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Quantum Mechanics and Atomic Physics 1

by: Mrs. Peter Toy

Quantum Mechanics and Atomic Physics 1 PHYS 3220

Marketplace > University of Colorado at Boulder > Physics 2 > PHYS 3220 > Quantum Mechanics and Atomic Physics 1
Mrs. Peter Toy

GPA 3.96

Steven Pollock

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Steven Pollock
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This 108 page Class Notes was uploaded by Mrs. Peter Toy on Friday October 30, 2015. The Class Notes belongs to PHYS 3220 at University of Colorado at Boulder taught by Steven Pollock in Fall. Since its upload, it has received 26 views. For similar materials see /class/232092/phys-3220-university-of-colorado-at-boulder in Physics 2 at University of Colorado at Boulder.


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Date Created: 10/30/15
SJP QM 3220 Ch 2 part 1 Page 1 Once again the Schrodinger equation 2 2 in 610090 h6 xat 6t 2m 6x2 V x t which can also be written II llXI if you like M once again assume V VX no t in there We can start to solve the PDE by SEPARATION VARIABLES Assume hope wonder if we might nd a solution of form I Xt uX pt Grif ths calls uX E llX but I can t distinguish a quotsmall wquot from the quotcapital 1 quot so easily in my handwriting You ll find different authors use both of these notations E Q at ux dt So Note full derivative on ri ht hand side 62111 W d2ux lt g 6x2 dx2 So Schrodinger equation reads with g E I and 2 u l x 2h m uquot Vu Now diVide both sides by I u p 17 W i quotx Vx I50 2m M06 EH function function of of 3 This is not possible unless both sides are constants ConVince yourself that is the key to the quotmethod of separation of variablesquot Let s name this constant quotEquot 2 Note units of E are 20r7 or simply VX either way check it s Energy 2m39dist time S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 SJP QM 3220 Ch 2 part 1 Page 2 So 1 thitE t 2 2 liuquotx Vxux E39ux 4 These are ordinary ODE s m Equation 1 is about as simple as ODE s get Check oeiEth b any constant it s a linear ODE lst order linear ODE is supposed to give w undetermined constant right This is quotuniversalquot no matter what VX is once we nd a 11X we ll have a corresponding Px 1 ux t Joe mun But be careful that uX depends on E h2 2 uquotx Vxux E39ux 2m T d2 this is L This is the quottime independent Schrodinger equationquot dx You can also write this as 1 E 39 ux which is an quoteigenvalue equationquot A h2 d2 H quotHamiltonianquot operator i72 Vx 2m dx In general I 1106 E 6 has many possible solutions f eigenfunctions eigenvalues 111X 112X unX may all work each corresponding to some particular eigenvalue E1 E2 What we will nd is not any old E is possible if you want uX to be normalizable and well behaved Only certain E39s the En s will be ok Such a uEX is called a quotstationary state of Hquot Whv Let s see S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 SJP QM 3220 Ch 2 part 1 Page 3 Notice that I nXI corresponding to url is 1Pquot x t unxe39iE 39h 4 go back a page 2 uquot m time dependence 2 So mos t for the probability density It39s not evolving in time it s quotstationaryquot 2 1E tlh quot l Conv1nce yourself Because e If you think back to de Broglie39s free particle whee Aeikxeiwr with E ha it looks like we had stationary states I nXI with this same simple time dependence eimt with 0 Equot h This will turn out to be quite general If you compute the expectation value of any operator Q for a stationary state the Edh in I multiplies the elmh in 1 and goes away This is assuming the operator Q depends on X or p but not time explicitly A A h 6 Q f 1 xtQx I nxtdx 1 6x In noquot 111 again no time dependence Stationary state are dull nothing about them that s measurable changes with time Hence they are quotstationaryquot gtNot all states are stationarv iust these sgecial states the I 39s And remember Hun x Enun x 4 Time Indep Schrod eq n 2 andal llincx En Pncx This is still the Schrodinger Equation after plugging in our time solution for pt S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 SJP QM 3220 Ch 2 part 1 Page 4 Check th out though for a state 1 Xt f I IinPndx f PEn1Pndx In stationary q En E Pndx E constants come out this is normalization of integrals So the state 1 quothas energy eigenvalue Enquot and has expectation value of energy E amp ltf12gtf1 1 12111n fw lvndx imsy E 3115101 dx givesE 19 I S0 of ltH2gt ltHgt2 E Ej 0 Think about this 7 there is zero uncertainty in the measurement of H for this state Conclusion 1 is a state with a de nite energy E no uncertainty that s really why we picked the letter E for this eigenvalue quot 6 11 Remember H IJ 171 is linear so at If 1 1 and 1 are solutions so is a I 1 b I z But this linear combo is not stationary It does not have any quotenergy eigenvaluequot associated with it S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 SJP QM 3220 Ch 2 part 1 Page 5 The most general solutions of IfllP ihaaiqjgiven V Vx is thus I 1Fgeneralx t ZCnIPnOCJ 2 cnunxeiEnth t EH amp an energy I I eigenstate Simple not stationary time dependence It39s a quotmixed energyquot for eaCh term energy eigenstate or quotstationary statequot constants you like real or complex Just so long as you ensure I is normalized Any every physical state can be expressed in this way as a combination of the more special stationary states or quoteigenstatesquot If you measure energy on a stationary state you get En de nite value if you measure energy on a mixed or general state we need to discuss this further Hang on S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 Ch 1 notes partl l of 26 Quantum Mechanics Quantum Mechanics Introductory Remarks QM is a new and absolutely necessary way of predicting the behavior of microscopic objects It is based on several radical and generally also counterintuitive ideas 1 Many aspects of the world are essentially probabilistic not deterministic 2 Some aspects of the world are essentially discontinuous Bohr quotThose who are not shocked when they first come across quantum theory cannot possibly have understood it quot Humans have divided physics into a few arti cial categories called theories such as classical mechanics nonrelativistic and relativistic electricity amp magnetism classical version quantum mechanics nonrelativistic general relativity theory of gravity thermodynamics and statistical mechanics quantum electrodynamics and quantum chromodynamics relativistic versions of quantum mechanics Each of these theories can be taught without much reference to the others Whether any theory can be learned that way is another question This is a bad way to teach and view physics of course since we live in a single universe that must obey one set of rules Really smart students look for the connections between apparently different topics We can only really learn a concept by seeing it in context that is by answering the question how does this new concept fit in with other previously learned concepts Each of these theories nonrelativistic classical mechanics for instance must rest on a set of statements called axioms or postulates or laws Laws or Postulates are statements that are presented without proof they cannot be proven we believe them to be true because they have been experimentally veri ed Eg Newton39s 2quotd Law Fm ma is a postulate it cannot be proven from more fundamental relations We believe it is true because it has been abundantly verified by experiment Actually Newton39s 2quotd Law has a limited regime of validity If you consider objects going very fast approaching the speed of light or very small microscopic atomic then this quotlaw begins to make predictions that con ict with experiment However within its regime of validity classical mechanics is quite correct it works so well that we can use it to predict the time of a solar eclipse to the nearest second hundreds of year in advance It works so well that we can send a probe to Pluto and have it arrive right on target right on schedule 8 years after launch Classical mechanics is not wrong it is just incomplete If you stay within its wellprescribed limits it is correct 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes part1 2 of 26 Each of our theories except relativistic Quantum Mechanics has a limited regime of validity As far as we can tell to date QM relativistic version is perfectly correct It works for all situations no matter how small or how fast Well this is not quite true no one knows how to properly describe gravity using QM but everyone believes that the basic framework of QM is so robust and correct that eventually gravity will be successfully folded into QM without requiring a fundamental overhaul of our present understanding of QM String theory is our current best attempt to combine General Relativity and QM some people argue quotString Theoryquot is perhaps not yet really a theory since it cannot yet make many predictions that can be checked experimentally but we can debate this Roughly speaking our knowledge can be divided into regimes like so A small A QM Relativistic nonrelativistic QM 1 size Mechanics Relativistic nonrelativistic Mechanics b1g 0 c speed In this course we will mainly be restricting ourselves to the upper left quadrant of this figure However we will show how nonrelativistic QM is completely compatible with non relativistic classical mechanics We will show how QM agrees with classical mechanics in the limit of macroscopic objects In order to get some perspective let s step back and ask What is classical mechanics CM It is most simply put the study of how things move Given a force what is the motion So CM studies ballistics pendula simple harmonic motion macroscopic charged particles in E and B fields etc Then one might use the concept of energy and conservation laws to make life easier This leads to new tools beyond just Newton39s laws eg the Lagrangian L and the HamiltonianH describe systems in terms of different but still conventional variables With these CM becomes more economical and solving problems is often simpler At the possible cost of being more formal Of course what one is doing is fundamentally the same as Newton39s Fma 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 3 of 26 The equations of motion are given in these various formalisms by equations like n d8L 9L 9x 17quot ii i0 or or Fma clth 9x 9px If you39ve forgotten the Lagrangian or Hamiltonian approaches its ok for now Just realize that the general goal of CM is to nd the equation of motion of objects Given initial conditions final xt analpt position anal momentum as a unction 0 time Then you can add complications E g allow for more complicated bodies which are not pointlike ask questions about rotation introduce the moment of inertia and angular moment Lrxp move to manybody systems normal modes etc QM is about the same basic thing Given a potential what is the motion It39s just that QM tends to focus on small systems Technically systems with small action I At lt21 h And the idea of quotmotionquot will have to be generalized a bit as we shall see soon Having just completed CM your initial reaction may be quotbut size doesn39t matterquot After all neither L nor H cares about size and CM often deals with so called quotpoint objectsquot Isn39t a point plenty small Unfortunately it turns out that in a certain sense everything you learned in 2210 anal 3210 is WRONG To be a little more fair those techniques are fine but only if applied to realworld sized objects As I said above there s a regime of validity Size doesn39t matter up to a point but ultimately CM breaks down if you try to apply the 3210 Lagrangian or Hamiltonian formalism to an electron in an atom or an atom in a trap or a quark in a proton or a photon in a laser beam or many other such problems you will fail big time It39s not just that the equations are wrong You can t patch them up with some clever correction terms or slight modifications of the equations like relativity does at high speeds The whole MIND SET is wrong You cannot ask for x t anal pt It39s not well defined Point particles do not exist Particles have a wave nature and waves have a particle nature There is a duality in the physical world which is simply not classical So we must start from scratch and develop a whole new framework to describe small systems There are many new ideas involved Some are formal and mathematical some are rather unintuitive at least at first I will try to motivate as much as possible and we ll study plenty of examples Quantum mechanics comes from experiment Feynman says that the one essential aspect to learn QM is to learn to calculate and we will basically follow this idea 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 4 of 26 QM is great fun very weird sometimes mysterious Philosophers still argue about what it all means but we will take a quotphysicist s viewquot mostly Issues of interpretation can come later As a colleague of mine once explained it s kind of like trying to learn Swahili slang First you must learn a new language and then you must learn a new culture and only then can you nally begin to truly understand the slang The Postulates of Quantum Mechanics The laws axioms postulates of Classical Mechanics are short and sweet Newton39s Three Laws You might add quotconservation of energyquot if you want to extend CM to include thermodynamics You can add two more postulates that the laws of physics are the same in all inertial frames and the speed of light is constant to extend CM to include special relativity The laws of classical electricity amp magnetism which I might argue still falls under an umbrella of Classical Mechanics are similarly short and sweet Maxwell39s equations plus the Lorentz force law Alas there is no agreement on the number the ordering or the wording of the Postulates of Quantum Mechanics Our textbook Grif ths doesn39t even write them down in any organized way They are all in there but they are not welllabeled and not collected in any one place Grif ths sometimes indicates Postulate by putting the statement in a box Quantum Mechanics has roughly 5 Postulates They cannot be stated brie y when stated clearly they are rather longwinded Compared to Classical Mechanics quantum mechanics is an unwieldy beast 7 scary and ugly at rst sight but very very powerful As we go along I will write the Postulates as clearly as I can so that you know what is assumed and what is derived Writing them all down now will do little good since we haven t yet developed the necessary vocabulary I will begin by writing partially correct but incomplete or inaccurate versions of each Postulate just so we can get started Later on when ready we will write the rigorously accurate versions of the postulates Don39t worry if these seem rather alien and unfamiliar at rst this is really the subject of the entire course we re just making our rst pass getting a kind of overview of where we re heading So let39s start 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 5 of 26 Postulate 1 The state of a physical system is completely described by a complex mathematical object called the wavefunction 1P psi pronounced quotsighquot At any time the wavefunction l11x is singlevalued continuous and normalized The wave function l11x is not quotthe particlequot or quotthe position of the particlequot it is a mathematical function which carries information about the particle Hang on In this course we will mostly be restricting ourselves to systems that contain a single particle like one electron In such a case the wavefunction can be written as a function of the position coordinate iquot of the particle and the time 1P l117 t Often we will simplify our lives by considering the rather artificial case of a particle restricted to motion in 1D in which case we can write 1P 1Px t We may also consider a particular moment in time and focus on just l11x In general l11x is a complex function of x it has a real and an imaginary parts So when graphed it looks something like Rem l IHIPP X X In fact it can look like anything so long as it is continuous and normalized Definition A wavefunction is normalized if f lWXtl2 dx 1 There are many different ways to write the wavefunction describing a single simple spinless particle in 1D at some time IJX ltIgtp and others to be explained later Here x is position and p is momentum If the particle has spin then we have to include a spin coordinate m in addition to the position coordinate in the wavefunction 1 Ilfm If the system has 2 particles then the wavefunction is a function of two positions 1 1 72 Postulate 2 has to do with operators and observables and the possible results of a measurement We will just skip that one for now 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 6 of 26 Postulate 3 has to do with the results of a measurement of some property of the system and it introduces indeterminacy in a fundamental way It provides the physical interpretation of the wavefunction Postulate 3 If the system at time t has wavefunction1Fxt then a measurement of the position X of a particle will not produce the same result every time I xt does not tell where the particle is rather it give the probability that a position measurement will yield a particular value according to lIIJOc tl2dx Probability particle will be found between X and XdX at time t An immediate consequence of Postulate 3 is f xt 2 dx Probparticle will be found between x1 and x2 Since the particle if it exists has to be found somewhere then Probparticle will be found between 700 and 00 1 Hence the necessity that the wavefunction be normalized f I Ijxtl2 dx 1 This QM description is very very different from the situation in classical mechanics In classical mechanics the state of a oneparticle system at any given instant of time is determined by the position and the momentum or velocity 713 So a maximum of 6 real numbers completely describes the state of a classical singleparticle system Only 2 numbers X and p are needed in 1D In contrast in QM you need a function To specify a function you need an in nite number of numbers And it s a complex function so you need 2 X 00 numbers In classical mechanics the particle always has a precise definite position whether or not you bother to measure its position In quantum mechanics the particle does not have a de nite position until you measure it The Conventional Umpire quotI calls em as I see 39emquot The Classical Umpire quotI calls em as they Equot The Quantum Umpire quotThey ain39t nothing till I calls 39emquot In quantum mechanics we are not allowed to ask questions like quotWhat is the particle doingquot or quotWhere is the particlequot Instead we can only ask about the possible results of measurements quotIf I make a measurement what is the probability that I will get suchand such a resultquot QM is all about measurement which is the only way we ever truly know anything about the physical universe 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 7 of 26 Quantum Mechanics is fundamentally a probabilistic theory This indeterminacy was deeply disturbing to some of the founders of quantum mechanics Einstein and Schrodinger were never happy with postulate 3 Einstein was particularly unhappy and never accepted QM as complete theory He agreed that QM always gave correct predictions but he didn t believe that the wavefunction contained all the information describing a physical state He felt that there must be other information quothidden variablesquot in addition to the wavefunction which if known would allow an exact deterministic computation of the result of any measurement In the 6039s and 7039s well after Einstein39s death it was established that quotlocal hidden variablesquot theories con ict with experiment Postulates l and 3 are consistent with eXperment The wavefunction really does contain evthing there is to know about a physical system and it only allows probabilistic predictions of the results of measurements The act of measuring the position changes the wavefunction according to postulate 4 Postulate 4 If a measurement of position or any observable property such as momentum or energy is made on a system and a particular result X or p or E is found then the wavefunction changes instantly discontinuously to be a wavefunction describing a particle with that definite value of X or p or E Formally we say quotthe wavefunction collapses to the I r iml cur r J39 to the I quot If you39re not familiar with this math terminology don39t worry we39ll discuss these words more soon If you make a measurement of position and find the value X0 then immediately after the measurement is made the wavefunction will be sharply peaked about that value like so 2 llpl I ll After measurement Before measurement V l Xe The graph on the right should have a much taller peak because the area under the curve is the same as before the measurement The wavefunction should remain normalized Postulate 1 states that the wavefunction is continuous By this we mean that lI Xt is continuous in space It is not necessary continuous in time The wavefunction can change discontinuously in time as a result of a measurement Because of postulate 4 results of rapidly repeated measurements are perfectly reproducible In general if you make only one measurement on a system you cannot predict the result with certainty But if you make two identical measurements in rapid succession the second measurement will always con rm the first 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes part1 8 of 26 QM is infuriatingly vague about what exactly constitutes a quotmeasuremen quot How do you actually measure position or momentum or energy or any other observable property of a particle For a position measurement you could have the particle hit a uorescent screen or enter a bubble chamber For a momentum or energy measurement it39s not so clear More on this later For now quotmeasurementquot is any kind of interaction between the microscopic system observed and some macroscopic manyatom system such as a screen which provides information about the observed property Postulate 5 the last one describes how the wavefunction evolves in time in the absence of any measurements Postulate 5 The wavefunction of an isolated system evolves in time according to the Schrodinger Equation where V VX is the classical potential energy of the particle which depends on the physical system under discussion One s first reaction to Postulate 5 is quotWhere did that come fromquot How on earth did Schrodinger think to write that down We will try to make this equation plausible coming soon and show the reasoning that lead Schrodinger to this Nobelprizewinning formula But remember it s a Postulate so it cannot be derived We believe it is true because it leads to predictions that are experimentally verified 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 9 of 26 Statistics and the Wavefunction Because QM is fundamentally probabilistic let39s review some elementary statistics In particular to start let39s consider random variables that can assume discrete values Suppose we make many repeated measurements of a random discrete variable called X An example of X is the mass rounded to nearest kg or height rounded to the nearest cm of a randomlychosen adult We label the possible results of the measurements with an index i For instance for heights of adults we might have X125 cm X2 26 cm etc no adult is shorter than 25 cm The list X1 X2 Xi is the called the spectrum of possible measurement results Notice that Xi is not the result of the im trial the common notation in statistics books Rather Xi is the i3911 possible result of a measurement in the list of all possible results N total of measurements ni times that the result Xi was found among the N measurements Note that N E ni where the sum is over the spectrum of possible results not over the N I different trials In the limit of large N which we will almost always assume then the probability of a n part1cular result Xi 1s fractlon of the tr1als that resulted 1n Xi The average of many repeated measurements of X expectation value of X rix l l sum of results of all trials n x number of trials N 212x i The average value of X is the weighted sum of all possible values of X x 2 xi Again this is called the expectation value of x even though you might e g NEVER find any particular individual whose height is the average or quoteXpectedquot height We can generalize this result to any function of X x2 2 P x ltrxgt 2 mo 2quot The brackets means quotaverage over many trialsquot We would call this the quotexpectation value ofx A measure of the eXpected spread in measurements of X is the standard deviation 0 defined as quotthe rms average of the deviation from the meanquot quotrmsquot rootmeansquare take the square average that then squareroot that 0 J ltx x2gt O2 is called the variance 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 10 of 26 Let us disassemble and reassemble The deviation from the mean of any particular result X is Ax x x The deviation from the mean is just as likely to be positive as negative so if we average the deviation from the mean we get zero ltx x 0 To get the average size of AX we will square it first before taking the average and then later squarerootit 0 ltx x2gt It is not hard to show that another way to write this is a x2x2 There are times when this way of finding the variance is more convenient but the two definitions are mathematically equivalent Proof ltxltxgt gt geemin 2xr2xiltxgtltxgt2n o2 a 2ltxgt2xsz gt221 ltx2gt 2ltxgt2ltxgt2 x2 x quot 1 o2 XZgt x2 Now we make the transition from thinking about discrete values of X say X l 2 3 to a continuous distribution e g X any real number We define a probability density pX I pX dX Prob randomly chosen X lies in the range X gt XdX In switching from discrete X to continuous X we make the following transitions pXidx 2R 1 gt jpxdxl x2lel gt xfxpxdx1 Please look again at these equations on left and right think about how they quotmatch up and mean basically the same thing We39ll use both sides throughout this course From Postulate 3 we make the identification l11x t 2 dx pxdx and we have x fxl llx2dx So in QM the eXpectation value of the position X of a particle with given wave function 1P is given by this simple formula for ltXgt It39s the quotaverage of position measurementsquot if you had a bunch of identically prepared systems with the same wave function 11 More generally for any function f fX we have fx ffxyxl2 dx 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl ll of 26 Grif ths gives an example 1 l of a continuous probability distribution Let39s redo that example just slightly modi ed to help make sense of it Take a look at it rst though A rock released from rest at time t0 falls a distance h in time T 1 2 1 2 x 2 gr h 2 gT A move is taken as the rock falls from t0 a T at 60 frames sec resulting in thousands of photos of the rock at regularlyspaced time intervals The individual frames are cut out from the lm and then shuf ed Each frame corresponds to a particular X and t and a particular dx and dt dx might show up visually as a smear since the rock moved during the short time that picture frame was taken All frames have the SAME dt but different frames have different dx39s dxdt gt gt dx g t dt We can de ne the probability distribution in space px and the probability distribution in time 1t pxdx Prob frame chosen at random is the one at X a xdx 1tdt Prob frame chosen at random is the one att a tdt Here39s a little picture that might help a dxe dx h I I I I I I gt I I I I quotI I I I I I I I gt a e k t dt dt To be precise I should really be writing Ax instead of dx and At instead of dt In the end I39ll take the limit At 90 Notice all the dt39s are the same size but the dx39s start out short and get longer and longer Now 1tdt dtT that is 1t constant lT Convince yourself That39s because any random frame is equally likely to be at any given time early middle late So the probability 1t needs to be constant But why is it lT That39s to ensure that the total probability of the frame being somewhere between 0 and T is exactly one T 1T 1 f Etdt fdt Tl 0 T0 T Each frame is equally likely and the probability of grabbing one particular frame is proportional to UT It is also proportional to dt if the frames are all longer there are fewer overall and the probability scales accordingly Convince yourself 8272008 University of Colorado Michael Dubson mods by S Pollock x TtdxdtlTgt But we know T 2h g and t 1 2X g see our kinematics equations above So pXlTgt VgZh 1 g 2xg 2 hx That s what Grif ths got thinking about it slightly differently Check out his derivation too The key formula in this problem is IO dx dt It is vital to remember that when using this formula X and t are not independent The X is the X which corresponds to the particular t and dX is the interval in X which corresponds to the dt of that quotframequot 7tdt pXdX gt pX By the way you might be uncomfortable treating dXdt as though it was just a fraction AXAt But we often quotpull apartquot dXdt and writ things like di fx gt dX fXdt or dt 1 dx dxdt This makes sense if you remember that E lim E dt Atgt0 At To physicists dXdt really is a tiny AX dX divided by a tiny At dt 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes part1 13 of 26 Complex Number Review Wave functions 1P are in general complex functions So it39s worth a quick review of complex numbers since we39ll be dealing with this all term i 1 ii 1 gt i 1i gt i4 1 Any complex number 2 can always be written in either Cartesian form Z 0739 Polar form Z A619 lm You can visualize a complex number by thinking of it a point in the complex plane A 39lt This picture also matches up with one of the most important 8 theorems of complex numbers Euler s relation J Re e 9 cosH isinB Which can be proven w a Taylor Series expansion if you like This means that Rez x A cosG Imz y A sine Again look at the picture above do you see the connections The complex conjugate of Z is called 2 xiy 1m which is also 2 Ae ia Note that Z zfkxiyx iyxziyx ixyy2 Re x2 y2 purely real 2 We therefore call Z quotmodulus of Zquot or quotamplitude of zquot and de ne it as Izl 1x2y2 A Note that z z z2 Also notice that z2 z z 7 z2 z z Squaring complex numbers does NOT always yield a real result and in general is quite different than multiplying by the complex conjugate ie the square of a complex number is DIFFERENT from the square of the amplitude of that number 8272008 3 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 14 of 26 Here s a useful fact 21 22 21 22 e ee where Z1 22 are any 2 complex numbers This means in particular that 6M4 6 39 equot which in turn can be used to derive various trig identities like e g that cosab cosacosbsinasinb just look at the real part of the equation 396 396 Also if 21 Alel 1 22 1426l 2 then it is very quick and easy to nd the product Z1Z2 A1142elw1 62 One more useful fact about complex numbers Any complex number 2 written as a complicated expression no matter how messy can be turned into its complex conjugate 2 by replacing every i with i so e g 5 6i 7i 5 6i7i Z Z 39 yyw 2i3em 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes part1 15 of 26 Classical Waves Review QM is all about solving a wave equation for 1pxt But before learning that let39s quickly review classical waves If you39ve never learned about waves in an earlier physics class take a little extra time to be sure you understand the basic ideas here A wave a selfpropagating disturbance in a medium y A wave at some moment in time is described by y fx displacement of the medium from its equilibrium position X Claim For any function yfx the function yxt fxvt is a1dimensional traveling wave moving rightward with speed v If you ip the sign you change the direction We will prove the claim in a couple of pages but rst let s just make sense of it Example 1 A gaussian pulse y fx Ae39xzZ jz If you are not familiar with the Gaussian function in the above equation stare at it and think about what it looks like It has max height A which occurs at x0 and it has quotwidthquot 0 Sketch it for yourself be sure you can visualize it It looks rather like the form shown above A traveling gaussian pulse is thus given by yxt fxvt fie H392 quot202 Note that the peak of this pulse is located where the argument of f is O which means check the peak is where xvt0 in other words the peak is always located at position xvt That39s why it39s a traveling wave Such a wave is sometimes called a traveling wave packet since it s localized at any moment in time and travels to the right at steady speed x Example 2 A sinusoidal wave y fx As1n2ui This one is probably very familiar but still think about it carefully A is the amplitude or maximum height Y I The argument changes by 27 exactly one quotcyclequot t0 I whenever x increases by L That39s the length of V I the sin wave or quotwavelengthquot of course Now think about the traveling wave l X yxt fxvt try to visualize this as a movie the A wave looks like a sin wave and slides smoothly to the right at speed v Can you picture it 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 16 of 26 Review of sinuosoidal waves 2n For s1ne waves we de ne k 7 7 wave numberquot k has un1ts quotrad1ansmeterquot k is to wavelength as angular frequency 00 is to period T Recall or much better yet rederive a 2n T 2713 angular frequency radssec Remember also frequency f cycles time l cycletime for 1 cycle lT In the previous sketched example the traveling sin wave yX A sinkX gt yXy A sinkXvt Let s think about the speed of this wave v Look at the picture when it moves over by one wavelength the sin peak at any given point has oscillated up and down through one cycle which takes time T one period right That means speed v horizontal distance time 2 T 2 f So the argument of the sin is 271 A x t kx vt7x t2nE kx out Don39t skim over any of that algebra Convince yourself this is stuff we ll use over and over Summarizing for our traveling sin wave we can write it several equivalent ways x t y A 39 k A 39 7 7 xt s1n x vt s1n2 L Asinkx wt The argument of the sign changes by 27 when X changes by 2 or t changes by T The wave travels with speed v A T nk We ll use these relations all the time Please check units to make sure it s all consistent Technically this speed v ook is called the phase velocity because it s the speed at which a point of constant phase like say the zero crossing or first peak or whatever is moving Soon we will discover for some waves another kind of velocity the group velocity Never mind for now I said that fXvt represents a traveling wave 7 it should be reasonable from the above pictures and discussion but let s see a formal proof 7 next page 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 17 of 26 Claim yXt fX i vt represents a rigidly shaped quotdispersionlessquot traveling wave The upper sign gives you a LEFTmoving wave The sign is what we39ve been talking about above a RIGHTmoving wave Proof of Claim Consider such a traveling wave moving to the right and then think of a new moving coordinate system X39y39 moving along with the wave at the wave39s speed v y I y Here Xy is the original coordinate system And X y is a new moving coordinate system V traveling to the right at the same speed as the X wave A k r X Let s look at how the coordinates are related Look at some particular point the big black dot It has coordinates Xy in the original frame It has coordinates X39y39 in the new frame But it39s the same physical point Stare and convince yourself that xX39vt and yy39 That39s the cordinate transformation we re after or turning it around X39Xvt and y39y Now in the moving x 39 y frame the moving wave is stationary right Because we39re moving right along with it It39s very simple in that frame y In this frame the Xy axes are running y away from us off to the left at speed v y39fx39 but never mind v The point is that in this frame the wave is simple y39fX39 at all times It just sits there X If y39fX39 we can use our transformation to rewrite this y39y X39Xvt giving us yfXvt This is what I was trying to prove this formula describes the waveform traveling to the RIGHT with speed v and fixed quotshapequot given by f 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 18 of 26 In classical mechanics many physical systems exhibit simple harmonic motion which is what you get when the displacement of a point object obeys the equation F kx or d2xt dt2 If you have a bunch of coupled oscillators like a rope or water or even in free space with oscillating electric elds you frequently get a related equation for the displacement of the medium yXt which is called the wave equation w2xt Hopefully that looks pretty familiar 8 v2 91 2 I If you re curious go back to your mechanics notes it s likely you spent a lot of time deriving and discussing it 92 In Just one spat1al d1mens1on think of a string that equation is 8732 x Theorem Any 1D traveling wave of the form yXt fX i Vt is a solution of the wave equation above Proof We are assuming yXt flt where Xt Xvt and we re going to show no matter what function flt you pick that this yXt satisfies the wave equation This is just the chain rule and I used the fact that 1 9x d 9x d 9x Please make sense of where I write partials and where I write full derivatives Now do this again 2 2 9 y 9 1 dijzc 1 Once againusing 1 d d d 9x d 9x 9x2 3 Similarly we can take time derivatives again using the chain rule v i here I used the fact that g v you see why that is 91 d 9t d 91 And again repeat the time derivative once more 2 2 8733 v vi Vi vV2LJ 2 9t 9t d d d 9t d d d 2 Using 1 and 2 to express dig two different ways gives what we want dzf i d 2 122 9t2 9x2 That last equality is the wave equation so we39re done Again ANY lD traveling wave of the form fX i vt solves the wave equation and the wave equation is just a very basic equation satisfied by MANY simple linear systems built up out of coupled oscillators which means much of the physical world 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 19 of 26 2 2 11 where quotEquot can be Eiy or E72 here and V 1 80 c the speed of light 3E8 ms So we re saying that EM waves do NOT have to be quotsinusoidal wavesquot they can be pulses or basically any functional shape you like but they will all travel with the same constant speed c and they will not disperse or change shape in vacuum Example 1 Maxwell39s equations in vacuum give Example 2 A wave on a 1D string will satisfy 772 where y represents the displacement of the T v2 9t rope and X is the position along the rope and the speed is given by v m masslength So here again on such a string wave pulses of any shape will propagate without dispersion the shape stays the same and the speed is determined NOT by the pulse but by the properties of the medium the rope it s tension and mass density 92y 1 92y x2 Superposition Principle If ylXt and y2Xt are both separately solutions of the wave equation then the function yly2 is also a valid solution This follows from the fact that the wave equation is a LINEAR differential equation Look back at the wave equation write it separately for yl and y2 and simply add We can state this a little more formally if 92y 1 92y A 7 770 gt L xt 0 8x2 V2 2 Y Here we are defining a linear operator L which does something to FUNCTIONS A 92 1 92 L 1 2 7 2 9x v 9t The key properties for any linear operator are that 101 3 21 L 1 L 3 2 and L cy cLy for any constant c Reminder Functions are things which take numbers in and give out numbers like fX y Here X is the quotinput numberquot and y is the quotoutput numberquot That s what functions ARE Now we have something new which will reappear many times this term we have an operator which takes a function in and gives a function out yx gx Here yX was the input function the operator operates on this function and gives back a different function out gX 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 20 of 26 Wave Nature of Light and Interference Because of the superposition principle waves add just as you would expect That is if you send two waves quotdown a stringquot they just add or cancel as simply as ytotal y1y2 This leads to constructive and destructive interference one of the de ning characteristic properties of waves Following you will nd some notes from a freshman course reviewing interference in case you ve forgotten For instance how is that you get an interference pattern from two slits The math and physics here will apply directly in quantum mechanics because particles also exhibit wavelike behaviour In general wavelike effects with light are difficult to detect because of the small wavelength of visible light 400 nm violet gt 700 nm red The problem is even tougher with particles it requires very special circumstances to demonstrate the wave nature of matter So in many situations light behaves like a ray exhibiting no obvious wavelike behavior Light passing through hole in wall gt wavefronts M gt l gtgtgtgtE gt A tiny hole D z 2 wavebehavior big hole D gtgt 2 raybehavior Newton late 1600 s did not believe that light was a wave since he always observed ray behavior Wavelike behavior was not clearly observed until around 1800 by Young Wavelike behavior of particles was not clearly observed until around 1925 in Davisson and Germer s experiment using a crystal of nickel as a quotgratingquot 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 21 of 26 Review of ConstructiveDestructive interference of Waves Consider 2 waves with the same speed v the same wavelength 2 and therefore same frequency f c 2 traveling in the same or nearly the same direction overlapping in the same region of space If the waves are in ph they add gt constructive interference V If the waves are out of phase they subtract gt destructive interference x If wave in nearly the same direction add subtract WW 7 7 plane ane spherical wave same 2 f Hu en39s Princi le Each oint on yg P P speed c I c a wavefront of given f 2 can be considered to be the source ofa gt spherical wave gt lt A A c wall with in nitesimal hole To see interference of light waves you need a monochromatic single A light source which is coherent nice clean plane wave This is not easy to make Most light sources are incoherent Qumble of waves with random phase relations and polychromatic many different wavelengths 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 22 of 26 Young39s Double slit experiment 1801 monochromatic plane wave 7 H Id gt A lt 2 slits screen What do you expect to see on the screen If Intensity 1 you believe light is a ray then you expect to see 2 bright patches on the screen one patch position X of light from each slit on screen d But here is what you actually see A series of bright and dark fringes wave interference W x How do we explain this Consider the 2 slits as 2 coherent point sources of monochromatic light Two sources are coherent is they have the same wavelength 2 and therefore the same frequency f and they emit peaks and troughs in sync in phase Each slit source emits light in all forward directions but let us consider only the parts of the waves heading toward a particular point on the screen 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 23 of 26 If the screen is far away L gtgt d then the rays from the two slits to the same point on the screen are nearly parallel both heading in the same direction at the same angle 6 to screen d d 39 6 The ray from the lower slit has to travel further by an extra distance d sine to reach the screen This extra distance is called the path difference When the path difference pd is one full wavelength or 2 full wavelengths or an integer number of wavelengths then the waves will arrive inph at the screen There will be constructive interference and a bright spot on the screen pd dsinQ m m0l2 constructive interference But if the path difference is 12 wavelengths or 32 wavelengths etc then there will be destructive interference at the screen and the screen will be dark there pd dsinQ m 12t m0l2 destructive interference Notice that the formula pd d sine is NOT a definition of path difference It is a formula for path difference in a specific situation namely when the screen is quotat infinityquot The definition of path diff is pd distance to one source 7 distance to other source A plot of brightness intensity vs angle position on the screen I intensity 1112 11 m m2 39 39 39 sinese d Kd ZKd Maxima at angles where sine m 2 d s 6 rads Recall sine s 6 rads if 6 ltlt 1 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes part1 24 of 26 Young s experiment was the first real proof that light is a wave If you believe that light is a ray there is no way to explain the destructive interference seen on the screen In the ray view when you hit a screen with two rays the brightness of the 2 rays M adds and you see a bright spot there It is impossible to explain destructive interference of two light sources unless you admit that light is a wave The same thing happened in the 1920 s with Davisson and Germer s experiments with electron diffraction here with the even more bizarre conclusion that forces you to admit that electrons are quotwavesquot too Single Slit Diffraction quotDiffractionquot interference due to infinitelymany Slit lled with imaginary sources packed infinitely close via Huygen s Huygen39s sources lt Principle Huygen39s Principle says that a slit that K is illuminated by a plane wave can be consider to be filled with an array of coherent point sources U gt Iloooooooll Consider the light from just two of the infinitelymany sources one at the top of the slit and one exactly in the middle of the slit 6 When the path difference between these two sources and the screen is 12 wavelength that E D A D2 g is when isin i then the light from D Z to screen 2 2 g aymsme 0 these two source interfere destructively and 8 no light from those two sources illuminates the screen at that particularly angle 6 But notice that all the sources can grouped in pairs with each pair s members D2 apart The light from all the sources the entire slit cancel in pairs and D A D2 there is no light at the position on the screen at the angle 6 such that Esme 00000000000000 MM The angle sine sin is the first intensity minimum on the screen 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes part1 25 of 26 The intensity pattern on the screen looks like this I intensity 39 sin 6 s 6 K D 0 A D 2 D 2 D The angular width of the quotcentral maximumquot is 0 2A D Notice that in the limit D gt 2 slit width becomes as small as the wavelength of light the central maX becomes so broad that we get spherical wave behavior H I gtgt U ml V V Dgt Diffraction Grating A diffraction grating is an array of many na1row slits with a uniform interslit spacing d A grating with quot500 lines per cmquot has a slit separation of d 15 71 0002cm 002mm 20m A typical diffraction grating has thousands of slits With exactly the same argument we used in the doubleslit case we see that maximum brightness occurs when pddsin9 m m0l2 The maXima occur at the same angles as with a double slit of the same d but the peaks are much sharper and much brighter 8272008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes partl 26 of 26 As N the number of slits increases the width of each peak decreases and gets brighter Why 2 slits I intensity gtd lt 39 39 39 sin 0 s 0 d 0 Kd 2 d 3 d Nslit grating 1 1st order 2quotd order 3rd order d I I I I 3 e d 0 Kd 2 d 3 d With just 2 slits when we are M the maximum at the angle 6 A d then the waves from the two slits are m in phase and we have nearly complete constructive interference and nearly maximum brightness But with Nslits when we are near the angle 6 A d any two adjacent slits are nearly in phase but the next slit over is a little more out of phase and the next one over is even more out of phase With many slits if you are just a bit off the special angle for maximum brightness the phase differences among the slits quickly add up and gives destructive interference Another nice feature of the grating is that with many slits for the light to get through the pattern on the screen is brighter than in the doubleslit case 8272008 University of Colorado Michael Dubson mods by S Pollock SJP QM 3220 Ch 2 part 2 Page 1 The In nite Sguare Well Let s make a choice for VX which is solvable that has some approximate physical relevance VX0for0ltxlta VX oo elsewhere A r 0 a It39s like the limit of as this gets big height VX and this gets small a distance where it rises x r Classically A d F div is 0 in the middle free but big at the edges Like walls at the edges x Like an electron in a small length of wire 00 I a I free to move inside but stuck with large force at the ends prevents it from leaving Given VX we want to nd the special stationag states and then we can construct m physical state by some linear combination of those S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 SJP QM 3220 Ch 2 part 2 Page 2 Recall we re looking for unX such that IE Iflu x E u x and then lI ct 1t ce 7 or dropping quotnquot for a sec and then at last 2 u x Vxux E ux lgt lI gmmlm t Z c lp x t Inside the well 0 lt X lt a and VX 0 so in that region d2 2 E I E 2 ux E k2ux Here I simply de ned k 2 2 E h2 It39s just shorthand k E m so E ikz h 2m However I have used the fact that E gt 0 you can t ever get E lt Vmin 0 Convince yourself I know this 2quotd order ODE and its general solution is uXAsinkXBcost g ueikxBeikx Postulate I says uX should be continuous Now outside 0 lt X lt a VX gt 00 This is unphysical the particle can t be there So uX 0 at X 0 and X a This is a BOUNDARY CONDITION uX0A 0B 10soB0required uXaA sinka0 But now I can t set A 0 39cause then uX 0 and that s not normalized So sinka0iekn7Iawithn 123 Ah ha The boundary condition forced us to allow only certain k s Call them k M a nz zhz Zma2 h2 Then s1nceE ikz we get E 2m Note negative n s just redef1ne quotXquot it s not really a different state A sin kX is the same function as A sin kX Note n 0 no good because sin 0X 0 is not normalizable S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 SJP QM 3220 Ch 2 part 2 Page 3 Thus our solutions the quotenergy eigenstatesquot are u x Asink x Asin n 1 2 3 0 lt x lt a a and 0 elsewhere 2 2 and E n2 hzj 2ma For normalization fun x2dx 1gt A Z f sin2 dx 1 0 a 2 ConV1nce yourself then A 2 71s required a So u x 2 sin for 0ltXlta a a Note signphasequot out front is not physically important If you multiply unx by eie you have a wavefunction with the same luxl it39s physically indistinguishable So eg unx and funx are not quotdifferent eigenstatesquot n1 n2 n3 Energies are En These are E1 4E1 9E1 excited states 4E1 n2 nl This is ground state Energy is quantized due to boundary condition on U Energies grow like n2 Lowest energy is not 0 You cannot put an electron in a box and have it be at rest S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 SJP QM 3220 Ch 2 part 2 Page 4 Key properties to note many of which will be true for most potentials not just this one Energy eigenstates unX are I quotevenquot or quotoddquot with respect to center of box nl is even n2 is odd this alternates oscillatory and higl1er energy 4gt more nodes or zero crossings Here un has nl intermediate zeros a 0 ifn m I orthogonal 1 u xum xdx 6 lifnm check this if you don t know why sin forms orthogonal states work it out a I complete Dirichlet s Theorem the basis of quotFourier seriesquot says ANY function fX which is 0 at X 0 and X a can be written uniquely fx Zc u x 6 can always nd the cn s 2 Recall u JsianC here a a Fourier s trick nds those cn39s given an fX If fx 2 c zp xthen do quotthe trickquot Multiply both sides by quotpsiquot fx1px 2 c zp x1px Then integrate so ffxlpxdx Ec fw xwux x o n o This is am H our 111 is real so isn39t important in this examp e Note that only ONE term the one with nmi contributes in this sum All others Vanish This tells us ipping left and right sides cm ffx1p xdx 4 this is how you gure out cm s o The above features are quite general Not iust this problem S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 SJP QM 3220 Ch 2 part 2 Page 5 And last but not least this was all for unX but the full timedependent stationary states 5 i T h I are u xe or wr1t1ng it out 2 rianrthZmaz i Wm Fsin e h z a K a 39 this particular functional form is speci c to quotparticle in boxquot This is still a stationary state full solution to quotSchrod in a boxquot with de nite energy En These are special particular states quoteigenstates of Hquot The most general possible state of any quotelectron in a 1D boxquot then would be m linear combo ofthese 10061 2 c lp x t 2 C u xeiiEnth You can pick m cn39s you like even complex and this will give you all possible physical states You might choose the cn s to give a particular I Xt0 that you start with using Fourier s trick If lI c0 Ec u is given nd the cn s and then the formula at top of page tells you I Xt So given initial conditions we know the state at all times That s the goal of physics For other potentials game is same nd unX s and corresponding En s then form a linear combo at t0 to match your starting state and let it evolve 1pxt Ec u xe39i5quot h S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 SJP QM 3220 Ch 2 part 2 Page 6 One last comment If 1pxt Ec u xe39iEquot39h then normalization says fll lpdx 1 But when you expand I and I all cross terms vanish after integration fa xum xdx 5m leaving only the terms with nm which integrate simply Work it out fll lpdx 1 tells you 1 2 2 fonC uxu xeiEm39Equot39hdx a Note this collapses the whole integral to 5m thus emm 39 gt 1 26266 2 2 Similarly try it use Hwn Enwn to get H C 4 convince yourself 2 Cr Interpretation which we39ll develop more formally 1pxt Ec u xe39iEquot39h Ec lp om n says quotthe state I is a linear combination of the special stationa states the I n squot The cn39s carry information H E Cr 2E which looks like 211E lcnlz Probability that this particle39s energy would be measured En 2 l is just quotconservation of probabilityquot Cr so 2 This is a central concept in QM we ll come back to it often S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 Spin 12 Recall that in the Hatom solution we showed that the fact that the wavefunction I r is singlevalued requires that the angular momentum quantum nbr be integer 0 l 2 However operator algebra allowed solutions 0 12 1 32 2 Experiment shows that the electron possesses an intrinsic angular momentum called spin with K 12 By convention we use the letter s instead of for the spin angular momentum quantum number s 12 The existence of spin is not derivable from non relativistic QM It is not a form of orbital angular momentum it cannot be derived from 1 r X The electron is a point particle with radius r 0 Electrons protons neutrons and quarks all possess spin s 12 Electrons and quarks are elementary point particles as far as we can tell and have no internal structure However protons and neutrons are made of 3 quarks each The 3 halfspins of the quarks add to produce a total spin of 12 for the composite particle in a sense T Nx makes a single T Photons have spin 1 mesons have spin 0 the deltaparticle has spin 32 The graviton has spin 2 Gravitons have not been detected experimentally so this last statement is a theoretical prediction Spin and Magnetic Moment We can detect and measure spin experimentally because the spin of a charged particle is always associated with a magnetic moment Classically H a magnetic moment is de ned as a vector u associated with a loop of current The direction of p is perpendicular to the plane of the current loop i righthandrule and the magnitude is u iA i11r2 The connection between orbital angular momentum not spin and magnetic moment can Q be seen in the following classical model Consider a particle with mass m q In charge q in circular orbit of radius r speed v period T i1 V 3 i HiA qV xrl T T 21tr 27 2 7122010 Dubson Phys3220 langularmomentumlLpr mvr so vrLm and u q gr 2L m So for a classical system the magnetic moment is proportional to the orbital angular momentum it l orbital The same relation holds in a quantum system In a magnetic eld B the energy ofa magnetic moment is given by E 139 uz B assuming B B2 In QM LZ hm Writing electron mass as n1e to avoid confusion with the magnetic quantum number m and q 7e we have uz where m Z The quantity HE E 26 h is called the Bohr magneton The possible me energies of the magnetic moment in B B2 is given by Emb uz B uB Bm For spin angular momentum it is found experimentally that the associated magnetic moment is twice as big as for the orbital case Fl i g spin We use S m instead of L when referring to spin angular momentum This can be written u i m 2uBm The energy ofa spin in a eld is Espin 2uB Bm m me ilZ a fact which has been veri ed experimentally The existence of spin s 12 and the strange factor of 2 in the gyromagnetic ratio ratio of ii to was rst deduced from spectrographic evidence by Goudsmit and Uhlenbeck in 1925 Another even more direct way to experimentally determine spin is with a StemGerlach device next page 7122010 Dubson Phys3220 This page from QM notes of Prof Roger Tobin Physics Dept Tufts U SternGerlach Experiment W Gerlach amp 0 Stem Z Physik 9 349252 1922 F vgm pw 132 De ection of atoms in zdirection is proportional to zcomponent of magnetic moment uz which in turn is proportional to L2 The fact that there are two beams is proof that s 12 The two beams correspond to m l2 and m 712 If 1 then there would be three beams corresponding to m 71 0 l The separation ofthe beams is a direct measure of Hz which provides proof that uz 2 MB m The extra factor of 2 in the expression for the magnetic moment of the electron is often called the quotg factorquot and the magnetic moment is often written as uz g uB m As mentioned before this cannot be deduced from nonrelativistic QM it is known from experiment and is inserted quotby handquot into the theory However a relativistic version of QM due to Dirac 1928 the quotDirac Equationquot predicts the existence of spin s 12 and furthermore the theory predicts the value g 2 A later better version of relativistic QM called Quantum Electrodynamics QED predicts that g is a little larger than 2 The g 7 12 2010 Dubson Phys3220 factor has been carefully measured with fantastic precision and the latest experiments give g 20023193043718i76 in the last two places Computing g in QED requires computation of a infinite series of terms that involve progressively more messy integrals that can only be solved with approximate numerical methods The computed value of g is not known quite as precisely as experiment nevertheless the agreement is good to about 12 places QED is one of our most wellverified theories Spin Math Recall that the angular momentum commutation relations L2LZ0 LiLji 1Lk ijkcyclic were derived from the definition of the orbital angular momentum operator 1 r X The spin operator S does not exist in Euclidean space it doesn39t have a position or momentum vector associated with it so we cannot derive its commutation relations in a similar way Instead we boldly postulate that the same commutation relations hold for spin angular momentum SZSZ 0 Si 8 ihSk From these we derivejust a before that Z Z 3 Z 39 S Isms h ss1smsgt 2h Isms s1nces12 SZ smsgt hms s ms i hlsmsgt sincems ss 1212 Notation since s 12 always we can drop this quantum number and specify the eigenstates of L2 LZ by giving only the n1s quantum number There are various ways to gt gt write this Ismsgt lmsgt gt gt IT 1 These states exist in a 2D subset ofthe full Hilbert Space called spin space Since these two states are eigenstates of a hermitian operator they form a complete orthonormal set 7122010 Dubson Phys3220 SS within their part of Hilbert space and any arbitrary state in spin space can always be written as all bli Grif ths notation is x ax bx Matrix notation Note that 1 0 If we were working in the full Hilbert Space of say the Hatom problem then our basis states would be In ml ms Spin is another degree of freedom so that the full speci cation of a basis state requires 4 quantum numbers More on the connection between spin and space parts of the state later Note on language throughout this section I will use the symbol S Z and SK etc to refer to both the observable quotthe measured value of SZ is h 2 quot and its associated operator quotthe eigenvalue of SZ is h2 quot The matrix form of S2 and S Z in the mZgt basis can be worked out element by element n l T Eh T Recall that for any operator A Amn m A ltTSZT Tsz 0 etc Ts s 1 0 1 0 sz 2 s 1 4 0 1 2 0 1 Operator equations can be written in matrix form for instance Tgt ZITgt 3 28 113 28 We are going ask what happens when we make measurements of S Z as well as SK and i 0 etc 2 z SZ Sy using a StemGerlach apparatus Will need to know What are the matrices for the operators SK and Sy These are derived from the raising and lowering operators 7122010 Dubson Phys3220 SSXiSy s 3 1 SK ssX isy sy s s To get the matrix forms of S S we need a result from the homework sms h ssl mml sms hquot ssl mm l For the case s 12 the square root factors are always 1 or 0 For instance s 12 s ms 1 s ms 1 m l2 gives ssl mml 1 iL 1 Consequently 2 2 2 S L th S T 0 and ST ll SL 0leadingto T S T 0 T S L h etc and 0 1 0 0 S h S h Notice that S S are not herm1t1an 0 0 1 0 S and Sy S S yields h 0 l h 0 i S S These are herm1t1an of course a h 0 l 0 i 1 0 Often written S 56 where 6X 6y 5 52 are called the Pauli spin matrices Now let39s make some measurements on the state alT bli Normalization 1 3 lal2 lbl2 1 a Suppose we measure SZ on a system in some state b Postulate 2 says that the possible results ofthis measurement are one ofthe SZ eigenvalues h2 or h 2 7122010 Dubson Phys3220 Postulate 3 says the probability of nding say h 2 is lt0 In of this measurement which found h 2 the initial state collapses to 2 Z Prob nd h2 x gt lbl2 Postulate 4 says that as a result But suppose we measure Sx Which we can do by rotating the SG apparatus What will we nd Answer one of the eigenvalues of Sx which we show below are the same as the eigenvalues of SZ h 2 o r h2 Not surprising since there is nothing special about the zaXis What is the probability that we nd say Sx h 2 To answer this we need to know the eigenstates of the Sx operator Let39s call these so far unknown eigenstates TXgt and IslXv Grif ths calls them and How do we nd these We must solve the eigenvalue equation S x 0 h2 a a 7t h2 a 7 wh1ch can be rewr1tten 0 In 11near h2 0 b b h2 7 b algebra this last equation is called the characteristic equation x Alx where 7 are the unknown eigenvalues In matrix form this is This system of linear equations only has a solution if 7 h 2 7 h 2 Det h 2 7 0 So M hzf 0 2 A hz h2 L As expected the eigenvalues of Sx are the same as those of SZ or Sy Now we can plug in each eigenvalue and solve for the eigenstates 28 an 28 2 a an 2J 2 gowehavelwx and luv g 7122010 Dubson Phys3220 1 Now back to our question Suppose the system in the state I ll 0 and we measure Sx What is the probability that we nd say Sx h 2 Postulate 3 gives the em 06 a Question for the student Suppose the initial state is an arbitrary state b and we recipe for the answer 2 2 2 1 J2 Prob nd SX h2 Tm Tu gt l2 measure Sx What are the probabilities that we nd Sx h 2 and h 2 Let s review the strangeness of Quantum Mechanics 1 Suppose an electron is in the Sx h2 eigenstate TXgt Ifwe ask What is the value of Sx Then there is a de nite answer h 2 But if we ask What is the value of S Z then this is no answer The system does not possess a value of S 2 If we measure S Z then the act of measurement will produce a de nite result and will force the state of the system to collapse into an eigenstate of 82 but that very act of measurement will destroy the de niteness of the value of Sx The system can be in an eigenstate of either Sx or 82 but not both 7122010 Dubson Phys3220 T ne ForanSm a 62me MltW 0m mu M M 61mm 4 MW WW Yum A OWWNQ M V A 1 9W4 Y T95E amp y T Am L Ming T152 1 41 E V M MA 4m WWW mam whoa E gt W Mum o4 TVsexy um wae E F DS F kw gt ma 5 Mm v m a WM LE Ljpuwuxe m 1790 a c a quotta H979 W hm LMmez BWM 4k ltgtlt gte mt WNW KW Mm amtF mscmtc 9W9 7 vavu Q WAXWW VA S 1quot km WWHWMMWMX Am Idx 441 W S I k IMWKgKH IJxe x SKK TN Zea st 4 Ax 48 693 q siSf AIR 3 p24rgts a 3r 4 39lt 3 1 H H m x 0lt fu olt r3 ltnrroltw 9 3 Jo 3 1 u f lt53 g 4 m n h 4ault1lt13 ltwm 2mu wf nmq EuTx wgw r 3 lt L r Emmch wk bk 5 m w Zak Rim a 333 w M i 55 Min 2 J 5 A 1 4b Jefi FG 4 Sf 543312 35 WL S SPF 153 menm zzz qqquotZV 3434 0 A if mix 5 2 L S E in be 5 gig 33m F z PMNJLMQ 12 Ev Amy 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1393397 V mWW Wm m 5 Eff91 TMPWLW m n X XL A Wm gum 4AM Wyn ampG f y 7 191 A 43 2391 Wm Wm H xJ a ngwm D39vn 37M E Vw NW W mm m w ilk LA W mz rm Wu km W2 w W A me MM 1 1 mx Wm m SJLQ MMW o mm M M7w3ymm 0 1 1me me MWaWM wk bug r 39 HIM WNW JAVAAide W I Gg w quot gm fag m a 235 E thm ow ed W W A E WVWV pw k WWW T T MAIN M vb WW Wiam mwmg VkaWu M a W w W mel mwm Aw the if w W7 KRm bamny H yes fxfxxw ru 7 SUTI Kym hmmm vms hatKAPka kKLe39kquot 53 3210 GMQH The Harmonic 05c qwf J A Manj in uent a Cl ssm We F quotka Hocke39s Law 7539 Is 7 Harmonic Oscillqwr v00 an VCX 3 ikXL Now when Imus l 9 x VUO wnk A mlnlmum look 391 94 4 Human lKe VHS 41 96151 for qu X to 7395 nkjslcqlb q V6 Common S39l39IUq39Iwl a Classicanj J th Asm wf Ecosw for Helm0c bunk fw 3 J3 So v 2 MW X A defan bunn 39lj I 507 n 7912 yon f9rldJ T R N So 0 5 ogr c 6 H 39 an39HM Hmoqlc OSC I Q70 A A afar Scrarcne x Qij an 00 w Sal os 390 1 ue 395 n i t 1 3 w Ive as 325 u K L m w x 439 x E M 37041303 31M m c E t 71mg prondenct 53 3am QM AJX 3U Mk 6 03 We win mi 7A4 94 w 1135157 Mon quot30 J 5800 quotLunch ceaa nwwaquot Wt 309 find 3an cer39lqln E s quot94 En will WarJ 7 CortmsronJMj an X 9W 5 WIJVC 50 jun INC m 7k box we My 4 x375 5 Back 0 conufdacl ly w o S39lemau INHC wnk Jlsae lt Cnffjj In Hus 561 0 ani odor 005 w 7M2 quotX 7am mam n Ami 1 Solve 7quot are 7r5cK Rfi lrijt Trick FH For hach X x1 Jammvcs So 1 391 1 11 J iL439ltgtlt 0E mwx0 3 LMWX U Lm I X S n check N no mill r o f l a UQJP39IUMIMJ 1 MWX mwx XH A6 25 8c 2 3 me 3quot1396er 036 1 V nq7j 13 399 do 7055 A an fl j s c NOW and track it can a 318 L An aide SET 3310 QM Numcucal Ano krr Mealor 70 Know dfi Now lo solve 4 39 333 E VxHX not yew Knowmj E Even sollns ave an E Wt A Similar Sinanon a Jus7 H o vans For Harmop aghha arj 1 00 has Vans af en 97 m mquc 3051 73 1 q q Swag1m 79855 e f lop l39m look fdf I u a 0 7kg wen 511 I u Le 10 T we nonmach Lner39 Pick ox quotSuryan E Tiny I US 55 10 7e you L4 O 34quot0 39 E anuf l 39 Us 4 9 QMH0 3910 comrwc 018 H US Adm u39g u39o 39w cowuvc H hi W E NOW urea QUOSS K In S7Pf 6 E if M S39IQHS 7o LION 2F 7 he m Iquot 1 Conq Urc C sun 0 mac Q q 31 If M ow lawn you nacJ leSSn CUIVOTUff low Q JaII SIS 3140 QM 28 6 I For natal 50quot0 Sun wnL 030 0 quot39quot 1 we nomJ39H 5mm junk mug So Lawra b 7o rig an E numfrlcauj quotslogquot ram 71W 03 k tf fun E will y 792 q Sal39n W ICL bdumes we on In x In 11 m9 7 can coaFl Mud and 71w 75 army s or Sumc lnj fo1n 395 9056 ujo 1w Lave 10 feel an 397 wn do cw VM39S Ulric anqu39r lc 5a for Mntx Mi C MI W can LC quotsakequot awnnu 53 3210 QM 9 L mw39x AYSU xe he wonder M x c 6 ET Maggt 50 col LL quotSWJen 1 qu use 75C love 11 fume Mnlsol of Fro enl39uj I 71g 9 kX Go i Q X 1quot 41 X1 4 In sec WA Barrens For nova I m Jouj 70 Skio 75 aljcldacallj 79110quot Exercise sklr 7 tin fuochc J If you msisz 2L0 aux 57 nomdlzq le 7M flai 71quot Series f AU Mei 70 pg 1 mus LC a fmnc order folynomal Ami 7 a will Aq o ot n F M1 3312 If E quotNK S on sfccml VAUCSJ name En Quirk 1 lnkjgr 1 1 7 mam 1 611 391 lion quot E 21 Pajama I IN dv Tlquot k x wanton 7M7 com are Daljnbmiqbece CriffMS f5 531 70110 Gm 2w 0 ubr l a Name ML wt P Y 39 and Ca around 7411 1110 no quot0 3quot A A N M u L 1 390 WWquot quot333x ng 5 E 2 quott 1 QJWI 0 Non hug lm1J LqSICuHD 397 579 DU 007 I o 3U 70 nommjdc 39 39 JXtak 3 u 1 m 1 Ii 3993 X 6 J Equot it t 2 L9 Cent s m7 395 oIJcr foiynomlql Van could J05 Venfj 7M7 an tag fa usc 51 a 1 1 L 3 u39icx tmw x Ham v E un t 1m bmcmLU 7M5 jive 1106 Nine defenclcme IS 51171916 39Enf 57q ilamu 3mm 319 J Un d C c 1 You can 7ch 01m an com mnlm of 75ch 70 walla cmj fan39tl39v 1 1 q wen 747eJ 7h 7mc french C 1 st S 2 C0 Mnxcquot cn Ht 53H 0 bSuJa39I Ivns 39 V VCA 0f 01 oscillvalj ML 3lKO QM 22 O H397Ik In LOX 000 AFC n0 2 to 395 Mn lenj 0A uncx 4mm 1x 3 Inquot 39 Or rko anal 3 m I Comrlue Any a Cam lac aloud Vaq found quotRR U E Cn Un X 1 7 W39n39lskcj futM G7 n a 39 EVqu of cch 39LOWC S clqu 3 9 Wyn j MI 1 JISCrHC jlows Ian n 31Hquot quot1 3 0 0395 YOM mnnW IsNIquot q mas on a 5u a7vm sluran mg n Hr I y 39 XVIMX I For Man n manor 4va quotwk WW 1 all r 57113 4 45ch 0er 1 75 clamsch W ffX funt7lon Nun Hun 14551019 E jou quotTun rovn t an Xx k quot E max U E can Le anyvlznj Luv JIK n L 4 j H 9 lmje n n SOL3 393 J05 I yon quotQWMJC owe 540001ij 531 QM 311 cl212 Truck 2 for Solwv 7 HMquot 056 vt u M 115 39TTICR S W47 our 73mg 1 Wood MWquot 50 Cam I w n K we 13 C00 wms om 0 be mm jcma 7quot jam C n IMujmt H71 now 3739 753 Lens for q SlmJM Wm 3910 JOJNS I th Jew ththM m 3390 394 7 SWIJ munHy m 70 qun39lum HM 75009 50 5 WW lequotquot j 391 W2 also nd VS Some quotelegant mc ols 754 MC Jeff j ccqua 0 Q 50 CRJOJ 755 Iri ckfgn e ma a 13 fun N Cm Jam 15 75b 5 SCIHOJ Cz n WE VQ 7v4ylf a fmmxu EM wAf Q c x f Course 1 O For nmet rs J CC cEL SLJ o 7h mmf39rs u 7 rcwrbc 7 quot ngnm quot hf 1 39 4 NY m chr rMWr quot J Grants of A 307 1 ISn39 57 I because 0 of My X S 47 L375 53 cm 3210 42x LeT S dc at 0 new oftmor mm m 0 A A A Q J rmwlt 11 KIHHIJ IA 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5M 3 NHL I Usquot 01 4432 ram Gunny So a 4 GAO z 71w 393 q twmiadi two A a on u 9 H Eu 0 0 4ERw M or 3ct4u7 E FtWV a4 4 of H wnl Cljcnuauc Eiw A So 44M Is an tljcnf39MmW a Clven H 1 70an Cl 032 clfiuan eljmfvnnlan chn vqlvr 57 61M 3110 126 NW4 jbm why I 1 Mme Manly A 3930 M7 1 claw joq l39 Jl7 E iw qq 4 39 A So now wt see AQ 25 Of a or a 7yag ttwme 06M 50PM Jl9ren7 CI eanveg J39FIjmfw39nlwquot H A 0 3 quot0015113 Ofefq lor 5601 I7 4 36 E 1i N u i u a lowcnv 39 Mg E z W 1L0 art quotIaJlgr orCIq39IOls terms E c qnjt In 51 Kw New for Crux 3 1yoq a a 6 a M you Crane 0 7 7C W Wnl E m kw 70 am la 75 IAI IQ UID 0671 at 10 3quot ncj owe cno g39j 60 VAN 4 71mmquot 7M7 5amp7 E am am A lowquot 75a 71 Ml tmvm 0 V00 yakL I O 7513B Baal In 55 7 quot IS 0 Lonvm 57m Ola SucL an q U0 0 None Hun o quotS39aIVCSU 7k 565 n39 J u57 n07 M39N rCS f aan fien 75c 039 0 414 1 40 0 531 CM 3120 219 a quot gt 39 1 i 3 o OK oCX 0 q mw 0 mkw JX 7533 m Mouth 70 SoIVC I See Quizmks arjun trH taxi11 v 000 739 A m works far any 3 n a PICK A 70 nolmnlllt bl CV4 SC X 1X 393 o 1 X711 30 Mac 1 M q mw X C 397 Sills5 S 5357 l7 m 0 71 717 you ll cc E0 foray E0 A L HUo EO M0 1 WQQLHo 39 E0 MO 71quot quotkillsquot 04 0 o 3 E0 Ho A An n35 J a U0 give a 37411241 7 HI Vt wnL E E01494 myth quot 7 70 animal Jain unm Aquot en New wk 6 miMW Luz5 normqjilt C cls 00f Gladia fb gt 53f QM 3ampu10 228 Codie We Have Missed Some Sal n Ill Way 7 M1717 7quot 5 q x W H CA is n11 JISCOVFH39J A arsenal affllavlunj 06 q 7o 4 7 Foryu mklj no 1 11m Wm anquot doc Joule 52 Am 0 5M 0 a q 53193 J not US 7Aerc o 5 a UOwc37 q Lu we 7kg 011161 lowc7 57am alf u j w arlf psk MS C070 7 HK ya amongII 7 Snncj fee a 9 HX NC also Ac qn 6191 Ich 70 00 PIWb S Math and JX Jmn See quot3900 RV 73 leak 7 11 0 3970 I I u H 9770 1151 lFroLCnIuj at lalclfr 09814 le we J faunJ 4 7M S M Iwnatj 57a n x J Wu C Mzt In jenuab 092mm Vw a quotLam 514795 J wr L Jilscrnc enujie5 BU 67 s move on 70 q auburn Jiffucm ftoLlem when we re 1333 boml SJP QM 3220 Ch 2 part 3 Page 1 m m Particle Consider VX 0 No PE just a free particle We talked about this at the start de Broglie39s ideas Now we can tackle it with the Schrodinger Equation There is some funny business here but clearly this too is an important physics case we m to describe quantum systems that aren39t bound Z Lu39xx Eux That looks easy enough m De ne k so u x k2ux JZmE h I can solve that 2quotd order ODE by inspection ut x A6 36le The label quotkquot identi es u different k s different u39s There is no bounda condition so k thus E is not quantized Free particles can have m energy And lI kxt Aeikx Beiikxkqam Let s de ne a E h as usual with E hzk2 2m ikxr t rikx t Rewr1t1ng lPkOC t A2 k Be k this is a fn ofxrvt this is a fn ofxVt rightrmoving wave le m i g wave speed Vwk speed wk The speed of a simple plane wave is also called quotphase velocity Go back to our chapter 1 notes on classical waves to see why vu w Eh hszZm hk Here v k k hk 2m Is that right What did we expect I m thinking p hk so v g m sli min A h a For the functlon lPkxt A6 2 quot you get pIPk ja IPk thk z x So this I k is an eigenfunction of with eigenvalue hk Just exactly as our quotde Broglie intuitionquot says The I k at the top of the page with k gt 0 only is a miX of k and k which means a miX of right moving and left moving plane waves hk But wa1t What39s w1th that funny factor of 2 then Why d1d I get v 2 We re gomg to In need to consider this more carefully coming soon S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 2 Summav So we have solution lPkxt A2 x ZmE h kxiwt where k i Note 2 values of k for each E one plus one minus and a E h hkz 2m comes from the usual free particle energy basically pA22m 2 because l1116 i A l1 06 k hlkl The speed phase 2m These waves have A velocity Small 7 faster speed h 2m There s that funny quot2quot we saw above again Classically I expected hk 2E v g without the factor of 2 m m m Re wk kgt0 or klt0 gt 4 So the Funny v factor of 2 curiosity is one issue we need to understand Also here39s a wave function which is not renormalizable Wkxt 2dx A21dx yikes Conclusion from this This I k is not representing a physical object There no free quantum particles with definite energy But not to worry these I k39s are useful even essential We can form linear combos that are physical So these plane wave solutions are eigenfunctions of momentum they are eigenfunctions of the freeparticle Hamiltonian ie if V0 they are easy to write down they have a simple form That s all good news They are not normalizable that s bad news I would consider them to be idealizations of physical systems a kind of limit which cannot be actualized but can be approximated of an ideal free particle with definite momentum And as we will see by the methods of Fourier transforms they provide the mathematical basis for constructing perfectly physical freeparticle states S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 3 Since k is a continuous variable combining I k39s requires not a sum over k but an integral over k l m l m I Consider l11 xt 7 k 1P xt dk 7 k e W39Ek39Uk gt min klt gt mid stuck this in for This plays the role Sum over all This is the plane wave the later convenience of cn before it s a k39S 0r eigenstate of energy for a free function of k 1321111016 This looks a little scary but it s just the continuous analogue of our old familiar quotFourierquot formula general 111W WHOC t 2 c lqummmlkx t where we basically replace an gt kdk This I has many k s momenta energies superposed 1 IfI give you 111g x0 7 f ke kxdk went 4 7m then you can figure out pk like Fourier s trick and given pk you then know from equation at top of page I at all future times So we need to review quotFourier s trickquot for continuous k ikx Note TkX0 e simple enough Suppose I give youfX and ask quotwhat pk is neededquot so that fx f kka xdx This is our problem Given TgenX t0 find pk Because once you39ve got it you know I at all times S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 4 The answer is Plancherel s Theorem also known as the quotFourier Transform 1 k E f f xequotkxdx Note the minus sign in the exponential here 2 2 Recall that iffx E c sinL x then Fourier s trick gave c ffx sin dx a a a at It39s rather similar the coefficients are integrals of the desired function with our quotorthogonal basisquot functions I pk is the quotFourier transformquot of fX here I If g here I X t0 is normalizable to m with I Xt will be too as will pk I You can quotbuildquot almost M FUNCTION g you want this way Digression 7 Some examples of Fourier transforms fX Griffiths p 62 has quotw1dthquot 1s m 2a then k ffwkiikxdxis easy 1 1 sin ka you get k ET Do it try it yourself k W gt na is quotwidthquot roughly pk N constant spread out fX localized Ifa gt0 Need many momenta to build a narrow wave packet 390 m quotWW If a gt 00 x spread out 2 V V One sharp momentum broad wave packet where is it S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 5 fx Digression continued width N 20L Let s do another fx Ae39xzAquot 1 Here x 239 de can be done anal icall uess what39s on the next homework 1271 f yt y g x2 E kxudx A 2 ribE dx m K m x A l rkz rkz T k 7 e e e a e quot T T That little trick on the right is called quotcompleting the square you need to DO that algebra yourself to see exactly how it works it s a common and useful trick i2dx39 1 m x dx B I llltt39 39 39k dx 39 39X as1caye1ngx 210714 215 ie 243 you then get k 1r Ae39kzquot 9quotquot 2 width Wu Once again na1row in f wide in p and Vice versa Iffis centered around X0 fx Ae39lx39x 24 work it out This will add well really multiplyU a phase eikx to pk So the phase of pk does carry some important information but not about momentum itself uzze oryouw at oesap ase eIUXmutipying X o P If h d h k 1 1 f d7 S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 6 So here39s the central idea instead of llX Pure eikx which is not normalizable and thus not a physical state we will start with a more physical llX a quotwave packetquot wx Mk Ak which 4 D hasa A A Fourier V V v v transform I k AX I k02m0 Any llX can be quotbuiltquot like this the math of Fourier transforms yields the following in general I pk may very well turn out complex but might also be real If llX is quotmostly sinusoidalquot with wavelength N X0 like above then pk will be centered around k 2n X0 If llX is localized with size AX then pk is localized with size Ak Z l AX Given I X t0 pk I X t is determined for all times Claim If we start with a simple wave packet A otk I WW 7 k V As time goes by the ripples in the llX wave packet will move with phase velocity mk but the envelope itself which we interpret as quotwhere the particle is locatedquot moves with d a different velocity quotgroup velocityquot 7 S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 7 DIGRESSION on phase and group velocitv and the 39 quot of the speed puzzle Let s look more at this last statement about phase and group velocity E h d hk In our case a free particle a 7 7k2 so 7w 7 no funny factor of2 present h 2m dk m This resolves both our quotfree particle issuesquot hk l The packet moves With Udmml 7 m 2 The I X t is perfectly normalizable And something el happens different k s move with different speeds so the ripples tend to spread out the packet as time goes by higher k s move up front SO 7 AV gt So AX grows With time Let me go back and make a casual proof of the quotclaimquot above about the time development of a wave packet 1 Pgexr 7 f ke dk 390 hkz Recall m is itself dependent on k a 7in this case ofa free particle it s just a consequence ofE pA22m Let s consider a wave packet where pk is peaked near k0 So we have a quotreasonably welldefined momentum this is what you think of when you have a E particle like an electron in a beam If you had many k s each k travels at different speeds packet spreads not really a quotparticlequot So this is a special case but a commonpractical one So if k s are localized consider Taylor expanding mk mkmk0m39k0kk0 This should be a fine approximation for the m k in that integral we need to do at the top of the page Let s define k E k k0 as a better integration variable so dk39 dk but mmk0m39k0k S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 8 1 m pgmog 2 f kyk0exkxerxwknterw knk I ka imam 1 y ik39x riw39kDk39t e 2 my k0e e knxrwut 1 EH This is a function of X m o t These are ripples f k0 my H D klk a function of This is the timeevolving envelope X 7 nok0 t which travels at speed do They travel with vphase nok0 That s the group velocity 0 390 7 hko So the quantum free wave packet moves with 39U a 0 s 7 In Just as you39d expect classically The ripples travel at vphase but l w l2 don39t care about the ripples really Wave packets do quotspread outquot in time we could compute this turns out it spreads faster if starts out narrower END OF DIGRESSION Which showed how group and phase velocities are different how wave packets DO behave classically and how they represent physical particles At this point we re armed to solve a variety of general quantum mechanics problems Given VX and wt0 find I X t That s huge gt Bound states will generate discrete En39s and un s this is the story we started the term with 7 like the harmonic oscillator or the infinite square well The time dependence is especially interesting when you start with mixed states gt Free particles will become quotscattering statesquot This is what we ve just gotten started We39ll talk more about this soon at which point we ll go back and tackle a few more simple problems to get familiar with the P39s for scattering states S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 9 But first let s talk about the INTERPRETATION of our Fourier transform what does pk mean It plays an important role in Quantum Mechanics and will lead us to even more powerful ways of thinking about wave functions We ve already danced around it pk is the momentum distribution it tells you about the particle in momentum space it s aprobabz39lity distributionfor momentum Let s see how this works To begin consider again the general math formulas for Fourier transform and inverse 1 W 1 m I x 7 k el kxdk 69 k 7 x 239 de fwm m m f Now consider the Fourier transform of a 6 function ie let fX6X so from above 1 m 1 k 7 5 x 6de 7 The last ste is ure math inte atin 6 X ives m m pp grgg you the rest of the integrand evaluated at the point where the argument of 6X vanishes But now the inverse of this just read it off from the Fourier formulas above says 1 1 x 5 x 7 k el kxdk 7 el kxdk Let me name this relation 1 f127 2 It s strange the integral on the right looks awful It oscillates forever is it really defined Well it s a little dicey you do need to be careful how you apply it but basically if X is NOT 0 the oscillating integrand gives zero think of real and imaginary parts separately the area under a sinusoidal function is zero but if X is zero you integrate 1 over all space and get infinity That s the delta function OK armed with this identity consider any normalized wave function lI X for which m 1 l11x11 xdx 1 Let s write 1P X 7 ke kxdk which ou can L m y ALWAYS do so here pk is just the usual Fourier transform of 1PX We ll put this formula for lI X into the normalization integral twice once for lIJX and once for lIJgtIltX Just be careful to change dummy indices for the second one Lj f kkmdk f k39eik39xdk jdx 1 That s a triple integral ack but check it out the only Xdependence is in TWO places the two eXponentials So let s pull that X dependence aside and do the Xintegration first m 1 1 6 de 7 kdk 7 k dk 1 L V2 lg 1271 If Note the 7 sign in the ikX term came from the compleX conjugation Follow the algebra here don t take my word for anything S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 10 But I can DO that integral over X just use our cute integral 1 to simplify Let me first take integral 1 and rewrite it ipping X s and k s 5k 2L felikxdx l Convince yourself treat X and k as pure symbols 7r But this is precisely what I have in my triple integral except instead of k I have k k Substitute this in getting rid of the dX integral which eats up the Sqrt27c terms too ff5k39 k k k dk dk 1 It s easy to glaze over this kind of math Pullout a piece of paper and DO it the steps are not hard and you ll use these tricks many times Check that you see what happened to the 275 s and the integral over X There s still a double integral over k and k but with the 6k k it s easy enough to do I love integrating over 6 s Let s integrate over k and collapse it with the 6 giving f k kdk 1 Ah 7 this is important We started with a normalized wave function lI X and what I just showed is that the Fourier transform pk is ALSO normalized But there s more Consider now p fm 1P x ilmx lx Play the eXact same game as before but with that 4 1 x derivative ddX inside All it will do will pull out an ik convince yourself everything else looks like what I just did and we get 12 fhk k kgtdk Now deBroglie identifies hk with p So just stare at this until you realize what it s saying I would eXpect from week 1 that p p Prob densitypdp So here it is apparently other than annoying factors of h since p and k are essentially the same thing p hk we know what the Probability density for momentum is it s just mac we To get the h s right I would define CIgt p k again with p hk I ll let you convince yourself that CIgtp is properly normalized too 1 flt1gt plt1gtpdp and p fplt1gtpqgtpdp We ve just done a lot of math so let s pull this all together and summarize S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 11 Bottom line We d gotten used to thinking about lI x as the wave function When you square it you learn the probability density to nd a particle at position x Expectation values tell us about OTHER things any other operatormeasurable you want lI x contains ALL information about a quantum particle Now we have the gt7 Fourier transform pk or equivalently CIgt p idle This function depends on p It contains the SAME information as lI x It just lives in momentum space instead of position space It s a function of p 39 CIgtp is normalized just like lI x was lCIgtpA2 tells the probability density of finding the particle with momentum p CIgtp is the wave function too There s nothing special about x CIgtp is called the momentum space wave function 39 We can compute expectation values in momentum space to learn about anything just like we could before Let s now consider this in more detail We already know p p61 pltIgtpdp In fact it shouldn t be hard to convince yourself that for any function of momentum fp fp f fplt1gt pgtlt1gtpdp How about ltxgt Here again start with the usual old formula in terms of lI x and play the same math game as on the previous pages expand lI x as a Fourier transform you get a triple integral do the 6function trick and try this it s not so hard you find x flt1gt p hiiqup Look at that for a second There s a lovely symmetry here A A 9 In pos1tion space we ve used x x and p 4118 Now we re see1ng that x A 9 A in momentum space x my and p p P It s almost perfect symmetry except for a minus sign but don t forget the inverse Fourier transform has an extra minus sign in front of the ikx which is the source We will discover there are other spaces we might live in Energyspace comes to mind but pretty much anyoperatoryoulikespace can be useful too The momentumspace example will be the first of many But for now we move back to our central problem solving the Schrodinger equation We ve done two bound states and the free particle Next we will consider particles with positive energy so they re still free but where V is not just simply 0 everywhere This is still a lot LIKE the free particle it s a free particle interacting with things This is getting to be real physics like the beam of protons at LHC interacting with nuclei We ll still stick to one dimension your beam can only go forward and backward you can only transmit or re ect 7 no scattering off at 27 degrees just yet And at the same time we will tackle other bound state problems too to keep the math simple we ll consider examples with piecewise constant Vx like in the picture on the next page S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 12 Back to our problem in general of solving the TISE Schrodinger s equation 2 2 Iflu x E unx or writing it out 7721400 Vxux Eux 2m dx 2 u x h71E Vxux 6 So u and u need to be continuous at least ifV is nite VX VX I Eltv A E KE EgtV I gt x If E gt Vx that means uquot k2 u Classically KE gt 0 De Bro lie 2 2mE V 21sz g p2 k a 2 E V i h 2m 2m usual uquot k2 u gt curvature is towards axis This is a Wiggly function If k2 is constant u A sin kx B cos kx or A39 elkx B e39lkx Big KE 9 Big k2 9 more Wiggly smaller 7 Classically Big KE 9 faster speed 9 less likely to be found there This is very different it corresponds in QM not to 7 but amplitude lt well usually but there are exceptions So as a specfic example F smaller k 9 longer 7 A 1 I expect u x S OW fast higher prob 9 big M W Vx X v v v v v X l L Big k 9 small 7 low prob 9 small l M S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 13 What if E lt V Classically this never happens negative KE VX classical turning point a 2m In QM u x E Vx ux 2 so u K39 u IfE lt V de ne Kquot 2m E This is not Wiggly curvature is away from axis If K2 constant constant V in a region e4er is good W e39Kxis good uXCe39KXDeKX ifxlt0 ifxgt0 X The other signs eg e X for X gt 0 blow up at 00 no good So in the quotclassically forbiddenquot regions can have uX i 0 But u gt 0 exponentially fast So eg VX ux sinusoidal ere em E T Ld X X T d quotPenetration depthquot h1l2mV E If VgtgtE rapid decay S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 14 With just these ideas you can guesssketch ux in many cases u u continuous always if V E nite EgtV 9 sinusoidal Bigger E 7 V 9 smaller 7 and usually smaller l w l E lt V 9 decaying exponential Bigger V 7 E 9 faster decay Example Finite square well continuous spectrum free Vx Discrete spectrum bound Very rapid decay exponential moo Ground state even about center Turns out you can prove if Vx is even VxVx then ux must be sinusoidal even or odd so luxl2 is symmetric less rapid decay more quotleakingquot lSt excited state 1 node Let s work this out more rigorously to check ok Let Vx 0 a lt x lt a Note I shifted the V0 outside zero wrt Grif ths Assume 0 lt E lt V0 ie bound state S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 15 Let s start with even solutions for uX VX We39ve solved for uX in each region already V a a X Region Q No good blows up J2 V E uxCeKquot 39quotwi 11s Region No good looking for even solution right now Region 0 CeKX R eglon uxAcoskxBs39 kx withk Same C 39cause looking for even solution Bounda conditions Continuity of u at Xa umx IXZM uHx IHz Continuity of u at Xa u39m x IXZM u H x I z Conditions at X a add no new info since we ve already used fact that u is even coefficients We have 3 unknowns E and A and C The energy eigenvalue We have 2 boundary conditions and normalization so we can solve for 3 unknowns Continuity of u says C6 Acos ka Continuity of u says 7Ce K Ak sin ka Dividing these 7 ktan ka 12 E I E tanTma VOT A nasty transcendental equation with w variable E S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 16 Can solve that transcendental equation numerically or graphically 2 E Let zska mz a 2 h so gg 2mV0 E z Let z0 hz Q So we want tanz z0 z2 1 1st sol39n 1 another one I see there are solutions for 2 which in turn means E remember 2 and E are closely related by de nition if you know one you know the other They are discrete There is a nite number of them No matter what 20 is there39s always at least solution Look at the graph If V0 gt 00 solutions are near 712 3 712 h2 2 hznz 722 72 Zma 2m2a which is exactly the even solutions for our 00 well with width g GivingE 12 or 32 Knowing z E k and K our BC s give A in terms of C and normalization xes C I leave the odd solutions to you The story is similar and you will get as your transcendental equation cotka This time it s not always the case there s even a single solution The well needs to be deep enough to get a 1st excited state but even a shallow well has w bound state S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 17 What about E gt V0 This is the continuous region also called the quotscattering statesquot because u will be nonzero off to 00 it s like a particle comes in interacts with our potential well and is affected scattered Lots of physics in this example like quotshooting electronsquot at some object it interacts with Griffiths treats the square well I ll do a simpler example E The STEP POTENTIAL V0 Think of wire w junction Higher voltage on right lower quot quot left 0 4 x Region Region We could build a wave packet send it in from left with E gt V0 and solve for time dependence That s the right way to make this physically realistic But our approach will be simpler Look for uX which solves SE Q which is physically interpreted as quotsomething coming in from the leftquot ZmE hz MHOC Ce m Dewquot with k39 4 Remember our interpretation of free particle solutions from awhile back the Aeikx term incoming wave traveling rigl1twards with p hk u1x Aeikx 32quot with k Be39ikx represents a moving wave but it s in region I so it must physically represent re ected wave in steady state remember Ceikvx represents quottransmitted wavequot going right in region H We set D 0 because otherwise we d have quotincoming wave from the righ quot and our physics condition was quotwaves coming from leftquot Boundary Conditions Continuity of u and u at the origin gives quot100 x0 quot1100 L64 i100 x0 3100 x0 S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 18 What about normalization These are not normalizable u39s like the free particle you must make a quotwave packetquot if you want it to be normalized But wait what are we after Well given A amplitude entering I want to know T quotTransmission Coefficientquot and R quotRe ection Coefficientquot how much of an incident wave moves on and how much quotbouncesquot Intuitively BA will tell us about R and CA will tell us about T Really R and T are quotthe physicsquot we re after We need to be careful I said quottell us aboutquot but it s not quotequalquot we39ll have to come back to R and T in a minute Let s first compute these ratios though Boundary condition on uX at X0 gives A B C Because e k390 1 BC on u gives k A B k C Now divide by k add the two equations 2 C SI t h k 2AC 1k k 77 o ge c ec or A 1kvk 2 Put this back into lst equation A B C WA which then means again check that you follow this algebra it s not hard and worth doing B 2 1 k k A1k k 1k k k E V0 Wit 7 k E OK so we know what BA and CA are NM we can talk about interpreting these coefficients of nonnormalizable quotplane wavesquot S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJ39P QM 3220 Ch 2 part 3 Page 19 WWW m 7 1 L M evulves wnh ume mummy Sn prutablhly aws m and em amps regun a Inlam th 0 X lx randmsnunenems m ed rubab1hlycunml quot n lh39x X Pmblltx ltb pxry dz Yeu need m du me algebm m prev equauun whereFPrub density vyz Let s rewme me rst dawned equauun as 7 Dr mlegaung a a b 7Ibelalaelb mte emmmup curmuthala 7 cunenl lalb mabregun Jusmke eleeme charge 39hename amt es geem m am 2 aw N t that x 7 72 391m 7 71m 7 e 0 2m Wax m ax Furaplanevave mat 3W 1 w 5 171p kl m which mans 253m fur a planB wave M ms xskey luuks Me i 7 v Sn uwufpmbablhly m planewavexs11km4Az pmm 39 hrgemw WMFM elemnew Thnkufelemcmmmtwherejiv K M M spee e e emeaemcy Vary slmllar M plays me rule er eharge dansny hare s Fuuucknaken umM Dubsun wnh mm 1 Andersun furlypese mg Fall 2m SJP QM 3220 Ch 2 part 3 Page 20 So back to our scattering problem with particles coming in from the left and hitting a quotstep upquot potential but with EgtV E0 A gt Jim V0 2 RA B gt a C Jtransmitted etc If J quotprobability currentquot then Jm R re ection coefficient J f which is just lBlz lA 2 do you see that from the formula me for J we got on the previous page W k39 mmx T Transmission coefficient J m which is NOT lClz lA2 it is J 2 Jr W k Think a little about the presence of the k k term there I might not have expected it at first thinking maybe T CZAZ but when you look at our probability current formula you realize it makes sense the waves have a different speed over there on the right side and probability current depends not JUST on amplitude but ALSO on speed We solved for BA and CA so here 1 2 R 7 kw convmce yourself then that R T 1 l ConV1nce yourself 4 k T X 1 4 k E V0 Note If V0 gt 0 I T gt 1 ConV1nce yourself and then check from the formulas above that R gt 0 and T gt 1 This makes sense to me there s no barrier in this limit kl On the other hand if E gt V0 I claim E gt 0 again do you see why and in THIS limit the formulas above tell us that R gtl and T gt0 Again this makes sense in this limit it is quotforbiddenquot to go past the barrier Classically ifE gt V0 R 0 QM gives something new here S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 3 Page 21 Other examples Potentlal barrler V i 39 same 2 sun my Pain but can compute R T E lt V0 T is not 0 even though classically it would be This is tunneling E gt V0 Classically Tl but QM some re ection S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 20 SJP QM 3220 Ch 2 part 2 Page 7 The Harmonic Oscillator Many objects experience a classical force F k X VX Hooke39s Law This is the harmonic oscillator and VX 12 k X2 Note that almost any VX with a minimum will look at least apprOXimately like this at least for small X so it s physically a very common situation Classically Xt A sin mt B cos mt for Harm Osc withmaldm soV12mm2X2 X T a de ned quantity but itte you period T 2n 03 So let s consider the Quantum Harmonic Oscillator As before separate X and t and look for solutions not 0 h 1 Tu Kx Emwzxzmx E ux 6 these u s will give us stationary states In riEth then if you nd such a u 1px t uxe gives the time dependence Just like before we will nd that if we insist ux gt 0 ie our quotboundary cond then we will nd only certain E39s called En will and the corresponding unX will be unique So just like in the b0X we ll have I nXt39s each n corresponding to a stationary state with discrete energy This differential equation is a 2quotd order ODE but that quotX2quot term makes it hard to solve There are tricks See aside neXt page Trick 1 For large X X2 dominates so if 1 1 iu x E rmwzxzm mwzxzu 2m 2 A Solution check it s not familiar mwxz ux Aei 2 2 Be 2 Note 2 undetermined constants for 2quotd order ODE nice T This B term is very nasty as X gt oo toss is as unphysical NeXt comes trick 1 part 2 which is to quotfactor outquot this large X behavior S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 SJP QM 3220 Ch 2 part 2 Page 8 Let s now assume hope wonder if ux hxe39mw2xZ2h2 and hope that maybe hx will be quotsimplequot Indeed use the love this name Method of Frobenius tLv hx a0 a1x azx2 Plug in see what happens For now I m going to skip this algebraically tedious exercise and skip to the punchline If you insist that ux stays normalizable you nd that series for hx needs to m it must be a nite order polynomial And that will happen if and only if E takes on special unu 1 values namely E n ha where is an integer If E is this then the polynomial is quotnth ortherquot and the hnx functions that come out are Hermite Polynomials See Grif ths p 56 U1 x U3 x 1 quotMAX 1 7mw2x22h2 1 7 7 th E 7h u1x e w1 1 2 a Note here h1xl basically the stuff out front is just to normalize 1 ma i 2 ma mwzxzth 3 x 7 7 7xe w1th ener E 7hw quot2 7m Vg h gy 2 2 and check out what s in front of the exponential besides all the constants there s my lSt order polynomial the rst Hermite polynomial You could just verify if you want just be TAKING the derivatives that at least for these 2 h 1 lSt 2 examples 7M Hx 7mw2x2unx E u x 2m 2 Remember these give unx time dependence is simple Stationary states as a function of time look like 1p x t u xe You can then form any combination of these to build any quotparticle in a wellquot state and 7iEnth the time dependence follows 11 Ec u xe39iEquot39h S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 Ch 2 part 2 Page 9 Observations Like particle in box unx are I even or odd I oscillatory with un having n l zero39s I orthogonal really orthonormal funxumxdx 6m I Complete AA fn can be expanded via Fourier39s trick as 2 cnun x nl if fn vanishes properly at oo anyway I Energy of each un is discrete grows like n here I Lowest energy is not 0 it39s 12 ha This means you cannot quotstopquot a mass on a quantum spring v L K MLJJ 111 This dashed curve is the classical px function Note that classically E can be anything but given E you quotturn aroundquot at 12 kx2 max E For small n unx2 doesn39t look classical at all but for large n it sort of does if you quotaverage over short Wigglesquot Here39s a quotrea quot picture from Shankar39s text g 72 page 202 N Myll S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 SJP QM 3220 Ch 2 part 2 Page 10 Aside Another method to know of you might call it Trick 2 Numerical solutions 2 How to solve u h71E Vxux w1thout yet knowrng E Even solutions Guess an E use a similar situation or just units For Harmonic Oscillator ha has units of energy so maybe just try it as a starting guess call it E Let u0 l u390 0 4 at at origin since I m looking for the even solution u0 may not be 1 but we39ll normalize later Now pick a quotstepsizequot a tiny We39ll quotstep alongquot in X nding uX point by point I Use SE to tell you uquot0 u 0 E V0u0 I Use u390 us 7 uO to compute us I Use uquot0 u39gsiu39101 to compute u39s Now repeat stepping acfoss X in stops of a If u starts to blow up you need more curvature at start so raise E and start again If u blows down you need less curvature lower E For odd solutions start with u0 0 u 0 1 again u390 may not be 1 but we ll renormalize later Otherwise it s the same game So basically you pick an E numerically quotshootquot from the origin and keep xing E till you get a solution which behaves well at large X In this way you can plotcompute uX and nd the energies for essentially any VX In this sense you don t have to feel bad that only a few VX s yield analytic solutions for unX and En All V39s can be quotsolvedquot numerically m Aside S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 SJP QM 3220 Ch 2 part 2 Page 11 Trick 3 for solving the Harmonic Oscillator This trick is way out there I would never have come up with it but it s cool and turns out to be more general than you can imagine right now It39s the basis for a similar trick to understand angular momentum in 3D and then spin and moving on to quantum eld theory So it s worth learning It will also teach us some quotoperator methodsquot that are deeply central to QM So enjoy this quottrickeryquot The math is fun Here again is the 1D Schrodinger equation we re studying hi 1 7p2 mwx2 u Eu withp f ofcourse 2m 1 Bx For numbers c2 b2 ic b ic b 1 so th1s tempts us to rewr1te the equatlon on left as 271 mwx ip mwx m But this isn39t right because 13 operates on any X39s it hits Still inspired by my idea of rewriting the equation let39s de ne new operators fl 1 B mm thmw i f A 1 2mwfc a E Vthw i These are just de nitions and the names in and El will make more sense in a sec We introduced the constants out front just for units Now try writing what I had before aside from some constant um mw ij7 mane 1 A A A A A A Mp2 mum2 imwxp 17x The last terms looks like zero since for numbers X p 7 p X 0 but not for operators We call A3 All the commutator of A and B My expression has in it What39s that do S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 SJP QM 3220 Ch 2 part 2 Page 12 What does mean It39s an operator so consider acting it on m old function fx AA AA AA hB ha Mph10Cwpmfxfif eeexf 13x 13x thix operates onxf Bx A A h please check this for yourself 16 pf ff 1 B B but 3709 f x The term xai explicitly cancels leaving x x Since fx is totally arbitrary this is try for ANY f we say 3C A 7 l P I This is very important ih it s not zero we39ll talk much more about this OK back to our problem looking again at that product of El and 52 we have A A 1 A2 A2 imw 7 h dial 2hmwp mama 2hmw l 1 A 7Hl ha 2 A A A 1 or Hhwaia E Now go through yourself and convince yourself that if instead we had written down that product in the opposite order we would have gotten A A A 1 H hw aha 7 note the other s1gn for the 12 term at the end H g 2 3552 And indeed we can easily compute the commutor of El and it which you could do directly by writing out it and in in detail but it s easier to use the expressions just above A A 1 A 1 1 A 1 a a i i i 7 l Kim 2 Km 2 1 OK but wh have we done all this Here comes the crux 1 Suppose ux is a solution to H u E u ie suppose we knew the u s which is what we39re after remember Let s consider acting H on 52 which in turn is acting on u Why You39ll see S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 SJP QM 3220 Ch 2 part 2 Page 13 1903 human gm ThisisjustH A A 1 A h A 7 wa 7 2a A A A 1 hwa a 7 01 2 A AA 1 ahw aai1 iu W z Comm slide Hm I used 1 21 11 past aw right a 1 A h A A 1 h a waa7 5 mu But this isjust H gain Look over all the steps these tricks are not hard but require some practice This is quotoperator methodsquot we ll play this sort of game more often later in the course OK one more step now use H u Eu Thus IIampu 21E hm u or flaw E hm am So fan is an eigenfunction of H with eigenvalue E hm Given u I found a new different eigenfunction with a dszerent eigenvalue Now you try it this time with H a u I claim you ll get E hm CALM So now we see the w of 5 or 5 they generate new solutions different eigenvalues and eigenfunctions 5 quotraising operatorquot because it raises E by ha 5 quotlowering operatorquot because it lowers E by ha They are quotladder operatorsquot because E changes in steps of ha S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 SJP QM 3220 Ch 2 part 2 Page 14 Now for crux 2 If you apply amp7amp7ampamp7 u you create a state with E mhw m of these You can always pile up enough as to get a negative energy but there s a theorem that says E can never be lower than the minimum of VX which is 0 ACK This is bad unless there is a bottom state uo such that a uo 0 That39s the ONLY way out If there was no such quotbottom statequot then we could always nd a physical state which is impossible one with negative energy Note that 0 quotsolvesquot the S Eq it s just not interesting And from then on applying more as doesn39t cause any new troubles flfliuo 5370 0 OK so amp7u0x 0 Let s write this equation out in detail because we know what 5 is mwx u 0 Vthw dx 0 That s no problem to solve See Grif ths or just check for yourself u0x Ae mmz 2h works for any X And it s a 1st order ODE so with one unedtermined coefficient A here that s the most general solution We can even find A pick it to normalize 110X use the handy relation fe39xzdx 17 1 AermwzZh So 1400cK 7m 1 This solves SEjust plug it in On the right you ll see E0 pop out E0 aha Indeed there s a very cute operator trick to see this directly 19 Eouo gt ham a 12 0 Eouo Ihix killx n 1 so Ehw Mo Eou0 S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 SJP QM 3220 Ch 2 part 2 Page 15 OK we found the quotbottom of the ladderquot But now at uo gives another state call it u1 3 with E1 E0 ha Ehw although you may have to normalize again A 1 And in general climbing back up the ladder u x A a u0x w1th E n Ehw Although you must normalize each one separately see Griffiths or homework Question could we have missed some solutions in this way Might there be a 11x which is not quotdiscoveredquot by repeated applications of 5 to uo Fortunately no If there were 21 12x would still have to give 0 for some n and thus there d be a quotlowestquot 11x but we found the unique lowest state already Grif ths has a cute trick to normalize the states see p 47 48 He also has an elegant trick to prove f u xu xdx 5m see page 49 But let39s leave this for now Bottom line By quotFrobeniusquot or quotladder operatorsquot we found all the stationary states 1 unX With E n Ehw In general potentials VX give quotbound statesquot with discrete energies But let s move on to a qualitatively different problem where we re not bound S Pollock taken from M Dubson with thanks to J Anderson for typesetting Fall 2008 Ch 1 notes part2 l of 10 A Brief History of Modern Physics and the development of the Schrodinger Equation quotModernquot physics means physics discovered after 1900 ie twentiethcentury physics 1900 Max Planck German tried to explain blackbody quotBBquot radiation that s radiation from warm objects like glowing coals using Maxwell39s equations and statistical mechanics and found that he could not He could only reproduce the experimentallyknown BB spectrum by assuming that the energy in an electromagnetic wave of frequency f is quantized according to EEMWM nhf where n l 2 3 and h Planck s constant 66gtlt10 34 SI units Planck regarding this as a math trick he was baf ed by its physical signi cance 1905 Albert Einstein motivated in part by Planck s work invents the concept of a photon though the name quotphotonquot came later to explain the photoelectric effect A photon is a quantum packet of electromagnetic radiation with energy E hfhw Note the de nition of h E 2Ll05x103934 J s 66x103916 eV s Called quothbarquot n 1911 Ernest Rutherford New ZealandBritain shows that an atom consists of a small heavy positivelycharged nucleus surrounded by small light electrons But there is a problem with the classical theory of this nuclear atom An electron in orbit about a nucleus is accelerating and according to Maxwell39s equations an accelerating charge must radiate give off EM radiation As the electron radiates giving energy it should spiral into the nucleus 1913 Niels Bohr Danish a theorist working in Rutherford s lab invents the Bohr model This is essentially a classical model treating the electron as a particle with a de nite position and momentum but the H Jr model has two nonclassical ad hoc assumptions 1 The angular momentum of the electron is quantized L nh a 2 The electron orbits determined by l are stable quotstationaryquot do 14quot Ha not radiate unless there is a transition between two orbits and then the atom emits or absorbs a single photon of energy hf I Ef Ei I Classically an electron in an atom should radiate and spiral The predictions of Bohr model match the experimental spectrum of inward as it loses energy hydrogen perfectly 8152008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes part2 2 of 10 It is important to remember that the Bohr model is simple useful and wrong For instance it predicts that the ground state of the Hatom has angular momentum L h when in fact the ground state of the H atom sstate has L 0 The Bohr model is a semi classical model meaning it combines aspects of classical and quantum mechanics Semiclassical models are frequently used by physicists because they are heuristically useful easy to understand and often give correct results But they must always be used with extreme care because the microscopic world is really purely quantum We insert classical mechanics into the microscopic world not because it is correct but because it is convenient 1922 Louis de Broglie French proposes waveparticle duality Theory and experiment indicate that waves sometimes act like particles photons Perhaps argues de Broglie particles can sometimes act like waves For photons E1 hf hi According to Special Relat1v1ty and Maxwell39s equations light of energy E carries momentum p c E Hence p1 Y hk c De Broglie argues that the same equations apply to particles and introduces the idea of matter waves de Broglie relations Ehfhw phk De Broglie39s hypothesis provides a nice explanation for Bohr s quantization condition L 11 assuming that an integer number of wavelengths fit in one orbital circumference the condition for a standing wave we have nk2nr gt 1393 2n Lrp r nh 13 231 Soon there was indisputable experimental verification of the photon concept and of the de Broglie relations In 1923 the American Arthur Holly Compton observes the Compton Effect the change in wavelength of gammarays upon collision with electrons This effect can only be explained by assuming that gammarays are photons with energy E1 hf and momentum p h 1 A 39 Then in 1927 Americans Davisson and Germer diffract a beam of electrons from a nickel crystal experimentally verifying that p hk for electrons Late in 1925 Erwin Schrodinger then Professor of Physics at Zurich University gives a colloquium describing de Broglie39s matter wave theory In the audience is physicist Peter Debye who called this theory quotchildishquot because quotto deal properly with waves one has to have a wave equationquot Over Christmas break Schrodinger begins developing his equation for matter waves n5 8152008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes part2 3 of 10 1927 Erwin Schrodinger Austrian constructs a wave equation for de Broglie39s matter waves He assumes that a free particle potential energy V 0 is some kind of wave described by wxt Aexpikx ot Recall Euler s relation c 3 cos 6 i sin 6 so expikx mt coskx not isink x ut Initially Schrodinger works with a complex wavefunction purely for mathematical convenience He expects that in the end he will take the real part of 1P to get the physically quotrealquot matter wave l 2 p2 The energy ofth1s free part1cle 1s all k1net1c so E Emv In According the de Broglie relations this can be rewritten hk ha 1 2m Schrodinger searches for a wave equation that will reproduce this energy relation He notes that 611 0 k t P d at atexp1 x m in an aw a aim e ik wt ikl11 ik 11 k2 IJ ax ax xp x 1 6x2 2 so 2 2 2 2 2 the trial equation ih h a I leadsto ih in h k2 hm h k at 2m 6x 2m 2m This looks promising To describe a particle with both KE and potential energy V VX Schrodinger added in the term V41J producing nally 6111 h 62111 1h 2 6t 2m 0x Vx11 The Schrodinger Equation is really an energy equation in disguise When you look at the SE you should try to see E KE PE For a particle with frequency f energy E h f and wavelength 2 momentum p hK in a potential V this equation appears to correctly predict that 2 p iisz Emt KE PE 2 V wh1ch accord1ng to de Broglle1s hm V m 2m This quotderivationquot is merely a plausibility argument Schrodinger immediately used the equation to solve for the energies in a hydrogen atom and found that he got the right answer for the energy levels This gave him con dence that the equation was correct 8152008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes part2 4 of 10 The physicist Paul Dirac famously asserted that the Schrodinger Equation accounts for quotmuch of physics and all of chemistryquot It is probably the most important equation of the 20Lh Century Its effect on technological progress has been much much greater than the more famous equation E mc Schrodinger was quite puzzled by the nature of the wave function What is the physical meaning of lI xt He wanted to think of it as some kind of physical matter wave like an electromagnetic wave Ext But this interpretation could not explain a host of experimental results such as that fact that a particle with a large extended wave function is always found at one small spot when a position measurement is made It was German theorist Max Born who late in 1927 proposed that the wave function is a kind of information wave It provides information about the probability of the results of measurement but does not provide any physical picture of quotwhat is really going onquot Bohr Heisenberg and others argued that questions like quotwhat is really going onquot are meaningless Humans live at the macroscopic level excellently described by classical mechanics and our brains evolved to correctly describe macroscopic classical phenomena When we ask quotwhat is going onquot we are really asking for an explanation in terms that our brains can process namely a classical explanation The microscopic world is fundamentally different from the classical world of large objects that we inhabit and our brains39 internal models simply don t apply at the level of atoms There may be no hope of understanding quotwhat is really going onquot in atoms because our brains are not built for that job All we can know are the results of measurements made with macroscopic instruments This view that the wave function provides probabilistic information but not a physical picture of reality is part of the quotCopenhagen interpretationquot of Quantum Mechanics socalled because it was largely developed at Bohr s research institute in Copenhagen Einstein de Broglie Schrodinger himself and others were very dissatisfied with this view and they never accepted the Copenhagen interpretation Nobel Prizes for QM Many of the pioneers of QM eventually received Nobel Prizes in Physics 1918 Max Planck concept of energy quanta 1921 Albert Einstein photon concept and explanation of photoelectric effect 1922 Niels Bohr Bohr Model 1926 James Franck and Gustav Hertz FranckHertz experiment showing quantization of atomic levels 1927 Arthur Compton Charles Wilson Compton effect Wilson cloud chamber 1929 Louis de Broglie waveparticle duality 1932 Werner Heisenberg Uncertainty Principle and Matrix formulation of QM 1933 Erin Schrodinger and Paul Dirac Formulation of QM 1937 Clinton Davisson and George Thomson experimental discovery of electron diffraction 1945 Wolfgang Pauli exclusion principle 1954 Max Born interpretation of the wave function 39 Ernest Rutherford received the 1908 Nobel prize in chemistry for experimental investigations of radioactive decay but never received the prize for discovery of the nuclear atom 39 Albert Einstein never received the prize for either Special or General Relativity 8152008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes part2 5 of 10 8152008 University of Colorado Michael Dubson mods by S Pollock Ch 1 nules panz 6 M m MaxPlannk Gaman 185871947 Albert Emgem GamaanwAssrAmEnmn 187971955 ansmax Umversny ufculumdu Mmhael Dubsunmnds by s Fulluck Ch mulesme 7 um EmestRu39herfurd NewZealandrE nush 187171937 NxelsEuhrDamsh188571962 ansmax Umversxly ur Culumdu Mxnhael Dubsun muds by s Fulluck Ch lnotes part2 8 of 10 Lows de BrogheFrench 189271987 Erwm Rudolf Josef Al Exander Schrodinger Ausman 188771961 8152008 Umversxty of Colorado Mahael Dubson mods by s Pollock Ch 1 notes part2 9 of 10 Comment on numbers and scales QM involves solving partial differential equations or matrix equations or operator equations which can at times be mathematically formidable but in many cases understanding the simple relations described in the sections above and knowing the basic units and numerical constants is enough to give you a sense for a lot of physics Simple relations to remember All waves AU v wavelength frequency speed ofwave For light vc 3XlO8 ms Light Ehv energy of photon is Planck s constant times frequency ha same just convert frequency v to angular frequency 00 by v 27100 deBroglie relations quotmatter wavesquot p h 2 p is momentum lambdawavelength of the matter wave k THIS the quotwave numberquot Einstein E nc2 only for objects at rest otherwise use E2 7262 mzc4 It39s amazing how much back of the envelope physics you can do with freshman physics plus these Common numbers and constants You can always look up constants and you should locate your favorite collection but having a few common constants at your fingertips is really pretty handy Still eg 72105Xl034 J s is a number which I often forget it s so small The combination of this constant times the speed of light he 2000 eV Angstroms roughly is somehow a simpler number for me to remember And this combination occurs often 39 An Angstrom is 103910 m or 01 nm a very convenient distance in atoms since that happens to be the typical size of atomic orbits 39 An eV is not a standard SI energy unit but this one I always seem to remember from EampM leV 16X103919J is atypical atomic energy scale It39s the energy gained when an charge quotequot like a proton or electron moves through a potential difference of 1 Volt You use eV a lot in practical atomic problems Remember from EampM that electrical potential energy of two charges a distance r apart is 4611i near 2 which tells us using qe that has units of energydistance the exact SAME units as he 6 47130 Sure enough the following combination of constants which is thus UNITLESS a pure number 1137 appears over and over in physical problems This is called the quotfine structure constant of naturequot Another combination which you often see is mac2 511 keV where m is the electron mass and c the speed of light By Ech this must have units of energy If you solve a problem and the answer does NOT have such a combination of constants you might quotmultiply and dividequot by some power of c or 7 to GET this combination so you can then just quotrememberquot the answer I can t tell you how much calculator grief this has saved me over the years 8152008 University of Colorado Michael Dubson mods by S Pollock Ch 1 notes part2 10 of 10 Sligl1tly silly Example How many photons does an ordinary lightbulb emit each second This is a highly complex and illdefined problem but we can make a simple crude estimate by making some simplifying assumptions 1 Assume light from a bulb is visible in the middle of the spectrum yellow You would then have to go look up the wavelength of visible light it s about 6000 Angstroms 2 Assume that all the energy is going into visible light ok assumption for a uorescent bulb You would then have to go look up the power of a bulb though you can probably guess that s about 100W for a bright one Each individual photon has energy Ehv and so the total power 100 Js must be producing a total of 100 Js hv Joulesphoton That s it that s how many photonssec should come out Now vc which is handy because F6000 Angstroms was what I looked up And leV 16x103919 J converts us to a more quotphotonicquot energy scale So the answer is 100 Js 1 6x103919JeV 2 hc and then multiplying top and bottom by 2 313 I get 2007 1 6x10 M he and now Iput in A6000 Angstroms and he 2000 eV Angstroms to get about 1022 photons each second No wonder we never notice or care that they39re discrete that s a huge number Slightly less silly Example What energy should electrons have if you want to do an experiment to demonstrate their quotwave nature quot7 I am thinking that quotshowing wave naturequot means generating some sort of interference and for THAT you need to have a wavelength comparable to a quotslit sizequot The smallest slit I can imagine is ONE atom wide imagine bouncing electrons off a crystal they can bounce off of one atom or its neighbor and those two quotsourcesquot will interfere Thus without doing any hairball calculations I would roughly want Melectron atomic size 1 Angstrom or so DeBroglie then tells me I want the electron to have momentum p h A And the energy associated with that momentum will be the good old freshman kinetic energy 2 formula E p22m So E 2h That s it I could plug in numbers mhz I But I can also multiply top and bottom by cA2 and divide top and bottom by 2 pi 2 to get 4 2 h 2 E this is the SAME formula but now has combinations of constants I remember me This I can do in my head 702 is about 10 so Ihave without a calculator E 47502 2000 eV AngstromA2 2 511 keV1 AngstromA2 That s about 40410 6 eVA2 1000 1000 eV 160 eV Anything LOWER than that will have a smaller momentum or LONGER wavelength and will diffract even better So that s an upper bound You don t need a high voltage supply for this This is the basic physics of the Davisson and Germer experiment that first showed that deBroglie was not off his rocker 8152008 University of Colorado Michael Dubson mods by S Pollock SJP QM 3220 Formalism 4 Let s now start the leap from 1D to 3D ID gt 3D x gt 7 xayaz 11106 gt 1M 2 o 2 3A flwool dx 1 9 MM d r allx all Space volume K element 777 dxz dyz dzz 994 2 a2 9 311 12 62 a2 a2 2m 2m3x2 2m 2m Simplest example Particle in a 3D box surprising good approximation for describing conduction electron in a metal z 0 0 lt xy z lt a V x y Z 00 elsewhere 3 TDSE 991p mg 31 First Separation of variables l117 t 1117 p0 a y s 99W E W rpm X a 2 2 2 2 i w if if Ew 2m dx dy dz Page 3D l6 M Dubson typset by J Anderson mods by S Pollock Fall 20 SJP QM 3220 Formalism 4 Next separation of variables again Search for solutions of form I xyz W1 X 39 W2 y 39 W3 2 Plug into TISE and divide thru by 1111 w 1113 1 Whips gt 32 1 d21p1xzpfx M2lady2211231p1 dxz WI etc hz w39x1p39239xw39339x E 2 W105 1020 1P3Z EH EH EH fx gy hz f g h must each be a constant otherwise cannot always add up to constant regardless of x y Z wi39ltxgtE 2m wx 1 2m My 2 etc E1 E E3 E Note E1 E2 E3 not independent must sum to E 2 gt 11500 k1p1x likewise for xyz x 7 define k 2mE1 h2 Have broken our 3D problem into three lD problems 2 2 2 nxmc E1 nxhzz a 2mg 1D solution gt 10106 2 sin nx 123 1 n n 2 2 1P2 gsiny7y E2 h n j etc a a 2 2 2 Em nx ny nz 2 2 EE1E2E3 h 2m E V wltxyz2 a a a a Page 3D l7 M Dubson typset by J Anderson mods by S Pollock Fall 20 SJP QM 3220 Formalism 4 States identi ed by 3 quotquantum numbersquot nx ny nZ all 2 1 n0 not allowed since gt 11 0 Dirac Notation nx ny nz e annynz hznz Ground state 111 has energy 372 3 8 2ma But there are three lstexcited states 112 121 211 all with the same energy 121222s6s These 3 states are linearly independent one state cannot be written as a linear combo of the other two Linearly independent states with same eigenvalue are called degenerate states Doesn t arise in 1D usually Note symmetry in Xyz in V7 potential Degenerate states often arise from symmetries Also note we can write if jx 7y 72 A j if a2 2m a7 x 2 1 lJnxnynz is simultaneously an eigenfunction of all 4 operators jx 7y jZ wEwnxznyznzzsq mwE1nxzsw etc Remember from our formalism chapter when can two different operators have simultaneous eigenfunctions When they commute A A h2 2 So for our particle in a 3D box HwyQ 0 gt states of definite jx and 7y simultaneously are allowed Sure We know what those are eg 1 Illnxnynz What s next We ll start thinking about angular momentum Page 3D18 M Dubson typset by J Anderson mods by S Pollock Fall 20


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