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# Quantum Mechanics I PHYS 6210

Utah State University

GPA 3.6

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Physics 6210Spring 2007Lecture 5 Lecture 5 Relevant sections in tezrt 12714 An alternate third postulate Here is an equivalent statement of the third postulate Alternate Third Postulate Let A be a Hermitian operator with an ON basis of eigenvectors aim The only possible outcome of a measurement of the observable represented by A is one of its eigenvalues aj The probability for getting aj in a state W1gt is D Pr0bA aj Z lti7aj wgt 27 i1 where the sum ranges over the D dimensional space of eigenvectors i ajgti 1 D with the given eigenvalue aj Note that if the eigenvalue 1k is non degenerate then Pr0bA aj ltjW1gt 2 Note also that ltjW1gt is the component of W1gt along Let us prove the probability formula in the case where there is no degeneracy Allowing for degeneracy is no problem think about it as an exercise Let x be the characteristic function of aj We have exercise Then in the state W1gt we have PTOMA aj ltfltAgt W WM W WWW Because the state vectors have unit norm we have 1 ltwwgt Zwmmw ZWWP J J For now we state it in a form that is applicable to finite dimensional Hilbert spaces We will show how to generalize it to infinite dimensional ones such as occur with a particle in a little while Physics 6210Spring 2007Lecture 5 that is the probabilities Pr0bA 1 add up to unity when summed over all the eigen values This indicates that the probability vanishes for nding a value for A which is not one of its eigenvalues We can write the expectation value of an observable so that the probabilities feature explicitly As usual we have Ali aili ltiljgt 6zj If a system is in the state W1 we compute by inserting the identity operator twice 7 good exercise ltAgt MAW Zwmwmmw iJ39 2 2 WW 239 We see that the expectation value is just the sum of all possible outcomes weighted by the probability for each outcome 7 just as it should be Note the special role played by normalized eigenvectors If the state is an eigenvec tor of some observable A with eigenvalue 1 then the probability for getting a is unity Eigenstates are states where you know one or more observables with certainty Here7s a good exercise for you Show that every state vector is an eigenvector of some Hermitian operator Therefore at least in principle every state can be determined by a suitable measurement The third postulate makes very specific predictions about what are the possible out comes of a measurement of a given observable In essence all the physical output of quantum mechanics appears via this postulate l have to emphasize that the predictions are always of a probabilistic nature Of course some probabilities are l or 0 so this does not mean that one can never make any statements with certainty Let us apply this new form of the third postulate to a couple of simple examples coming from the spin 12 system Consider the state 1 What is the probability for finding say Sy to have the value ihQ Since W1 l5y7gt H iH i lSy7igt H ilgt is orthogonal to lSy ltSy7 lSy 07 2 Physics 6210Spring 2007Lecture 5 this probability is zero In the same state what is the probability for nding 52 to have the value h2 We have i SS2727 lltz ly gtl l l And so forth SternGerlach revisited Let us now derive from our model of the spin 12 system some of the salient features of the Stern Gerlach experiment View our beam of spin 12 system as a large number of identical systems When we pass the beam through an apparatus 5G2 and keep only the spin up beam we have a large collection of particles all in the state 52 gt If we pass the beam through another 5G2 apparatus we find all the particles are still spin up In our model this result is simply the statement that the probability for finding 52 h2 in a the state W1gt lSZgt is unity llt527l gtl2 ltSzal527gtl2 1 Now consider passing the ltered beam through an SGI apparatus We know that the beam splits in half In our model this is seen by computing the probability distribution for SI when W1gt lSZgt We get 2 2 1 lltSzziWJgtl lltSzzilSmgtl Now we come to a crucial aspect of the use of our model to describe nature Suppose we take all the particles that had SI h2 and pass them through an 5G2 apparatus what does our model predict The key observation is that the beam of particles that are going to be passed through 5G2 are now all in the state i The measurementpreparation ltering process using SGI has determined a new state W l5x7gt l527gt l52rgt vector for the system To verify this we just pass the SGI filtered beam through another SGI apparatus and see that the particles have SI h2 with probability unity This means that they are in the corresponding eigenstate 51 gt So when we pass this beam through 5G2 we find the 50 50 probability distribution miw Michal We now want to spend a little time understanding this sort of result in general 3 Physics 6210Spring 2007Lecture 5 Compatible and Incompatible Observables Let us generalize the unusual result found in the Stern Gerlach experiment as follows Suppose we have two observables A and B If we measure A we will get one of its eigen values say a For simplicity only we assume that this eigenvalue is non degenerate The state of the system is now the eigenvector a where Aa aw If we then measure the observable B we get an eigenvalue say b and the system is in the state b where B W 5W for simplicity we also assume that b is a non degenerate eigenvalue The probability for getting a is now Rah1H2 WNW Unless there is a real number 0 such that a ei b this probability is less than unity When do we have a ei b This condition means that A and B have a common eigenvector Note that there are no common eigenvectors for the spin operators 7 ex ercise As long as A and B do not have all of their eigenvectors in common there will be situations where knowing one of the observables with certainty will preclude knowing the other with certainty Such observables are called incompatible Under what conditions will two observables be compatible Evidently compatible observables are represented by operators that have a common basis set of eigenvectors so that the argument just given above cannot apply This implies that the observables must commute AB E AB iBA 0 To see this let us denote by a b the common set of eigenvectors which form an ON basis A az a azwz B az b azbgt7 ltazb a7bgt 6aa6bb We have ABM b bAa b bow b thz b BAa b Since the operator A B maps each element of a basis to the zero vector and since every vector can be expanded in this basis we conclude that A B must be the zero operator To see this use the Schwarz inequality 7 exercise W b 2 S lta altbb equality iff hi constb Physics 6210Spring 2007Lecture 5 So if a pair of Hermitian operators have a common basis of eigenvectors then they must commute It is a fundamental result from linear algebra for nite dimensional vector spaces that the converse also holds If two Hermitian operators commute then they admit a common ON basis of eigenvectors Thus compatible observables are represented by commuting operators Physically it is possible to have states in Which the values of compatible observables are determined With certainty 7 unlike different components of the spin By the same token incompatible observables are represented by operators that do not commute For incompatible observables there Will exist states in Which their values cannot be determined With statistical certainty We see then that the unusual feature of nature embodied in the Stern Gerlach experiment is encoded in the algebraic structure associated to the observables by the commutator of the corresponding linear operators Physics 6210Spring 2007Lecture 12 Lecture 12 Relevant sections in tezrt 21 The Hamiltonian and the Schrodinger equation Consider time evolution from t to 15 6 We have U6JIeiHOAO As usual the unitarity of U implies that Ht is Hermitian z39e it represents an observable 7 the Hamiltonian There is one signi cant difference between the spatial momentum and the Hamiltonian however The spatial momentum is de ned once and for all by its geometrical nature in the Schrodinger picture The Hamiltonian depends on the details of the interactions within the system and with its environment Thus there can be many useful Hamiltonians for say a particle moving in 3 d but we always use the same momentum operator in the Schrodinger picture Let us extract the Hamiltonian in a different way We have that Ut 6150 Ut e tUt t0 Substitute our result above for Ut 615 divide both sides by E and take the limit as E a 0 thereby de ning the derivative of U We get exercise dU t t mgLQ2H Um dt Let us turn the logic of this around Let us suppose that given a self adjoint Hamiltonian Ht we can de ne Utt0 by the above differential equation When solving the differ ential equation an initial condition will have to speci ed in order to get a unique solution The initial condition we need is that Ut0t0 1 Thus we can say that given a Hamiltonian the time evolution of the system is determined By focusing attention on H rather than U we get a considerable advantage in our ability to describe physical systems Indeed we shall always de ne the dynamics of a system by specifying its Hamiltonian Note that it is much easier to give a formula for the energy of This supposition can be proved rigorously when the Hamiltonian doesn7t depend upon time One will have to make additional hypotheses in the more general case but we won7t worry with those technical details 1 Physics 6210Spring 2007Lecture 12 a dynamical system than to explicitly display its dynamical behavior lndeed rarely will we be able to explicitly compute the time evolution operatorJ The relationship we just derived between the time evolution operator and the Hamil tonian is an abstract version of the Schrodinger equation To see this simply apply both sides of this operator relation to an arbitrary state vector representing the initial state of a system at time to We have d1057750 hi i dt l 7t0gt HtUt7t0l 7t0gt This is d which is the traditional form of the Schrodinger equation in terms of abstract vectors Perhaps you would be more familiar with its coordinate wave function form in the case where the Hamiltonian is of the kineticpotential form for a particle P2 a 117 VX 2m We then get u i P2 a h 2 u u HWQC W f VXW1gt V f VW W95 To see this you should verify that mm gl wm eraM in a a and using the definition that We also have d 8 hi t hi at W dtiw gt i atwcc gt So the Schrodinger equation is after taking components in the position basis 252 2 s u 8 a lt2mV 11xt i thE LJt This does not mean U does not exist of course but rather it means the dynamical evolution of the system is suf ciently complicated that no simple formula will suf ce to describe it 2 Physics 6210Spring 2007Lecture 12 Solving the Schrodinger equation for the state vector at time t given any initial state vector is equivalent to determining the time evolution operator You see that the Schrodinger equation simply captures the fact that the Hamiltonian is the generator of time evolution To see what that time evolution actually is one needs to get informa tion about the solutions to the Schrodinger equation But a key observation here is that the solutions are ultimately determined by the choice of Hamiltonian Determining the Hamiltonian is the key step in making a model of a physical system Formal solutions to the Schrodinger equation It is possible to give formulas for the time evolution operator analogous to the exponen tial form of the spatial translation operator There are 3 cases to consider First suppose that H doesn7t depend upon time Then we are in an identical setting mathematically speaking as with the spatial translations We have Ut to W403 You can easily check that this operator satisfies the operator version of the Schrodinger equation including the initial condition Second suppose that H H05 but that for any times t t we have that lHt7Htl 0 Then it is not hard to check that z 1 dt H t Utt0e hfto H Note that this formula includes the previous result as a special case To check this result just note that one can manipulate the operators as if they were ordinary functions since all the different operators Ht commute Finally suppose that H H05 but that the operators at different times do not commute This case is somewhat harder and we shall take a crack at it much later For completeness let me just say that the resulting evolution operator is given in terms of the time ordered exponential i z Ulttyt0 Tei fto dtHt For a formula see your text We won7t be using this last case for a while so we defer its discussion Physics 6210Spring 2007Lecture 12 State vector evolution when 0 From now on let us focus on the common case where 0 We have seen how to build Utt0 in this case Let us have a look at how state vectors evolve in time in the Schrodinger picture Given H let us denote its orthonormal basis of energy eigenvectors by Any state can be expanded in this basis particularly the initial state WW ZltEji zt0gtiEjgt J The state at time t is given by w 6th Zltijogtijgt J39 ZltEji zt0gt57t7t0EjiEjgt 139 As a good exercise you should check directly that this formula gives the solution to the Schrodinger equation matching the initial state W1 t0gt So the effect of time evolution can be viewed as a transformation of the components in the energy basis ltEjiwz t0gt ltEjl 7t0gte t t0Ej This is a very important result If you want to find the state vector at time t in the Schrodinger picture given the initial state vector you must perform the following computations 1 Find the energy eigenvectors and eigenvalues 2 Expand the initial state vector in the energy basis 3 The components of the state vector at time t in the energy basis are the original components times the phase ef t tO You can now see why the energy eigenvectors are so important Finding them is the key to understanding time evolution Stationary states Suppose the Hamiltonian does not depend upon time If the initial state vector for a system is an energy eigenvector HlEkgt EkiEkgtz 4 Physics 6210Spring 2007Lecture 12 so that the energy is known with certainty at t to then the this state vector changes in a trivial fashion as time progresses This can be seen in 2 ways Firstly simply apply the time evolution operator Ultttogtwlttogtgt WWW reman You can also check this result by using the 3 step process described above exercise The energy eigenvectors change only by a phase factor hence all physical predictions probability distributions will not change in time Thus energy eigenvectors are states in which the system exhibits stationary behavior Such states are naturally called stationary states Nothing ever happens in a stationary state Often times we prepare the initial state by a filtering process based upon measurement of one or more commuting observables If these observables commute with the Hamilto nian then one can arrange that the preparedfiltered states are energy eigenvectors thus stationary states Let me point out a paradox here Surely you have heard of the process called spon taneous emission You know where you put an atomic electron in an excited state and even if you do notthing to the atom it will decay to a lower energy state And I know you have learned about energy levels of atoms in a previous course in quantum mechanics The energy levels are eigenvectors of the Hamiltonian for the atom ie stationary states Now the paradox appears how can an excited state of the atom 7 a stationary state 7 decay into anything if nothing every happens in a staionary state Think about it Conservation of energy The components of a state vector along the energy eigenvector basis determine the probability distribution for the energy in that state Time evolution amounts to multiplying these coef cients by phase factors In general this changes the vector in a non trivial way this causes probability distributions for various observables to change in time However since the probabilities for energy arise via the absolute values of the components you can easily see that the probability distribution for energy does not change in time still assuming that the Hamiltonian is time independent This is how energy is conserved in quantum mechanics Of course in classical mechanics we usually consider conservation of energy to be characterized by measure the energy initially ii measure the energy at some later time iii energy is conserved if these results are the same One can do that here as well but if you measure the energy initially you will then have preparedfiltered your system state to be an energy eigenvector Then of course the probability distribution is quite simple As we have seen if the system is in an energy eigenvector at one time it remains 5 Physics 6210Spring 2007Lecture 12 there for all time Thus you Will always get the same result for any subsequent measurement of energy That is certainly much like the classical way of vieWing conservation of energy However as we have seen in a stationary state all probability distributions are time independent ln non stationary states Where the dynamical evolution is non trivial one cannot say With statistical certainty What the initial energy is 7 one has a non trivial probability distribution What one can say is that While other probability distributions Will in general change in time energy compatible observables being the exceptions the energy probability distribution Will not This is What conservation of energy means in quantum mechanics Physics 6210Spring 2007Lecture 37 Lecture 37 Relevant sections in teth 57 Electric dipole transitions Our transition probability to first order in perturbation theory is 47904 yWA a Paan4fjwwmm mmwinv ommtmmi mhw where a 12W th137 is the ne structure constant and W2 2 Nwwgtgmwn is the energy per unit area per unit frequency carried by the EM pulse characterized by the vector potential in the radiation gauge with frequency components AW We now want to analyze the matrix element which appears This factor re ects the atomic structure and characterizes the response of the atom to the electromagnetic wave Let us begin by noting that the wavelength of the radiation absorbedemitted is on the order of 27rcwfz N 1076771 while the atomic size is on the order of the Bohr radius N 1078771 Thus one can try to expand the exponential in the matrix element a 1A a a lt lefsz 5 Z wfl che anizlizmigtlt lefsz ipfwfi zn X5 P izlizmigt The first term in this expansion if non zero will be the dominant contribution to the matrix element Thus we can approximate 7 I l A I Ti A a A a lt lef7mf 5 Z wfl chemeylzymo lt lefsz 5 anizlizmigta which is known as the electric dipole approzrimation Transitions for which this matrix ele ment is non zero have the dominant probability they are called electric dipole transitions We shall see why in a moment Transitions for which the dipole matrix element vanishes are often called forbidden transitions This does not mean that they cannot occur but only that the probability is much smaller than that of transitions of the electric dipole type so they do not arise at the level of the approximation we are using If we restrict attention to the electric dipole approximation the transition probability is controlled by the matrix element ltnf lf mff fni li To compute it we use the fact that a 23110 mg m 1 Physics 6210Spring 2007Lecture 37 and that Ho nz lzmgt En nz We get a m a a lt f7 lfszlplnizlizmigt 91107 famleHO 7 HOXlniz 1mm imwfiltnf7 lfszlenizlizmigt39 Now perhaps you can see why this is called a dipole transition the transition only occurs according to whether or not the matrix elements of the component along 6 of the dipole a moment operator qX are non vanishing Selection rules for Electric Dipole Transitions We have seen that the dominant transitions are of the electric dipole type We now consider some details of the dipole matrix elements lt lefszlq 39Xlnizlizmigtl In particular we derive necessary conditions on 1 and m such that the dipole matrix element is non zero and hence electric dipole transitions can occur These conditions are usually called selection rules for the first order electric dipole transitions Transitions which do not obey these selection rules are usually called forbidden transitions Of course they are only forbidden insofar as our approximations are valid The forbidden transitions may very well occur but they will be far less likely than the first order electric dipole transitions being considered here The selection rules we shall derive are determined solely by the angular momentum properties of the unperturbed stationary states Thus the selection rules rely upon the fact that the stationary states can be chosen to be orbital angular momentum eigenstates which requires that the atomic potential V0 be a central potential V0 see below On the other hand the selection rules do not depend upon any further properties of this potential Digression Rotational Symmetry Here we brie y explain how angular momentum conservation is tied to rotational sym metry Recall the unitary rotation operator Physics 6210Spring 2007Lecture 37 a On positionmomentum eigenvectors lxgt we have 292 lRm m WWW me where R02 6 is the 3 d orthogonal transformation rotating about the axis 1 by the angle 6 From this it follows that exercise and The last relation can be seen using the spectral decomposition of V02 or by verifying this relation in the position basis If the potential is rotationally invariant 26 is spherically symmetric 26 depends only upon the distance from the center of rotation 26 describes a central force then afgh39LvltXgteW Vltlegt The Hamiltonian is then rotationally invariant a 2 76hL P ml 239 A r P2 e 92 Vle6 Vle By considering an infinitesimal transformation it easily follows that 11 E 0 This is just infinitesimal rotation invariance But it also means that angular momentum is conserved 7 its probability distribution is unchanged in time This is because stationary states can be chosen to be angular momentum eigenstates Thus we see how a symmetry corresponds to a conservation law For us however the key thing here is that for a central potential the energy eigenvectors can be chosen to be also angular momentum eigenvectors Selection rules involving m We now consider restrictions on m and mf needed so that the three components of X have non vanishing matrix elements Our main identity arises because X is a vector operator we have that Rh iFLEZJ kRk 3 Physics 6210Spring 2007Lecture 37 In particular XLZ iihY YLZ ihX ZLZ 0 These formulas simply give the in nitesimal change of the position vector under rotations exercise and are the infinitesimal versions of the formulas given above You can easily check them explicitly From these identities we have 0 lt fzlfszliZszllnizZizmigt mi mfhltnfzlfszlzlnizlizmigtz so that the 2 matrix element vanishes unless ml m f Next we have miimfhltnflfmlelnilimigt ltnflfmflXL2Hnilimigt iihltnflfmlelnzlimigt and miimfhltnflfmleanlimigt ltnflfmflYL2Hnilimigt ihltnflfmlelnilZmigt from which it follows that either mi 7W 17 or ltnflfmlelnlimZgt ltnflfmlelnlimZgt 0 Thus for the X and Y matrix elements to be non vanishing we must have mf m i 1 In short no electric dipole transitions occur unless Am 0i1 lf Am 0 then only radiation with polarization with a component along 2 will stimulate a transition in this approximation lf Am i1 then polarization in the x 7 y plane will stimulate transitions Likewise these are the polarizations that feature in the respective emission processes that is the emitted radiation will have this polarization structure Selection rules involving 1 We have seen how the electric dipole transitions require Am 0 ii We can get se lection rules involving 1 by playing a similar game as above but now using the commutator L2 m Mag RLZ 4 Physics 6210Spring 2007Lecture 37 Take the initial final matrix element of both sides of this equation and use the fact that vectors de ning the matrix element are L2 eigenvectors You Will find that this identity implies exercise 2mm 1gtziltzi1gtH1flt1f 1gt7ziltzi 1gti2ltnf1fmf2mzmgt 7 0 Therefore if ltnlefszlilmzli7migt 3quot 0 then lfaf 1 11 1 7 1fo 1 7 11 12 0 This condition can be factored into the form exercise zf 112 71Hltzf 71 71 7 0 Keeping in mind that l is non negative you can see that the first term vanishes if and only if 2 lf 0 The second term vanishes if and only if lf 7 2 i1 We conclude that the transition is forbidden unless Al i1 or lf lz 0 In fact the second case is excluded an explicit calculation easily shows that the dipole matrix element actually vanishes if the initial and final states are zero angular momentum states To see this simply recall that each of XY Z is a linear combination of l l spherical harmonics lf lf l 0 then the angular integrals in the inner products vanish by orthogonality of l l spherical harmonics With l 0 spherical harmonics exercise To summarize the electric dipole selection rules are Al i1 Am 0 i1 These conditions are necessary for a transition to occur given our approximations These selection rules are compatible With an interpretation in terms of emission and absorption of photons by the atom Using this point of vieW the photon Will have frequency w lwal interpretable as conservation of energy and the photon carries angular momentum xih or as one says the photon must have spin l by conservation of angular momen tum lndeed if the photon carries spin l then our previous discussion of addition of angular momentum coupled With an assumption of conservation of angular momentum for the atom photon system imply that the angular momentum of the atom after the tran sition must be l l i 1 exercise The first possibility doesn7t occur in the electric dipole approximation Physics 6210Spring 2007Lecture 37 This is all true but it is a mistake to think that this picture can be obtained from our treatment of an atom in a radiation field There are two reasons why our current description is inadequate First we have not treated the electromagnetic field as dynamical 7 we have simply postulated its form Because the electromagnetic field configuration is specified once and for all there is no way to describe emission andor absorption of energy and angular momentum from the electromagnetic field which is not allowed to change This is re ected in the fact that the Hamiltonian we have been using does not give the total energy of the combined system of electron and electromagnetic field rather it just gives the non conserved energy of the electron Similarly the angular momentum f X gtlt 13 that we are speaking of is not the total conserved angular momentum of the charge and electromagnetic field but rather just the unconserved angular momentum of the charge alone To include the electromagnetic field in the bookkeeping of conservation laws we must include the electromagnetic degrees of freedom into the system and include suitable terms in the Hamiltonian to describe the dynamics of these degrees of freedom and their coupling to the charge This leads us to the second dif culty with our previous treatment Our model was semi classical since the charge was given a quantum treatment but the electromagnetic field was given a classical treatment It does not appear possible to have a consistent theory of charges and electromagnetic fields in which the former are described via quantum mechanics and the latter treated via classical physics This was realized early on in the history of quantum mechanics What is needed then is a method for incorporating electromagnetic degrees of freedom into the system using the rules of quantum mechanics It was I think Dirac who first showed a way to quantize the electromagnetic field and then consider a quantum dynamical system of charges and fields Thus QED was born A healthy dividend was paid for this quantum description of electrodynamics one could now explain spontaneous emission which is the phenomenon where an atom or other quantum system in an exited bound state will spontaneously emit radiation and drop to a lower energy state 7 even in the absence of a perturbation Physics 6210Spring 2007Lecture 1 Lecture 1 Relevant sections in tezrt 11 What is a theory Students often consider quantum mechanics to be rather strange compared to other theories they have encountered eg classical mechanics electromagnetic theory And I guess it is a little unsettling to find position and momentum being represented as dif ferential operators particles being described by complex waves dynamics being defined by the Schrodinger equation measurements characterized by probabilities and so forth In particular the rules of quantum mechanics at first sight seem rather strange com pared to the rules of Newtonian mechanics essentially just Newton7s 3 laws But a lot of the intuitive appeal of Newton7s laws comes from years of familiarity Unfortunately the strangeness of quantum mechanics largely bred from lack of familiarity often leads to the feeling that the subject is very dif cult Of course the subject is not easy nor should it be it is one our most advanced descriptions of nature But even if the basics seem a little dif cult it is only because one is having to use new mathematical models for familiar physical structures After all wasn7t it a little unsettling when you first started using vectors in a systematic way to describe displacements velocities and so forth Can you remember that far back So I would like to begin by emphasizing that in many ways quantum mechanics is a theory like any other To do this I must give a very coarse grained description of what a theory is in general and quantum mechanics in particular Of course the devil 7 and the physics 7 is in the details Essentially a physical theory is a set of rules 2e postulates that can be used to explain andor predict the behavior of the world around us 7 in particular the outcome of experiments Which experiments can be explained how to characterize the physical ingredients in these experiments what information needs to be specified in advance etc are part of the rules of the theory By the way keep in mind that a theory can never be said to be true but only that it agrees with experiment It is always possible that the next experiment will falsify the theory And it is possible that more than one theory will be able to explain a given set of results Insofar as these results are all we have experimental access to the theories are equally good Usually though there is one theory that explains the most results with the simplest postulates This theory is usually considered the best So while classical mechanics can be used to explain a wide variety of macroscopic observations quantum mechanics can also explain these results and a host of other results from microscopic physics atomic physics nuclear physics etc that cannot readily be explained using classical mechanics It should be emphasized that the word theory has a number of connotations and this 1 Physics 6210Spring 2007Lecture 1 can be confusing For example quantum mechanics is a theory but the use of quantum mechanics to model the hydrogen atom as say a non relativistic electron moving in a fixed Coulomb field is also a theory in some sense 7 a theory of the hydrogen atom Clearly these two notions of theory have somewhat different logical standings in the sense that one can build a variety of theories of the hydrogen atom by adding eg spin orbit coupling special relativity finite size nucleus etc within the framework of the theory of quantum mechanics Given this slightly confusing state of affairs I will try but may fail to call quantum mechanics a theory and I will call models the various theories built from quantum mechanics of eg the hydrogen atom What are some successful physical theories There are many of course Some ex amples are Newton7s theory of matter and its interactions valid at large length scales weak gravitational fields and small velocities classical mechanics Maxwell7s theory of the electromagnetic field and its interaction with charged sources Einstein7s theory of gravitation and of course the theory of matter and its interactions at small length scales which we call quantum mechanics along with its descendant quantum field the ory One confusing feature of all this is that theories really come to us in overlapping hierarchies For example using the classical Maxwell theory of electrodynamics we can create a theory of atoms as bound states of charges These theories are ultimately incorrect being classical A correct theory of electromagnetic phenomena in general and atoms in particular arises via quantum electrodynamics in which the theory of quan tum mechanics better quantum field theory is melded with Maxwell7s theory and then used to build a theory of interacting charges atoms etc Thus we can discuss the physical theory called quantum mechanics and using the framework defined by this theory we can build theories of various physical phenomena eg crystalline solids The theory of the phenomena can be wrong without quantum mechanics being wrong or perhaps one is unable to build a satisfactory theory of the phenomena owing to a failure of the parent theory quantum mechanics Similar comments apply to Einstein7s theories of relativity and the various physical theories that are formulated in the context of Einstein7s relativity Here again we are drawing a conceptual distinction between the idea of a theory and a parcticular model built within the confines of that theory After all these general philosophical sounding statements it is time to get down to business What does it mean to have a set of rules that can be used to explain7 andor predict the outcome of experiments Of course useful theories are necessarily somewhat intricate but taking a very coarse grained view of things the structure of a generic theory can be characterized in a pretty simple way Basically one can view a theory as a way of describing observables states and dynamics Below we will describe each of these three As you know mathematics has for some centuries now been the language and tool of choice in building a physical theory Consequently the initial job in this course will be to 2 Physics 6210Spring 2007Lecture 1 give a mathematical representation for these 3 basic ingredients Observables Measurable aspects of the experimentally accessible world are the observables Any theory is to provide a means of assigning a mathematical representation to the observables By specifying the observables we are going a long way toward specifying the kinds of physical situations that our theory is meant to cover Quantum mechanics postulates a universal ground rule for observables they must be self adjoint operators on a Hilbert space The way in which we implement this rule may vary from physical model to physical model For example using the theory of Newtonian mechanics we can build a model of a par ticle in which some important observables are position momentum angular momentum energy etc In fact our usual model of a Newtonian point particle supposes that all observables can be viewed as functions of position and momentum which we can call the basic observables Mathematically the basic observables for a single particle in Newtonian theory are represented by 6 numbers at y zpzpypz that is the observables are func tions on the six dimensional phase space Normally these 6 numbers are actually viewed mathematically speaking as a pair of vectors The behavior of a particle is documented by monitoring the behavior of its observables and we build our theory using the mathemat ical representation eg vectors for these quantities Other quantities like mass electric charge and time are in some sense observable too but in Newtonian mechanics these quantities appear as parameters in various equations not as quantities which one measures to document the behavior of the system In other words while we may consider the way in which the position of a particle changes we normally don7t include in our model of a particle a time varying mass or electric charge Of course more sophisticated models of matter may attempt to give a better or more fundamental description of mass and electric charge in which these quantities become observables in the sense described above As another example consider the electromagnetic field as it is described in Maxwell7s theory Observables are of course the electric and magnetic field strengths Also we have polarization of waves energy density momentum density etc All electromagnetic observables in Maxwell7s theory are built from electric and magnetic field strengths which are the basic observables of the theory Mathematically they are represented as vector elds Another measurable quantity is the speed of light 0 Like mass and charge in Newtonian mechanics this quantity appears as a parameter in Maxwell7s theory and is not something that can change with the configuration of the system We don7t use the term observable for the speed of light in the language we are developing As my final example consider the theory of bulk matter known as statistical mechan 3 Physics 6210Spring 2007Lecture 1 ics This theory has a number of similarities with quantum mechanics For now let us note that some of the observables are things like free energy entropy critical exponents etc All of these quantities can be computed from the partition function which is in some sense the basic observable Of course statistical mechanics is normally built from classical andor quantum mechanics in which case the partition function itself is built from more basic observables By the way temperature is of course a measurable quantity But it plays the role of a parameter in statistical mechanics in a canonical ensemble and so in this case temperature is treated as a parameter just like mass and electric charge in classical electrodynamics So we see that different physical phenomena require different kinds of observables and different theories use different mathematical representations for the observables One of our two main goals this semester is to get a solid understanding of how quantum mechanics represents observables Let us remark that in all of our examples 7 indeed in most theories 7 the observable called time enters as an adjustable parameter It is normally not modeled as an observ able in the same sense as we model say the energy of a particle In quantum mechanics time is not an observable in the sense described above States A physical system can be in a variety of configurations that is it can display a variety of possible values for its observables When we speak of the state of the system we are referring to a mathematical object which can completely characterize the outcomes of all possible measurements of the observables of the system Thus the mathematical representation of state is intimately tied to the representation of observables Let us illustrate the notion of state with our previous examples In the Newtonian mechanics of a particle we can determine the state of the system by measuring the basic observables the coordinates and momenta Indeed all other observables are functions of these observables and the time In electromagnetic theory the state of affairs is quite similar In source free electro dynamics the electric and magnetic fields at one time determine all electromagnetic observables and the state of the system In these two examples the notions of state and observables are closely intertwined which gives the impression that they are really the same thing But this is not always the case In statistical mechanics we have a more subtle way of defining the state To be explicit let us focus on classical statistical mechanics of a system of particles From the point of view of classical mechanics the underlying basic observables can be viewed as the coordinates and momenta of all the particles making up the system the phase space and specifying a point in phase space determines the state of the system In statistical mechanics the 4 Physics 6210Spring 2007Lecture 1 state is specified by giving a probability distribution on the phase space which is a single function rather than a point in phase space as is done in Newtonian mechanics Mom the probability distribution one can compute all observables of statistical mechanics We see that given the state of the system all observables are determined but the way we specify the state can be rather different To get a better handle on the distinction between states and observables you can think as follows The state of a system re ects the way it has been prepared which normally is a re ection of the particular initial conditions used Given a particular preparation procedure performed by various measurements and or ltering processes the system will behave in a particular 7 indeed unique 7 way as re ected in the behavior of its observables A physical model for a system principally involves an identification of the observables needed to describe the system This is done once and for all The states of the system represent various ways the system can be started off 7 and can be adjusted by experimental procedures Dynamics The measured values of observables of a system will usually change in time Normally a theory will contain a means of describing time evolution of the system that is a dynamical law or a law of motion Assuming that we use a time independent mathematical model for the observables we can view dynamics as a continuous change in time of the state of the system according to some system of equations This way of formulating dynamics is what is often called the Schrodinger picture of dynamics and a famous example of the dynamical law is provided by the Schrodinger equation In statistical mechanics the state of the system is determined by a probability distribution on phase space This distribution evolves in time according to the Liouville equation In classical mechanics and electrodynamics the state of the system is known once one speci es the values of the basic observables For example if you give the positions and velocities of a Newtonian particle at an instant of time these quantities will be uniquely determined for all time by a dynamical law 26 Newton7s second law In these theories one can therefore think of dynamics as a time evolution in the value of the observables according to some system of equations F ma Maxwell equations One important aspect of dynamics that is usually incorporated in any theory is a very basic notion of causality In the Schrodinger picture given the state of the system at one time the dynamical law should determine the state uniquely at any other time Granted this you see that the state at a given time along with the dynamical law which is part of the specification of the theory will determine the outcomes of all measurements at my time To summarize A theory requires 1 A mathematical representation of observables 5 Physics 6210Spring 2007Lecture 1 2 A mathematical representation of states and a prescription for determining the values of the observables 7 the physical output of the theory 7 from any given state 3 A speci cation of a dynamical law which tells us how to extract physical output as a function of time Our goal this semester will be to see how quantum mechanics takes care of l 2 and 3 and uses them to build models of a number of physical systems A Word of Caution One prejudice that arises from say classical mechanics that must be dispelled is as follows In classical mechanics knowing the state is the same as xing the values of all observables at one time So if we know the state of a Newtonian particle at one time we know the values of its coordinates and momenta and every other observable Other theories may be set up differently and this kind of result need not apply For example in quantum mechanics and in classical statistical mechanics the state of the system will provide probability distributions for all observables One may completely determine specify the state by assigning values to some of the observables the energy of a simple harmonic oscillator is 5 ergs with probability one but this may leave some statistical uncertainty in other observables As we shall see for example roughly speaking specifying the position of a particle will completely determine the state of the particle in quantum mechanics This state will allow for a large statistical uncertainty a very broad probability distribution for momentum Likewise specifying the energy of a particle will in general imply a statistical uncertainty in the values of position and momentum SternGerlach experiment We now describe an experiment conducted by Stern and Gerlach in the early 19207s It gives us a valuable demonstration of the kind of phenomenon that needs quantum mechanics to explain it It also provides an example of what is probably the simplest possible quantum mechanical model As you probably know this experiment involves the property of particles known perhaps misleadingly as their spin which is an intrinsic angular momentum possessed by the particles Note though at the time of the experiment neither intrinsic spin nor quantum mechanics was very well understood Our goal in studying this important experiment is to introduce the basic rules of quantum mechanics in what is probably the simplest possible mathematical setting The Stern Gerlach experiment amounts to passing a beam of particles through a region with a magnetic field which has the same direction everywhere but a varying magnitude Recall that the classical potential energy of interaction of a magnetic moment M with a magnetic field B is 7M B Thus the force as opposed to the torque exerted on the 6 Physics 6210Spring 2007Lecture 1 magnetic moment is non zero if the magnetic eld varies in space We have F VM B If the magnetic eld varies only in one direction then the force is parallel or anti parallel to that direction depending upon the orientation of the magnetic moment relative to the fixed direction of B Thus the force experienced by the magnetic moment allows us to measure the component of M along the direction of B we simply have to observe which way the particles are de ected Of course this reasoning was purely classical so we are initially con dent it will work with macroscopic objects but it can be justifiedexplained quantum mechanically For now we just note that the correct quantum model of the interaction of a magnetic moment with a magnetic eld does indeed use the potential energy function shown above and since the atom is su iciently massive its motion can be well approximated using a classical mechanics description Here we have a nice example of how classical macroscopic reasoning gives us important clues to modeling a quantum mechanical microscopic system Stern and Gerlach passed a beam of silver atoms through such an apparatus These atoms have a magnetic moment thanks to the electron spin and so are de ected by the apparatus Based upon a classical model of the atoms magnetic moment as coming from motion of the charge distribution and given that the atoms in the beam have random orientations of their magnetic moments one expects a continuous de ection of the beam re ecting a continuous spread in the projections of the magnetic moment along the inho mogeneity axis lnstead what was observed was that the beam splits into two parts The explanation for this phenomenon is that the magnetic moment vector M of the atom is due to the presence of an electron which carries intrinsic angular momentum S 7 its spin 7 and hence a magnetic moment Here M is proportional to S The electron is in an atomic state which does not have any orbital angular momentum so its motion about the atom does not provide a contribution to the magnetic moment In this initially bizarre explanation the electron can be in one of two spin states relative to any given direction More precisely the projection of the spin of an electron along any axis can only take two values ih2 Such particles are said to have spin 12 an intrinsic state independent property of the electron If each atom is randomly selected one expects that these two alternatives occur with 50 50 probability and this is what is observed Half the beam gets de ected in each direction To explain this discreteness 7 indeed two valued nature 7 of an angular momentum is one challenge faced by any putative theory But there is much more Using Stern Gerlach apparatuses we can measure any component of the magnetic moment 7 equivalently the spin vector 7 of a spin 12 particle by aligning the SG magnetic Of course there are 47 electrons in a silver atom However 46 of them are in a state with no net angular momentum and hence no net contribution to the magnetic moment 7 1 Physics 6210Spring 2007Lecture 1 field direction along the axis of interest then passing the beam of particles through and seeing which way the particles are de ected corresponding to spin up or down along that direction Let us therefore try to model the behavior of the spin vector S in such an experiment ignoring all the other degrees of freedom that the atoms might have Thus the atom is modeled as a spin 12 particle Let us call an SG apparatus that measures the spin along an axis characterized by a unit vector 11 SGn Thus the apparatus SGT measures S n The empirical fact is that if you measure S 11 you always get ih2 Let us pass a beam of spin 12 particles through SGT and keep say only the particles that de ect according to S 11 having the value 9 If we pass this ltered beam through another such SGn lter device we see that 100 of the beam passes through We say that we have determined the spin along 11 with certainty for all the particles in the filtered beam We model this situation by saying that all the particles in the ltered beam are in the state S n We can say that we have prepared many particles all in the same state by passing a beam of particles through an SG apparatus and only keeping those de ected up or down Suppose we pass a beam through the apparatus 5G2 and only keep one spin projection We now have many electrons prepared in the state 1521 Let us try to pin down the value of SI that these electrons possess Pass the beam all particles in the state 52 through another Stern Gerlach apparatus SGI Particles are now de ected according to the projection of their magnetic moments or spin vectors along the x direction What you find in this experiment is that the beam splits in half This is perfectly reasonable we have already decided that any component of the spin has just two projections along any given axis Since there is nothing special about the x or 2 directions we should get similar behavior for both In the 5G2 filtered beam we did not prepare Sr in any special way so it is not too surprising that we get the beam to split in half Let us continue our investigation as follows We have passed our beam through 5G2 and kept the spin up particles We then pass these spin up particles through SCI let us focus on the beam that gave h2 for the SI measurement Therefore roughly half of the beam that entered the SGI apparatus is kept and we now have 14 of the original particles left in our prepared beam After this filtering process we can if we like verify that a repeated filtering process with apparata SGI keeps all the beam intact evidently the state of the particles could be represented by 51 Now we have a beam of electrons that have been measured to have the following properties 1 52 is h2 2 Sr is h2j Given 1 and 2 above it is reasonable to Of course one naturally prefers to write the state as something like 52 1 511 but we shall see that this is not appropriate We could now go and measure Sy in this doubly filtered beam you will find that half the beam has spin up along y half has spin down exercise But let us not even bother with this Physics 6210Spring 2007Lecture 1 believe that the electrons all have de nite values for 52 and SI since we have ltered out the only other possibilities This point of view is not tenable Suppose you go back to check on the value of 52 Take the beam that came out of 5G2 with value h2 and then SGI with the value h2 and pass it through 5G2 again You may expect that all of the beam is found to have a value h2 for 52 but instead you will find that beam splits in two This is despite the fact that we supposedly filtered out the spin down components along 2 So if you measure 52 and get say h2 and then you measure it again you will get h2 with probability one assuming no other interactions have taken place If you measure 52 and get say h 2 then measure SI and then measure 52 the final measurement will be ih2 with a 50 50 probability This should get your attention the values that you can get for the observable 52 in two measurements depends upon whether or not you have determined the value of Sr in between the 52 measurements Given this state of affairs it is hard to make sense of the classical picture in which one imagines the electron to have given definite values of all its observables 69 Sr and 52 One sometimes says that the measurement of SI has somehow disturbed the value of 52 This point of view is not incorrect but is not a perfect description of what is going on For example as we shall see the quantum mechanical prediction is unambiguously independent of the way in which we make the measurements Nowhere do we really need to know how the SG devices worked Moreover the disturbance in 52 due to the SI measurement is not a function of how carefully we make the SI measurement that is one cannot blame the strange behavior as coming from some experimental error the measurements can ideally be perfect and we still get the same result The fact of the matter is that one shouldn7t think of observables such as 52 and SI has having given fixed values that exist in the object of interest This may be philosophically a bit sticky and psychologically a bit disturbing but it seems to be quite alright as a description of how nature actually works If all this seems perfectly reasonable to you then you probably don7t understand it too well Our macroscopic experience with matter just doesn7t give any hint that this is the way nature works Electrons and other elementary particles are not like tiny baseballs following classical trajectories with tiny spin angular momentum arrows attached to them and there is no reason experimentally to believe that they are It is a purely classical prejudice that a particle has definite values for all observables that we can measure Try to think this way what is a particle It has mass total spin charge etc and other intrinsic real properties that do not change with the state of the particle Based upon experiment one may want to assign other observable properties such as position energy orbital angular momentum spin component along an axis to the particle But according to experiment 9 Physics 6210Spring 2007Lecture 1 these properties change With the state of the particle and cannot be Viewed as existing in the particle independently of the measuring process Which changes the state As it turns out according to the quantum mechanical explanation of this sort of phenomenon all you are guaranteed to be able to assign to a particle is probability distributions for its various observables Our next task is to build up the quantum mechanical model of the spin 12 system using the rules of quantum mechanics 96 Physics 6210Spring 2007Lecture 9 Lecture 9 Relevant sections in tezrt 16 17 Momentum wave functions We have already indicated that one can use any continuous observable to de ne a class of wave functions We have used the position observable to this end Now let us consider using the momentum observable We define as usual Plpgt plm p E R 1100 dplpgtltpt 1MP ltpl gt7 PW pwp and so forth The interpretation of the momentum wave function is that W1pl2 dp is the probability to find the momentum in the range 1 dp In other words 12 Pr0bP 6 ab dpl ltpgtl2 Note that a translation of a momentum wave function is a simple phase transformation exercise 1 mm ltpeaplwgt Mama Physically this means that a translation has no effect on the momentum probability dis tribution exercise Exercise use the momentum wave function representation to check the unitarity of Ta A very useful and important relationship exists between the position and momentum generalized eigenvectors To get at it we study the scalar product yelp which can be viewed as the position wave function representing a momentum eigenvector It can also be viewed as the complex conjugate of the momentum wave function representing a position eigenvector This complex function of It must satisfy for each 1 ltx 7 elm mm e i6pltlpgt This implies to first order in 6 exercise at pltlpgt Of course these properties eg the spectrum of P ought to be derived directly from the de nition of the translation operator This can be done but we won7t do it explicitly here 1 Physics 6210Spring 2007Lecture 9 The solution to this equation is gulp constepz The constant can be determined by the normalization condition 00 6mpw4mw acmcw 700 Using the Fourier representation of the delta function 00 700 we see that exercise 61 cwel p m Thus we have recovered a familiar result from wave mechanics the position space wave function for a particle in a state such that the momentum has with probability 1 the value p is a complex plane wave with wavelength Because the absolute value of the wave function is unity the particle has an equal probability of being found anywhere think uncertainty relation Note also that since the energy of a free particle of mass m is P2 7 this wave function describes a free particle with energy p22m Because 1 I p x 767 lt l gt 27 we see that the momentum space probability amplitude for a particle in an idealized state with a definite position is also a plane wave For an arbitrarily well localized particle all momenta are equally likely uncertainty relation again With these results in hand we can give an explicit relation between the position and momentum bases 96gt 1 dplpgtltplgt dpe fmlm and 00 1 0 i1 mmmmmm mmwma Of course plane waves are not normalizable but we have already discussed this subtlety 2 Physics 6210Spring 2007Lecture 9 If we set ltW1gt7 1211 ltPW1gt7 we get exercise 1 00 1 7 d 1 1W M100 W 1109 and 1 00 Jim 11p M100 date h Thus the position wave functions and momentum wave functions are related by Fourier transforms Note that exercise 00 00 lt l gtgt dx wltxgt gtltxgt dpwltpgt gtltpgt 00 00 In the momentum representation the momentum operator is a multiplication opera 77 tor PM ltplPl gt pltpl11gt mXp While the position operator is a differentiation operator 1 0 1ng Xwltpgt i mdxe h We 1 00 i 7 77px lwdme h ltgt 7L 9 l 00 i 77 771 iapx27rh food e h W3C WWI i8 Expectation values Expectation values in the position and momentum representation are easy to compute Using the notation ltW1gt7 1211 ltPW1gt7 we have in particular the following lt VQXM gtdfOWltxM2dp u f7dpMpt 2 mwcv wct 1MP wmmwmmgt 3 Physics 6210Spring 2007Lecture 9 You should prove these as a very good exercise Particle in 3 dimensions The generalization to a particle in 3 dimensions is done by essentially tripling our previous constructions Here we brie y describe this generalization Afterward I will show you how to view in general this way of adding degrees of freedom77 to a quantum mechanical model In 3 d we now have position vectors X with components Xi X Y Z and momen tum vectors P with components PZ PIPyPZ Each pair is represented by seif adjoint operators exactly as before We demand that they have the following canonical commutation relations XiXj PlPj 0 XiPj may We have position generalized eigenvectors ix and momentum generalized eigen vectors p XZiXgt x PAP pz39ipgt These form a generalized basis d3xxgtltx Id3ppgtltpi Here it is understood that the integrals run over all of positionmomentum space The seif adjoint momentum operators generate unitary transformations corresponding to 3 d translations 1 Ta ci a39P Taix ix a The canonical commutation relations re ect the fact that translations are commutative operations I TaTb TbTa e ltabgt39P Ta 13 Note that the commutation relations allow us to choose a basis of simultaneous gen eralized eigenvectors of the position or momentum The relation between the two bases is i 1 l plx ltX Pgt We The position wave functions and momentum wave functions are defined as usual by taking the components of a state vector along the corresponding basis WK ltXWIgt7 WP ltpW1gt Physics 6210Spring 2007Lecture 9 We then have Xiwx wx awltxgt wxx and i h 8 X 1MP 787Tpr PMP pMp The probability distributions for nding position momentum in a volume VV are Pr0bXEVVd3x1Jx2 Pr0bPEVVd3p11p2 Physics 6210Spring 2007Lecture 31 Lecture 31 Relevant sections in tezrt 51 52 Example nite size of the atomic nucleus One improvement on the simple particle in a potential model of an atom takes account of the fact that the atomic nucleus is not truly point like but instead exhibits finite size structure in both its mass and charge distributions It is possible to use perturbation theory to get a quick look into the effect on atomic spectra of this feature Of course this effect is one of a myriad of small corrections that need to be added to the atomic model Let us model the nucleus as a very massive uniform ball of total charge Ze with radius r0 As a nice exercise in elementary electrostatics you can check that the potential energy Vr 76 for such a charge distribution takes the form 2 izi forr 2m 7 V r 2 3 forrSm I do not know if a closed form solution to the eigenvalue problem for the Hamiltonian P2 H 7 V 2m is known but I doubt that such a solution exists We well treat the potential energy due to the finite size To of the nucleus as a perturbation of the usual Coulomb potential To this end we write V0 V00 BO where Z 2 V00 776 for 7 gt 0 and 2 2 BO 670 732 for 037 gm 0 for 7 2 To The idea is then that since the unperturbed energy eigenfunctions are non trivial over a range corresponding to the Bohr radius a for the given Z as long as To ltlt 1 we expect that the effect of the perturbation B will be small We will make this more precise in a moment Physics 6210Spring 2007Lecture 31 Recall that the energy eigenstates ln l m have position wave functions given by ltrln7 l7 mgt E 11nlm7 797 Rnlrylmgz 7 where the Ylm are the usual spherical harmonics and 3 12 z 2 H7171 L 27 211 27 R 7 7 M L 7 n10 7111 2nnll3 6 ml n l lna7 with E2 1 7 m electron mass Zmee2 and Lg is the associated Laguerre polynomial See the text for details Note that a is the Bohr radius for the hydrogenic atom with a given Z it sets the scale for the atomic size Let us consider the first order correction to the ground state of the atom due to this perturbation We choose the ground state for simplicity and since the ground state energy is non degenerate The shift in the ground state energy is given by AE lt100lBl100 Using the fact that 573 2 11100 7 V 471113 2 2 L 3 r E m 7 This integral can be computed explicitly Try it I did it using Maple I am going to give you the answer after it has been simplified with the assumption that To ltlt 1 which we find that exercise 7 0 2Z 2 AE drr2 63 0 70a 721 11 is needed in any case to ensure that the perturbation is suf ciently small to render our approximation scheme viable In this case we get 2 2 N 2Z6 r0 7 4 r0 2 r0 AE N 5a 7 ElEgroundl 7 Z ltlt1 I got this expression by having Maple perform a Taylor series expansion Thus the effect of the finite size is to shift the energy upward and the scale of the shift is determined by 77 To get a feel for the size of the correction let us consider a couple of examples For hydrogen we note that the charge radius of a proton is on the order of 10 15 m and that the Bohr radius is on the order of 107mm so that the the perturbative correction is on 2 Physics 6210Spring 2007Lecture 31 the order of one part in 1010 Very small There are physical systems in which the nite size of the nucleus has a more pronounced effect For example one can consider muonic hydrogen consisting of a proton and a 7 muon Muons are like electrons only much more massive For muonic hydrogen the Bohr radius is smaller by a factor of 1072 leading to an energy shift on the order of 1 part in 1076 At the other extreme consider a muonic lead atom a negatively charged muon bound to a lead nucleus There we have that both To and a are on the order of 10 15 m Hence here the finite size effect is as important as the Coulombic effect t Note however that our perturbative approximation is no longer valid since the effect of the perturbation is not small so this is at best a qualitative statement Let me finish this example with a few comments about the perturbative approxi mation to the ground state To compute it we need to compute the matrix element 11 l mlBll 0 0 n 31 1 Since the perturbation commutes with angular momentum it is clear that this matrix element will be non zero only if 1 m 0 We then have exercise To nlmlBl10 0 ammo0 dr ran0rR10rBr Assuming that m ltlt 1 we can approximate R7110 Rnl0 We then get exercise Z62 2 ltn7 la 17 07 WTORHO0R1006106 WLO which can be used to compute the approximate ground state wave function as a superposi tion of unperturbed l 0 m hydrogenic stationary states Note that this superposition will lead to a spherically symmetric ground state wave function which is non vanishing for 7 lt r0 Thus there is a non zero probability for finding the electron or muon within the nucleus Degenerate Perturbation Theory We now consider the case where the unperturbed eigenvalue is degenerate that is 0 i eigenvalue E720 These eigenvectors span the degenerate subspace D which is a finite there are d linearly independent eigenvectors i 12 d for the unperturbed dimensional vector space sitting inside the full Hilbert space of state vectors Degeneracy is associated with a symmetry of the unperturbed in this case Hamiltonian The full Hamiltonian with the perturbation included will typically not have all the symmetry of the unperturbed Hamiltonian Thus the true eigenvalues that are approximated by the unperturbed eigenvalue will usually not all be degenerate Put differently as the pertur bation is turned on by mathematically varying A from 0 to 1 some of the unperturbed eigenvectors with the same unperturbed eigenvalue become eigenvectors with a distinct eigenvalue so that the degeneracy can be lifted by the perturbation One says that the energy levels split as the perturbation is turned on 3 Physics 6210Spring 2007Lecture 31 Consider for example an atom modeled as a particle moving in a central potential lt7s excited states are degenerate because the unperturbed Hamiltonian H0 is rotationally invariant In particular since H0 commutes with E it is easy to see that all states differing only by their m values must have the same energy In detail if ln l mgt is an eigenvector of H0 then so is Liln l Thus all such states will have the same energy Suppose this atom is put in a uniform electric field Stark effect so that the perturbation is VeEX This potential breaks the rotational symmetry to just that about E so that the degen eracy is lifted States constructed as outlined above will have differing energies But now there is a subtlety in the perturbative account of this phenomenon In the unperturbed system any basis for the degenerate subspace could be used to define the unperturbed states But in the perturbed theory the energy eigenvectors that are no longer degenerate cannot be linearly superimposed to get eigenvectors This means that in fact the perturbation theory must select a preferred basis of unperturbed eigenvectors namely the ones that the correct eigenvectors collapse to as A a 0 We will see this happening in what follows The problem is now that in perturbation theory one begins with the unperturbed states But which states do we choose The results of degenerate perturbation theory take care of this as you will see 1 Physics 6210Spring 2007Lecture 30 Lecture 30 Relevant sections in te1t39 39 51 Bell s theorem cont Assuming suitable hidden variables coupled with an assumption of locality to determine the spin observables with certainty we found that correlation functions must satisfy lltA 1B 2gt ltA 1B 3gtl S 1 ltA 2B 3gt We now show that quantum mechanics is not compatible with this inequality We compute the expectation value of the product of the two observers7 measurements in units of h2 using quantum mechanics 2 ltAltmgtBlt 2gtgt lt51sggt nial 31m an Again this quantity is the correlation function of the two spins With 2 chosen along n1 this quantity is easily computed exercise WW1 1 2 2l gtlt l i lt W2 2l gt l gtgt 722cos6 4 where 6 is the angle between in and n2 To get the last equality we assume that m and m are in the x z plane with 2 along and in Using 6 to denote the angle between m and m we have ltl7lt7D 2 2ltl7gtl7gtgt lt l 7l lt7DiC056522Sin652zll gtl gt 7 0089 Of course the result is geometric and does not depend upon the choice of coordinates Thus defining 62 nz nj Bell7s inequality 7 if it applied in quantum mechanics 7 would imply lcos613 7 cos612l S 17 cos 623 which is not 13111le Thus quantum mechanics is not consistent with all observables having local definite values based upon some unknown hidden variables On the other hand if reality is such that all observables for the individual particles are compatible and locally defined with QM just giving an incomplete statistical description then this To see this just let n1 point along y let n3 point along it and let n2 lie at 45 from it or y in the x y plane so that 612 7r4 623 613 7r2 l Physics 6210Spring 2007Lecture 30 inequality should be valid experimentally speaking Assuming of course that the correct description can be obtained using some hidden variables A as described above Experiments to check the Bell inequality have been performed since the 19607s Many regard the Aspect experiment of the early 19807s as definitive It clearly showed that the Bell inequality was violated while being consistent with quantum mechanical predictions The experiment actually used spin l particles photons arising from atomic transitions rather than spin 12 systems but the ideas are the same as described above Approximation methods We now begin studying various approximation methods in quantum mechanics Ap proximation methods have a practical and a conceptual value On the practical side we use such methods to get useful approximations to wave functions energies spectra as well as transition probabilities and other dynamical quantities On the conceptual side we shall see that some of our most cherished ways of thinking about energy levels and dynamics stem principally from the point of view of approximation methods The need for approximation methods arises from the simple fact that almost all realistic physical systems one wants to study are too complicated for explicit analytic solutions to be available This is no less true in classical mechanics So for example while we can analytically handle the hydrogen atom when modeled as a charged particle in a Coulomb field we cannot handle helium or more complicated atoms in the same fashion 7 let alone dynamical processes involving the interaction of these atoms with electromagnetic fields In fact even more realistic models of the hydrogen atom including things like spin orbit coupling hyperfine interaction finite size of nucleus etc are not exactly soluble Thus the only way we can understand these systems is to find methods of approximation We shall study two of several possible approximation techniques First we shall look at what is usually called time independent perturbation theory TlPT which gives ap proximate solutions to eigenvalue problems But this is also called stationary state pertur bation theory since one is usually studying the eigenvalue problem for the Hamiltonian Then we shall study time dependent perturbation theory TDPT which is designed to give approximate solutions to the Schrodinger equation For the most part I am going to explain the results of the theory with essentially no derivations Then we will look at some important applications Time independent perturbation theory This approximation method is designed to approximate the eigenvalues and eigen vectors of a given observable This observable is usually the Hamiltonian whence the 2 Physics 6210Spring 2007Lecture 30 alternate name stationary state perturbation theory but the techniques and results are not restricted to just the Hamiltonian any observable will do The basic idea is that one is attempting to view a given observable of interest as in some sense close to a simpler well understood observable One then approximates the eigenvalues and eigenvectors of the given observable in terms of those of the simpler ob servable For example one could be interested in the energies of an anharmonic oscillator with Hamiltonian P2 1 2 2 4 Hi7mw X X 2m 2 Assuming that the anharmonicity described by Q is suitably usmall one can usefully approximate the eigenvalues and eigenvectors of H in terms of those of the harmonic oscillator Let us now make this more precise We suppose that the observable of interest H admits eigenvalues and can be expressed as HK where V is a small perturbation of H0 69 its matrix elements in the basis of eigenvec tors of H0 are small compared to the eigenvalues of H0 For simplicity we assume that H0 has discrete spectrum We assume that all the operators in question are su iciently well behaved such that the eigenvectors and eigenvalues of H can be obtained via a l parameter family of operators beginning with H0 HA H0 AV H0 H0 H1 H For each value of A we have HlEgt ElEgt The idea is that if the effect of V is small compared to that of H0 then one should be able to approximate the eigenvectors and eigenvalues of H by expanding them in a power series in A keeping just the first term or so and then evaluating at A 1 This is equivalent to approximating the eigenvectors and eigenvalues using an expansion in powers of the matrix elements of the potential which can be taken to be AV Thus we assume that EMA E730 AES HE Eno Engtlt0gt MEngtlt1gt mm The plan is to solve the equation H0AVlEngt0 MEngtlt1gt mm gt E720 A321 A2322 lEngt0 AlEngt1 AZlEngt2 The parameter A is not essential it is just a convenient means of bookkeeping as you will see Physics 6210Spring 2007Lecture 30 order by order in A to derive the corresponding perturbative approximations to the eigen vectors and eigenvalues Let us note that eigenvectors are only determined up to multipli cation by a scalar This is of course what allows us to normalize them and view them as state vectors Therefore when we solve the eigenvalue problem perturbatively we will still need to normalize the result At zeroth order it is easy to see by inspection that the relevant equation is 0 Hm t E h t which just says that the zeroth vector and scalar are the unperturbed eigenvector and eigenvalue This should come as no surprise The higher order equations constitute a tri angular77 system of linear inhomogeneous equations Each set of equations depends only upon the results of the lower order equations So one can successively solve these equa tions from lower to higher order In the end any perturbative correction can be expressed completely in terms of the unperturbed eigenvalues and eigenvectors Of course the idea is that for a su iciently small perturbation a good approximation can be obtained by just sticking to relatively low orders The details of the solution process which constitutes a very nice application of linear algebra techniques can be found in your text Here I will just state the results for first order perturbation theory 0Q corrections and show you how to use them As it turns out the solutions of the rst order equations are rather simple when the unperturbed A 0 eigenvalue is non degenerate We begin with that case If Ego is non degenerate then we have E721 Vnn E 0ltEanlEngt07 V jEngt1 Z l gtlt0gt7 kin EhO E120 where Vim 0ltEhlVlEngt039 The rst order approximation to the energy eigenvalue is then 0 En m E72 Vm The un normalized eigenvector is to first order Engt Engtlt0gt 2 3 lam Vi EEo kin n k 4 Physics 6210Spring 2007Lecture 30 Thus the correct eigenvector is a superposition of the old eigenvectors which form a basis If we wish to approximate the state of the system using this vector we must normalize it It is a straightforward exercise to see that to rst order in perturbation theory lETlgtn0Tmalized V anEngt7 where Z 7 2 lenl2 he E720 13713 You can easily see that Zn can be interpreted as the probability for getting E780 when measuring H0 in the state where H is known to have the value En with certainty Put differently we can say that Zn is the probability for finding the state lEngt0 when the system is in the state In general assuming that at least one of the Vlm 31 0 we have that Zn lt l Physics 6210Spring 2007Lecture 19 Lecture 19 Relevant sections in tezrt 26 Charged particle in an electromagnetic eld We now turn to another extremely important example of quantum dynamics Let us describe a non relativistic particle with mass m and electric charge 1 moving in a given electromagnetic eld This system has obvious physical significance We use the same position and momentum operators in the Schrodinger picture X and 13 although there is a subtlety concerning the meaning of momentum to be mentioned later To describe the electromagnetic field we need to use the electromagnetic scalar and vector potentials ftTft They are related to the familiar electric and magnetic elds EB by 18 23 The dynamics of a particle with mass m and charge 1 is determined by the Hamiltonian Eiv i VgtltT 1a 2 H p 7 imam q Xtgt This Hamiltonian takes the same form as the classical expression in Hamiltonian mechanics We can see that this is a reasonable form for H by computing the Heisenberg equations of motion and seeing that they are equivalent to the Lorentz force law which we shall now demonstrate For simplicity we assume that the potentials are time independent so that the Heisen berg and Schrodinger picture Hamiltonians are the same taking the form H e 1 13 gm m 7 2m 0 q For the positions we get exercise d in e 1 m H e 1 1 mm dt T m 7 T m e We see that just as in classical mechanics the momentum 7 de ned as the generator of translations 7 is not necessarily given by the mass times the velocity but rather a di q a 130 m7 21020 c As in classical mechanics we sometimes call P the canonzeal momentum to distinguish it from the mechanical momentum m 13 714TXt Physics 6210Spring 2007Lecture 19 Note that the mechanical momentum has a direct physical meaning while the canonical momentum depends upon the non unique form of the potentials We will discuss this in detail soon While the components of the canonical momenta are compatible Pia 07 the mechanical momenta are not I qui Ming k HZHj i mja 7 8162 i zhEEZjkB Thus in the presence of a magnetic field the mechanical momenta obey an uncertainty relationl This is a surprising non trivial and quite robust prediction of quantum mechan ics In particular if the field is uniform then two components of mechanical momentum will obey a state independent uncertainty relation rather like ordinary position and mo mentum Can this prediction be veri ed As you will see in your homework problems this incompatibility of the mechanical momentum components in the presence of a magnetic field is responsible for the Landau levels77 for the energy of a charged particle in a uniform magnetic field These levels are well known in condensed matter physics The remaining set of Heisenberg equations are most simply expressed using the me chanical momentum Starting with n2 a H7 X 2mq using the commutation relations between components of the mechanical momentum above and using X2 Hj ih gI we have exercise c imam q 130300 i es x Bow 7 Ma x 1120 Except for the possible non commutativity of ll and g this is the usual Lorentz force law for the operator observables The Schrodinger equation Dynamics in the Schrodinger picture is controlled by the Schrodinger equation If we compute it for position wave functions then we get exercise h a 2 0 i N e 9m mt q ltatgtwltmgt mama 1 C 2 Physics 6210Spring 2007Lecture 19 The left hand side represents the action of the Hamiltonian as a linear operator on position wave functions We have in detail 1 E h a a 2 magi Avw ltVAgtwl HQ A2q lw H LJ if 2m me As you may know one can always arrange by making a gauge transformation if necessary to use a vector potential that satisfies the Coulomb gauge V A 0 In this case the Hamiltonian on position wave functions takes the form 722 7 9 c a 2 Hw vgwe Avwm A2q l Some typical electromagnetic potentials that are considered are the following i The Coulomb field with k E7 which features in a simple model of the hydrogen atom the spectrum and stationary states 45 1107 should be familiar to you We will soon study it a bit in the context of angular momentum issues ii A uniform magnetic field E where a 1 a b 0 A EB gtlt at The vector potential is not unique of course This potential is in the Coulomb gauge You will explore this system in your homework The results for the stationary states are inter esting One has a continuous spectrum coming from the motion along the magnetic field but for a given momentum value there is a discrete spectrum of Landau levels77 coming from motion in the plane orthogonal to To see this one massages the Hamiltonian into the mathematical form of a free particle in one dimension added to a harmonic oscillator this is the gist of your homework problem iii An electromagnetic plane wave in which 0 ffTocoslfikct7 1910 0 Of course this latter example involves a time dependent potential This potential is used to study the very important issue of interaction of atoms with a radiation field maybe we will have time to study this toward the end of the semester 3 Physics 6210Spring 2007Lecture 19 Gauge transformations There is a subtle issue lurking behind the scenes of our model of a charged particle in a prescribed EM field It has to do with the explicit appearance of the potentials in the operators representing various observables For example the Hamiltonian 7 which should represent the energy of the particle 7 depends quite strongly on the form of the potentials The issue is that there is a lot of mathematical ambiguity in the form of the potentials and hence operators like the Hamiltonian are not uniquely defined Let me spell out the source of this ambiguity You may recall from your studies of electrodynamics that if axT define a given EM field EH then the potentials 6147 given by 4577715 4KVf define the same E for my choice of f ftf Because all the physics in classical electrodynamics is determined by E and H we declare that all potentials related by such gauge transformations are physically equivalent in the classical setting In the quantum setting we must likewise insist that this gauge ambiguity of the potentials does not affect physically measurable quantities Both the Hamiltonian and the mechanical momentum are represented by operators which change their mathematical form when gauge equivalent potentials are used The issue is how to guarantee the physical predictions are nonetheless gauge invariant Let us focus on the Hamiltonian for the moment The eigenvalues of H define the allowed energies the expansion of a state vector in the eigenvectors of H defines the probability distribution for energy and the Hamiltonian defines the time evolution of the system The question arises whether or not these physical aspects of the Hamiltonian operator are in fact in uenced by a gauge transformation of the potentials If so this would be a Very Bad Thing Fortunately as we shall now show our model for a particle in an EM field can be completed so that the physical output of quantum mechanics spectra probabilities are unaffected by gauge transformations For simplicity only we still assume that H is time independent and we only consider gauge transformations for which 0 The key observation is the following Consider two charged particle Hamiltonians H and H differing only by a gauge transformation of the potentials so that they should be physically equivalent Our notation is that if H is defined by ab then H is defined by the gauge transformed potentials 4 gt I K WW It is now straightforward to verify see below that if satisfies HlEgt E lEgt7 4 Physics 6210Spring 2007Lecture 19 then I a W efltXgtEgt satis es HlEgt ElEgt Note that the eigenvalue is the same in each case The operator 6fO is unitary and this implies the spectra of H and H are identical Thus one can say that the spectrum of the Hamiltonian is unaffected by a gauge transformation that is the spectrum is gauge m varz ant Thus one can use Whatever potentials one Wishes to compute the energy spectrum and the prediction is always the same To be continued Physics 6210Spring 2007Lecture 25 Lecture 25 Relevant sections in tezrt 36 37 Position representation of angular momentum operators We have seen that the position operators act on position wave functions by multipli cation and the momentum operators act by differentiation We can combine these two results and using spherical polar coordinates 7 6 b get a useful position wave function representation for the angular momentum operators We have mow ltzgtgt i sin 1559 e cowcoswt we as Lywr 6 b 7 cos e589 7 cot Qsin 215 4M7 6 b h 121103 67 67 You can see that L2 is particularly simple 7 it clearly generates translations in b which are rotations about the z axis of course The other two components of E also generate rotations about their respective axes They do not take such a simple form because spherical polar coordinates give the z axis special treatment Combining these results we have in addition L2wltnen gt eh 1 1 2 78 78 39 68 6 sin 6 5 M 9ltsm 9 w gtgt You may recognize that this last result is up to a factor of ihgrg the angular part of the Laplacian This result arises from the identity see text L2 r2122 7 X 13 13 where r2 X2 Y2 Z2 so that exercise 1 E27 P2wltr6 gtgt ehgvgwww eh2lt L2 83 gaww gt Thus we get in operator form the familiar decomposition of kinetic energy into a radial part and an angular part Orbital angular momentum eigenvalues and eigenfunctions spherical harmon ics A good way to see what is the physical content of the orbital angular momentum eigenvectors is to study the position probability distributions in these states Thus we consider the position wave functions 39Wm ltf lzmgt 1 96 Physics 6210Spring 2007Lecture 25 corresponding to orbital angular momentum eigenvectors These are simultaneous eigen functions of L2 and L2 so they satisfy 2 szlml mlhwlmlz L 11sz where 7 on general grounds 7 lltl1gth2wzm 0l1 2 and ml7l7l1l71l We note that the angular momentum eigenfunctions will always involve an arbitrary multiplicative function of the radius 7 This is because the angular momentum differential operators only take derivatives in the angular directions What this means physically is that the states of definite angular momentum will always have a degeneracy This should not surprise you just specifying the angular momentum of a state of a particle is not expected to completely determine the particles state We now argue that the half integer possibility does not occur for orbital angular mo mentum First we note that it is quite easy to find L2 eigenfunctions in spherical polar coordinates since L2 7ih8 Evidently mm 7 mm 6W lmmediately we see that ml can only be an integer 7 otherwise 1mm will not be a continuous function Now discontinuous wave functions are not inherently evil Indeed there are plenty of discontinuous functions in the Hilbert space But these functions will fail to be differentiable and hence will not in the domain of the momentum and angular momentum operators This is a contradiction since of course we are trying to construct the angular momentum eigenfunctions which by definition are in the domain of the operators Thus we can conclude immediately that for orbital angular momentum we can only have at most 10l2 and ml7l7lll7ll As we shall see all of the indicated values do in fact arise By the way in your homework assignment you will find a problem which gives an alternative argument for this result Orbital angular momentum eigenvalues and eigenfunctions spherical harmon ics Having solved the L2 equation we now must solve the L2 equation which is an ordinary For example consider the classical motion of a particle with vanishing angular momentum The motion is such that position and momentum vectors are parallel but are otherwise arbitrary Physics 6210Spring 2007Lecture 25 differential equation for flmlO 6 1L2 7 ml 10 3990 ill flml m m 65111 6 flml 1flml The solutions of this equation are the associated Legendre polynomials 131mlcos 6 and the angular momentum eigenfunctions are thus of the form lmlrz 67 flml7nYlmlgz 7 where the Y17m6 b are the spherical harmonics and the functions flmlO are the integra tion constants77 for the solution to the purely angular differential equations See your text for detailed formulas for the spherical harmonics Note that all non negative integer values are allowed for l As discussed earlier the functions flmlO are not determined by the angular momentum eigenvalue problem Typically these functions are xed by requiring the wave function to be also an eigenfunction of another observable which commutes with L2 and L2 69 the energy in a central force problem In any case we will assume that 00 2 2 0 drrmmml 1 This way with the conventional normalization of the spherical harmonics 7r 2 d6 d gt sin2 lmlt6 ltzgtgtYz mlw ltzgtgt Six5mm 0 0 l l we have that 00 7r 27f ltl7m l7mlgt 0 dr 7020 d6 0 d 78in2 Qi l ymg y 9 1lemlr 9 5w5mlm2 For a state of definite angular momentum ll mlgt we see that the angular dependence of the probability distribution is completely determined by the spherical harmonics The radial dependence of the probability distribution is not determined by the value of angular momentum unless other requirements are made upon the state Addition of angular momentum Two spin 12 systems We now will have a look at a rather important and intricate part of angular momentum theory involving the combination of two or more angular momenta We will primarily focus on the problem of making a quantum model for a system consisting of two distin guishable spin 12 particles ignoring all but their spin degrees of freedom Again the idea is simply to combine two copies of our existing model of a spin 12 system The tech nology we shall need has already been introduced in our discussion of the direct product 3 Physics 6210Spring 2007Lecture 25 construction We shall take this opportunity to review the construction in the context of the problem of combining 7 or adding 7 two spin 12 angular momenta For a system of two spin 12 particles 69 an electron and a positron we can imagine measuring the component of spin for each particle along a given axis say the z axis Obviously there are 4 possible outcomes exercise Having made these measurements we can denote the states in which these spin values are known with certainty by 152 gt 152 gt 152 gt 152 gt 152 7gt 152 gt 152 7gt 152 gt Here the first factor of the pair always refers to particle 177 and the second factor refers to particle 2 We view these vectors as an orthonormal basis for the direct product Hilbert space of states of 2 spin 12 particles We thus consider the 4 d Hilbert space of formal linear combinations of these 4 basis vectors An arbitrary vector W1gt is given by lwgt alSagt lSagtaJSmgt Sz 7gta7l527 igt l527gta77l5277gt l527 7gt Here the scalar multiplication is assigned to the pair as a whole but by definition it can be assigned to either of the factors in the pair as well If you wish you can view the scalars Iii as forming a column vector with 4 rows the squares of these scalars give the various probabilities for the outcome of the 52 measurement for each particle Other bases are possible corresponding to other experimental arrangements 69 SI for particle 1 and Sy for particle 2 Physics 6210Spring 2007Lecture 35 Lecture 35 Relevant sections in tezrt 56 Fermi s Golden Rule First order perturbation theory gives the following expression for the transition prob ability 2 Pi a ni 31 11 4 Vm sin2 T E e E02 2h We have seen that the energy conserving transitions if there are any available become dominant after a suf ciently large time interval Indeed the probabilties for energy non conserving transitions are bounded in time while the probability for energy conserving transitions grow quadratically with time for as long as the approximation is valid Here large time means that the elapsed time is much larger than the period of oscillation of the transition probability for energy non conserving transitions 27th T 2 En Note that the typical energy scale for atomic structure is on the order of electron volts 0715 This translates into a typical time scale T N 1 s so large times is often a very good approximation In the foregoing we have been tacitly assuming the final state is an energy eigenstate coming from the discrete part of the energy spectrum If the final state energies lie in a continuum at least approximately we get a qualitatively similar picture but the details change We shall see that the transition probability at large times still favors energy conserving transitions but it will only grow linearly with time because the width of the probability distribution about such transitions is becoming narrower with time We can see this by supposing the probability is a continuous function of AE En 7 EZ see discussion below and considering the large 15 limit via the identities hm sin2x a oo 0a mm 5m am Then we have exercise 27139 2 P L a 111 31 11 w 6AEt tgtgt T This implies that in this case quasi continuum of final states the transition rate g is constant at late times d 2 Em a m 7e 11 gim mm tgtgt T 1 Physics 6210Spring 2007Lecture 35 a result which is one version of Fermi s Golden Rule An example of a continuum of final states appears in the Auger ionization of a Helium atom Here two electrons in the 28 unperturbed stationary state make a transition thanks to their electrostatic repulsion to a state in which one electron is in the 18 state and the other is ejected Another example would be elastic scattering of an electron off of an ion In such situations there is a continuum or near continuum of nearly degenerate states that can serve as the state denoted by In this case one will have a density of states77 factor The quantity pEdE is the number of states with energies between E and E dE The transition rate from the initial state into states with energy E E S is then d 27139 27139 am a s dEnpltEngtflvnn26ltEtEngt fMEDle lz 8 This is probably the most commonly used version of Fermi7s Golden Rule We will consider an example of this a bit later Harmonic Perturbations We now consider perturbations which depend upon time Usually we can Fourier ana lyze the time dependence and view the perturbation as a superposition of many perturba tions that each depend harmonically on the time Therefore the next thing to consider are such harmonically varying perturbations As a good example which we shall consider later consider the effect of an electromagnetic wave incident upon an atomic electron If the wave is nearly monochromatic then one can model the time dependence of the associated perturbation as harmonic Suppose then that the perturbation is of the form Veiwt vigilwt where V is some time independent operator and w is a given frequency Note that there are now two time scales in the problem that associated with the perturbation and that naturally defined by the difference in unperturbed energies We again suppose that at the initial time t 0 the system is in an eigenstate of H0 We ask what is the probability for finding the system in the state at time t n 31 Using our previous results we have that exercise 1 En Ez h h The integral is the sum of two terms where each is of the form encountered in the time t 2 7g n 0 dt vniezwt Vlieilwtelwnit 7 wni independent perturbation case except that now the frequency in the exponents is shifted via tan a tum i w 2 Physics 6210Spring 2007Lecture 35 This allows us to use our previous analysis for the late time transition rates provided the frequencies are modi ed as shown above in order to handle the current situation In detail after taking the absolute square in the above formula for the probability we have 3 terms two direct terms and a cross term Each has a vanishing denominator on resonance that is when the initial and final states are such that tom i to 0 As before the limit as the denominator vanishes for each of these terms is finite and grows with time Thus as before the principal transitions at late times are such that one of these frequency combinations vanishes For a given initial and final state only one of these two options i can occur The direct terms have the square of the growing probability while the cross term only involves this quantity linearly The cross term can therefore be ignored compared to the direct terms In summary at large enough times the only transitions with appreciable probability will be those for which either tom to 0 or tom 7 to 0 Of course only one of these two situations can occur for a given choice of frequency to and for a given initial and final state In the first case we have tomw0ltgtEnEZihw and in the second case we have tom7w0ltgtEnEZhw So using the unperturbed energy observable to interpret these results we have the follow ing situation When En E 7 Tito we speak of stimulated emission since the perturbation has from the unperturbed point of view caused the system to emit a quantum of energy hw When En E Tito we speak of stimulated absorption since the perturbation has from the unperturbed point of view caused the system to absorb a quantum of energy RM To get a formula for the late time transition rates we simply have to make the appro priate shift tan a tom i to in our previous formulas In the case of a quasi continuum of final states we get d 2 W a y n E muWm 7E m lVLl26En 7E 7 7w 3 Physics 6210Spring 2007Lecture 35 This is yet another version of Fermi7s Golden Rule Note that since 2 2 Mil vzivm vjnwjnr will the transition rate per unit energy for i a n is the same as that for n a i a phenomenon often called detailed balancing Physics 6210Spring 2007Lecture 20 Lecture 20 Relevant sections in tezrt 26 3 Gauge transformations cont Our proof that the spectrum of the Hamiltonian does not change when the potentials are rede ned by a gauge transformation also indicates how we are to use our model so that all probabilities are unaffected by gauge transformations We decree that if W1 is the state vector of a particle in an EM field described by the potentials a5 11 then W fomlw is the state vector of the particle when using the gauge transformed potentials 6147 Note that this is a unitary transformation Let us now see why this prescription works For a particle all observables are functions of the position and momentum operators Here momentum means either canonical or mechanical The position observable is represented in the Schrodinger picture by the usual operator X no matter the gauge Any observable function G of the position has an expectation value which does not change under a gauge transformation MGme ltwle fXGltXefXl1Jgt ltleXgtl11gt The momentum operator is where things get more interesting The mechanical momentum is a gauge invariant observable But it is represented by an operator which changes under a gauge transformation Indeed we have n 57 gl 1 gig1 W However it is straightforward to check that exercise CfXWgt 5f gt Put differently we have the operator representing the mechanical momentum 7 which is a gauge invariant observable 7 transforming under a gauge transformation as a unitary transformation a a 6fX eifX Any function of the position and mechanical momentum will have a similar transforma tion law In particular the Hamiltonian can be expressed as exercise 1 H 7H gas 2m 1 Physics 6210Spring 2007Lecture 20 so it follows that exercise HefXl1Jgt egfltXgtlegt that is I I H 6fXHeifX The physical output of quantum mechanics is not changed by a unitary transformation of the state vectors and a unitary similarity transformation of the observables This is because the expectation values will not change in this case ltwl0lwgt WOW where Wgt UM 0 UOUT It is now easy to see that if you compute the expectation value of any function of the mechanical momentum you can use the state W1 and operator fl or you can use the vector W and operator lT and get the same answer In this way one says that the physical output of quantum mechanics is suitably gauge invariant Different choices of potentials lead to unitarily equivalent mathematical representations of the same physics It is not hard to generalize all this to time dependent gauge transformationsf ft Here we simply observe that if Wat is a solution to the Schrodinger equation for one set of potentials then exercise wwbwWXWmgt is the solution for potentials obtained by a gauge transformation defined by f Thus one gets gauge invariant results for the probability distributions as functions of time This result also shows that position wave function solutions to the Schrodinger equation transform as MW e ew 2 under a gauge transformation Aharonov Bohm effect The Aharonov Bohm effect involves the effect of a magnetic field on the behavior of a particle even when the particle has vanishing probability for being found where the magnetic field is non vanishing Of course classically the Lorentz force law would never lead to such behavior Nevertheless the AB effect has been seen experimentally You will explore one version of this effect in a homework problem Here let me just show you how technically such a result can occur Physics 6210Spring 2007Lecture 20 The key to the AB effect is to cook up a physical situation where the magnetic field is non vanishing in a region from which the charged particle will be excluded and vanishing in a non simply connected region where the particle is allowed to be Since the magnetic field vanishes in that region we have that VXIT0 In a simply connected contractible region of space such vector fields must be the gradient of a function In this case the potential can be gauge transformed to zero and there will be no physically observable in uence of the magnetic field in this region However if the region is not simply connected it need not be true that A is a gradient 26 pure gauge As an example relevant to your homework we study the following scenario Consider a cylindrical region with uniform magnetic field magnitude B along the axis of the cylinder You an imagine this being set up via an idealized solenoid Outside of the cylinder the magnetic field vanishes but the vector potential outside the cylinder must be non trivial In particular A cannot be the gradient of a function everthere outside the cylinder To see this we have from Stokes theorem Kw u C S where O is a closed contour enclosing the cylinder and S is a surface with boundary 0 so that the right hand side is never zero if the ux of B through S is non zero which it isn7t in our example But if I is a gradient then the left hand side vanishes exercise 7 contradiction In fact the vector potential can be taken to be exercise iBR A R 2 2T 69 rgt where R is the radius of the cylinder 7 gt R is the cylindrical radial coordinate and 69 is a unit vector in the direction of increasing cylindrical angle Since I is necessarily not gauge equivalent to zero it can affect the energy spectrum 7 and it does Angular momentum introductory remarks The theory of angular momentum in quantum mechanics is important in many ways The myriad of results of this theory which follow from a few simple principles are used extensively in applications of quantum mechanics to atomic molecular nuclear and other subatomic systems The mathematical strategies involved have a number of important generalizations to other types of symmetries and conservation laws in quantum mechanics The quantum mechanical theory of angular momentum leads naturally to the concept of intrinsic spin Just as we saw for spin 12 a general feature of angular momentum in 3 Physics 6210Spring 2007Lecture 20 quantum mechanics is the incompatibility of the observables corresponding to any two components of spin The nature of this incompatibility is at the heart of Virtually all features of angular momentum Just as linear momentum is intimately connected With the notion of translation of the physical system so angular momentum is deeply tied to the theory of rotations of the physical system being considered We shall use this geometric interpretation of angular momentum as the starting point for our discussion Physics 6210Spring 2007Lecture 10 Lecture 10 Relevant sections in tezrt 17 Gaussian state Here we consider the important example of a Gaussian state for a particle moving in l d Our treatment is virtually identical to that in the text but this example is suf ciently instructive to give it again here We de ne this state by giving its components in the position basis z39e its wave m my exp 7 You can check as a good excercise that this wave function is normalized function 1 ltwlwgt dx ltwlxgtltxlwgt dx WM Roughly speaking the wave function is oscillatory with wavelength 2 but sitting in a Gaussian envelope centered at the origin The probability density for position is a Gaussian centered at the origin with width determined by d Thus this state represents a particle localized near the origin within a statistical uncertainty specified by d Let us make this more precise and exhibit some properties of this state As an exercise you can check the following results ltXgt WXW mum 0 so that the mean location of the particle is at the origin in this state Next we have ltX2gt WWW 00 d2l xl2 3 Thus the dispersion in position is d2 37 z39e is the standard deviation of the probability distribution for position Next we have hd 77w hk i die 7 00 ltmwmw mwm 00 telling us that on the average this state has the particle moving with momentum his and h 22 2hk 00 d2 ltP2gt Wm 00m menopwx g 1 Physics 6210Spring 2007Lecture 10 so that the momentum uncertainty is 722 AP2 7 lt gt 2d2 Thus the momentum uncertainty varies reciprocally with d relative to the position uncer tainty The product of position and momentum uncertainties is as small as allowed by the uncertainty relation One sometimes calls W1gt a minimum uncertainty state Because the Fourier transform of a Gaussian function is another Gaussian it happens that the momentum probability distribution is a Gaussian 2312 ltpiwgt dx ltplxgtltxiwgt 00 l die 700 V 27th 7 77 ink d Mag 2209 gt You can see that this Gaussian is peaked about the expected momentum value as it aim4x should be and that its width varies like ld ie reciprocal to the position uncertainty as expected In summary the Gaussian state we have defined corresponds to a particle which on the average is moving and has a Gaussian spread of position and momentum values such that the uncertainty product is minimized One can use such states to model macroscopic objects Just put in reasonable values for d and k and you will find that the quantum uncertainties are suf ciently small to be negligible Combining systems in quantum mechanics In our generalization from a particle moving in l d to a particle moving in 3 d we tacitly took advantage of a general quantum mechanical scheme for combining systems to make composite systems In particular we viewed a particle moving in 3 d as essentially 3 copies of a particle moving in l d We do this sort of thing all the time in physics For example if we have a satisfactory kinematic model of a particle and we want to consider a kinematic model for 2 particles we naturally try to use two copies of the model that we used for one particle As another example we have presented a model of a spin 12 system what if we want to describe 2 spin 12 systems Or what if we want to describe a particle with spin 12 All these situations require us to know how to combine quantum mechanical models to make composite models Here we outline the scheme using our generalization from a l d particle to a 3 d particle as illustration Later we will have occasion to use this construction again Physics 6210Spring 2007Lecture 10 The tensor product Suppose we wish to combine two systems described by Hilbert spaces H1 and Hg with observables represented by linear operators 14131 and 14232 which are of course only de ned on their respective Hilbert spaces We can build up a Hilbert space for the combined system as follows We rst consider the set consisting of all ordered pairs of vectors from each Hilbert space Typical elements are denoted l117Xgt1l11gt lgtltgt7 W 6 H1 lXgt 6 Ha These vectors are called product vectors Physically a product vector W1gt lxgt is a state of the composite system in which system 1 is in state W1gt and system 2 is in state Product vectors admit a natural scalar multiplication inherited from that existing on H1 and H2 CW1 ICl gt lXgtEl11gt Clxgt It is possible to de ne a notion of addition on special product vectors We have W W W W I W W W and W W W W W W W Obviously this does not de ne an addition operation on all product vectors For additions that are not of the above special form we simply define W W W W to be a new element of our set We then simply take the set of all formal linear combinations of product vectors to be the set used in the new composite Hilbert space This set is a vector space called the tensor product of H1 and H2 and is denoted by H1 8 H2 Our text uses the terminology direct product77 for this construction A basis for the tensor product can be obtained by taking all product vectors built from a basis for each of H1 and H2 exercise If lcigt are a basis for H1 and are a basis for H2 then every W1gt 6 H1 8 H2 can be written as W Eagle lfjgt ivj Note that not every basis for the tensor product is going to be built from product vectors Note also that the dimension of the product space is the product of the dimensions of the individual Hilbert spaces In other words if m is the dimension of H1 and n2 is the dimension of H2 then H1 8 H2 has dimension 111112 To see this simply note that there 3 Physics 6210Spring 2007Lecture 10 are H1712 scalars in the array of numbers llj in the above expansion of an arbitrary vector in the product basis The tensor product space is to be a Hilbert space The scalar product is defined as follows For product vectors we have ltay lm5gt ltalvgtlt l5gt For general vectors we expand them in a basis of product vectors and then define the scalar product via the usual linearityanti linearity of the scalar product exercise Let us note that since the space of states must be a vector space this forces us to consider as state vectors all possible normalizable linear combinations of the product state vectors Such linear combinations will not be interpretable as saying that each subsystem is in a definite state Rather in such linear combinations each subsystem will have various probabilities to be in various states Moreover there will exist correlations between the probability distributions for each subsystem This is the essence of quantum entanglement it is responsible for very striking physical behavior of composite systems compared to their classical analogs Hopefully we will have a little time to discuss this further at the end of the semester As an example consider a system consisting of 2 spin 12 systems eg an electron and a positron with all degrees of freedom except for spin ignored A product basis for the Hilbert space of states could be the four combinations of li Where the first vector denotes SZ eigenvectors of particle l and the second denotes SZ eigenvectors of particle 2 These vectors represent states in which 52 for each particle is known with certainty A vector such as 7 7 represents a state in which neither particle has statistically certain values for 52 W lgt lgtlgt lgt You can easily check that all the constructions above can be implemented in an identical fashion for the space of bras In particular we have ltal n ltmgt W As an example let us consider a particle moving in 2 d as built from a product of 2 particles labeled by x and y moving in l d A basis for the particle moving in x is and a basis for the particle moving in y is The corresponding product basis is then lac A general vector in the product space can be written as W d296l96gt lygtltl ltylWgt 4 Physics 6210Spring 2007Lecture 10 We de ne Way lt94 ltylW1gt You see that vectors are characterized by their position wave functions Which are complex Valued functions of two variables These wave functions must be square integrable as you can see by setting w dxdywmm w W dxdywmxxl ltyt and computing exercise ltwlwgt dgx um Of course With unit normalization one can intepret W1y 2 to be the probability density for nding position at any Physics 6210Spring 2007Lecture 15 Lecture 15 Relevant sections in tezrt 22 CompatibleIncompatible observables in the Heisenberg picture Let us also note that the notion of compatibleincompatible observables is left undis turbed by the transition to the Heisenberg picture This is because the commutator trans forms as exercise mmmmwMumBmmm w Note any two operators related by a unitarysimilarity transformation will have this commutator property Thus if A and B are incompatible in the Schrodinger picture they will be incompatible at each time in the Heisenberg picture Note also that the commutator of two observables in the Schrodinger picture 7 which is i times another observable 7 makes the transition to the Heisenberg picture just as any other Schrodinger observable namely via the unitary transformation A UMu Unitary transformations in general Let us note that while our discussion was phrased in the context of time evolution the same logic can be applied to any unitary transformation For example for a particle moving in one dimension one can view the effect of translations as either redefining the state vector leaving the operator observables unchanged W wyA A or equivalently as redefining the observables with the state vectors unchanged A a TJATa Welt1t Note in particular that the position and momentum operators change in the expected way under a translation exercise 7 you played with this stuff in the homework X m pnp Heisenberg equations We saw that the conventional Schrodinger equation is really just a consequence of the relation between the time evolution operator and its infinitesimal generator in the context of the Schrodinger picture mgwmt wwmt ihwlttgtgtHlttgtwlttgtgt 1 Physics 6210Spring 2007Lecture 15 Given a Hamiltonian this equation is the starting point for investigating quantum dynam ics in the Schrodinger picture We can now ask what is the analog of this in the Heisenberg picture In the Heisenberg picture dynamical evolution occurs through the operator observables which for simplicity we assume to be time independent in the Schrodinger picture We have At Ultt0AUtt0 Differentiating both sides and using our basic differential equation for Utt0 we get At 7Ultt0HtAUtt0 Ultt0AHtUtt0 7Ul05t0HtUtt0Ultt0AUtt0 UWEt0AUtt0Ultt0HtUtt0 l iAlttgtHHlttgti Here we have introduced the Heisenberg picture version of the Hamiltonian Hmc Ultt0HtUtt0 If the Schrodinger Hamiltonian is time independent as is often the case then we have exercise 8H HHmOf Ht H 7 0 at This is pretty important to keep in mind The equation d 1 HA 5 i dt lt ih is the Heisenberg equation of motion for the Heisenberg operator At Given a Hamiltonian iAt7 HHeistl it is a differential equation that in principle can be solved to find the Heisenberg operator corresponding to an observable at time t given initial conditions 1050 A Given At one gets the time dependence of probability distributions in the usual way The outcome of a measurement of the observable represented at time t by At is one of the eigenvalues 12 The probability for getting 1 assuming no degeneracy at time t is Pai7t lltai7tlwzt0gtlg lltaizt0 Ut7t0lwzt0gtl27 where lai t is the eigenvector of At with eigenvalue 1 and l11t0 is the state vector for the system Physics 6210Spring 2007Lecture 15 The Heisenberg equation can make certain results from the Schrodinger picture quite transparent For example just by taking expectation values on both sides of the equation using the single time independent state vector it is apparent that exercise d 1 3AM mltAHigt Note that here the notation 14 H means the observable corresponding to the commutator which is well de ned in either picture Similarly it is easy to see that operators that commute with the Hamiltonian assumed time independent for simplicity so that HHeis H at one time to are constants of the motion d 1 if i dt O ih Thus conserved quantities satisfy AtH UlAt0HU 0 At At0 A You can also see this result directly If an observable commutes with H at one time t to say then it will not change in time since Ac UiltmogtAlttogtUltmogt UiltmogtUltmogtAlttogt Ace Evidently a conserved quantity has a time independent probability distribution since the operator in the Heisenberg picture does not change in time Functions of observables We have seen that the time evolution operator defines via a similarity transformation the Heisenberg operators at time t At UlAU If we have an observable that is a function of A FA say we have of course FAt UlFAU It is important to note that one can also express this as FAt FAt E FUlAU To see this we use the spectral decomposition FM ZFWWWL a 3 Physics 6210Spring 2007Lecture 15 so that UTFltAgtU Z FltagtUUagtltaw Z Fltagtwalttgtgtltalttgtw FltAlttgtgt Heisenberg Picture Dynamics of a Particle in a potential For a particle with position at one time say If 0 X and momentum P these are spatial vectors made of operators 7 we are going back to our original notation we consider a Hamiltonian of the form P2 H Note that this operator does not depend upon time so it is both the Schrodinger and Heisenberg Hamiltonian In particular we have that E Ht 110 H 2 VXt 2m This result can be viewed as a mathematical version of the conservation of energy in the Heisenberg picture Let us compute the Heisenberg equations for Xt and momentum Pt Evidently to do this we will need the commutators of the position and momentum with the Hamiltonian To begin let us consider the canonical commutation relations CCR at a fixed time in the Heisenberg picture Using the general identity AlttgtBlttgt UmtogtAlttogtBlttogtUltttogt we get exercise Xit7Xjtl 0 MWPADL Xit7Pjtl M5 In other words the Heisenberg position and momentum operators obey the OCR at any fixed time Note that this means the uncertainty relation between position and momentum is time independent 7 a fact you can also prove in the Schrodinger picture exercise It is now straightforward to compute ih XilttgtH Em so that dX Z t 1 dt HP 0 Thus one relates the momentum and velocity of the particle a result that is a bit more tricky to establish in the Schrodinger picture To compute the Heisenberg equations for the momentum we need to compute exercise Ii0711 Plt VXtl 4 Physics 6210Spring 2007Lecture 15 Probably the simplest way to do this is to use Iat VltXlttgtgtl Um one vltxgtiUlttogt 7mm mgmw enigma Here we have used 8V PVltXgti wagon which can be veri ed by checking it on the position eigenvector basis using the de nition of Pl as the generator of in nitesimal translations good exercise All together we get dPZt 8V dt egocc Using the Heisenberg equations for X05 we can write the Heisenberg equation for P05 as d2Xit 239 dtg F ltxlttgtgt where Filtxlttgtgt igm can be viewed as the quantum representation of the force at time t in the Heisenberg picture This is a quantum version of Newton7s second law From this result it is tempting to believe that a quantum particle is actually behaving just like a classical particle After all the basic observables obey the same equations of motion in the two theories Of course this is not true if only because it is not possible to know both position and momentum with statistical certainty in the quantum theory In the next section we will take a closer look at this issue As a elementary example let us consider a free particle in l d V 0 The Heisenberg equations are dXt Pt dPt 777 707 dt m dt Here X0 X and P0 P are the operators we discussed previously in the Schrodinger picture The momentum is evidently a constant of the motion the momentum probability distribution for a free particle is time independent The position probability distribution changes in time 69 Physics 6210Spring 2007Lecture 3 Lecture 5 Relevant sections in tezrt 12 13 Spin states We now model the states of the spin 12 particle As before we denote a state of the particle in which the component of the spin vector S along the unit vector n is ih2 by 5 n igt We de ne our Hilbert space of states as follows We postulate that H is spanned by 5 nigt for any choice of n with different choices of n just giving different bases Thus every vector W1gt E H can be expanded via l gtalsn7gtailSnrgt lt1 We define the scalar product on H by postulating that each set 5 nigt forms an 07 thonormal basis ltSnilSnigt1 ltSnlSnigt 0 Since every vector can be expanded in terms of this basis this defines the scalar product of any two vectors exercise Note that the expansion coef cients in 1 can be computed by 1i ltS nil11gt This is just an instance of the general result for the expansion of a vector W1gt in an orthonormal ON basis i 12 n where the ON property takes the form ltiljgt 5ij We have exercise lwgt 2 cm a MW i We can therefore write w Z ligtltilwgt i If we choose one of the spin bases say 5 n igt and we represent components of vectors as colurnns then the basis has components ltSnilSngt ltSnilSnigt More generally a vector with expansion w alsngt a4sn gt l Physics 6210Spring 2007Lecture 3 is represented by the column vector ltSnilwgt if The bra lt l corresponding to W1gt has components forming a row vector Msme a at Linear operators Our next step in building a model of a spin 12 system using the rules of quantum mechanics is to represent the observables SISySZ by self adjoint operators on the foregoing two dimensional vector space To do this we need to explain how to work with linear operators in our bra ket notation A linear operator A is a linear mapping from H to itself that is it associates to each vector W1gt a vector AliO This association mapping77 operation is to be linear Aalagt bl aAlagt bAl gt If you think of vectors as columns then a linear operator is represented as a square matrix I will explain this in detail momentarily As you know if you take a row vector and multiply it on the left with a column vector of the same size you will get a square matrix that is a linear operator More generally given a ket lozgt and a bra lt l we can define a linear operator via A lagt lt l What this means is AW WWW You can easily check as an exercise that this is a linear operator This operator is called the outer product77 or tensor product77 operator You can easily see that the sum of two linear operators defined by A BMW Ali1H Eli1 is a linear operator as is the scalar multiple MWO GAR1 Thus the set of linear operators forms a vector space Moreover you can check that the product of two operators defined by ABMW ABW 2 Physics 6210Spring 2007Lecture 3 is a linear operator Thus the set of linear operators forms an algebra It is not hard to see that every linear operator can be written in terms of an orthonormal basis ONE as A ZAijligtltjlz i where Aij MAW are called the matrz39a elements of A in the basis provided by To see this simply expand the vectors W1gt and Al gt in the ONE Aw Z ligtltilAW1gt Z ligtltilAljgtltjl gt ZAMW i ij ij A very important example of this is the identity operator defined by 1W W7 VW lt has the decomposition good exercise 1 2w lt2 This resolution of the identity77 is used all the time to manipulate various equations Don7t forget it As a simple example you can use 2 to view the expansion in a basis formula as a pretty trivial identity w 1w Z w w Z ligtltil gt I The array Aij is in fact the matrix representation of the linear operator A in the given basis To see how this works we consider the action of a linear operator on a vector and see the familiar rules of matrix multiplication coming into play when we expand in an orthonormal basis Watch ltilAl gtltilZAjkljgtltkl11gt jk ZAjkltiljgtltkWgt jk ZAikltkl gt k More generally any scalar of the form ltozlAl gt is called a matrix element 3 Physics 6210Spring 2007Lecture 3 The final line shows how the ith component of ARM 7 the ith entry of the column vector representing Al gt 7 is given by matrix multiplication of the array AZk with the column vector We can equally well see how matrix multiplication gets de ned via the product of linear operators Consider the matrix elements of the operator AB ltilABlkgt ZltilAljgtltlelkgt j ZAz ijk 139 Given a linear operator A we recall the notion of eigenvectors and eigenvalues They are solutions A and lgt to the equation AlAgt MM The zero vector is not considered an eigenvector Note that the eigenvector lgt correspond ing to an eigenvalue A is not unique since any scalar multiple of the eigenvector will also be an eigenvector with the same eigenvalue However note that such eigenvectors are not linearly dependent It may or may not happen that there is more than one linearly inde pendent eigenvector for a given eigenvalue When there is one says that the eigenvalue is degenerate Note also that the number of eigenvalues and the dimension of the eigenspace can range from 0 to the still nite dimension of the Hilbert space Finally let us note that a linear operator A on H also defines a linear operation on the dual vector space z39e the space of bras This operation is denoted W m WA To de ne lt11lA we should tell how it acts as a linear function on kets the de nition is MilUM lt lAl gt As an exercise you should check that lt11lA so de ned is indeed a linear function z39e a bra and that A is a linear operation on bras Selfadjoint operators A linear operator A on a finite dimensional Hilbert space defines an operator Al called the adjoint of A It is de ned by demanding that for all kets we have ltwm gt lt lAlwgt Note that this is equivalent to de ning Al as the operator whose matrix elements are the complex conjugate transpose of those of A ATM Aji 4 Physics 6210Spring 2007Lecture 3 If AlA we say that A is self adjoin or Hermitz an It is a standard result from linear algebra that a Hermitian operator has real eigenvalues and that eigenvectors corresponding to distinct eigenvalues must be orthogonal See your text for the elementary proofs An extremely important theorem from linear algebra says that a Hermitian operator always admits an orthonormal basis of eigenvectors Typically we will use a notation such as AW ailigt to denote the eigenvalues and eigenvectors Physics 6210Spring 2007Lecture 28 Lecture 28 Relevant sections in tezrt 39 Spin correlations and quantum weirdness The EPR argument Recall the results of adding two spin 12 angular momenta The fact that the total spin magnitude is not compatible with the individual spin observables leads to some somewhat dramatic consequences from a classical physics point of view This drama was already noted by Einstein Podolsky and Rosen EPR in a famous critique of the completeness of quantum mechanics Much later Bell showed that the basic classical locality77 assumption of EPR must be violated implying in effect that nature is truly as weird as quantum mechanics makes it out to be Here we give a brief discussion of some of these ideas The original EPR idea did not deal with spins but with a pair of spinless particles We shall in a moment following Bohm deal with a spin system But it is worth first describing the EPR argument which goes as follows Consider two spinless particles characterized by positions 1152 and momenta p1p2 It is easy to see that the relative position 172 and the total momentum P P1 P2 commute so there is a basis B of states in which one can specify these observables with arbitrary accuracy Suppose the system is in such a state Suppose that an observer measures the position of particle 1 Assuming that particles 1 and 2 are well separated there is no way this experiment on particle 1 can possibly affect particle 2 This is essentially the EPR locality idea Then one has determined particle 27s position with arbitrary accuracy 7 without disturbing particle two Thus particle 27s position is known with certainty 7 it is an element of reality Alternatively one could arrange to measure 191 By locality the value of m is undisturbed and is determined with arbitrary accuracy One concludes that given the locality principle particle 27s position and momentum exist with certainty as an element of reality But of course quantum mechanics prohibits this situation Thus either quantum mechanics is incomplete as a theory unable to give all available information about things like position and momentum or the theory is non local in some sense because the locality idea was used to argue that a measurement on particle 1 has no effect on the outcome of measurements on particle 2 The loss of this type of locality was deemed unpalatable by EPR and so this thought experiment was used to argue against the completeness of quantum mechanics In fact the correct conclusion is that quantum mechanics is in a certain sense non local Physics 6210Spring 2007Lecture 14 Lecture 14 Relevant sections in teth 21 22 TimeEnergy Uncertainty Principle c0nt Suppose the energy is discrete for simplicity with values Ek and eigenvectors Any w Zcm k Assuming this is the initial state at t t0 the state at time t is given by Ultttogtwgt ch EW tONEu k state can be written as Let us use an observable A to characterize the change in the system in time which is after all what we actually do Let us denote the standard deviation of A or H in the initial state W1gt by AA or AE From the uncertainty relation we have in the initial state AAAE ltAHgt Recall our previous result which relates time evolution of expectation values to commuta tors we get ltAHgt 2 1 d 7 7 A 2 dtlt gt Th erefore h d AAAE 2 2 dt If we want to use A to characterize the time scale A for a significant change in the system we can do this by comparing the rate of change of the average value of A to the initial uncertainty in A AA At d7 WWW With At so de ned we then have AtAE 2 So the shortest possible time scale that characterizes a significant change in the system is given by AtAE N h Of course if the initial state is stationary 7 that is an energy eigenvector then AE 0 which forces At a 00 which makes sense since the physical attributes of the state never change Physics 6210Spring 2007Lecture 14 The time energy uncertainty principle is then a statement about how the statistical uncertainty in the energy which doesn7t change in time since the energy probability dis tribution doesn7t change in time controls the time scale for a change in the system In various special circumstances this fundamental meaning of the time energy uncertainty principle can be given other interpretations but they are not as general as the one we have given here lndeed outside of these special circumstances the alternative interpreta tions of the time energy uncertainty principle can become ludicrous What I am speaking of here are things like the oft heard You can violate conservation of energy if you do it for a short enough time or The uncertainty of energy is related to the uncertainty in time We shall come back to these bizarre sounding statements and see what they really mean a little bit later For now beware of such slogans As a nice example of the time energy uncertainty principle consider the spin precession problem we studied last time Recall that we had a uniform static magnetic field B Bl along the z axis The Hamiltonian is B H 52 me We studied the time dependence of the spin observables when the initial state was an SI eigenvector It is not hard to compute the standard deviation of energy in the initial state lSzgt Using 2 H2 7 2mc AE 3 17w 2mc 2 we have exercise so that we expect a significant change in the state when 1 iwAt N l 2 Thus the frequency w controls the time scale for changes in the system as you might have already ascertained from 69 the probability distributions 7L cos2 t Pr0bSI rims g Heisenberg picture Let us now see how to describe dynamics using the Heisenberg picture in which we encode the time evolution into the operator representatives of the observables rather than in the state vectors The idea is that time evolution is mathematically modeled by allowing 2 Physics 6210Spring 2007Lecture 14 the correspondence between physical observables and self adjoint operators to change in time To see how this works is straightforward given our previous work in the Schrodinger picture It is useful to use a very explicit notation We denote the physical observable by A We denote its time independent self adjoint operator representative by A We have ltAgttlt tlAl11tgtlt ltt0lUltat0AUtatol1Itogt If we define At Ultt0AUtt0 then we can view time evolution as occurring mathematically speaking through the time dependent identification of a self adjoint operator At to the observable A ltAgtt lt ltt0lAtl1Itogt We call A the Schrodinger picture operatorobservable and we call At the Heisenberg picture operatorobservable You can see that at t t0 the Schrodinger and Heisenberg representatives of A are the same In the Heisenberg picture the unit vector representing the state of the system is fixed once and for all Wt W050 while the operator observables evolve in time In the Schrodinger picture the states evolve in time while the operator observables are held xed As you know one of the most basic predictions of quantum mechanics is that the set of possible outcomes of a measurement of A is the spectrum of its operator representative In the Schrodinger picture we thus consider the spectrum of A to get at the possible outcomes of a measurement of A In the Heisenberg picture we have to evidently consider a different operator at each time to see what are the possible outcomes of a measurement of A at each time This looks bad If in the Heisenberg picture the operator representing A can in principle be different at different times we are in danger of saying that the possible outcomes of a measurement of A 7 which is a fixed set for all time in the Schrodinger picture 7 is different at different times For example perhaps it is possible that a particle could at one instant of time have the whole real line to move on while it could only move on some subset of the real line at some other time This would be a serious inconsistency Wweshallalwzws restrict our attention to observables A which do not involve the time t explicitly in their definition What I mean here is that we shall exclude from our discussion observables like the energy multiplied by the time It is easy to allow for such observables but it might be confusing at first glance Your text shows how to generalize our results to the case of explicitly time dependent observables 3 Physics 6210Spring 2007Lecture 14 Of course there is no inconsistency You can easily check that if laigt are the eigenvectors of A Alaigt ailaigtz then lailttgt 2 WWW are eigenvectors of At with the same eigenvalue Mal75 Mal0 In fact it is not hard to prove that the spectrum of At is identical to the spectrum of A Thus the possible outcomes of a measurement of A are the same in both pictures What does change between the two pictures is the mathematical representation of the states in which the observable A is fixed with statistical certainty As you know the states where A are known with certainty are the eigenvectors of the operator representative In the Heisenberg picture we then get in general a different eigenvector at each time This is exactly what is needed to get the proper time evolution of probability distributions PTOMA 1239 lltaitl t0gtl2 lltailUt7t0l1Itogtl2 lltail tlgtl2 where the last expression is the Schrodinger picture formula for the probability This result on the eigenvectors changing in time can lead to confusion so let me belabor the point a bit The state vector in the Heisenberg picture is the same for all time but the basis of eigenvectors of an observable will in general change in time If you start your system off in an eigenvector of AGED this means that the observable A is known with certainty at time t t0 and we have W105 l t0gt lat0gt At some other time t the state vector is still lat0gt but this is no longer a state in which A is known with certainty since that state vector would be latgt not lat0gt One sometimes summarizes this situation with the slogan In the Schrodinger picture the basis vectors provided by the observables are fixed while the state vector evolves in time In the Heisenberg picture the state vectors are held fixed but the basis vectors evolve in time in the inverse manner This is an instance of the active vs passive77 representation of a transformation 7 in this case time evolution This is true for any two operators related by a unitary transformation A lt gt UlAU 4 Physics 6210Spring 2007Lecture 11 Lecture 11 Relevant sections in tezrt 17 21 Product observables We have seen how to build the Hilbert space for a composite system via the tensor product construction Let us now see how to build the observables Let A1 be an observable for system 1 and Bg an observable for system 2 We de ne the corresponding observables for the composite system as follows Consider the following linear operators on product vectors A1 Iiagt i gtz41iagt i gt7 I B2iagt W W 132 We extend the definitions of these operators to general states by expanding the state in a product basis and demanding linearity of the operator Thus defined these operators are Hermitian Moreover they commute good exercisesl A1 II B20 which means that one can still determine with statistical certainty the subsystem observ ables within the composite system For example if we are looking at 2 particles it is possible to ascertain both particle 17s position and particle 27s momentum with arbitrarily good statistical accuracy Likewise when using this contruction to build a model for a particle moving in 3 d from our l d model we end up with all the position variables com muting among themselves and all the momentum variables commuting among themselves Of course the commutation relations obtained in the subsystems appear in the combined system Exercise Show that if 14131 01 then A1 1731 81 01X 1 thus explaining the generalization of the canonical commutation relations we obtained for a particle in 3 d Please take note that it is conventional in many references to simply drop the 81 or I factors when denoting the extension of the subsystem operators to the composite system Indeed it is common to also drop the symbol entirely Also note that not every observable of the system is of the above form As usual any self adjoint operator can represent an observable Thus there will be observables that refer 1 Physics 6210Spring 2007Lecture 11 to the composite system as a whole 69 the total energy and are not observables of either of the subsystems alone Returning to our example of a particle moving in 2 d We have position operators for each of the two particles X 8 I and I 8 Y For example we have X Il7ygt X Ilgt lygt QCW lygt Let us see how these operators act on position wave functions YWMI WWW ylt7ylwgt mam where we used Yl7ygt yl7ygt 5 lt96le ylt7yl Similarly we can de ne the momentum For example you can check as a nice exercise that P1wltxygtltltxi ltyigta Ilwgt gum Hint PI acts on via infinitesimal translations As an example of an observable that can only be de ned on the whole system consider the total energy function for two particles moving in l d equivalently one particle moving in 2 d 1 2 2 It acts on position wave functions as exercise 2 2 2 w t 97214967yH V967y 967y Dynamics the Schrodinger picture We now have enough tools to formulate quantum dynamics Dynamics are character ized by the final postulate Time evolution is a continuous unitary transformation We have of course just studied an example of a continuous unitary transformation Transla tion by an amount a In that application a is not identified with time though The idea of dynamical evolution is that measureable aspects of the system are changing in time In For us time will be modeled in its Newtonian form as a non dynamical a priori way of ordering events valid in any inertial reference frame In particular in this framework time is not treated as an observable whose value depends upon the state of the quantum system but instead as a special kind of parameter Of course there are many interesting issues here but we will not be able to explore them further in this course We simply assume that some suitable standard of temporal reference has been chosen once and for all 2 Physics 6210Spring 2007Lecture 11 quantum mechanics the characterization of the system at any given time can be viewed as the totality of probability distributions for all observables at that time Time evolution therefore ought to correspond to a time varying change in the probability distributions As we have noted already all probability distributions can be computed by taking expec tation values of suitable observables eg characteristic functions etc Therefore time evolution can be defined in terms of the evolution of expectation values The rules of the game say that given an observable A the expectation value is to be computed via ltAgtlt lAl1Igt where W1 is a unit vector in Hilbert space representing the state of the system and A is the operator representing the observable We now consider doing this as time varies At each instant of time we need to have a system of vectors and operators that can be used to make physical predictions via say expectation values Our strategy will be to assume we have a single Hilbert space for all time but we allow the mathematical identification of states and observables to vary in time To describe mathematically how expectation values change in time we then have a number of options Two such options are often convenient First we can let the operator representing the observable change in time with the vector W1 fixed for all time This point of view of dynamics is called the Heisenberg picture and will be discussed later Another way of viewing dynamics known as the Schrodinger picture is based upon letting the vector W1 change in time while holding the operator representatives of the observables fixed in time There are infinitely many different pictures intermediate to these two We shall look first at the postulate on dynamics from the point of view of the Schrodinger picture The Schrodinger picture The time evolution operator Much as we did with spatial translations we assume that the state vector at time t denoted by W1 t is related by a continuous unitary transformation from the state at any earlier time to Therefore we write Wat Ut7t0l 7t0gt Here liJt0 can in principle be any unit vector U is unitary so that the state vector remains normalized as time progresses ltwtwtgt ltwtoUiUwtogtltwtowtogt1 It is not hard to show that a bounded operator on a complex Hilbert space vanishes if and only if all its diagonal matrix elements vanish exercise Since l11t0 is arbitrary we see that the state vector remains normalized if and only if Ul U71 that is U is unitary Given UWJO U lcnso 3 Physics 6210Spring 2007Lecture 11 we naturally associate the inverse transformation as time evolution from t back to to so that U71ltt7t0 U050 Finally we assume that time evolution from to to t can be viewed as time evolution from to to t1 and then from t1 to t with to lt t1 lt t so that Ut7t1Ut17t0 Ut7t0 The operator U is called the time evolution operator The principal issue when studying dynamics using quantum mechanics is to uncover the nature of the time evolution operator for the given system Generally speaking this is not easy to do if one proceeds in a direct fashion After all the time evolution operator has to be a pretty complicated object given that it knows how to evolve any initial state A deep principle of physicsmathematics going back to Lie is the following When considering continuous transformations it is always easier to work with infinitesimal transformations This is because the finite transformation is completely characterized by the infinitesimal transformation which is a much simpler structure than the finite transformation This is one of the reasons that you usually find dynamical evolution whether in Newtonian dynamics uid dynamics electrodynamics etc expressed in terms of differential equations Thus we consider the infinitesimal generator of time evolution viewed as a continuous unitary transformation The infinitesimal generator of the unitary time evolution will be an observable called the Hamiltonian in analogy with classical mechanics where the Hamiltonian is the generating function of a canonical transformation corresponding to motion in time Normally the Hamiltonian represents the energy which is conserved provided H does not depend upon the time lndeed just as we defined the momentum as the generator of spatial translations we de ne the Hamiltonianenergy as the generator of time translations One subtlety which occurs here unlike what occurs with spatial translations is that the Hamiltonian may be time dependent that is it may be a family of operators paramterized by the time This demands a slightly more sophisticated analysis than we used for the spatial translations Physics 6210Spring 2007Lecture 7 Lecture 7 Relevant sections in tezrt 16 Observables with continuous andor unbounded values We are now ready to turn to the quantum mechanical description of a non relativistic particle We shall de ne a spinless particle as a system that is completely characterized by the position and linear momentum which are the basic observables in this model This means that all observables are functions of position and momentum While it is possible that the position and momentum variables take a discrete set of values as angular momentum and 7 often 7 energy do there is currently no experimental evidence of this We therefore create a model in which these observables can take a continuous unbounded set of values Evidently we need to de ne a Hilbert space that admits self adjoth operators with a continuous unbounded spectrum Neither of these features are possible on nite dimensional vector spaces and so here we are forced into the in nite dimensional setting 2e spaces of functions This leads to some mathematical subtleties that we need to be wary of I shall not try to be perfectly rigorous in our discussion since that would take us too far a eld But I will try to give you a reasonably fool proof 7 if somewhat formal 7 treatment First let me give you a recipe for dealing with the situation Then let me give you a avor of the underlying mathematics which makes the formal recipe work and also which shows where it can become tricky Formal recipe We want to de ne an observable A with a continuous unbounded set of values a E R say We postulate the existence of a self adjoth linear operator again denoted A and a continuous set of vectors fagt such that Afagt afagt We say that A has a continuous spectrum These vectors are to be orthonormal in the delta function sense ltafagt 6a 1 You can think of this as a continuum generalization of the usual orthonormality expressed via the Kronecker delta The vectors fagt are to form a basis that is they provide a continuous resolution of the identity 11daagtlta 1 Physics 6210Spring 2007Lecture 7 so that we can write Wwwwn Using quotation marks to indicate where the underlying mathematics is considerably more subtle than the words indicate we can say the following We interpret the eigenval ues a as the possible values of an outcome of the measurement of A the vectors lagt are states in which A has the value a with certainty The scalar KaliO lgda is the probability for finding A to have the value in the range 1 a do in the state In particular the probability that the observable A is found to have a value in a region 11 a2 is given by PMEMw AWWWWR Thus in this setting we call Kali1H2 the probability density for A in the state 11 You can see that given the operator A with continuous spectrum the abstract state vector is completely characterized by the complex valued function wownwww wnL 11a represents the components of the vector W1gt along the basis provided by lagt Its the continuous analog of the column vector representation This function 1a ltalwgt is called the wave function associated to the observable A The Hilbert space can thus be identified with the set of all square integrable functions 11a Of course you have seen this before in the position and momentum representations for a particle Subtleties A glimpse of the real story For the most part using the above formalism you can treat operators with continuous spectra much the same as we did linear operators on finite dimensional Hilbert spaces It is worth warning you however that there are mathematical subtleties that can come into play To begin with note that the eigenvectors lagt cannot really be elements of the Hilbert space since they do not have a norm 7 6a a is not defined To see this concretely in a familiar example consider the familiar momentum operator to be justified soon acting upon the Hilbert space of square integrable functions of l variable h d 191415 lts eigenfunctions are of the form We N 6 2 96 Physics 6210Spring 2007Lecture 7 with eigenvalues No matter how you choose H 00 dac MW 7 00 This means that 16 does not actually correspond to an element of the Hilbert space This dif culty can be traced back to the fact that 1 has continuous spectrum Our formal prescription delta function normalization etc works if we pick H ik where k is a real number If the spectrum of an observable is unbounded then another dif culty arises not all elements of the Hilbert space are in the domain of the operator This dif culty 7 which can only occur if the Hilbert space is infinite dimensional 7 arises whenever the spectrum is unbounded irrespective of whether the spectrum is discrete or continuous To illustrate this return to the momentum operator It is clear that say a square wave pulse defines a normalizable wave function that is an element of the Hilbert space of a particle However the derivative of a square wave is not defined at the corners so one doesn7t get a function after applying the operator 7 let alone a square integrable function We say that the square wave function is not in the domain of the momentum operator Likewise it is easy to construct wave functions that are not in the domain of the position operator For example you can easily check that at W15 m is a square integrable normalizable function of x However the position operator X acts on such functions via 30496 WW so that in our example 2 at X 7 W 362 1 which is not square integrable 26 not an element of the Hilbert space for a particle In a more rigorous treatment of observables with continuous and or unbounded spec trum one handles these issues as follows One accepts the limited domain of unbounded operators One adds to the definition of self adjoint operators the requirement that the op erator and its adj oint have the same domain The domain of an observable with unbounded Since we don7t have any eigenvectors how do we define spectrum Roughly the spec trum of an operator A is the set of scalars A such that the linear operator A 7 AI has no inverse Physically such a wave function describes a particle that is known with certainty to be somewhere in the region in which the square wave function is non vanishing For example this would be the state of a particle after it has passed into a particle detector of finite size Physics 6210Spring 2007Lecture 7 spectrum forms a dense subset of the Hilbert space This means that every vector can be arbitrarily well approximated relative to the Hilbert space norm by elements in the domain If the spectrum is discrete if unbounded an orthonormal basis of eigenvectors for the Hilbert space can be found inside this dense domain If the spectrum is continuous the eigenvectors are not in the Hilbert space but are generalized eigenvectors They live in the vector space of distributions which are linear functions on a dense subspace of the Hilbert space Thus the generalized eigenvectors have a well defined scalar product77 with elements from a dense subset of the Hilbert space but not with the whole Hilbert space or among themselves It can be shown that a self adjoint operator with continuous spectrum admits a delta function normalized set of generalized eigenvectors which can be used to define a generalized basis much as indicated in our formal treatment above If all this mathematical detail leaves you a little dazed don7t feel bad I only gave you vague hints as to how the story goes I did not attempt to give a complete coher ent description Still now you at least know where some of the subtleties lie In what follows all of our formal manipulations will have a justification within the more complete mathematical formulation that l hinted at above Position Operator We have already mentioned that a particle is defined as a system in which the only relevant observables are position and momentum How do we implement this idea in quantum mechanics We need to define self adjoint operators representing position and momentum As mentioned earlier we do this so that the spectrum is continuous and unbounded ranging over all real numbers We begin with the position operator which can be constructed formally as we did the generic operator A above We will denote by X the position operator and the spectral values will be denoted at so that we have the generalized eigenvectors We have on a suitable dense domain X Xl The position wave function is defined by 00 2 we WM dx WM WW 00 A vector is completely specified by its position wave function and vice versa The position wave function is a continuous version of a column vector A linear operator maps vectors to vectors and so defines 7 and is defined by 7 a linear mapping from wave functions to wave functions This mapping can be computed via 1Z1z a A11z E 4 Physics 6210Spring 2007Lecture 7 In particular the position operator maps functions to functions via Mic X ltle11gt WCW WW Here we have used The wave function LJ is the probability amplitude for position This means that W420 is the probability density for position This means that b Pr0bX e 1 b dark160 a Note the unit vector condition implies normalization of the wave functions 1 ltwiwgt dx ltwlxgtltxiwgt dx WM Finally we note that the generalized eigenvector lzgt represents a state in which the particle has definite position Formally the wave function for this state is then 11196 969 WWl Of course this wave function is not normalizable 7 except in the delta function sense Try it Remarks Note that all the foregoing results are in fact valid for any observable with continuous spectrum We simply select one and use it to represent position and adapt our notation accordingly Also note that as is the tradition when presenting material at this level we have been completely cavalier with domains The idea is that the formal manipulations given above will make sense provided the domains are suf ciently restricted These restrictions are usually not explicitly made since we are only trying to understand the formal structure of things When performing concrete computations one must usually pay more attention to such things But its amazing how far one can get without doing so Physics 6210Spring 2007Lecture 25 Lecture 26 Relevant sections in tezrt 36 37 Two spin 12 systems observables We have constructed the 4 d Hilbert space of states for a system consisting of two spin 12 particles We built the space from the basis of product states77 corresponding to knowing the spin along 2 for each particle with certainty General states were however not necessarily products but rather superpositions of such How are the observables to be represented as Hermitian operators on this space To begin let us consider the spin observables for each of the particles Call them g1 We de ne them on product states via 1agt W GM W and 2agt W W W Here the operators g are the usual spin 12 operators acting on a two dimensional Hilbert space that we have already discussed in some detail If 04 is an eigenvector of spin along some axis then so is 04 5 for any This means that if we know the spin component along the chosen axis with certainty for particle one then we get an eigenvector of the corresponding component of 61 as we should The same remarks apply to particle 2 The action of 61 and 62 are defined on general vectors by expanding those vectors in a product basis such as we considered above and then using linearity to evaluate the operator term by term on each vector in the expansion Sometimes one writes g1 1 QI to summarize the above de nition The two spin operators g1 and g2 commute exercise and have the same eigenvalues as their 1 particle counterparts exercise In this way we recover the usual properties of each particle now viewed as subsystems Total angular momentum There are other observables that can be defined for the two particle system as a whole Consider the total angular momentum defined by 51 53 You can easily check that this operator is Hermitian and that Sky 51 iheklmsmz 1 Physics 6210Spring 2007Lecture 25 so it does represent the angular momentum Indeed this operator generates rotations of the two particle system as a whole The individual spin operators g1 and 62 only generate rotations of their respective subsystems Using our general theory of angular momentum we know that we can find a basis of common eigenvectors of 82 and any one component say SZ Let us write these as lsm5 where 82 sm5 88 lh2lsm5 Szlsm3 mhlsm3 Let us de ne lii lSzi lSzi This product basis physically corresponds to states in which the 2 component of spin for each particle is known with certainty In the following we will find the total angular momentum eigenvalues and express the eigenvectors in the product basis li i To begin with it is clear that eigenvectors of SZ are in fact the basis of product vectors since with m1 i m2 i Szlm17m2gt 512 S2zlm1m2gt 7711 m27blm17m2gt We see that m 71 0 l with m 0 being doubly degenerate exercise From our general results on angular momentum it is clear that the only possible values for the total spin quantum number are 5 01 From this we can infer that the m i1 eigenvectors must be 82 eigenvectors with s l but we may need linear combinations of the m 0 product eigenvectors to get 82 eigenvectors To see why the vectors l and l 7 7 must be also 82 eigenvectors one reasons as follows Our general theory guarantees us the existence of a basis of simultaneous 82 and SZ eigenvectors It is easy to see that the l and l 7 7 are the only eigenvectors up to normalization with m i1 since any other vectors can be expanded in the product basis and this immediately rules out any other linear combinations exercise Therefore these two vectors must be the 82 eigenvectors Because they have m i1 and we know that s 01 it follows that the l and l 7 7 vectors are 82 eigenvectors with s 1 To determine the linear combinations of the m 0 product vectors l 7 and l 7 that yield 82 eigenvectors we use the angular momentum ladder operators Si SI t iSy Sli Sgi If we apply S to the eigenvector ls17m1gtlgt7 we get exercise 1 l817m0gt S7l817m1gtS7lgt517527lgt 3lgtlgt 2 Physics 6210Spring 2007Lecture 25 The other eigenket l0 0gt must be orthogonal to this vector as well as to the other eigenkets l gt and l 7 7gt from which its formula follows exercise All together we find the total angular momentum eigenvectors ls mgt are related to the individual angular momentum product eigenkets by l11gtlgt l l170gt Em 7gt l 7 l171gtl7gt loogt 70 7gt l gtgt These vectors form an orthonormal basis for the Hilbert space so they are all the linearly independent eigenvectors of 82 and SZ The eigenstates with s l are called the triplet states and the eigenstate with s 0 is the singlet state Notice that by combining two systems with half angular momentum we end up with a system that allows integer angular momentum only A lengthier 7 but more straightforward 7 derivation of the eigenvectors ls m5gt arises by simply writing the 4 gtlt 4 matrix for 82 in the basis of product vectors l i igt and solving its eigenvalue problem This is a good exercise To get this matrix you use the formula 52 52 th ss It is straightforward to deduce the matrix elements of this expression among the product states since each of the operators has a simple action on those vectors Notice that the states of definite total angular momentum ls m5gt are not all the same as the states of definite individual angular momentum say l i igt This is because the total angular momentum is not compatible with the individual angular momentum For example 527 Slil Si t 5 2511521 51y52y Slzs2zlzslil 7 0 A couple of complete sets of commuting observables are given by 5253 S12 22 and 52 5282 Sz The eigenvectors of the first set are the product basis lm1m2gt l i igt representing states in which each individual spin angular momentum state is known with certainty The eigenvectors of the second set are given by lsm5gt representing states in which the total angular momentum is known with certainty A remark on identical particles Let us remark that there is yet another postulate in quantum mechanics that deals with identical particles These are particles that are intrincsically alike same mass spin electric charge etc Thus for example all electrons are identical though of course they can be in 3 Physics 6210Spring 2007Lecture 25 different states This does not mean that electrons cannot be distinguished literally since we can clearly distinguish between an electron here on earth and one on the sun These are two electrons in different position states But we view these particles as interchangeable in the sense that if one took the electron from the sun and replaced it with the one here on Earth putting them in the respective states when you weren7t looking then you couldn7t tell This intrinsic indistinguishability of identical particles opens up the possibility of having the states of multi particle systems re ect this symmetry under particle interchange This symmetry is modeled as a discrete unitary transformation which exchanges particles The postulate of quantum mechanics which can more or less be derived from relativistic quantum field theory is that particles with integer spin bosons should be invariant under this unitary transformation even under exchange and they should change sign odd under exchange if they particles have half integer spin fermions You can see that the total spin states of two spin 12 systems are in fact even and odd under particle interchange If the two particles are identical and no other degrees of freedom are present then one must use the anti symmetric singlet state only Of course real particles have translational degrees of freedom and the state will re ect that Using position wave functions to characterize these degrees of freedom one again can consider the symmetric and anti symmetric combinations Only the total state vector must have the appropriate symmetry For example consider the ground state for two electrons in a Helium atom The position space ground wave function is symmetric under particle inter change Thus the ground state must be a singlet Excited states can however by described by symmetric spin states if the position part of the state vector is anti symmetric under particle interchange Physics 6210Spring 2007Lecture 34 Lecture 34 Relevant sections in te1t39 56 Timedependent perturbation theory c0nt We are constructing an approximation scheme for solving d i magma eEEn EmtVnmtemt Vnm ltantlmgt For simplicity we shall suppose as is often the case that the initial state is an eigen state of H0 Setting l1b0gt ie taking the initial state to be one of the unperturbed energy eigenvectors we get as our zeroth order approximation 0 cnt on am We get our first order approximation 021 by improving our approximation of the right hand side of the differential equations to be accurate to first order This we do by using the zeroth order approximation of the solution cn0 in the right hand side of the equation So we now approximate the differential equation as d i thaw e E L Emet with solution 1 t i gnu of c5905 5712 g dte En EitVnZt 0 Successive approximations are obtained by iterating this procedure We shall only deal with the first non trivial approximation which defines rst order time dependent perturbation theory We assumed that the system started off in the formerly stationary state defined by H0 Of course generally the perturbation will be such that is not a stationary state for H so that at times t gt 0 the state vector will change We can still ask what is the probability for finding the system in an eigenstate of H0 Assuming that n 31 i this is l t i Pt a m 7r n O dte E EZtVmtl2 This is the transition probability to first order in TDPT The probability for no transition is in this approximation Pigti17Pigtni7En 1 Physics 6210Spring 2007Lecture 34 TDPT is valid so long as the transition probabilities are all much less than unity Other wise our approximation begins to fail The transition probability for i a n is only non zero if there is a non zero matrix element Vm connecting the initial and final states Otherwise we say that the transition is forbidden Of course it is only forbidden in the leading order approximation If a particular transition 2 a n is forbidden in the sense just described then physically this means that the transition may occur with very small probability compared to other non forbidden transitions To calculate this very small probability one would have to go to higher orders in TDPT Example timeindependent perturbation As our first application of TDPT we consider what happens when the perturbation V does not in fact depend upon time This means of course that the true Hamiltonian H0V is time independent too since we assume H0 is time independent If we can solve the energy eigenvalue problem for H0 V we can immediately write down the exact solution to the Schrodinger equation and we don7t need to bother with approximation methods But we are assuming of course that the problem is not exactly soluble Now in the current example of a time independent perturbation we could use the approximate eigenvalues and eigenvectors from time independent perturbation theory to get approximate solutions to the Schrodinger equation This turns out to be equivalent to the results we will obtain below using time dependent perturbation theory lt7s actually quite a nice exercise to prove this but we won7t bother Assuming V is time independent the integral appearing in the transition probability can be easily computed and we get exercise Pi a mi 31 11 4 Vm 2 sin2 T En 7 El 2h Assuming this matrix element Vm does not vanish we see that the transition probability oscillates in time The amplitude of this oscillation is small provided the magnitude of the transition matrix element of the perturbation is small compared to the unperturbed energy difference between the initial and final states This number had better be small or the perturbative approximation is not valid Let us consider the energy dependence of the transition probability The amplitude of the temporally oscillating transition probability decreases quickly as the energy of the final state differs from the energy of the initial state The dominant transition probabilities occur when En Ei i 31 11 For this to happen of course the states and must correspond to a degenerate or nearly degenerate eigenvalue When En EZ we have 1 Pi a m 7 nEn E lmF252 2 Physics 6210Spring 2007Lecture 34 Because of the growth in time our approximation breaks down eventually in this case Still you can see that the transition probability is peaked about those states such that En Ei with the peak having height proportional to t2 and width proportional to 115 Thus as the time interval becomes large enough the principal probability is for transitions of the energy conserving type En Ei For shorter times the probabilities for energy non conserving transitions are less negligible Indeed the probability for a transition with an energy difference AE at time t is appreciable once the elapsed time satis es N 1 AE as you can see by inspecting the transition probability This is just a version of the time energy uncertainty principle expressed in terms of the unperturbed stationary states One sometimes characterizes the situation by saying that one can violate energy conservation by an amount AE provided you do it over a time interval At such that AtAE N h All this talk of energy non conservation is of course purely gurative The true energy represented by H0 V is conserved as always in quantum mechanics when the Hamiltonian is time independent It is only because we are considering the dynamics in terms of the unperturbed system using the unperturbed energy defined by H0 that we can speak of energy non conservation from the unperturbed point of view You can see however why slogans like you can violate conservation of energy if you do it over a short enough time might arise and what they really mean You can also see that such slogans can be very misleading Physics 6210Spring 2007Lecture 5 Lecture 6 Relevant sections in text 14 Complete set of commuting observables Only in the simplest physical systems will the measurement of a single observable suf ce to determine the state vector of the system Of course the spin 12 system has this property if you measure any component of the spin the state vector is fixed up to an irrelevant phase factor More complicated systems need more observables to char acterize the state For example recall that the energy levels of a hydrogen atom can be uniquely specified by the energy orbital angular momentum magnitude one component of the angular momentum and the electron spin state Other choices are possible After measuring an observable and getting a particular outcome 7 an eigenvalue 7 the state of the system is the corresponding eigenvector It may happen that more than one state vector is associated with that measurement outcome Mathematically the eigenvalue is degenerate To pin down the state uniquely other observables must be measured Evi dently if we have a set of observables whose measurement uniquely determines the state vector 7 up to a phase then that state vector must be a simultaneous eigenvector of all the observables in question If we demand that the possible outcomes of measurements of this set of observables defines a basis for the Hilbert space then the observables must be compatible te their operator represntatives must all commute Such a set of variables is called a complete set of commuting observables 0800 For a hydrogen atom modeled as an electron in a Coulomb potential the energy orbital angular momentum magnitude one component of the orbital angular momentum and one component of the electron spin constitute a 0800 For a single spin 12 system any component of spin defines a 0800 Evidently a 0800 is not unique Given a 0800 A B we can label the elements of the ON basis it determines by the eigenvalues of the 0800 1a b These are states in which the elements of the 0800 are known with certainty to have the values a b The vectors satisfy lta b 1ab dad6bb Incompatible observables The uncertainty principle Incompatible observables are represented by operators which do not commute As we have seen incompatible observables are such that there exist states where both observables Two states that differ by a phase factor em will give the same expectation value to all observables 7 exercise This means that all probability distributions are insensitive to the phase factors and hence all physical predictions are as well 1 Physics 6210Spring 2007Lecture 5 cannot be determined with certainty We will now make this more precise and give a very general version of the celebrated uncertainty principle 7 probably better called the uncertainty theorem Given an observable A and a state W1 we de ne the dispersion ofA in the state W1 to be ltAA2gt I ltA ltAgt2gt WWW lt lAl gt2 ltA2gt ltAgt2 The dispersion also called the variance is a non negative real number which characterizes the statistical uncertainty of the observable A in the state To see that the dispersion is indeed non negative note that the expectation value of the square of any Hermitian operator 0 is a positive number ltwl02wgt alt TOW the right hand side is just the length squared of the vector CW1 which must be non negative Setting 0 A 7 AI and noting that A must be a real number exercise we conclude that AA2 2 0 Note the dispersion vanishes if and only if the state W1 is an eigenvector of A You can easily verify the if part Let me show you the only if77 part Write ltltAAgt2gt ltwlltA AgtIgt2iwgt ltwlltA MM 7 ltAgtIWgt WA ltAgt1W1gtH2 The norm of a vector vanishes if and only if the vector is the zero vector so if the dispersion 7lt 7lt vanishes we have A ltAgtIWgt 0 which is the eigenvector condition exercise Using the Schwarz inequality the following relation between the dispersions of 2 oberv ables can be established see your text for the proof ltAA2gtltA32gt 2 ltlAaBlgtl2 1 ll This is the general form of the uncertainty relation In a given state it relates the product of the statistical uncertainty of a pair of observables to the expectation value of the com mutator of the observables If the commutator vanishes or if its expectation value does in the given state then the uncertainty relation has no content Otherwise it provides information about the effect of the incompatibility of the observables In general this effect depends upon the state that is chosen You can see this from the fact that the expectation values occurring in the inequality above are defined by a given state This is 2 Physics 6210Spring 2007Lecture 5 important to keep in mind In certain cases 69 the position momentum relation to be studied later the uncertainty relation turns out to be state independent and hence it is much more dramatic 7 and famous For a spin 12 system it is straightforward to compute the uncertainty relation for various observables To do it you need the following commutation relations which you derive in your homework 17 Syl 3527 51752 ihsrz 52751 ihsy Let let W1 all bli lalg lblg l which is a general state vector We consider the uncertainty relation for SI and 52 We have exercise ltASZ2ltASI2 2 h4imab2 Of course if a or b vanish the state is an eigenstate of 52 and the uncertainty relation has no import Otherwise though you can see how the lower limit for the product of uncertainties varies with the choice of state For example if a l and b so that the state is an eigenstate of Sy we have 4 ltASZ2ltASI2 2 Physics 6210Spring 2007Lecture 23 Lecture 25 Relevant sections in tezrt 32 35 Spin precession as a rotation It is enlightening to return to the dynamical process of spin precession in light of our new results on rotations You will recall that a spin system with magnetic moment when placed in a uniform magnetic field E can be described by the Hamiltonian PM H7 where p You will recall that the behavior of the spin observables could be viewed as precession about g z39e a continuously developing in time rotation about an axis along We can now see this result immediately Let n be a unit vector along B so that H ipB S This means that the time evolution operator is Utt0 e itowB g This operator represents a rotation about 1 by an angle 3057150 which is exactly our previous result for the dynamics Note that while the physical observables are precessing with frequency MB the state vector itself has is precessing at half the frequency since eg it takes a 47139 rotation to get the state vector to return to its initial value It is possible to experimentally see this difference in frequencies thereby confirming the projective representation being used by a pair of spin 12 systems one of which propagates freely and one of which travels through a region with a magnetic field The latter spin will precess according to the time it spends in the magnetic field The two particles can be brought together to form an interference pattern The interference pattern depends upon the relative phase of the two particles If the magnetic field region is set up just right you can arrange for the second particle to change its state vector by a minus sign by the time it leaves the magnetic field region The interference pattern that you see confirms this fact See your text for details Angular momentum in general We can deduce quite a lot about the angular momentum f of any system just knowing that it is represented by 3 self adjoint operators satisfying the angular momentum commu tation relations To begin with it is clear that the 3 components of j are not compatible 1 96 Physics 6210Spring 2007Lecture 23 so that generally speaking one will not be able to determine more than one component with certainty Indeed the only state in which 2 or more components of f are known with certainty is an eigenvector of all components with eigenvalue zero 26 a state with vanishing angular momentum To see this suppose that la is an eigenvector of J1 and Jy then it is easy to see from Jan Jy th2 that la is an eigenvector of J2 with eigenvalue 0 You can easily see that this same argument can now be used to show that la has zero eigenvalue for all three components Thus if there is any angular momentum in the system at all at most one component can be known with certainty in any state When we consider states with a definite value for a component of f we usually call that component J2 by convention But it is important to realize that there is nothing special about the z direction one can find eigenvectors for any one component of spin 12 We next observe that the squared magnitude of the angular momentum 2 2 2 2 J JI Jy J2 is a Hermitian operator that is compatible with any component Ji To see this is a very simple computation J27 Jkl Z JllleJkl My Jliz 61kmJsz Jsz 0 l Lm where the last equality follows from the antisymmetry of 61km We will assume that J2 is self adjoint Consequently there exists an orthonormal basis of simultaneous eigenvectors of J2 and any one component of jusually denoted J2 Physically this means that while the 3 components of angular momentum are not compatible there exists a complete set of states in which the magnitude of angular momentum and one component of angular momentum are known with certainty Angular momentum eigenvalues and eigenvectors Of course given an observable represented as an operator the most pressing business is to understand the spectral properties of the operator since its spectrum determines the possible outcomes of a measurement of the observable and the generalized eigenvectors are used to compute the probability distribution of the observable in a given state In our case we have defined angular momentum as operators satisfying f ff Jh Jm ihelann The quantity in parenthesis is symmetric under Z lt gt m while 61km is anti symmetric when this interchange is performed This guarantees that each term in the double sum will be canceled by another term in the double sum 2 Physics 6210Spring 2007Lecture 23 Just from these relations alone there is a lot we can learn about the spectral properties of angular momentum We assume that each of the operators JZ and J2 admit eigenvectors Let us study the angular momentum eigenvalues and eigenvectors the latter being simultaneous eigenvec tors of J2 and J2 We write lm alabgt leabgt blabgt The possible values of a and b can be deduced much in the same way as the spectrum of the Hamiltonian for an oscillator can be deduced using the raising and lowering operators To this end we de ne the angular momentum ladder operators Ji J1 i Hy JI Of course these two operators contain the same physical information as J1 and Jy In terms of the ladder operators the angular momentum commutation relations can be expressed s exercise Jm Ji thi Ning 0 Jim lm From these relations we can see that the vector Jila bgt satis es exercise J2lt7ilazbgtgt Wham Lamb bi Wham Thus when acting on angular momentum eigenvectors eigenvectors of J2 and J2 the ladder operators preserve the magnitude of the angular momentum but increase decrease the 2 component by a quantum of angular momentum77 h To be continued Physics 6210Spring 2007Lecture 17 Lecture 17 Relevant sections in tezrt 23 Spectrum of the SH0 Hamiltonian cont It is shown in detail in the text that the eigenvalues of N are non degenerate and are non negative integers AEn012 The essence of the proof is to note that the number operator must have a non negative expectation value WNW ltwwaiawgt ltltwwaigtltawgtgt 2 o where WWW 0 ltgt IiiO 0 On the other hand since a lowers the eigenvalue of N by one unit there must be a unique up to normalization lowest eigenvalue eigenvector l0gt such that al0gt 0 One then easily infers that the spectrum of N is non degenerate consisting of the non negative integers Ningt nlm n012 Therefore the energy spectrum is purely discrete 1 En n hw n012 and the energy eigenvectors are labeled These results arise from the assumption that X and P are self adjoint operators on a Hilbert space or equivalently that a and al are adjoints of each other with respect to the Hilbert space scalar product Finally it is shown in the text that we have aw xH n 71gt a m H 1m 1gt ltmmgt 5m Note that the ground state77 l0gt has energy hw and satisfies al0gt 0 Energy eigenfunctions We have defined the simple harmonic oscillator and computed the spectrum of its Hamiltonian Now we explore some properties of the energy eigenvectors that is the 1 Physics 6210Spring 2007Lecture 17 stationary states Of course each of these vectors represents a state in which the energy is known with certainty to have the value En n These states also de ne probability distributions for all other observables in particular the position and momentum Let us consider the position probability distribution which is controlled by the position wave functions MW ltlngt It is easy enough to compute these functions For example consider the ground state wave function no it satisfies mow ltlal0gt 0 ma a g X um g x we we have that Since h d as 71015 i 0 This equation is easily solved and normalized to give a Gaussian exercise 2 1 i 2 10 u i 6 0 7114 0 7 where h 360 is a length scale set by the problem and as you can see it determines the width of the Gaussian Here we can see one way in which the quantum and classical mechanics regimes are related Of course a classical mechanics description of an oscillator implies that in the ground state the oscillator has position x 0 with certainty Quantum mechanics provides instead a Gaussian probability distribution about at 0 However provided that 0 is small the width of the Gaussian is negligible and the quantum description starts to coalesce with the classical description in this regard I used quotation marks about the word small since to has dimensions of length whether or not you consider it to be small depends on comparing it to some other length scale When we speak of macroscopic phenomena we usually are interested in length scales on the order of say centimeters masses on the order of grams and times on the order of seconds In such a regime to is indeed very very small But of course in a microscopic regime to can be appreciable The excited states 11 gt 0 are easily obtained from the identity exercise We lngt w Eb Physics 6210Spring 2007Lecture 17 so that 2 1 2 d n 67 Fugim g Z i o at e As you may know this formula represents up to the dimensionful constants one of the standard generating function77 methods for de ning the Hermite polynomials Thus is a Hermite polynomial times the ground state Gaussian See your text for detailed formulas Expectation values It is instructive to see how to compute stationary state expectation values and of course to see what they look like To begin we observe that stationary state expectation values of observables that are linear in position and or momentum will vanish ltanlngt 0 ltanlngt To see this just note that such observables are linear in the ladder operators and we have by orthogonality nlaln olt 71 0 nlalln olt 1 0 Another way to see that the expectation values of position and momentum vanish in stationary states is to note that exercise X olt RH P olt From which the result easily follows exercise On the other hand quadratic functions of position and momentum need not have vanishing expectation values For example in the ground state exercise x2 x2 lt0X20gt 30mm a2 ala aall0 Which gibes with our earlier comment about the width of the ground state Gaussian position probability distribution A similar computation for P2 shows that exercise E2 lt0lP2l0gt i 2x0 We see that the dispersions in position and momentum satisfy in the ground state 2 2 g ltAX gtltAP gt I which is in accord with the uncertainty principle but also shows that the ground state is a minimum uncertainty state The excited states are not minimum uncertainty states a straightforward computation reveals exercise AX2AP2 n 9252 3 Physics 6210Spring 2007Lecture 8 Lecture 8 Relevant sections in tezrt 16 Momentum How shall we view momentum in quantum mechanics Should it be mass times ve locity or what Our approach to the de nition of momentum in quantum mechanics will rely on a rather fundamental understanding of what is momentum To motivate our definition let me remind you that the principal utility of the quantity called mo mentum77 is due to its conservation for a closed system One can then understand the motion of interacting systems via an exchange of momentum Next recall the intimate connection between symmetries of laws of physics and corresponding conservation laws In particular symmetry under spatial translations corresponds to conservation of linear momentum In the Hamiltonian formulation of the classical limit of mechanics this cor respondence becomes especially transparent when it is seen that the momentum is the infinitesimal generator of translations viewed as canonical transformations In the Hamil tonian framework the conservation of momentum is identified with the statement that the Hamiltonian is translationally invariant that is is unchanged by the canonical transfor mation generated by the momentum We shall see that this same logic applies in quantum mechanics lndeed nowadays momentum is mathematically identified 7 by definition 7 as the generator of translations Let us see how all this works Having defined the position generalized eigenvectors which represent idealized states in which the position of the particle is known with certainty we can define a trans lation operator Ta via Talia lac 1 Since the span the Hilbert space this defines the operator Note that exercise TaTb TM T0 I Physically we interpret this operator as taking the state of the particle and transforming it to the state in which the particle has been moved by an amount a in the positive x direction Since we demand that this operator map states into states we must require that they stay of unit length 1 ltwlwgt ltwlTJTal gt It can be shown that this requirement for all vectors forces for all a which we suppress here TlT TTl Iltgt Tl T4 1 96 Physics 6210Spring 2007Lecture 3 We say that the operator T satisfying this last set of relations is unitary Note that a unitary operator preserves all scalar products exercise Note that for position wave functions we have good exercise Ta ltlTal1jgtlt ah 1496 i I So moving the system to the right by an amount a shifts the argument of the wave function to the left To see that this makes sense consider for example a particle with a Gaussian wave function exercise The unitarity of Ta is expressed in the position wave function representation via exercise wltx7agt2 As we mentioned above momentum is identified with the infinitesimal generator of WW2 1 translations Thus consider an infinitesimal translation T5 6 ltlt 1 We assume that Ta is continuous in a so that we may expand the operator in a Taylor series EI76PO The mathematical definition of the Taylor series of an operator needs a fair amount of discussion which we will suppress For our purposes you can just interpret the expansion as meaning that any matrix elements of the operator can be so expanded The factor of 7 has been inserted for later convenience Here P is a linear operator called the in nitesimal generator of translations The unitarity and continuity of T implies that P is self adjoint lndeed considering the 06 terms you can easily see exercise I n LH O U7PO PPh In fact the self adjointness of P is also suf cient for T to be unitary This can be seen by representing Ta as an infinite product of infinitesimal transformations 22m 67w T a hN lim I 7 N700 It is not hard to check that any operator of the form em with Al A is unitary exercise Thus P represents an observable which we identify with the momentum of the particle The canonical commutation relations We now consider the commutation relation between position and momentum We can derive this relation by studying the commutator between position and translation operators Technically the infinitesimal generator is 7P but it is a convenient abuse of terminology 7 and it is customary 7 to call P the infinitesimal generator 2 Physics 5210Spring 2007Lecture 8 and then considering the limit in which the translation is infinitesimal Check the following computations as an exercise XTElac ac 6lac 6 TEXlac Julio 6 Subtracting these two relations taking account of the definition of momentum and working consistently to first order in 6 we have exercise Xlt7Pgtxgt 7 7PgtXlxgt 7 m This implies 7 keep in mind that is a generalized basis 7 X P oil This relation along with the trivial relations XaXl 0 P7Pl7 constitute the canonical commutation relations for a particle moving in one dimension Note that these commutation relations show us that position and momentum are in compatible observables They thus satisfy an uncertainty relation 1 AX2AP2 2 1222 This is the celebrated position momentum uncertainty relation It shows that if you try to construct a state with a very small dispersion in X or P then the dispersion in P or X must become large Note also that the uncertainty relation shows the dispersion in position or andor momentum can never vanish However either of them can be made arbitrarily small provided the other observable has a suf ciently large dispersion Momentum as a derivative It is now easy to see how the traditional representation arises in which momentum is a derivative operator acting on position wave functions Consider the change in a position wave function under an infinitesimal translation We have ma 7 lt1 7 gnaw 7 we 7 e 7 we 7 6 Comparing terms of order 6 we see that Page 2 man 7 Of course you can now easily verify the position wave function representation of the dime die 08 We idac canonical commutation relations XPiwltxgt ii496

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