Quantum Theory I
Quantum Theory I PHYS 321
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Quantum Mechanics of Spin12 1 SternGerlach Experiment 11 Schematic diagrams for the SternGerlach experiment In the standard Stern Gerlach experiment as illustrated in Fig 12 particles are red through a gap between the poles of two magnets which are arranged to provide an inho mogeneous magnetic eld B Bz2 Detector Screen in 1quot g Figure 12 Stern Gerlach experiment for particles with identical magnetic dipole moments In the simplest realization 132 B02 where B0 is constant and in this case a classical analysis predicts that the particle follows a partly curving trajectory toward a detector screen resulting in a de ection d which is proportional to M where p is the magnetic dipole moment of the particle Despite the fact that the Stern Gerlach experiment measures magnetic dipole moment it is conventional to regard it as a measurement of a component of the particle s spin angular momentum This follows since the spin angular momentum of the particle S is related to the magnetic dipole moment via gq p 7 g S 12 where q is the charge of the particle m its mass and g the g factor for the particle The Stern Gerlach experiment indicates that for spin 12 particles such as the electron only two de ections where possible corresponding to 1 ML2 and S iii2 It is typical to consider this entire experiment as nothing other than a measurement of 1 and the Stern Gerlach apparatus as a merely a device for measuring 1 In the standard Stern Gerlach experiment there is no control over the spins of the particles prior to their entry into the region between the two magnetic poles the best assumption is that the spins are randomly oriented although the total magnitude of the spin is the same for each particle Then for spin 12 particles each is found to emerge at one of two locations on the screen as illustrated in Fig 13 with approximately half emerging in the upper beam for which 51 Hi2 and half in the lower beam for which Sz iii2 Note that any individual particle only emerges at one of the two detector locations and it is impossible to predict at which it will arrive Detector Screen E o 0 SZ aarnsoror r H 7 7 7 7 7 7 7 7 7 7 7 7 M 4 o 39 0 o E Z 2 Figure 13 Stern Gerlach experiment for particles with randomly oriented magnetic dipole moments The entire apparatus could be rotated through 90quot about the z axis so that the magnetic eld is now oriented along the y axis It follows that this would measure the y component of the particle s spin Sy The possible outcomes of the experiment cannot be changed by a mere rotation of the apparatus and for a spin 12 particle must be S h2 or Sy iii2 One can now conceive of a Stern Gerlach experiment in which the magnetic eld is oriented along an arbitrary direction 73 in which case the Stern Gerlach deVice is an apparatus for measuring Sn the component of particle spin along 73 Again the two possible outcomes are S h2 or S 752 A schematic description of any of these Stern Gerlach measurements requires a speci cation of the direction of the magnetic eld f1 and the two possible locations in which the particle emerges corresponding to S h2 or S 7252 The entire arrangements of magnets can be represented by a box labeled with f1 and from which emerge two particle trajectories each corresponding to one of the two outcomes This is illustrated in Fig 14 and is called an SG measurement 5 h2 5 752 Figure 14 Schematic diagram of an SG measurement The horizontal line on the left indicates the trajectory of particles red into the apparatus Those on the right are the trajectories corresponding to each of the two outcomes 12 Repeated SternGerlach experiments One can imagine a sequence of Stern Gerlach experiments performed on the same particle The particle is red into one Stern Gerlach apparatus emerges in one of the two trajecto ries and is then directed through another Stern Gerlach apparatus In such sequences it is possible to ask about the outcomes of later measurements given a knowledge of outcomes of earlier measurements in the sequence For spin 12 particles this amounts to thought ex periments the actual experiments are either extremely dif cult or have not been performed at all However one can show that the polarization states of individual photons can be described mathematically in a way completely analogous to the states of spin 12 particles such photon states are relatively easy to subject to sequences of polarization measurements that are analogous to the sequences of Stern Gerlach experiments to be considered 121 Repeated measurements of the same type Suppose that a particle is subjected two a succession of two SG2 measurements as illus trated in Fig 15 The experimental evidence indicates that any particle that emerges from the rst apparatus in the upper trajectory for which Sz Hi2 will emerge from the sec ond apparatus in the upper trajectory again giving Sz h2 with certainty Similarly a particle which emerges from the rst apparatus in the lower trajectory will always emerge from the second apparatus in the lower trajectory This correlation between the two succes sive measurement outcomes is independent of the particle s state prior to entering the rst apparatus Note that there are states prior to the rst apparatus such that the particle will never emerge in the lower or alternatively the upper trajectory the correlation between the two measurement outcomes however is always preserved 7 SZ 5 2 Si 7h2 8G2 l l l l l 52 752 never SGE SZ 2 never SGz 51752 Sz h2 Figure 15 Successive SG2 measurements It follows that if the rst measurement yields S h2 then if the only subsequent actions on the particle are measurements of S2 these will always yield Sz Hi2 A similar statement applies to S 752 In this sense one can say that after the rst SG 2 measurement the particle has a de nite value of S1 That is any subsequent measurement of S1 will yield just one of the values with certainty In this sense and only in this restricted sense the particle is in the de nite physical state for which S Hi2 De nite states of physical systems such as those described above are denoted by a ket an arrangement of brackets containing a label which indicates a certain outcome upon a particular measurement These have the form llabel where the contents of the ket are a label which is usually an abbreviated way of describing the particle s physical state From the description of sequences of 8G2 measurements we Hz ltgt measuring Sz yields S ML2 with certainty and 1 3 liz ltgt measuring Sz yields S 752 with certainty 39 Thus the particle which emerges from the rst apparatus in the upper trajectory is in state Hz However neither of these states necessarily applies to particles entering the rst apparatus since these can emerge with either outcome for S2 Note that the symbol within the brackets here 2 or 72 is a label which indicates the relevant measurement and outcome This can be generalized to situations involving successive SG apparati for some xed direction 73 It follows that the corresponding states of the spin 12 particle are H71 ltgt measuring Sn yields S ML2 with certainty 1 4 1 73 39 ltgt measuring Sn yields S 752 with certainty Schematically the physical meaning of the general states H71 and lift phrased in terms of future measurement outcomes is provided in Fig 16 Sn FL2 certainty l gt iiiiiiiiiiiii i never Figure 16 The meaning of the state H71 in terms of an input to SG 73 The measurement outcome Sn ML2 will occur with certainty A consequence of the relationships of Eq 14 is that if a particle is subjected to an SG measurement and emerges with Sn ML2 then its state immediately after mea surement is H71 Similarly if it emerges immediately after measurement with S 752 then its state immediately after measurement is lift An example is illustrated in Fig 17 Example Describe an experiment which is such that it prepares particles with certainty in the state H133 Figure 17 The meaning of the state H71 in terms of an outcome after SG A particle is supplied and if the measurement outcome is Sn ML2 then subsequently the state is Answer Suppose that you had particles which are in some sense randomly prepared Each is subjected to an SG ii apparatus Those that emerge with Sm ML2 are in the state H133 and those that emerge with S 752 are in the state 31gt Keep all of the former and discard any of the latter Note that this will not work if by some chance all the particles are in the state 1733 prior to entering the SG ii apparatus In this case more manipulations to be explained later are needed Thus there are in nitely many states available to a spin 12 particle corresponding to all possible directions 73 In fact it will emerge that every state available to a spin 12 is of the form H71 for some unit vector 73 Physically this means that for any particle there is some possibly unknown to the experimentalist direction f1 such that if the particle is subject to an SG apparatus the measurement outcome will be S ML2 with certainty 122 Repeated measurements of the different types To understand how the many possible states are related to each other consider successive Stern Gerlach measurements of different types For example suppose that the particle is rst subjected to an SG 2 apparatus and then to an SG ii apparatus as illustrated in Fig 18 SZh2 Szh2 SM 51 752 soz Sz h2 sci 51752 Sm 2 Figure 18 Successive Stern Gerlach measurements of different types Experiments show that if the particle emerges from the 8G2 apparatus in the upper trajectory then it can subsequently emerge in either but not simultaneously both of 10 the trajectories after the SG ii apparatus Unlike successive 8G2 measurements no nal trajectory is excluded Furthermore the probability with which it emerges in the upper trajectory is 12 and the probability with which it emerges in the lower trajectory is found to be 12 In terms of states7 one can say that FL 1 Sm 5 with probability 5 12 into SGzi 15 FL 1 Sm 75 with probability In general a particle can be subjected to an SGm apparatus followed by an SG apparatus where 731 and f1 are any two unit vectors Suppose that the particle emerges from SGm with Sm FL2 Thus the state immediately after the rst SG apparatus is 731 The experimental eVidence is that it will emerge from SG with S FL2 with probability 1 731 Thus the general rules are a 1 Sn 5 with probability 51 771 73 731 into SG 16 a 1 Sn 75 with probability 51 7 771 73 and FL i i i 1 g 1 Sn 5 w1th probability 51 7 m 71 7771 into SG A 17 Sn 7 with probability 1 771 77 Example Suppose that a particle is subjected to an SG ii measurement and emerges with Sm FL2 It is then subjected to an soy measurement List the possible outcomes and the probabilities with which they occur Answer If it emerges from the SG ii apparatus with S FL2 then its state after the rst measurement is 133 Using Eq 16 with 731 ii and f1 g one obtains outcome Sy FL2 with probability 12 and S 752 with probability 12 An unusual feature of quantum mechanics is eVident when one considers two Stern Gerlach measurements of the same type interspersed with a single Stern Gerlach measure ment along an orthogonal direction An example is illustrated in Fig 197 in which it is assumed that the particle emerged with S FL2 after the rst SG 2 measurement 11 8G2 1 h2 sz 752 Figure 19 Three successive Stern Gerlach measurements for the circum stance in which the particle emerges with S1 FL2 after the rst 8G2 measurement Applying the rules of 16 one can deduce that the probability with which the particle emerges with S ML2 after either of the latter SG 2 measurements is 12 Similarly the probability with which the particle emerges with S 752 after either of the latter SG 2 measurements is 12 Exercise Consider the arrangement of Fig 19 Suppose that the particle emerges from the SG ii measurement with Sm ML2 Show that the probability with which the particle emerges from the subsequent 8G2 measurement with S ML2 is 12 Repeat this for a particle that emerges with S1 752 is 12 Re peat this entire analysis for the case where the particle emerges from the SG ii measurement with Sm iii2 Exercise Repeat the argument of the previous exercise for the case where the particle emerges with S ML2 after the rst SG 2 measurement The implication is that it makes sense to speak of a particle as having a de nite value for S1 in the context where the only subsequent operations are 8G2 measurements but not where subsequent operations include SG measurements where f1 is distinct from 2 or 72 It is impossible to speak of a particle as having a de nite value of S1 for arbitrary general scenarios smh2 Sz52 SZ 752 S h2 Smh2 517 32 This gives 1 711 1 in l 1 tgt 5 tthzgt 5 tZl zgt 1 A 1 w A ltzgt 5 tl zgtl iiwtZ Sim2 and ignoring the global phase factor 5 the state of the particle at time t is equivalent to 1 1 Ilt 7 2 7 l lt gtgt l gt This is the same as the state H41 where f1 has spherical coordinates t9 7r2 and j wt The time dependence of the label f1 is clear in this case Note that this vector lies in the my plane and precesses about the z axis with an angular frequency 41 giant 72 The remaining step is to relate the time evolution operator to observables associated with quantities such as external magnetic elds The idea will be to consider derivatives of the state Clearly the notion of differentiating a timedependent ket will be crucial This is de ned via 1w clttgt12gt c4lttgt172gt gm 12gt 172 114323 In terms of representation in the l2gt 72 basis7 1 d6 1 E 11425 H if E E 11439 W Then 1w 0o Wlt0gtgtl rm M0 where Ut 1 is the inverse operator or matrix to It is straightforward to show that the operator A 005 Ut 1 is anti Hermitian7 ie Al 7A Thus the operator i l Ut 1 is Hermitian Thus means that 2 Mt 705 Wlt0gtgtl 2 1005 mail 11105 11440 and7 sincei 005 Ut 1 is Hermitian and therefore an observable7 it is reasonable to ask which observable this is Considerations of energy conservation lead to the conclusion that a plausible choice is that this is related to the energy or equivalently that the observable is the Hamiltonian This motivates the main rule which connects the evolution of quantum systems to factors such as magnetic elds that drive the evolution The time evolution of a quantum mechanical system that is isolated from other quantum mechanical systems is dictated by I d A 11441 mgwa Hwogt where H is the Hamiltonian Eq 11441 is known as the Schrodinger equation lt s usefulnesses here is that it frequently straightforward to construct the Hamiltonian given a knowledge of the physical circumstances and then to solve the Schrodinger equation to determine the evolution op erator7 It is seldom straightforward to arrive at the evolution operator without rst passing through this step Example For a spin 12 particle of mass m7 charge q and g factor7 g in a general7 possibly timedependent7 magnetic eld mwmoiaoyao2 the Hamiltonian is 7 7 3mm in Byt 9 Bzt bl gq 5 B205 BwT iBytgt 2m 2 BAT iByt z 39 The Schrodinger equation is represented in terms of vectors and matrices by mi 0 7 EL Bzt BAT 7 iByt 0 dt c 2m 2 BAT Hag2 Bzt c 39 Upon expansion this can be seen to yield two coupled rst order differential equations for cit 97 481 Solving the Schrodinger equation for timeindependent Hamiltonians 1n general the coupled differential equations yielded by the Schrodinger equation are too complicated to solve in closed form for the evolution operator However for timeindependent Hamiltonians there is a general solution that is straightforward Theorem 1f the Hamiltonian H is time independent then M amth 11442 Proof By Eqs 11440 and 11441 the evolution operator satis es 1 A A A 5 7U if U t 1 H 2 1 o o or equivalently it satis es the differential equation 1 A 72 A A 7U 25 iHU t d o h o Additionally Now consider the possible solution 00 eiilflt 39 Then 1 A d iU t i lHth dt dte g inth i 1 ng 7 FL Additionally A t e thh 70 f Thus Ut eiilAIth satis es the differential equation for the evolution operator and gives the correct initial value There is a unique solution to this type of rst order differential equation with a given initial value 0 Example Suppose that a spin 12 particle of mass m charge q and g factor g in the magnetic eld B B02 Determine the evolution operator and show that it is a rotation Find the axis and angle of rotation 98 Answer First7 the Hamiltonian is For convenience7 de ne Then Eq 11442 implies U05 eiifltE eiiw z tEZ eiiwffztZ39 This is a rotation about the 2 axis through angle wt This extends quite generally to any constant magnetic eld Consider a eld of magni tude B oriented along f1 nmzi mg nzz Then the eld is B B nmzi mg 7112 The Hamiltonian is A 5w H 7 Wm mm 71162 M A 7 7 where w iqu2m for a particle of mass m7 charge q and g factor7 9 Thus the evolution operator is 00 eiiwffntZ which is a rotation about axis f1 through angle wt We have thus arrived at For a spin 12 particle of mass m7 charge q and g factor7 g in a con stant magnetic eld of strength B in direction f1 the evolution oper ator is a rotation about 7 1 through angle wt where w iqu2m Since every timeindependent Hamiltonian has the form of that generated by a constant magnetic eld plus a possible multiple of the identity operator which is irrelevant for evolution we have classi ed every possible type of evolution generated by time independent Hamiltonians in terms of rotations ln fact7 this extends to neutral spin 12 particles where w 39yB where 39y is called a gyromagnetic ratio and is a property of a particle neutrons are an example These notions of evolution provide one of the basic concepts for nuclear magnetic resonance NMR In typical NMR scenarios there is an ensemble of particles in various quantum states The simplest case is that where each particle in the ensemble contains a single spin 12 particle It can be show that for this7 the state of the ensemble is 99 equivalent to a smaller ensemble each of Whose members particles is at any stage in the same state H71 for some direction 73 An NMR spectrometer s measurement outputs yield good approximations the expectation values Sm 2 and lt51 and these are gathered into an ordinary three dimensional magnetization vector M Sm i 21 lt51 2 It can then be shown that for an ensemble each of Whose members is in the state H71 the magnetization vector is M on where 04 is a constant Thus if the spin 12 particle is placed in a constant magnetic eld B Bl along some direction I the state rotates about this axis at a constant rate with an angular frequency w where w 39yB It follows that the magnetization vector also rotates about this axis with an angular frequency w This essentially produces a time varying magnetic eld which can induce currents in coils situated around the spin 12 particles This is the quantum mechanical manifestation of Larmor precession and the frequency with which it occurs is called the Lamar frequency 482 Evolution of energy eigenstates The evolution of energy eigenstates occupies a special place in the dynamics of quantum systems by virtue of the fact that their time evolution is particularly simple Additionally this is of importance since any state at an initial instant can be expressed as Cil gt1gt 62 l gt2gt where Mn and l gt2gt are the energy eigenstates and 01 and 2 complex coef cients Thus NW U W0 UCl l gt1gt 62 l gt2gtl c1Ul 1gt 20 l 2gt 11443 since the evolution operator is linear Thus the evolution of the state will be completely determined by the evolution of the energy eigenstates As an example consider a spin 12 particle in a magnetic eld B B02 The Hamilto nian is A m A H 7 01 where w iquO2m The energies and energy eigenstates of the Hamiltonian are given in Table 3 Suppose that the particle is initially t 0 in the state l gt1gt H2 Then The state at a later time t is W U M0 M W M0 57139 alt2 0 1 lt 0 elm2 eiiwtZ lt o lt gt Outcome Associated State E1 l gt1gt ligt l gt2gt 12gt Table 3 Table of measurement outcomes and states for measurement of the energy of a spin 12 particle with Hamiltonian H 72 Thus the time evolution of this energy eigenstate is l gt1gt H Wt e iEltE l gt1gt Exercise Show that the time evolution of the other energy eigenstate of a spin 12 particle with Hamiltonian H 71 is l gt2gt H Wt flight11 The results derived for this example are7 in fact7 completely general Theorem Eigenstate Evolution Suppose that is an energy eigenstate of a time independent Hamiltonian and the associated energy eigenvalue is E Then the eigenstate evolves as W H 1 e mith 1 11444 Proof Exercise For a particle initially in an eigenstate7 the state at a later time is identical except for a phase factor7 e iEith 1n this case7 this is a global phase and can be ignored Thus a system which is initially in a given energy eigenstate is subsequently in the same energy eigenstate 1n terms of measurement outcomes this means that the energy measurements will yield the same outcome energy eigenvalue regardless of the time at which the measurement is per formed Thus the energy is conserved7 a reasonable criterion for any quantum system which is isolated from other quantum systems and only subjected to constant outside in uences An immediate and general consequence of Eqs 11443 and 11444 is that a general state evolves as 01 41 2 42 a cle iEltE 41gt CgeiiEZtE 42 11445 101 49 Evolution of Expectation Values An ensemble of particles all initially in the same state initially t 0 and subject to the same Hamiltonian is such that each will at a later time be in the same state Wt 705 MO Suppose that the physical quantity A is measured at the same time on each of the particles The sample average of the outcomes will if the ensemble size is large enough approximate the expectation value A well However both of these will depend on t and it is of interest to determine the evolution of this with respect to time This is given by the following theorem Theorem Ehrenfest7s Theorem Consider an ensemble of identical particles each subject to the same Hamiltonian H The expectation vale of any observable A satis es d 81 2 A A 3 W lt1 t 5 W75 lt1 tl H714 Wt 11446 M where 5 accounts for the fact that the observable may depend on time explicitly Proof Starting with the left 1 d A a ltAgt e 3 mm A was d A 8A A d i am 14 lt 1 tlgl 1 tgtlt 1 tmdtNW5 However the Schrodinger equation implies that d 2 A amt 7Hwlttgtgt and this gives Thus and this completes the proof 0 102 Example Consider an ensemble of spin 12 particles7 each subject to the magnetic eld B B133 Denote the resulting Hamiltonian H7ampm Determine equations for the time evolution of Sm and lt51 Answer First 1 a ltsmgt Similarly Thus It can be shown that the differential equations satis ed by the expectation values of the spin observables of a spin 12 particle are the same as those satis ed by the components of the spin of a classical dipole in a magnetic eld Ehrenfest s theorem frequently leads to such conclusions 103 Introduction 1 Scope and Nature of Classical Physics The ultimate aim of classical mechanics is to predict the trajectories of objects given that the nature of the interactions between them and their environments is known and that the initial states of the objects are provided In introductory and intermediatelevel physics courses7 one often focuses on one or two objects and attempts to nd their positions as functions of time by using Newton s laws Example Consider a block with known mass that is free to move across a frictionless horizontal surface The block is connected to one end of a spring of known stiffness and the other end of the spring is attached to an immovable wall Figure 12 Spring and mass system on a horizontal surface viewed from above The dashed line indicates the equilibrium position of the center of the block Suppose that the mass of the block is m7 the spring constant is k and x denotes the displacement of the center of the block from equilibrium Newton s second law applied to an object that moves in one dimension gives Fnet ma 12 which gives 12 7km W 13 where t is time Via this route the problem of determining the trajectory ie the dependence of x on t of a physical system has been reduced to the math ematical problem of solving a second order differential equation Fortunately the type of differential equation that appears in Eq 13 can be solved fairly easily The general solution is mos 350 comm 0 sinwt 14 w where k w a 15 and the two constants z0 and 120 have the physical interpretation of the initial position and velocity respectively The graph of position as a function of time is sometimes called a trajectory Fig 13 illustrates an example of this for the block connected to a spring Vt Figure 13 Position vs time graph for a block connected to a spring With a little insight it is clear that such a plot of position vs time indicates oscillatory motion and in this sense the trajectory offers a fairly concrete description of the block s motion In the example of the spring and block system we see that Newton s laws eventually give an expression for the trajectory of the block All other physically interesting quantities such as velocity momentum and energy that pertain to the block can be derived from the equation for its trajectory Exercise Show that the velocity of the block as a function of time is given by vt 7wz0 sinwt 120 coswt and the total mechanical energy is given by 1 E 5 mw2x02 mv02 The general scheme of classical mechanics is Given the forces acting on an object of known mass Newton s second law can be applied to eventually determine a complete expression for the position of the object as a function of time Needless to say this is frequently only true in principle In practice the equations of motion analogous to Eq 13 for most systems are too complicated to admit known solutions In many cases involving just one object it is possible to make some progress in understanding the motion of a system by abandoning the quest for complete knowledge of the trajectory of an object and rather considering speeds and energies For situations involving many objects such as a gas it is uninteresting to focus on any individual object the detailed description promised by Newton s laws must be abandoned for descriptions involving averages that appear in thermodynamics and statistical mechanics Nevertheless in principle Newton s laws could be applied to determine the complete details of all the objects7 motions provided that the interactions between them are well known Newton s mechanics and the theories derived from it are tremendously successful in describing a vast array of physical phenomena However by the early 1900 s careful ap plication of Newton s laws contradicted the observed phenomena in a variety of physical situations typically dealing with physics at the atomic and molecular level Some of these such as blackbody radiation the photoelectric effect the Compton effect the stability of an atom atomic spectra and the Frank Hertz effect are often described in detail in sophomorelevel modern physics courses A coherent description which entailed radical de partures from Newton s mechanics of these and other physical phenomena emerged in the period from 1900 to the 1920s By the late 1920 s a new theory capable of describing all of these and more very accurately had emerged this became known as quantum mechanics or quantum theory 2 Scope and Nature of Quantum Theory Quantum theory is a general purpose theory which claims to provide a framework for understanding all physical phenomena in the same way that prior to 1900 Newton s laws were thought to describe the physical world That is it provides a set of mathematical rules which ultimately allow for a description of any physical system Example An ion of charge q can be placed in a one dimensional electrostatic potential which is described to a good approximation by V 77 26 2 z lt gt where k is a constant with units of Vm2 In classical mechanics the compo nents of the force exerted on the ion as a result of this are determined via 8V Fm qa 27 and it is easily seen that in this case the z component of the force is Fz 7km 28 This is mathematically identical to the situation of the block and spring and is merely another example of a simple harmonic oscillator In classical mechanics the task would be to determine the trajectory ofthe ion In quantum mechanics it will emerge that there is no sense in trying to do this In fact the concept of force is inapplicable and so is Eq 27 and everything that follows from it The key mathematical entity in the quantum mechanical description of this type of particle is the wavefunction m t which can produce complex values One example of a wavefunction for the ion in the above potential or any harmonic oscillator turns out to be as illustrated in Fig 24 m 0 Figure 24 One possible wavefunction for a harmonic oscillator at t O The means by which one can arrive at a sensible wavefunction for a simple harmonic oscillator are typically described in a sophomorelevel modern physics course What is important to note at this stage is that the wavefunction does not represent a trajectory of a particle as time passes In fact the wavefunction as illustrated in Fig 24 is that at one particular instant It therefore cannot resemble any kind of evolution of the ions physical state What then is its use in physical terms In general the state of a particle that is free to move in one dimension is described in quantum theory by a wavefunction m t This is useful for determining the outcomes and the associated statistics of measurements For example Probposition measurement gives outcome from 0 to 0 dm l Izo tl2 dz 29 and there are a variety of mathematical rules for predicting the outcomes of measure ments of other physical quantities statistically This is really the only interpretation of the wavefunction Quantum theory provides the mathematical rules which enable one to de termine in such cases suitable wavefunctions and to use these to determine the outcomes of measurements The wavefunction is then more of a mathematical tool than a means for visualizing the physical evolution that the trajectory offers In this sense quantum theory is more abstract than classical physics Quantum theory has been enormously successful in describing the physical world partic ularly at small scales Quantum theory features prominently in the description of amongst others 1 atoms and molecules and their interactions with light 2 subatomic particles high energy physics particle physics 3 light at the low photon levels quantum optics 4 the properties of solids including thermal and electrical properties solid state physics 5 superconductivity super uidity and 6 quantum information IV Quantum Mechanics of Systems of Particles with Spatial Degrees of Freedom One Dimensional Case 1 States and Measurements For many physical systems it is possible to measure position or momentum Important examples include harmonic oscillators and atoms Even in the case of physical systems restricted to one spatial dimension not only are there are in nitely many possible outcomes for position measurements but the range of outcomes is continuous Although the basic language and rules for describing these systems are the same as those for systems in which any measurement can only yield one of nitely many possible outcomes there are a variety of special techniques for dealing with the issues that stem from the existence of an in nite number of possible measurement outcomes Experience with classical descriptions of systems with one spatial degree of freedom would suggest that position and momentum measurements should be the cornerstone on which the physical description of a system is based To a limited extent this is true for quan tum mechanical systems however notions of position and momentum exist here primarily to establish connections to classical mechanical systems In fact in quantum systems posi tion and momentum produce problematical mathematical issues In classical mechanics it is possible to consider a particle whose state is such that it will yield one precise value for a position measurement eg z 35m In contrast for a quantum mechanical system it turns out that such a situation is accompanied by the fact that were the particle s momen tum to be measured all values for momentum foo lt p lt 00 are all equally likely There is no way to reconcile this with standard probability theory However it is possible to discuss the process of measuring energy without too many tricky mathematical complications Fortunately too energy measurements or measure ments of quantities closely related to energies are far more readily performed in typical laboratory situations For example much of the early evidence that supported the quan tum mechanical description of the hydrogen atom was derived from spectroscopic exper iments involving hydrogen In these the frequencies of the light emitted by the atoms is easily related to differences in hydrogen atom energy levels The key feature that provides for a reasonable discussion here is that the possible outcomes of an energy measurement are in nite in number but not continuous A schematic of such a measurement is pro vided in Fig lV11 In such cases the energy measurements can be labeled with integers n 1 2 The outcomes and the associated states can be listed as in Table lV11 The fact that these states are all associated with mutually incompatible outcomes of a single measurement implies that they must be orthogonal Also it is standard to require that they are normalized Thus lt il jgt 61739 N11 where the Kronecker delta symbol is de ned as 0 if 67 T 7 lV12 1 1f1 J 113 l I gt Measure E E1 E2 Figure lV11 Cartoon of an energy measurement with an in nite but dis crete number of outcomes Energy Associated State E1 l gt1gt E2 l gt2gt E3 l gt3gt Table lV11 Energy measurement outcomes and associated states For many quantum systems corresponding to particles moving in one dimension there is only one energy state for a given value of energy such systems are called non degenerate In such cases the energy states form a basis for the space of all kets This means that a n1 where on are complex numbers In terms of vectors the ket can be represented in the basis general ket can be expressed as M l gtngtln12via M n1 114 Special cases are the energy states themselves 1 0 l gt1gtlt 0 7 l gt1gtlt gt 0 7etc Clearly the in nite nature of these vectors will inhibit their effective use in calculations and it is better to work with the abstract format Bra vectors are de ned in the usual manner7 via l Igtl If M 291 n1 then lt 1 l MT 00 T nil 6 sag 1 81 c M IV15 1 n In terms of vector representations bras are represented by row vectors Thus 00 lt Pl Z 0 lt gtnl n1 lt gt cf 05 0 lV16 The inner product of two vectors can be computed using these tools and extensions of the methods available for two dimensional vectors Thus if7 l gt an n1 1 10 2 l gtngt n1 115 then ltltIgtl1 gtlt 1gtll1 gt b ml cm ml 1 17 Z Cm n1 m1 ZbZGn IV17 Note that in such calculations it is crucial to use distinct labels in each sum7 since7 in general Zzbnbicibzcz5363m 71 71 while Zzbmcn 5161 5102 5103 m 7L and thus V L V L m V L Thus we have arrived at If llt1gtgt bn l gtngt NO ion l gtngt IV18 then our prcn n1 The outcomes of energy measurements can be predicted using the similar rule to that for spin 12 measurements Thus for a particle in the state 11 PrEn llt nlilgtl2 IV19 116 Exercise Show that if 00 1 10 Z en l ngt n1 then PrltEngtlon12 One consequence of this result is that 00 10an 1 IV110 n1 and this is equivalent to the requirement that the state be normalized Any ket that represents the state of a physical system must be normalized 139 lV111 Exercise Show that the general normalization condition7 Eq lV1117 implies Eq lV110 for the state 231 on The fact that the energy states for a basis for all possible kets is equivalent to the completeness relation 00 Z l ngt lt gtnl I7 IV112 n1 In terms of vector representations 1 1 0 0 0 l gt1gtlt gt1lH 0 1 0 7 0 0 and repetition of this eventually yields Eq lV112 It should be apparent that typical states of these physical systems involve in nite sums of energy states basis vectors We seldom actually sum these so as to arrive at a single ket which represents the state of the system7 as is possible for spin 12 systems Mostly these in nite expansions are placeholders for later calculations Even so7 it may be 117 Even for spin 12 systems there are measurable physical quantities other than spin components The most important is the energy of the system The route to the corre sponding observable can be ascertained by considering the relationship between the energy and spin of a classical charged dipole When placed in a magnetic eld B the dipole has potential energy U 7 B 11369 and as this is the only contribution to the total energy the energy is E 7 B 11370 1n the absence of an external magnetic eld it is meaningless to speak of energy The route to an observable for energy in the quantum mechanical description will be to relate the magnetic dipole moment to the particle spin and to apply results for quantum mechanical measurements of spin to the problem For a classical particle of mass m charge q and g factor g the dipole moment is related to the spin via 9Q 7 s 11371 u M where S is the classical spin of the particle Thus i E 2mB s 11372 Suppose that the magnetic eld is oriented along the 2 direction Then B B12 and E 7 BZSZ 11373 1t follows that given a value for the component of the spin the energy of the particle can be computed via Eq 11373 Thus for a particle in the state 2 a measurement of the 2 component of spin gives S ML2 with certainty 1t follows that the energy of a particle in this state must be 5 9q E if B 7 2m 2 2 and an analogous inference applies to a particle in the state 72 These are the only two measurement outcomes possible and the measurement details are listed in Table 2 The corresponding observable called the Hamiltonian is constructed via the usual scheme and is A M 5 gq 1L H 325 lzgt ltZl 325 l zgt lt Zl FL 7 B15 12gtlt217172gtlt721 i 131 2m Similarly if a spin 12 particle of mass m charge q and g factor g placed in an arbitrary magnetic eld B B73 Bmzi By 1 B22 then the corresponding Hamiltonian is H i BS 2m 66 Outcome Associated State 7amp3 2 1373 2 Table 2 Table of measurement outcomes and states for measurement of the energy of a spin 12 particle of mass m charge q and g factor g placed in a magnetic eld B Bzz Using the expression for 7 given by Eq 11364 this implies the following The Hamiltonian for a spin 12 particle of mass m charge q and g factor g placed in the magnetic eld B is A 7 A H 7 M B s 11374 where B s BmSw Bysy 131 11375 Note that the Hamiltonian can be constructed from the classical expression for en ergy by replacing quantum mechanical variables spin components in this case with the corresponding observables Thus E Bs a 1111 gs 11376 2m 2m 34 Compatible and incompatible measurements According to the above construction if a spin 12 particle is placed in a magnetic eld along 2 then there are two states namely 2 and 72 each of which gives de nite outcomes for measurements of both 51 and E In this respect 1 and E are compatible measurements provided that the magnetic eld is along 2 However there is no state which will give de nite outcomes ie outcomes with certainty for measurements of both 51 and Sm In this respect 1 and Sm are incompatible measurements The question arises of whether it is possible to determine these facts given the associated observables A hint at the answer to this emerges from the following theorem Theorem Uncertainty Relation For an ensemble of particles each in the state 11 67 and any two observables A and B 1 A A AA AB 2 5 AB 11377 where the commutator of A and B is A B AB 7 BA 11378 Proof The following property of the inner product called the Schwarz inequality is crucial to this proof Let and lltIgtgt be any two states Then the Schwarz inequality states that llt9l gtl2 lt9l9gt PW Now consider the states M A e ltAgtf M M B 7 Mi M Then lt9l9gt M A7 ltAgtf A7 ltAgtf M MA M e M ltAgtfmgt e M ltAgtANMgt M ltAgt2Wgt MZM lt14ltAgtAl11gtiltlltAgtAWgtltJIlltAgt2fWgt ltA2gt e 2ltAgt MAM ltAgt2ltMMMgt ltA2gt i ltAgt2 mm Similarly m Thus the Schwarz inequality gives llt9l 1gtgtl2 AA2 AB2 which leaves AAAB The left hand side of the preVious equation is a complex number For any complex number 27 2 2 272 2 272 2 2722 2R 2 I 2 Z 2 M l M lmzl 2 2 2 2 Thus Hgt 2 1i l 2 4 272 22 2 In this case7 lt9ltIgtgt 2 ltmlt1gtgt7ltmlt1gtgt 7 1MltIgtgt7ltltIgtM1 Then lt9llt1gtgt M A7 ltAgtf B7 ltBgtf M ltM1A31Mgt7ltAgtltM1f31Mgt7ltBgtltM1Af1MgtltAgtltBgtltM1ff1Mgt ltM1AB M 7 ltAgt M 7 ltBgtltAgt ltAgt ltBgt ltM1AB M 7 ltAgt M and similarly Thus lt0ltIgtgtgt MM7ltAgtltBgt7ltMMA1MgtltBgtltAgt ltM1BA1Mgtl 1 5 lt gt 1 A A 5 1111431107 1 A A 5 ltAB1 gt which completes the proof 0 Commutators of observables occur repeatedly throughout quantum mechanics in some instances it provides a good starting point for establishing a quantum mechanical descrip tion of a physical system The following properties ofthe commutator are easily established A I Cl A B A 0 1137 aA Bl aA Al 1137 A Bl AAl 1137 where A B and 3 are any operators and 04 is any complex number Exercise Prove the above Example Suppose that a spin 12 particle is in a magnetic eld B B12 Determine 113 Answer First Eq 11374 implies that 111 7 3 2m 69 Thus Eq ll379b implies An important example involves commutators of the observables corresponding to dis tinct spin components Example Determine 1 Answer The de nition of the commutator gives Sm S27 gm 7 gm 2 One technique for computing this and which does not rely on matrix rep resentations would be to use A 5 5 525 1zgtltz175 ezgtltez1 A 5 A 1 5 Sm 5 1wgt ltw1 e 514M751 together with repeated applications of the inner product rule A more concrete approach uses matrix representations in the l2gt 72 basis Thus Z7 m Amgm quotS mgz 510 501501510 H5071 107510 071 752 0 1 52 0 71 Z 710 Z 1 0 5 o 1 2 710 0 72 mlt 0 H553 Thus 70 Similar derivations enable one to show that va Ayl 25 AZ mm 11380 l 17 ml m y An application to the uncertainties in measurement outcomes is as follows Example Suppose that an ensemble of spin 12 particles are each in the state H2 and that either Sm or Sy is measured Determine ASm and A52 and verify that the uncertainty principle is valid Answer First Thus ltSzgt lt72l zl2gt aeao 7 0 3 0 Then ltS gt Hi l2gt Maa 321 0 1 W o 1 Thus AS lts gteltsmgt2 E 2 Similarly Mfg Now the right hand side of Eq 11377 requires lt2l l z7 yl l5gt t l i532 l5gt E m i 2 2 7L2 2 Thus the right hand side of Eq 11377 gives 524 1n this case the un certainty principle reduces to an equality 1n the previous example the right hand side of the uncertainty principle returned a non zero value and this necessarily implied that for the state 2 measurements of Sm or Sy could not return particular values with certainty However for arbitrary states the right hand side of Eq 11377 is not necessarily zero Exercise Consider an ensemble in the state 133 Show that the right hand side of Eq 11377 returns 0 For the state of the previous exercise one can show that Am 0 Thus the uncertainty principle cannot be used to extract any information about Ay The following general the orem from linear algebra clari es such situations Theorem Two observables A and B satisfy A B 0 if and only if there exist a complete set of states l gtab which are simultaneously eigen states of both operators A l abgt a l abgt 19ml b 1 Proof Later 0 1t follows that if such a simultaneous set of eigenstates exist then they are the states that correspond to de nite measurement outcomes of both quantities A and B Applied to spin 12 this means that there are no states which will give de nite outcomes for measurements of both Sm and Sy since Em 7 O This is clearly a stronger statement than that offered by the uncertainty principle 72 Figure lV412 A trial solution for an eigenstate where 0 lt E lt V1 for a piecewise constant potential The trial solution demonstrates the depen dences of a and the spatial frequency on the difference between the total and potential energy in each region and satis es the matching conditions at ma However it does not satisfy the matching conditions at 5 and is thus does not represent an energy eigenstate for this potential In such cases the matching conditions severely constrain the possible values of E only certain very speci c values will work This leads to a discrete set of energy eigenvalues and energy eigenstates This is quite general and typically if there is a region in which V is lower than for the surrounding regions the set of energy eigenstates is discrete and often nite in number On either side of this region the wavefunction for energy eigenstates decays exponentially and the probability of locating it there is negligible compared to that of locating it within the energy well Such states are called bound states A wavefunction for such an eigenstate is illustrated in Fig lV413 5 Momentum representation of the particle state The position wavefunction z t of a state is a useful for describing position mea surement outcomes via the position probability density Pzt l llztl2 153 Figure lV413 An eigenstate where 0 lt E lt V1 for a piecewise constant potential This demonstrates the dependences of n and the spatial frequency on the difference between the total and potential energy in each region and satis es the matching conditions at ma and 17 which can be used in probability calculations to give 17 Pr a g m g b Pxtdz 04 An equivalent momentum probability density7 15197 t7 which is possibly a very different type of function from the position probability density for the same state7 would play the same role in determining the probability of momentum measurement outcomes via Pb Prltpa lt p lt pi Plt tgtdp pa The issue is to determine an expression for the momentum probability density by deter mining a momentum wavefunction7 Ip7t7 and to provide a prescription for relating the position and momentum wavefunctions to each other This is accomplished by considering 154 that for a particle in the state P mltpltp0ltW0NAmWHM MWWgt a Amwmwmmmw Pb Pa Pb Pa where the momentum wavefunction is de ned to be ltpl1 tgtl2dp 2 W197 t dp 1429775 ltpl1 tgt IV587 Thus the momentum probability density is 2 P t l 097 75 lV588 The momentum wavefunction can be determined from the position wavefunction bu invok ing the completeness relation for position states and the position wavefunction for momen tum states7 Eq lV158 Thus mwmmm w i mem ww 1mmmmmm AOo e imE IxJMz It is conventional to use A 1 271 giving7 ip7t iWh z tdz IV589 F f ln mathematical terms the momentum wavefunction is the Fourier transform of the position wavefunction It follows that the position wavefunction is the inverse Fourier transform of the position wavefunction7 Ixt p eiPmE ip7tdp IV590 155 Example Suppose that the position wavefunction for a particle at t 0 is 1 W where p0 is a constant with dimensions of momentum and a gt 0 is a dimensionless constant Determine the normalization constant A and the momentum wavefunction Plot the position and momentum probability distributions Answer Normalization is attained via ltI l1 gt 1 which implies that l llzl2dz AP d mammum 57139 A2 l WO aSPO A Fm Eq lV589 gives the momentum wavefunction Thus7 ignoring the t variable7 Thus 1 00 p if h 7 aimh dm 3 2a 19000 eiipm 12 2 dm vm Fm 00 napeh a El gapPo if p lt 0 apo E 1190 gimpPo ifp gt 0 The position probability distribution is with the momentum distribution given 2 P t l 17297 75 Plots of these are as illustrated 156 A notable feature in the example above is that as a decreases the position probability density narrows while the momentum probability density Widens Thus for such wavefunc tions as the position measurement tends toward giving a single precise value momentum measurements tend to give outcomes that are more and more evenly distributed over a wider range of values 157 Mg m 200 dem 34 Emir spam 1 M oscanfar SW02 I D s an Wag1 engIWJC MK W343 27le we proved E 2 ii 33 3 an m39cAQtsfa c M 21 EHXLJ a1 quot E Fco as apes 4 WM d2 amisssemakd b5 mo in qu es 33 30m 644 TtSE we wtmmd m Wuquot 2 Male K W0 4 J 0 a We 1de M5 0 m mowL mowa rww ask whw M De 0 lt5 mm ladder w New m SWL H17 39 J61 lt E x lt 310 Z NA 2 to 0 pm Kw agmz m e g 14w boa mu m w Far mg yak MAMqu W 53mm Mm 0 OIId EMA t Z Fir MAM lm W a wm em W aLal t M39W 3 0 PM W D EM QMAldMAgt has Chadx C m39lxl M r4gt v1 9 tam g zw wwlawcbm Now 1d H gt EMMA J39Wisrs neg 433mg nowmhcd 1m EMA E Jaw mm Now he m s a mm EMiA JWgt W 0 mcxa we haw mm 0 WNW 315 aw4 iaw395 N ch anex guKhFQ pureed v has yam ltEw45 7 f w a m wampalgt LEW53 4W WV 40 Mo ltW gt 20 ma walkech 39WDVJW ltW 1o owl lt3Lampampv3d 2 17 we 1gt 73quotme 5quot 03a quot Emin iy 30 M15 M31 m ladder BEN5 mu I 05 r5 9 cg u 9 3F E x i m7 5 on e n 4576 p iii 3 25 c u 2f E 3 Aid H Ac6ltm ham 0110 E1535 n2 d E m 302 VJ 0 3355 ijgu g AGO V643 ATSQ A 0ch M AC hm Em nCCZV ESQsQ Y mi am vi 9 3va 9 QB 3amp3 We cm W Wag 2 Mam wm moh39ous ma a ExaMpka Given lo 4M0 06x5 6 dwm vltgtm M 8mg JTIO 2hgt 9 44H 4 X 39Miwi ltbox ggt zxe W7zu 3709mm Wm expedam mm m W from a 3mm an am M ltxgt klWU 1 J5 lta2cgtwgt 0 ltxgt Ewam www EWPM Far 43 8 h cc Anew Sm Exerdsc t ltgtltgt mam M310 2 W E w O xi Hg e l ltw ka gtM am a 0 CUFF p wk lw W i 2 mm WM21131 lt gt1 4amp4 mt ups gym V h Mow lt0 4 Particles in piecewise constant potentials A piecewise constant potential is one which takes constant values over nite ranges and changes discontinuously at a nite number of points An example is illustrated in Fig lV47 V ma zb Figure lV47 A piecewise constant potential In this example V1 for x g ma V9 V20 f0raltltma V3 for 17 lt m where V1 lt V2 The task is to determine the energy eigenstates7 which satisfy the TlSE The strategy for solving this makes use of the fact that Whenever V is constant then the TlSE 7 52 82 3 Vm gtEz E gtEz reduces to 82 2 3 77 V 7 E gtEz IV478 This is a second order differential equation with constant coef cients and the solutions to it are well known It follows that it will be possible to obtain separate solutions in each region in which V is constant For the example of Fig IVA77 this will take the form7 gtE1m for x g ma for mu m g ma lV479 gtE3m for 5 z A A where the subscript indicates the energy E which is the same for all pieces and a label to indicate the region in which each piece applies Thus7 for x g ma 82 E1m 7 2m W 52 V1 E E17 147 and so on Taking this approach it is possible to nd solutions for any possible energy E However the solutions are constrained as follows At any boundary between the two regions of distinct constant potentials 1 the wavefunction must be continuous and 2 the spatial derivative of the wavefunction must be continuous In the example of Fig lV47 this would imply that E1lta E2lt l1gt d gtE1 7 d gtE2 dm zma dm zma with a similar condition at 17 These conditions together with the fact that gtEz a 0 as x a ioo will severely restrict the possible values of E The general form of the TlSE for each region is 82 E 2m 8302 W V 7 E Ej IV480 where j labels the region There are two separate cases to consider either gt E or V lt E and each gives a distinct type of solution to the differential equation 41 Solution for regions Where lt E If lt E then Eq lV480 takes the form 82 Ej 8352 7k2 Ej IV481 where k IV482 The solutions to Eq lV481 are gtEm Acos km B sin km lV483 where A and B are constants A key feature of this solution is that the spatial frequency k increases as the difference between E and increases Without considering behavior at the edges of the constant potential region solutions of this type exist for any value E In the context of the example of Fig lV47 for any value of E gt V3 such solutions exist in each region As an initial candidate consider the function plotted in Fig lV48 In each region the solution is an oscillating function as required by Eq lV483 The spatial frequency of oscillation is determined by Eq lV482 and the plot demonstrates this qualitatively 148 1 1 1 1 1 1 1 1 ma l7 gtE m m 17 Figure lV48 An initial trial for an eigenstate where E gt V2 for piecewise constant potential The solution demonstrates the dependence of the spatial frequency on the difference between the total and potential energy in each region but does not satisfy the matching conditions at the edges of the regions This does not satisfy the matching conditions at the boundaries of the regions This would require for the solutions of regions 1 and 27 E1lta E2lt l1gt d gtE1 d gtE2 dm dm 125151 121a To attain such a solution requires adjusting the amplitude and phase but not the spatial frequency of the the solution for region 2 It turns out that this xes the phase and amplitude of the solution in region 27 relative to those of the solution in region 1 The solution of Fig lV49 demonstrates this The matching conditions at 17 require that gtE2b gtE3b d gtE2 d gtE3 dm dm mmb mmb This xes the phase and amplitude of the solution in region 3 relative to those of region 2 and hence of region 1 This is illustrated in Fig lV410 A solution of this form can be 149 1 1 1 1 1 1 1 1 ma l7 gtE I a 17 Figure lV49 An improved trial for an eigenstate where E gt V3 for piece wise constant potential The solution demonstrates the dependence of the spatial frequency on the difference between the total and potential energy in each region and satis es the matching conditions at ma but not those at 17 found for any E gt for all regions The matching conditions at the region boundaries merely determine the relative amplitudes of the solutions in each region 42 Solution for regions Where gt E If gt E then Eq lV480 takes the form H2 Ej IV484 where H IV485 The solutions to Eq lV481 are gtEz A5 Be m IV486 where A and B are constants A key feature of this solution is that a increases as the difference between E and increases Without considering behaVior at the edges of the constant potential region7 solutions of this type exist for any value E 150 Figure lV410 A solution for an eigenstate where E gt V3 for piecewise constant potential The solution demonstrates the dependence of the spatial frequency on the difference between the total and potential energy in each region and satis es the matching conditions at ma and at 17 As an example consider the potential of Fig lV47 where V1 lt E lt V3 Then in regions 1 and 2 the solutions will have an oscillating form and the matching conditions can be applied as before Now in region 3 the solution must approach 0 as x a 00 This eliminates the term Ae m from Eq lV486 Thus in region 3 Be m The best approach to constructing a solution here is to proceed from right to left Again consider any trial value for energy E This determines a and k for each region The solution in region 3 determines gtE2 and its spatial derivative at 17 this in turn determine the amplitude and phase of 1132 which ultimately determines the amplitude and phase of E1 via matching conditions at ma This is illustrated in Fig lV411 Such solutions exist for any value of E gt V1 An important feature of this eigenstate is that the position probability density is non zero for all values of z gt 17 This is the region where V gt E and in a classical mechanical system the probability of locating the particle in this region is zero In a quantum system the particle is able to penetrate the burner A more interesting case arises when 0 lt E lt V1 Here the solution is an oscillating 151 Figure lV411 A solution for an eigenstate where V1 lt E lt V3 for piecewise constant potential The solution demonstrates the dependence of the spatial frequency on the difference between the total and potential energy in each region and satis es the matching conditions at ma and at 17 function in region 2 but is exponentially decaying in regions 1 and 3 ln region 1 gtE1x a 0 as x a 00 This eliminates the term Be m from Eq lV4867 giving gtE1z Ae m in region 1 and similarly gtE3m Brain in region 3 Here 2W e E 52 This is larger in region 3 than in region 1 Given a trial value of E7 the values of the Ms and k are xed The wavefunction can be constructed from the right to the left by initially choosing an arbitrary value for A and applying the matching conditions at L1 This will x the amplitude and phase of the oscillating wavefunction in region 2 An example for an arbitrary choice of E is illustrated in Fig lV412 There is no value for B in Hj E3 Brain3m in region 3 which can possibly satisfy the matching conditions at n Thus there is no solution for this particular value of E 152 Figure lV412 A trial solution for an eigenstate where 0 lt E lt V1 for a piecewise constant potential The trial solution demonstrates the depen dences of a and the spatial frequency on the difference between the total and potential energy in each region and satis es the matching conditions at ma However it does not satisfy the matching conditions at 5 and is thus does not represent an energy eigenstate for this potential In such cases the matching conditions severely constrain the possible values of E only certain very speci c values will work This leads to a discrete set of energy eigenvalues and energy eigenstates This is quite general and typically if there is a region in which V is lower than for the surrounding regions the set of energy eigenstates is discrete and often nite in number On either side of this region the wavefunction for energy eigenstates decays exponentially and the probability of locating it there is negligible compared to that of locating it within the energy well Such states are called bound states A wavefunction for such an eigenstate is illustrated in Fig lV413 153 ma l7 Figure lV413 An eigenstate where 0 lt E lt V1 for a piecewise constant potential This demonstrates the dependences of a and the spatial frequency on the difference between the total and potential energy in each region and satis es the matching conditions at ma and 17 154 Exercise Show that 11m A5 satis es 8 1 x 723 a p Ipm provided that 04 ipFL The wavefunction for a state with de nite momentum is identical to the complex repre sentation of a wave7 5 where k 27r as the wavenumber From this it follows that the wavefunction for a particle is exactly that of a wave where the momentum and wavelength are related by 7 lV159 p lt gt This is the deBroglie relationship It is often necessary to consider function of momentum such as the kinetic energy of a particle of mass m p2 K 7 2m In general these will be constructed in terms of a series such as 1419 a0 lip 12192 2am V L The observable associated with this is de ned to be A a0 ms 12132 Eamon IV160 71 The following theorem is useful Theorem Consider the operator in where n is any integer Then gal 23 M e em W IV161 Proof To follow 0 Thus any A ll H in 45 W 14 Determining Energy Eigenstates It is often useful to provide a quantum mechanical analog of a classical system Examples include quantum mechanical descriptions of free particles and harmonic oscillators The 136 task will be to use the classical description as a guide for determining the states and dynamics of the analogous quantum system In general any quantum mechanical system can be described in terms of energy eigen states these are the states that return speci c energy measurement outcomes with certainty Energy eigenstates satisfy A H l Egt E l Egt IVL62 where where H is the Hamiltonian operator E denotes a possible energy measurement outcome and him is the state that gives outcome E with certainty when the energy of the system is measured The importance of these lies in the fact that a general state can always be expressed as a superposition of energy eigenstates ie l I gt 20E l gtEgt E where CE are complex numbers Additionally the time evolution of a system initially in an energy eigenstate l gtEgt proceeds via W05 Utl1 0gt Ut l gtEgt eiilflth eiiEth This can be extended to superpositions in the general fashion W0 20E l gtEgt ZCEE iEth l gtEgt E E Thus the key step in describing quantum systems in one dimension is to determine the Hamiltonian and to nd its eigenstates This is informed by the energy for the corresponding classical system The total energy of a particle of mass m is 2 P E 7 V M 95 where p is the particle s momentum and Vm is the potential energy As for spin 12 systems the transition to the Hamiltonian is accomplished by replacing the physically measurable quantities z and p by observables ie Hermitian operators i and 13 Thus IV163 Eq lV162 gives an operator equation i W l Egt E w 2m In some circumstances such as the quantum harmonic oscillator it is possible to proceed directly from this algebraic equation and via a series of operator manipulations arrive at 137 expressions for energy eigenvalues and eigenstates However more often the only route for solving the energy eigenvalue equation is to translate it into an equivalent equation involving wavefunctions Let l gtEgt lt gtE Then according to Eqs lV153 and lV161 the eigenvalue equation becomes 1 82 Em 27 7 m 2m 3352 l or equivalently 52 82 Em 7 W Vm gtEx 7 E gtEz IV164 This is the time independent Schra39dinger equation TISE Thus the task of nding the energy eigenvalues and eigenstates has been reduced to that of solving a second order differential equation It is important to note that for any single solution gtEz the energy E is one single real number which applies for all ualues of m However there are many possible energies each corresponding to at least one wavefunction Any solution to the TlSE must be normalized ie 00 l gtEml2dm 1 700 and this typically constrains the possible solutions since it implies that gtEmgt0 as maioo lV165 Additional constraints are that in any regions where the potential is not in nite both gtEm and 3E must be continuous If the potential is in nite at any point it is possible that W may be discontinuous at that point 141 Particle in an in nite well The basic techniques for solving the TlSE are well illustrated by considering a particle of mass m which is restricted to the region 0 g x g L and which moves freely inside this region The potential is then 0 0 g as g L We oo otherw1se and when graphed resembles a well of in nite depth The impossibility of a position measurement returning an outcome beyond the well implies that Im0 for zlt0 and zgtL It follows that for any wavefunction at the boundaries of the well Inside the well 0 g x g L the timeindependent Schrodinger equation gives 5 82 Eltmgt 2m 8x2 E gtE and the corresponding boundary conditions are gtE0 gtEL 0 Thus the task is to solve the differential equation 82 m 7 QmE 8m2 7 52 subject to the given boundary conditions The general solution is gtEm Asin km B cos km where A and B are complex constants and k satis es k2 222E IV166 The rst boundary condition implies 0 gtE0 Asink0 BcoskO B while the second gives 0 gtEL Asin kL Bcos kL Asin kL the only non trivial solutions are that 717139 k f where n 1 2 3 Thus the possible energy values and states can be labeled by a discrete quantity namely n Substituting into Eq lV166 gives the possible energy values nzwz z E lV167 2mL2 l where the index n labels the possibilities This can be used to label the corresponding eigenstates ie l gtngt which corresponds to wavefunction IV168 i 7171391 gtnm Asin The constant A may be determined by applying the normalization condition lt gtnl gtngt 1 Exercise Show that the normalization condition for any in nite well energy eigen state implies that 2 1427 H L 139 and thus 1 2 1 PrSz h2 5 Similarly PMS4 52 1951 12 Now A A Z lt Zl t 21gt E and thus 4 2 1 Z PrSz iii2 W 5 31 Equivalence between mathematical and physical states The kets H71 and 1773 have well de ned physical meanings in terms of SG measure ments To recap the physical meaning of 12 or 72 is that when a particle in this state is subjected to an SG measurement the outcome S ML2 or 1 752 will occur with certainty An alternative physical interpretation is that if an SG f1 measurement yields outcome S ML2 or 1 iii2 then that state of the particle immediately after measurement is H7 or 72 These are illustrated in Figs 1128 and 1129 Thus in terms of the basis l2gt 72 the general states for which such physical meanings based on measurement outcomes can be inferred are those given in Eqs 11326 However the vector space will admit all superpositions of the form 11 0 l2gt 2172 where 0 and c are any complex numbers The question now is whether physical meanings of the type given above can be associated with such an arbitrary state in the vector space This amounts to asking whether every state of this form can be expressed in the form of those in Eqs 11326 1f 0 and c are unrestricted then this is impossible However normalized states ie w 1242 1 are the only possibilities that are consistent with the standard interpretation in terms of probabilities of measurement outcomes For these the following result is important Theorem 1f 0 Hz 1 c 72 satis es lClZ 16712 17 then there exist real numbers 0 1 4p such that 0 W 7 0 5 coslt2gt 0 w W5 7 c e e sinlt2gt 35 Proof Any complex number can be expressed as 2 rem where r and 9 are real It follows that 3 a em 6 hem where 17130473 E R Normalization Eq 11323 gives lcl2 lcilz 1 a2b2 1 and it is always possible to nd 9 such that Thus 0 em e w sin Q 2 and setting 4p Oz and j B 7 Oz proves the result o It follows that the most general normalized state can be expressed as ll 5W cos H2 51 sin 72 11327 The overall constant factor of the form a is called a global phase Exercise Consider the two states l 31ewcosltggt l2gtei sin 72 and M cosltggt l2gtei sm 72 Show that7 ifa particle in each state is subjected to a SG f1 measurement for any direction 73 the probabilities of the measurement outcome S h2 are identical for the two particles 36 Example Consider two particles in the states 1 7 1 H3 and him 5W 12gt 172 Each particle is subjected to a SG 2 measurement Determine the proba bility with which 51 FL2 Answer For a particle in state 7 Pr5z 52lltil1 gtlz For the particle in state Nil 7 PMSZ 32 lltz l1 1gtl2 lt w gt 1 z 7 1 giving 1 Pr SZ FL2 5 For the particle in state 1 32 7 Pr SZ h2 llt2lx112gt12 and 1 211 W i lt l2gt 5 giving 1 1 1 PS E2 W7W77 1 1 5 g 2 These are identical Thus the value of 4p never affects the probabilities of the outcomes of any measurements In this sense it is physically irrelevant and can be omitted lgnoring global phases7 the most general normalized state has form 11 cos 12 51 sin 72 11328 for some angles 9 and 1 corresponding to a direction 73 this is exactly the state H71 of Eq 11326 It follows that any normalized ket7 has a physical interpretation in terms of the outcomes of an SG f1 measurement for some speci c unit vector 737 which depends on 37 MI in the sense that the measurement will yield the outcome S ML2 with certainty In general for spin 12 particle Any normalized state 11 ie satisfying 1 has the physical interpretation that there is some measurement such that when it is performed on a particle in this state exactly one of the possible outcomes is attained with certainty In the case of spin 12 particles the possible measurements are Stern Gerlach measure ments of spin components for a given state 11 it is necessary to nd 9 and 1 so that the state can be expressed as H71 according to Eq 11326 The relevant measurement is then that of the spin component Sn which is equivalent to submitting the particle to an SG f1 apparatus The de nite measurement outcome which will result is S ML2 32 Bras Calculating inner products Determining probabilities of measurement outcomes in quantum mechanics hinges on the ability to calculate inner products of ket vectors While the de nition of Eq 11321 is adequate for this there is a more convenient technique which only uses standard vector and matrix operations Recall that for M a l2gt a7 l72gt lltIgtgt 13 Hz b 72 the inner product is ltltIgtl 11gt ba bia It would be convenient to be able to compute this directly from the column vectors repre senting the kets in the l2gt 72 basis M H a la H There is no previously de ned operation on such column vectors which will result in a scalar The basic idea will be to convert one of the vectors to a row vector and apply matrix multiplication to the two Mathematical Tools Row vectors and dual vector spaces It is conventional to represent any real vector in two dimensions u um l uy g via a column vector apparent that it will be necessary to perform in nite sums when calculating probabilities or expectation values of measurement outcomes The following example indicates that this is frequently possible to actually do this Example Suppose that for a quantum mechanical system the energies are 11 En and that the system is in the state 00 A M Z 37 1m nl where are the energy eigenstates Determine A7 the probability with which an energy measurement yield the outcome En for any n and the expectation value of the energy measurements Answer The normalization condition gives co 2 A lgt1 37L 1 n1 Thus 00 1 2 1lAl E 3 n1 1 2 1A1 97 n1 1 1 1 A 2 7 1 7 7 l l 9 l 9 92 The term in brackets is a geometric series and the following applies 1 1TT2 provided that M lt 1 Thus 1 8 111412 g 1A1 g 1719 9 Which leaves 118 and the most convenient choice is 9 3 A 7 8 Thus the normalized version of the state is 00 1 1 n1 Now the probability of any given outcome for an energy measurement is PrEn lt gtnl1 gtl2 To this end i 1 1 3m71 1 8 3m71 MW 18 Hi 3 48 me 81 DJ 3 02 3 3 swim OJ 3 H Thus quotU 1 E 3 V H OOlH The expectation value is ltEgt 2 En Pr En n1 E1 2 1 71 i l 1 2 89 7 E1 2 n 169 9 n1 1699 9 9 E1 1 8991729 7 E1 89739 Note that the expectation value could also be computed using the energy observable7 119 ie the Hamiltonian 00 n1 and E MI 11 IV113 In practice calculating using Eq lV113 is no easier than rst calculating the prob abilities of the outcomes and then summing as in the previous example Time evolution of any state follows the general rules Thus for a time independent Hamiltonian the evolution operator is Ut eiililth39 Unlike the cases of spin 12 and other nite dimensional quantum systems it is seldom possible to exponentiate this operator in closed form However for a general initial state Eng l gtngt the state at a later time is Wt Utl1 0gt eiililth Eon n1 Zoneithh n1 00 Z onaiEnth W IV114 n1 Again it is usually impossible to sum this to arrive at an expression involving just a single ket and coef cient 11 Position Measurements Discrete approach The general description of position measurements in classical mechanics assumes that any real number foo lt z lt 00 is a possible outcome of a position measurement An equiva lent quantum mechanical description of position measurements will surely mirror this with in nitely many possible outcomes possible However if this is so then the number of out comes will not be countable ie cannot be labeled by integers and thus the mathematics of countably in nite sequences of vectors that was developed in the previous sections cannot be translated immediately to describe position measurement outcomes A rst step is to consider position measurements with a nite precision Thus suppose that the measuring device has a resolution of Am For example standard meter sticks are marked in steps of 1mm and for these Am 1mm We can break the range of all possible position measurement outcomes into sections or bins each of whose length is Am The actual outcome can be represented by the value of the x at the midpoint of each bin This is illustrated in Fig lV12 Each bin can be labeled with integers j according to the scheme of Fig lV12 Alter natively each bin can be labeled using the value at center of the jth bin ie zj jAz This 120 llllllll Figure lV12 Discrete position measurement The axis is divided into bins of length Am The bins can be labeled by counting from the centermost bin centered at z 0 outwards Bins to the right will be labeled by positive integers j 12 and those to the left by negative integers j 7171 Note that the center of the bin labeled j is at zj jAz forms the basis for assigning position values for this type of discretized position measure ment If the particle is located in bin j during a position measurement then its position value is said to be mi Thus the possible outcomes for a classical position measurement are x1m0m1 A quantum mechanical description of this measurement begins with a description of the position measurement outcomes and the associated states These are listed in Table lV12 As with other types of measurements on quantum mechanical systems the state has Position Associated State x2 lm2gt x1 lm1gt m0 lz0gt m1 lz1gt m2 lz2gt Table lV12 Discretized position measurement outcomes and associated states the physical interpretation that if the position of the particle is measured then the outcome will be that the particle is located in bin j with certainty In terms of position values the outcome will fall in the range xi 7 Az2 xi Am2 and we refer to this by saying that that the outcome of the position measurement will be zj with certainty although this really just refers to the label at the center of the bin These states are referred to as position 121 states Outcomes that lie within distinct bins will not be confused and thus the position states are orthonormal7 ie The position states form a basis for the vector space consisting of all states and this is indicated by the completeness relation 1 M ltle f lV116 For a general state l Igt7 Z W W10 IV117 The complex numbers lV118 form the components of in the basis 7 lz2gt 7 lm1gt7 g 7 la 7 g 7 7 ie 00 ll Z xIzj m IV119 Again the state dictates the probabilities for discretized position measurement out comes Thus Prm lltl 1 gtl2 IV120 Exercise Show that if M We m then Pr 35 lIzl2 IV121 One consequence of Eq lV121 is that which is equivalent to the fact that the state must be normalized7 ie 1 lV122 122 The observable associated with position measurements is 32 Z 357 ljgtltmjl IV123 1700 and it can be shown that i mn lV124 Exercise Show that i mn The expectation value of z for an ensemble of particles in the state is determined via ltgt 1492 lI gt mijl 93 l 2 WMWM k7oo M8 1m I WV ml 2 WWW lkgt j7oo k7oo Z I WWQEH Z Mlmklggw joo k7oo Z Z wltzjgtWltmkgtzkltmzkgt 7700 k7oo co If 2 wltzjgtwltzkgtzk6jk 7700 k7oo which gives 00 2 m Z x 14sz IV125 j7oo Another important consequence is that the probability that the measurement outcome yields any value in the bins ranging from that labeled jlow to jhjgh is jhigh 2 Pr leow lt m lt mm Z l llzjl IV126 jj10w 123 Thus jhigh Pr Mow lt 90 lt may Z QBWQ W jjlow jhigh Z ltjli gtltjl1 gt jjlow jhigh Z 1490 WHO jjlow jhigh M Z ljgtltjl M IV127 jjiow and in this sense the operator jhigh Z ljgtltjl jjiow is associated With position measurements in the entire range zjlow lt z lt mjhigh 12 Position Measurements Continuous approach The resolution of position measurements can always be improved this amounts to decreas ing the size of the bins Am in the discretized version The mechanisms described for the discretized quantum description above apply Whenever Am 7 O ls there an underlying description When Am 0 In a thorough mathematical treatment there is such a descrip tion but the mathematical tools used in the discretized version of position measurements have to be modi ed substantially Nevertheless the basic rules for calculation have many features similar to those of the discretized version The set of all position states is l all real Note that although the labels inside the kets are real numbers they are still just labels and they are not explicitly involved in typical real number calculations Thus i25gt l36gt 7r V31 Additionally there is an important conceptual caution it is impossible to prepare a particle in any of these states Thus they do not have any reasonable physical interpretation in terms of the outcomes to position measurements The role of these states is to support superpositions from which quantities pertaining to measurement outcomes can be calculated ln superpositions over states with continuous labels the sums must be replaced by integrals Thus the general state will have the form M 124 IV128 W Oscillcd W SPQOWM Fm Kw MW 5le H 3 MM 0 daFm mm a 3 J77 9 in Cm or 8 W733 P 1km aj gtE We HM um My m eshbhak tesulb hr W awswew values W3 WC cum 0 Mb E 3 140 El gt Tug Er f resad r is39 a mltaawgt w EM 7km lt ltw mm L33 45145 Elt l gt 7 1m lt laawgt it E 94 mm s awm xzed Tum UM N awzmm 7w W 3 6 M ltwtwgtgto Tm E W a Tm er 2mm 3J2ch is rashdad as 0 a lmeosmLk Addi mu we W atmmd m 1511 0 Co a watZ 4M 0 MMch 80th M T SE 39039 E3112 This S H44 Lowest WWW WW degAWE OM is mild m mm Mm We can also ambush M Kim39s is 46 m5 swim wh d E4732 39Proz 01C qu35 L4 500 be mow goluHM Irw oO u39ck Ema1 m 059 M4 e WV21 Tum wi 2 N 535 39W Mo 5302 So 090 0 MmltLux 2355 ow ax 39 t 0 Lquot Z L 0 10 OLE e mwxMK mi 5 R EX dungmcIw OW 0 J ER X lt20 1 7 3 if erkuxZfZ M42 50 erzwzng 0W R Jo ZMmX 03E c wazzk max quot311 39M OWLCt t m6 L 1 39E Olga if it 6 WMXLLt J w N ZZN 1M dxxl ZM om 7 43 EWX 0 4mm Aka gwaWk a AX my Jul did 2 L o t v L I 7 MLU V a U b d 0 d1 A ZM 937 739 0 2 oi ZM XX JFJMJX 564 Q MWXZt WW J A V ngo mm a0 3 agmw Wt WM E 7 Thus L L i 45quot Jmox OLL E 0 2M 0ka 0 092 M K 0 out 4 17707 v 37 9amp1 A e waxt or 5w cwsb A39 dx Now My w an 16 MUJX1Z waa 1 Ex ma W u w a JVF O A6 4790 as x4700 wak z39c fs Mcmmlren39r a 0 0M3 Fog wmb 3 A O b 033 0 17 40 cmfcn r OKX go we 5 mg we sdh agv LU ux r 3MP Muobe M Mm Wsw 7cm MM 0 KM Inc M Man rewlf is SUMD 5 m aws ra rc MK mom EHRLO a W n h n E ma SO elm mam an add wa th ogem andc QAMW ales Mao 6 W AEEV WMW wm Zutdhw we know m Mb 60 a Co 6 waLZt 3H0 8 0V Wam hm439th 0 MD A A am 1 E X m1 05 2 X t Weiquot D at Wax M K1 Ewoiser W40 M NO MD axe m At39 Amway lt9 WWw wmiw BX a la M UX7tx So am an xme z 1 Z 3 x M xLzh x 1 E W We cm SIAM a quot J wa t8 We all MK use quw Sudx as W W be Wm M W E MW 3 amp39ampRgtlm t to Tim U hm8l ggt Jr 2r EM CEF Zuo M we 5 m WW MMLW Elhco The regul ris wwedxk 9Wlar 9mm 2 Tm we can am a Ladder 03 uid u I 6 quotr M W Z meHM 73 35417 gL yia w 5 E v Ws K1016 2505 states Mk4 EA wCM39IA AND ngt Tm n0tz s S h2 D 7 h 2 z m 8G2 s it 2 SzquotL2 Z sci sz h2 8G2 8G2 Sm 752 Sz iii2 91752 Figure 19 Three successive Stern Gerlach measurements for the circum stance in which the particle emerges with S h2 after the rst 8G2 measurement Applying the rules of 16 one can deduce that the probability with which the particle emerges with S h2 after either of the latter SG 2 measurements is 12 Similarly the probability with which the particle emerges with S 7h2 after either of the latter SG 2 measurements is 12 Exercise Consider the arrangement of Fig 19 Suppose that the particle emerges from the SG ii measurement with Sm h2 Show that the probability with which the particle emerges from the subsequent 8G2 measurement with S h2 is 12 Repeat this for a particle that emerges with S1 7h2 is 12 Re peat this entire analysis for the case where the particle emerges from the SG ii measurement with Sm iii2 Exercise Repeat the argument of the previous exercise for the case where the particle emerges with S h2 after the rst SG 2 measurement The implication is that it makes sense to speak of a particle as having a de nite value for SZ in the context where the only subsequent operations are S02 measurements However Whenever subsequent operations include SG f1 measurements where f1 is distinct from 2 or 72 it is impossible to speak of a particle as having a de nite value of S1 Thus is quantum mechanics it is meaningful to speak of a particle as having a de nite value for Sn only in certain very restricted contexts One cannot universally ascribe a de nite value of Sn to a particle 13 Maximal description of physical states It is possible to consider measuring more than one component of spin via successive Stern Gerlach apparati as illustrated in Fig 110 S h2 S 7 FL 2 y w soy Sy7FL2 SGzi Syh2 SGy Sm 752 5 752 Figure 110 Successive Stern Gerlach measurements in an attempt to mea sure orthogonal spin components Any spin 12 particle to which these are applied will yield values for Sm and Sy But this joint outcome can only be considered a measurement of a property of the particle if when repeated again to a system that has undergone no extra measurements or interactions it yields exactly the same value as it did previously Exercise Consider the double SG experiment as illustrated in Fig 110 and a spin 12 particle that emerges from the uppermost output beam Suppose that this is then reapplied to the same apparatus Show that it does not emerge from the uppermost beam with certainty The preceding exercise indicates that two SG apparati cannot be combined to jointly measure two orthogonal components of spin in the sense that the measurement outcome is not repeatable Thus in the context of SG measurements it is not sensible to label the physical state of a particle via Hm 73 where 731 and 731 are not parallel such a state would indicate that a combined SG 731 and SG f1 measurement would yield S ML2 and Sn ML2 with certainty However a general argument similar to that of the preceding exercise indicates that this is not possible Thus in terms of repeatable measurement outcomes the only physically meaningful states are H71 and 77 where f1 is any unit vector The collection can be simpli ed by noting that in terms of SG f1 measurement outcomes 72 is equivalent to 7 gt Exercise Show that 7 gt yields S 752 with certainty To do so de ne 7h 773 and determine the probabilities with which H731 yields Sn iii2 Thus the collection of all possible physically distinct states of spin 12 particles is where f1 ranges through all unit vectors Mathematical Tools Probability Quantum mechanics provides a means for determining the probabilities with which the various outcomes of a measurement can occur The intuitive motion of probability is established by considering a large number of identical measurements on copies of identically prepared systems Suppose that N spin 12 particles are all prepared in the same state and subjected to the same SG measurement If the the probability of attaining S ML2 is p then in a typical run of N such measurements the outcome Sn FL2 will occur approximately pN times although this is not guaranteed Deep theorems in probability such as the central limit theorem quantify such statements and these become more precise as N a 00 The general tools of probability theory can be illustrated by considering a generic mea surement with n distinct outcomes labeled 77117712 mn Denote the probability with which the outcome mi occurs by pi Following the basic notion of a probability representing fractional occurrences of a given outcome after many measurements it is required that the probabilities satisfy 0 g pi g 1 18 and that i319 1 19 i1 The collection of all pi is called a probability distribution Example An unbiased die can give one of the following outcomes after being rolled 1 23 4 5 6 Regarding a roll of the die as a measurement the outcomes are m1 1 m2 2 m6 6 For an unbiased die the probabilities with which each outcome occurs are identical thus p1 1692 16 It is easy to see that these satisfy Eqs 128 and 128 The mean of a distribution gives a notion of the average numerical outcome and is de ned via ltmgt Emmi 110 i1 ln physics and particularly in quantum mechanics this is known as the expectation value of the quantity which is being measured Example For an unbiased die the mean is 6 6 1 ltmgt Emmi 27mg 11 11 6 1 2m8123456 i1 1 8 7 5 It is important to note that the mean is an idealized quantity and that a given run of measurements will not necessarily yield the mean when averaged Suppose that the outcomes of an experiment involving N measurements are 31 32 33 3N This constitutes a sample of the probability distribution and the sample average is de ned as 2 1 m I 3 111 l H Typically for large N one expects that m z m Example For an unbiased die in one trial consisting of ten rolls the following outcomes occurred 2 4 4 3 3 2 1 4 4 6 The sample average of these is 33 which differs from the mean 35 Various theorems in probability theory quantify the extent to which a sample average approximates the mean Generally as N increases the probability with which the sample average will be within a given range of the mean increases the probability with which it is beyond the a certain range of the mean diminishes as 1N An important tool in quantifying such discrepancies and uctuations away from the mean is the standard deviation or variance of a probability distribution This is de ned as 112 and this quanti es the extent to which measurement outcomes typically deviate from the mean A general result that simpli es this calculation is Am m2gt 7 ltmgt2 113 where 71 ltm2gt 27712191 I i1 These notions can be applied to physical systems subject to the laws of quantum me chanics Example Consider an ensemble of spin 12 particles each in the state where 731 i Each particle is subjected to an SG ii measurement Determine Sm and AS Answer We can list the measurement outcomes and probabilities Outcome 51 Probability 5 1 m 7 1 2 P1 2 FL il m 7 2 2 P2 2 where we have used 1 PrSmFL2 5 1Th 1 Pr5m7h2 5 1773141 etc Thus V L ltSzgtZmiPi L391 7FL 1lt FL il 2 22 2 22 7 FL 772V Then A52 ltS gt7ltSzgtz Here V L 53gtZmlzpi L391 gf 1ltjgtz 2 22 2 22 52 1 Thus FLZ FLZ FL AS 777 Z 4 8 22 Phys 321 Fall 2007 Quantum Theory I Problems I Spin12 Kinematics I1 SternGerlach Experiment Suppose that a beam of particles7 each with mass M and velocity 39v Umii enters a region in which the magnetic eld is B B02 2 where B0 600 Teslam This region extends in the z direction for distance LB A detector is placed LD beyond the end of the magnetic eld region The setup is illustrated in Fig 1 Detector Screen Figure 1 Question 1 a Suppose that prior to entering the magnetic eld7 the 2 component of each particle s magnetic dipole moment has the same value7 uz Find an expression7 in terms of M and Hz for the acceleration of the particles while they are in the region with non zero magnetic eld lgnore all forces on any particle except that exerted by the magnetic eld 7 V Assume that the particles follow trajectories governed by classical mechanics Find an expression for the total de ection in the z direction in terms of M7 uz Um7LB and LD Verify that it is proportional to Hz 12 Repeated SternGerlach Experiments Two Stern Gerlach devices are arranged as illustrated The rst is oriented in the direction 73 i g and is such that the beam corresponding to Sn ML2 is blocked The second is oriented along the direction g Neither of its outgoing beams are blocked 5 h2 U Q 3 soy a Suppose that a spin 12 particle is subjected to the rst device Express in terms of H2 and 72 the state which describes a particle that will emerge from the rst device with 100 certainty b Consider a particle that emerges from the rst device and which is then subjected to the second Stern Gerlach apparatus List the possible outcomes of the second measurement Sy and the probabilities with which they will occur for a particle which emerges from the rst device with 100 certainty c Suppose that a large number of particles is subject to this pair of Stern Gerlach devices Consider only those which emerge from the rst device Determine the expectation value of the the y component of the spin 2 for these 13 Repeatable vs nonrepeatable measurements Consider a measuring device which contains an SG2 apparatus combined with a blocking device on the lower outgoing beam This is packaged within the dotted box as illustrated Suppose that a single particle is sent into the boxed device The idea will be to subject the same particle to repeated measurements 5 h2 U Q N This apparatus is a measuring device in the sense that it answers the question Is 51 ML2 for the particle entering the apparatus77 lf particle emerges from the dotted box then a reasonable answer could be yes However this is only reasonable because the measurement can be repeated and always yields the same outcomes in the sense described below Suppose that a particle which emerges from the dotted box is immediately subjected to the same device again Note it is important to realize that the same particle is subjected to both devices The situation is very different if one particle is subject to the rst device and another particle subject to the second device 51 h2 SZ h2 U Q N Any particle which emerges from the rst boxed apparatus ie answers the question with yes will emerge with 100 certainty from the boxed second apparatus ie will answer the question when repeated with yes with 100 certainty One can repeatedly add other boxed apparati like this and any particle which emerges from the rst necessarily emerges from all the others This is equivalent to repeatedly asking Is 51 ML2 for the particle entering the apparatus and always receiving the answer yes In this sense it is reasonable to say that S ML2 for this particle This is an example of a repeatable measurement Now consider the composite measurement7 involving two blocked SG measuring devices with different orientations as illustrated Sm h2 SZ h2 This composite SG apparatus inside the dotted box may appear as a single measuring device which asks the question Is 51 ML2 and S ML2 for the particle entering the apparatus and it may seem that the answer should be yes for a particle which emerges from the combined apparatus a This will only make sense if the measurement is repeatable That is7 one must take a particle which emerges from the dotted box and feed it back into a second copy of the entire dotted box It must emerge with 100 certainty after this repetition for the measurement to be repeatable Determine the probability with which a particle that emerges from the dotted box will do so again if it is fed into another copy of the dotted box ls this measurement repeatable Does it make sense to say that Sz ML2 and Sm ML2 for the particle b Based on this result why do you suspect that we have not tried to describe the state of a spin 12 particle in terms of more than one spin component 14 Kets and SternGerlach measurements a A Stern Gerlach apparatus is oriented in the direction f1 i 2 Express as a superposition of H2 and H2 the state which emerges from the measuring apparatus if it gave measurement outcome S Hi2 Repeat this for a particle emerging from the device if it yielded S iii2 Demonstrate that the two states are orthogonal T V A Stern Gerlach apparatus is oriented in the direction f1 g 2 Express as a superposition of H2 and H2 the state which emerges from the measuring apparatus if it gave measurement outcome S Hi2 Repeat this for a particle emerging from the device if it yielded S iii2 Demonstrate that the two states are orthogonal 15 Spin12 states and SternGerlach measurements Consider a spin 12 system prepared in the state l ll A39 H2 48 202 72 where A is a normalization constant a Apply the normalization condition to determine A b Suppose that you would like to subject this particle to measurement via a Stern Gerlach apparatus whose magnetic eld is oriented in some direction f1 so that the outcome of the measurement is Sn ML2 with 100 certainty Determine values for the spherical coordinate parameters 0 1 corresponding to 7 1 which will ensure this 16 Orthonormality of standard states Consider the two states associated with mutually incompatible measurement outcomes of an SG f1 experiment Show that these are orthonormal 17 Global phase in quantum states This is a special case of Townsend s problem 19 Consider a spin 12 system particle described by the state M ltc l2gt of 72 where lcl2 lalz 1 and 4p is any real number a Suppose that the particle is subjected to an SG f1 measurement Show that the proba bilities of getting the two possible measurement outcomes are independent of 4p b A consequence of this result is that any states that differ only by a global phase ie ll and W 11 where 4p is real are physically equivalent in the sense that they give identical predictions about measurement outcomes Show that the states 0 i 0 1 l ll coslt2gt lzgte s1nlt2gtl z and int2 9 W2 t9 Mg e cos 5 Hz e sin 5 lizgt with 9 and 1 real are physically equivalent 18 Orthonormal kets and SternGerlach measurements Sometimes you will be given a complete description of the state of a spin 12 particle corre sponding to the measurement outcome S ML2 but will then have to deal with circum stances in which the particle is not in the state H71 and the measurement actually returns Sn ill2 Mathematically this corresponds to a situation where you have complete knowl edge of H71 but really need to work with 77 The aim of this exercise is to learn how to translate between these First consider an example A Stern Gerlach apparatus is oriented in such a way that it produces Sn ML2 for the state represented by 3 42 a 7 pz 5 with certainty In one particular run of the experiment on a particle in a different state the measurement returns Sn ill2 You need to determine an expression for the corresponding state ie that which gives S ill2 with certainty so that you can predict what happens in subsequent SG measurements 1 lngt lzgt gt a One way to determine an expression for this state is via the relevant spherical angles 9 and 1 Determine these for the state H71 and use these to derive an expression for the state of the particle which gives S ill2 with certainty Verify that the two states are orthonormal Hopefully you noticed a pattern in the answers for this example Now consider a general state m c l2gt c 72 where lcl2 lalz 1 b Show that l1 2gt 0 l2gt 0 l72gt is normalized and orthogonal to l l Describe why this seemingly purely mathemati cal result is relevant to the example of part a 19 Repeated SternGerlach measurements Consider a SG 7 measurement followed by an SG measurement where 731 sin 01 cos 1113 sin 01 sin m cos 012 f1 sin 02 cos 1213 sin 02 sin 121 cos 022 where 917 11 and 02 12 are the spherical coordinate parameters for 731 and f1 respectively Suppose that a particle emerges from the SG 731 apparatus with S ML2 In class it was stated that7 in this case7 the probabilities for the outcomes of the SG measurement are Outcome Probability Sn h2 1 ma 2 S 752 17 ram2 The purpose of this exercise is to verify that these are consistent with the probabilities that are calculated using the ket formalism a Provide an expression7 in terms of 017 gt1 and using the basis states H2 and 72 for the state of this particle immediately after it emerges from the SG 731 apparatus b Use the ket formalism to determine the probabilities with which the outcomes of the SG f1 measurement occur and verify that they are consistent with those given above 110 Bra vectors and measurements a Suppose that a particle in the state Hm 7 i l2gt H lea W m is subjected to an SG g measurement Use 1 1 i 2 2 7 iz l ygt l gt l gt to compute ltyl and use this to compute the probability with which Sy ML2 b Suppose that 731 g 2 and a particle emerges from an SGm measurement with Sm Hi2 This particle is then subjected to an SG measuring device where f1 ii 7 g Determine expressions for Mil and 773 and use these to compute the probabilities with which the outcomes S Hi27 and S 752 occur 111 Measurement calculations using different orthonormal bases Suppose that a spin 12 particle is prepared in the state 3 4i A 3 7 42 Z 7 5 l gt 5 and is subsequently subjected to an SG ii measuring device The aim of this exercise is to calculate the probabilities with which the two measurement outcomes occur in two ways the rst uses the l2gt 7 72 basis and the second the ligt7 basis M liigt a Express the states corresponding to the two de nite outcomes S ML2 and S 752 in terms of the l2gt 7 72 basis and use this to compute the probabilities with which Sm Hi27 and S 752 occur b Express in terms of the 7 basis and use this to compute the probabilities with which Sm ML27 and S 752 occur Do the results agree with the previous part 112 Observables for spin12 Consider the observable corresponding to the z component of spin A h A A A A s 75ewxw7wx7w0 a Find the matrix representation of Sm in the l2gt 7 72 basis b Verify that Sm is Hermitian c Find the eigenstates and eigenvalues of Sm and express these in terms of the basis l2gt 7 l2gt The observable corresponding to the component of spin along the direction f1 is gltH H bd H mgt d Using the standard representations 0 i 0 H771 cos Hz e ZS s1n l7zgt 0 0 l7 gt sin H72 7 e ZS cos 72 nd the matrix representation of S77 in the l2gt 7 72 basis
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