### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Quantum Mechanics I PHYS 3810

TTU

GPA 3.88

### View Full Document

## 26

## 0

## Popular in Course

## Popular in Physics 2

This 48 page Class Notes was uploaded by Vernie Wehner on Wednesday October 21, 2015. The Class Notes belongs to PHYS 3810 at Tennessee Tech University taught by David Murdock in Fall. Since its upload, it has received 26 views. For similar materials see /class/225721/phys-3810-tennessee-tech-university in Physics 2 at Tennessee Tech University.

## Similar to PHYS 3810 at TTU

## Reviews for Quantum Mechanics I

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/21/15

Phys 3810 Fall 2009 Computer Calculation Notes 01 Setting Up the Problem The goal here is to calculate the radial wave function for a bound state of a particle trapped in some thing like a nite spherical square well which is basically posed in Problem 49 Of course with a computer we can solve for a more complicated binding potential than the simple square well And the exact77 solution will involve some messy combinations of spherical Bessel functions we want to at out solve it numerically for any Z First write out the DE for the reduced radial function uzr This is what we will solve for the eigenfunctions and eigenvalues 437 is easily rewritten as dzu 2mV 7 Z Z 1 e um EMU lt1 The boundary conditions1 are u0 0 and uoo 0 Satisfying these will give solutions for the possible energies E for each Z Of course what we really do is to try to satisfy uloo 0 by searching on E The rst thing to do is to get the equation into a form so that it can be programmed up easily and we can input and receive numbers that we can understand We will be dealing with atomic or nuclearisized systems but we certainly dont want to deal with a lot of incomprehensible numbers in scienti c notation We want to use sensible units These will depend on the scales in the problem but here we will solve a problem on nuclear scales for which the energies are or the order of MeV7s and the length scales are of the order of fermis 1 fm 10 15 In particular the bound particle will be a proton which has rest energy 9383 MeV For de niteness we7ll say that the depth of the well is 600 MeV and its radius is 400 fm It is most convenient to rewrite 1 as dzuz i 2m02Vr 6 1 7 2m02E 7 0 2 d7 he r2 he WU With the very useful fact which you are encouraged to show he 19733 MeV fm and mcz 9383 MeV the radial equation is now dzw 6 1 W 7 004819Vr 7 E T wr 0 3 1To be a correct radial wave function w also needs to be normalized but if we only want the energies we won t worry about that In using this equation you must use V and E in MeV and r in ferrnis Then the units of the bracket in 3 are frn Eq 3 is of the form dg u W 7 mama 0 4 but of course if we dont know E then Vr isn7t fully known It is known provisionally if were searching on At this point there are several ways to go Maple provides a way of solving the DE directly while giving the eigenvalue E and we can use this at worst searching on E by hand One can vary E until one gets a solution that is small at large 7 or equivalently which goes through a large change in sign at large 7 when E is varied For this we want to take as the two boundary conditions ul0 0 1 We can give the second condition because as the function ulx does not need to be nor malized in our problem u does not have a de nite value it varies in proportion to the normalization Since I know that W is something like a sine function 1 is a rea sonable choice Regardless we are still looking for a dramatic change in the sign of W at large 7 when we vary E 02 Maple Maple has some fairly convenient routines for solving differential equations and plotting the results I will give the steps which will let you do the most basic and clurnsy calculation 1 hope you will do more than that First off while I hope you will use a more exotic form for the binding potential later on the square well77 is good to start with But how to get it into a Maple staternent Yes could de ne a function using some ii 7 statement but Maple does include a step function The famous Heaviside step function often denoted 0 is de ned by 0 f 0 6m T M 11f 20 except that its not called 0 they reserve the Theta name for something else rather its called Heaviside X With this function our potential function Vz given in MeV with z the radial coordiante in ferrnis is 7600 gtk Heaviside40 7 X which as you can see will give 7600 MeV if z is less than 40 frn and 00 if z is bigger than 40 frn At very least we would like to get a plot of the function uzr that results from a particular choice of E Maple has the means to solve the radial DE with plots Phys 3810 Quantum Mechanics Notes David Murdock7 TTU May 47 2009 Contents CO 4 The Schrodinger Equation and Its Meaning 11 The Schrodinger Equation 12 Statistical Meaning of 1 13 Probability 14 Momentum 15 Uncertainty Principle The Timeilndependent Schrodinger Equation 21 Introduction 22 Stationary States 23 In nite Square Well Box 24 The Harmonic Oscillator 241 Algebraic Method 242 Analytic Method 25 The Free Particle 26 The DeltaiF unction Potential 27 Finite Square Well Formalism Introduction 32 Hilbert Space 33 Observables 34 Determinate States 35 Eigenfunctions of Hermitian Operators 351 Discrete Spectrum 352 Continuous Spectrum 36 General Statistical Interpretation 37 The Uncertainty Principle 371 Minimum Uncertainty Wavepacket 38 EnergyiTime Uncertainty 39 Dirac Notation The Schrodinger Equation in Three Dimensions 41 Schrodinger Equation in Three Dimensions 42 Spherical Coordinates Separation of Variables 43 The Angular Equation 44 The Radial Equation 3 0 CONTENTS 45 Example Spherical Box 37 46 TheH Atoml 39 461 Introduction 39 462 The Radial Equation and Asymptotic Behavior 39 463 The States of the H atom 41 464 The H atom spectrum 43 47 Angular Momentum 43 471 Algebraic Approach to Angular Momentum 44 472 Analytic Approach to Angular Momentum 45 473 Spin 45 474 Electron in a Magnetic Field 45 MultiParticle Systems 47 51 The Schrodinger Equation for Two or More Particles 47 511 Central Forces Relative and CM Coordinates 47 52 Helium 47 Chapter 1 The Schrodinger Equation and Its Meaning 11 The Schrodinger Equation Classical mechanics is all about nding the trajectory of a particle of mass m when it is subjected to a force Fr t The object is to nd the position as a function of time rt One way or another that s what we did in classical mechanics We can not do this to describe motion at the atomic level because it is essentially impossible to nd such a trajectory function What we do nd is the wave function for the state of the particle The wave function z t is complexivalued For now we will work with one space dimension And without further ado the wave function is given by the Schrodinger Equation 8 11 52 821 mi 8t 7 3m2 W Essentially we will spend the rest of this chapter unraveling what it tells us and the rest of the course nding solutions to it in different cases of physical interest 12 Statistical Meaning of kl The statistical interpretation of 1 is due to Born and sez that l Iz tl2 gives the probability of nding the particle at point x at time t l Iz tl2 Probability of nding particle between z and z dz at time t See the discussion of the text for different attempts to get the meaning of this Though the search for understanding of QM continues to this day the consensus is that nature at its most basic level is probabilistic and we have to accept that facti and learn how to deal with it 5 6 CHAPTER 1 THE SCHRODHVGER EQUATION AND ITS MEANHVG 13 Probability Relations for a discrete probability distribution Nj A probability distribution for a continuous variable pz satis es 7 co 00 pa pm dz pm dz 1 ltfltzgtgt fltzgtpltzgt dz and the standard deviation of z is 02 E Am lt2gt ltgt2 If our wave function z t is to give a sensible probability distribution it must be normalized that is it must satisfy 00 l llmtl2dx 1 00 An essential idea of QM shown in a text example is that of the probability current One can show that with E 811 8 1 2 J t E 7 1 7 I i 7 2m lt 8m 8m then the rate of change of the probability of nding the particle between a and b is dPab J tint d m m 14 Momentum An expectation value of a physical quantity is the average value we get for that quantity if we make repeated measurements of it for particles prepared in the same 11 It is lt35 1zl llzt2dm The nearest thing to a velocity or rather its expectation value that we have in QM is to take the time derivative of We can then relate this to the momentum lt1 by multiplying by m We nd dlt gt 8 1 h 8 7 i 74 7 7 p 7m dt 2h Il 8m dm 11 8m Ildm and comparing the expectation values of z and p leads us to the associations z lt gt z factor operator pH 15 UN CERTAIN TY PRINCIPLE 7 When we want other physical quantities we will build them up from x and p Then the expec tation value of the general physical quantity is ltQltm7pgt m m As an example of this7 the expectation value of the kinetic energy7 T p22m is in 2sz m I dm An interesting result which is as close as we can come to Newton s second law in QM is a case of Ehrenfest s Theorem shown in one of the prololems7 dltpgt 7 8V W 7 lt7gt 15 Uncertainty Principle While we will do a full derivation of the famous Uncertainty Principle in Chapter 37 we can give a preview of what it will mean when applied to the quantities z and p CHAPTER 1 THE SCHRODHVGER EQUATION AND ITS MEANHVG Chapter 2 The Time Independent Schrodinger Equation 21 Introduction In this chapter we present the timeiindependent Schrodinger equation the thing people refer to wrongly as the Schrodinger equation We will consider the meaning of its solutions in terms the measurement of the energy of a particle and then we will go on to solve practically every oneedimensional problem you will need to solve Along the way we will pick up many mathematical techniques and develop our understanding of the wave functions of quantum mechanics 22 Stationary States We want the nd the solution z t for various physical situations Even without 3 dimensions we still have a functions of two variables to nd We will look for solutions that are products of a space function and a time function 1490775 MMW Putting this form into the timeedependent Schrodinger equation one gets 1d 52 1 12 77 7777 V z f dt 2m 1 dz l Here the left side depends only on t and the right side depends only on m It follows that each side must be equal to a constant which call E for energy The equation for ft then gives df iE 4E1 7 i if t l d h f gt f e where we should have a constant out in front but that will be handles later in the sz part and the z equation gives 2 2 7772V E 21 Eq 21 is called the timeindependent Schrodinger equation We can t solve it unless the potential function Vm is speci ed But when we do the solution will be of the form We 75 MME iEth 22 9 10 CHAPTER 2 THE THVIEJNDEPENDENT SCHRODHV GER EQUATION Some comments on our separated solution to the Schrodinger equation 1 We have found stationary states7 meaning that since the form of the solution gives m tl2 WW2 there is no time dependence to the probability density This implies that is constant and ltpgt 0 2 The states we nd this way are states of de nite total energy The total energy operator7 called the Hamiltonian is found from the classical expression by p2 A hz 82 HzpVm gt H7Vz so that the time independent Schrodinger equation is written compactly as P1 E The expectation value of the total energy operator is H fwdz Elzplzdz E and one can show that TH 0 Every measurement of the total energy will return the value E for a particle in this stationary state 3 The general solution to the time dependent SE is a linear combination of separable solutions our solutions of the form given in Eq 22 really give us all we need to solve any problem The general solution has the form um t i cnzpqm iEth 23 n1 where the conditions of the problem will give us the coef cients on 23 In nite Square Well Box We now go on to solve the timeiindependent SE for simple potentials Yep7 simple7 right We should understand why we are solving such problems In nature particles don t move in the simple potentials that we will consider here At best7 a real life potential might be close to one that we ll solve7 as with the harmonic oscillator Then we might say that our simple potential is a toy model But by solving these problems we will get some valuable mathematical experience Also7 since there are very few problems in QM which permit exact solutions7 we might be able to use one of these examples as a starting point for a more precise calculation 0 f0lt lt Vm 1 izia The simplest potential is otherwise as illustrated in Fig 21We want to use the TlSE to nd valid solutions for sz Outside the well it is clear that the particle can t be found7 so we expect to get 1 0 for z lt 0 and z gt a For the region 0 2 x g 017 with the de nition which we will make often V QmE h 23 HVFINITE SQUARE WELL BOX 11 Vx an Figure 21 In nite square well7 of width 1 it makes sense7 because we fully expect the energy values to be positive the Schrodinger equation can be written as 12 7 7k dmz w which has trig functions as solutions The condition that the wave function be continuous so 10 0 and zMa 0 leads to a condition on k and gives the solution 11z Asin a and energy eigenvalues zwzhz n7 2ma2 n123 24 and the normalization condition gives the full solutions may sin 25 We note 0 The solutions are alternately even and odd about a2 c As we increase in energy and in n the successive states have one more node 0 The wave functions satisfy woman dm 0 if m y n and so woman dm 6m and we say the din s are orthonormal A generalization of the orthonormal basis of our conventional vectors 12 CHAPTER 2 THE THVIEJNDEPENDENT SCHRODHV GER EQUATION The din s are complete meaning that any function satisfying the boundary conditions can be expressed as a linear combination of them 00 2 00 mm fWZMM EZmE 1 n1 We can nd the on using Fourier s Trick multiply both sides by mz and integrate The result is WWWMW It follows that the general stationary states of the in nite square well is 00 x t Onsin eiin27r2h2ma2t n1 2 a l mm on gO s1n m0dz This shows for this one example at least that knowing the solutions for the stationary states gives the most general solution to the Schrodinger equation where The meaning of the coef cients in the this expansion for a general state is that 0an gives the probability that a measurement of the energy of the system gives the value En One often says loosely that 0an is that probability that the system is in the nth state but as Grif ths notes the particle is in state 11 The en s must satisfy 00 23an 1 n1 and one can also show that the expectation value of H for the state 1 is 00 ltHgt Z loanEn n1 24 The Harmonic Oscillator The next simplest choice for the potential Vm is the spring potential77 of elementary classical physics It is of enormous importance for quanutm physics as well because if a potential function has a minimum somewhere in the neighborhood of that minimum one can always approximate the potential as a quadratic function having the form V kmz plus a constant which we can ignore z being the displacement from the minimum De ning w the angular frequency of the motion in the classical problem as the TlSE for 11 is 24 THE HARMONIC OSCILLATOR 13 which applies for all z though we expect the wave functions to be small but not zero in the regions where the classical motion is not allowed The solutions must approach zero as x a ioo so that they can be normalized The differential equation 26 can be solved we ll do it but it s not especially easy and requires us to think about power series In fact there are two different ways of solving it The second way involves the important method of ladder operators and uses cleaver algebraic tricks as opposed to standard methods of solving DE s In many courses this special method would be presented later on but Grif ths sez Let s do it now So we ll go though both methods We ll certainly know a lot about the harmonic oscillator when we re done 241 Algebraic Method Using the momentum operator the Schrodinger equation for the harmonic oscillator can also be written its mwale Ew We make the de nitions of the operators ai E ip mum 27 1 V 2 mw and in a proof which you will nd in the book Grif ths shows that the Hamiltonian can be written in terms of these operators as H w 0141 7 w aura Then he shows that if 1 is an eigenstate of H with eigenvalue E then the states 144 and 171 are also eigenstates of H and have eigenvalues E w and E 7 w respectively Because of these properties the operator 1 is known as a raising operator and a is known as a lowering operator Since the energy of all the states must be positive and so there must be a state 110 for which Li10 0 Now resurrecting the de nition of a which involves a derivative operator thru the momentum operator we get a much simpler DE for the lowest state 110 1110 7 mm H W0 for which the normalized solution is 28 mwgt14 Mmz 7 5 25 wow WE The energy of this state is 14 CHAPTER 2 THE THVIEJNDEPENDENT SCHRODHV GER EQUATION so that we can get all the states by successively applying the raising operator to this state The nal result is 1 7 n with E l h 29 man wow n 71 2 w Along the way in derivations of these results some important relations were used7 which we note down here The commutator of two operators A and B is A7 B E AB 7 BA and we have the very important commutator7 Mal m The raising and lowering operators relate the individual states by awn Wilma aw xqu 210 These relations can be useful when evaluating certain integrals involving the HO wave functions The integrand may involve z or p operating on a wave function We can write this in terms of 1 and a operating on the wave function and then use Eq 210 along with orthonormality to do the integrals very easily An example is found in Example 25 242 Analytic Method Now go back to original Schrodinger equation for the HO potential and try to get the solutions the conventional hard way There are reasons for doing this even though we can in principle get all the solutions from the results of the last section First7 Eq 29 can be unwieldy to use secondly the experience in solving this problem will be useful when we solve the more dif cult problem of the H atom later on Eq 26 can be cleaned up a little by introducing the dimensionless variable mw E E m 211 Then it can be shown that this DE becomes 12 T52 EziKliJ 212 where K This still isn t an easy equation We ve already gotten one solution but that doesn t help much We use some inspiration At large x large 5 52 has got to be much bigger than K so the DE of 212 behaves like 2 d dig z 1 large m and for a little guring7 this has the solution we Ae W 24 THE HARMONIC OSCILLATOR 15 there is a solution with positive exponent but it is not permitted since that function blows up at large This suggests that if we pull off a factor of 5 522 and de ne we 2 new then we will get a friendlier equation for 715 In fact the equation that we get is dzh dh 7 7 2 7 K 7 1 h 0 d5 565 lt gt which looks worse than the original Science takes a giant leap backwards At this point we have to set up a power series for 715 715 Z Lg5quot 73970 Substituting and doing some algebra gives a recurrence relation between the coef cients what we re normally after when use a power series 2j 1 7 K a 392 a39 7 J 1J 2 7 which in fact does determine the solution before we apply the boundary conditions One then realizes that if this power series has an in nite number of terms the solution will be unsuitable for a wave function which must vanish at large 5 The series must truncate and from this it follows that K is speci ed by 2n17K0 for n123 and this gives the possible values of the energy En n w which is the same answer as before gotten in a totally different way We can say that the reason that the energy is quantized is that for all the improper values the wave function found from the Schrodinger equation will behave improperly Reassembling the pieces to get the wave functions we note that from the recursion relation either a07 0a10 or a00a17 0 For the rst the wave function is even in z and for the second it is odd in m The polynomials 715 are simply related to some wellistudied functions in math called the Hermite polynomials of which the rst few are H01 111125 11124872 H38537125 and one can show that the normalized wave functions are mm 1 1M5 Ry4 m Hn5e 24 213 with 5 given by Eq 211 16 CHAPTER 2 THE THVIEJNDEPENDENT SCHRODHV GER EQUATION 25 The Free Particle Now we turn to a case which might seem even more simple than those we considered a particle moving freely that is Vm 0 everywhere In classical mechanics a particle would move at constant velocity under these conditions The TlSE for this case is simple 52 12 777 E 214 2m dmz w which is a bit more simply written as 121 2 w 7 7k 1 215 where k x2mEh This is the same DE as we had for the interior of the box There the solution was written in terms of sin and cos but here we prefer to use complex exponentials Thus Aeikz Beiikm But we have no boundary conditions so there s no reason to discard either of the terms here It helps our understanding to replace the time dependence in this solution Thus x7t Aeikzit Beiikzt and now from our experience with wave solutions we see that the rst term is a complex wave moving to the right and the second is a wave moving to the left The difference between the two terms is the sign associated with k which was de ned as positive but if we now allow k to be positive or negative we can write down one of the terms for simplicity So our solution for a wave traveling in some direction is 2 i v2 E has t Aelltkwtgt Wlth k iTm 216 Trying to identify the particle velocity for this wave gives a small problem The speed of the quantum wave is the ratio of the t coef cient to the z coef cient which gives E U uantum q 2m whereas from the classical formula E mv2 we have 2E Uclassical K The resolution of this paradox will be given shortly A more serious issue is that the wave function we have cannot be normalized The solution in 216 gives 00 CO tqukdz AF ldmoo x 00 00 which means that this wave functions cannot describe a true physical state So there is no such thing as a free particle moving with a de nite energy However the solution found here is of enormous use to us mathematically so we forge ahead 25 THE FREE PARTICLE 17 We can get a proper physical state if we sum over solutions of the type in 216 Rather we do an integral these functions over k weighted by a function we ll call Mk Mam gtkeikm tdk where we put a Um in front for later convenience Such a function can localized in space and thus normalizeable but it carries a range eigenvalues k and thus it surely carries a range of speeds regardless of the precise meaning of speed In particular at t 0 the wavefunction we ve constructed is z 0 may dk and if you had a healthy math background you d say that m 0 is the inverse Fourier trans form ofiJk Or equivalently gtk is the Fourier transform of m 0 The Fourier transform is of great importance in theory of wave and also in QM which is also all about waves ls the mathematical means by which we alternate between the coordinate representation of a wave and the frequency representation of a wave In QM we might say it is the means by which we go from a coordinate representation of a wave function to a momentum representation of a wave function A function and its Fourier transform x are related by Plancheral s theorem which is 1 00 i m 7 1 00 7i m agp m Fkek dz Fk m fme k dz 217 Note the different sign in the exponential in the two transforms Of course these integrals have to exist for particular choices of x and there are lots of juicy mathematical issues But we push onward From this theorem then we get 100 Imoe ikm dm ln words this gives the momentum wave function from initial t 0value of the coordinate wave function We see from working a few examples that if m 0 is thin then the momentum wave function is broad and viceversa This again brings us close to the content of the Uncertainty Principle but we return to the question of the velocity of the wave We did nd that for our wave Ilkz t the velocity of its wiggles is given by m Uwave 2m but these are not wiggles in probability which what is actually measured The velocity we need to be concerned with is velocity of wiggles in the probability and this is called the group velocity vgmup The velocity of the wave is properly called the phase velocity As a concrete example one can consider a wave packet which has the shape of a modulated but nite wiggle The group velocity is the velocity of the modulating envelope If the rapid 18 CHAPTER 2 THE THVIEJNDEPENDENT SCHRODHV GER EQUATION wiggle comes from a narrow distribution of frequencies centered on we with wavenumber k0 then whereas the phase velocity of the pure we wave is w hko Uphase E k0 one can show that the group velocity is dw k Ugroup R k0 As vgroup is what corresponds to the classical velocity then we have Ugroup Uclassical 239Uphase thus resolving the confusion 26 The Delta7Functi0n Potential So far the problems we ve solved have had either all localized or bound states square well harmonic oscillator or else like the free particle were unlocalized or scattering states These simple cases were not realistic in the real world a potential will go to zero at in nity Such a potential might give both bound and scattering states The 67function potential is the extreme case of a nite square well which we will also solve shortly This makes the answer simple at the expense of dealing with some slightly strange mathematics Recall the properties of the delta function roughly speaking one has 635 0 ifzy O oo ifm0 and 7 7 a fa so that 7 a fa6z 7 a Delta functions can be made into proper objects when they are treated as distributions a gen eralization of our usual functions The de nition can be extended to threedimensional integrals We consider the potential energy function Vm 7046z that is the function is a negative spike of strength 04 Watch out the units are subtle here since 04 must have units of Energy Length We will want to solve for all the quantum states both for E lt 0 and E gt 0 But how to solve the DE when we have such a strange function as 6z present It s not as scary as it seems we just need to break up the problem and use one special trick First de ne 7 72mE H 7 h which is OK because E is negative With this de nition the TISE for all z except z 0 becomes d2 11 azw dmz 26 THE DELTAiFUNCTION POTENTIAL 19 for which the general solution has the form Ae m Beer but to keep the function from blowing up for z a ioo we have to choose the terms with the right behavior This leads to Bem m lt 0 1M Fe m m gt 0 Now we don t have a DE right at z 0 but we can solve for B and F and the energy using boundary conditions an normalization First sz must be continuous at z O This give F B and we can write the wave function simply as we BM The other condition one might think of applying is the continuity of dipdz everywhere but in fact here it does not need to be continuous because of the behavior of V at z 0 The trick we use is to integrate the respective terms of the Schrodinger equation from x 5 to z 5 hZ 5 12 5 5 7 7 d V d E d 2 is M w is we as w z where we let 5 a 0 After a few steps for example the right side gives zero one nds 7 27710413 dm 0 dm 7 0 22 Here the left side is the difference in slopes of 1 between the right side of the origin and the left side A little more work gives us the value of the energy 2 777104 E 7 2h2 Finally we can get B from normalization of 1 One can show B LEW There is just one bound state for this potential The nite square wells which we ll solve later will in general have one or more Now we solve for the scattering states E gt 0 Again with the de nition k x2mE the Schrodinger equation is 12 7 7k dmz w which has solutions which can be expressed as trig functions or complex exponentials We choose the latter getting I I Aelkz Belkm and for now we can t rule out either term for the limit z a 00 And note that again these are non normalizable states so that some care is needed We will write 1 Aeikm B57111 m lt 0 z Femz Ge lkm m gt 0 20 CHAPTER 2 THE THVIEJNDEPENDENT SCHRODHV GER EQUATION We have to solve for these coef cients but we don t need to solve for the energy because any positive value is permissible Continuity of 1 gives A B F G If we do the same trick as before and integrate the Schrodinger equation over a small interval around the origin one can show F7 G A1 2m 7 B17 2m where e but this is only two equations to nd the four unknowns A B F and G The condition of normalization for localized states does not help us here We actually have to make a choice of coef cients depending on the experiment we are considering We have seen that the term with etikm corresponds to a wave traveling to the right e ikm gives a wave traveling to the left We choose to consider particles a whole lot of them in fact which are incident from the left They interact with the potential and afterward have probabilities of continuing to travel to the right or to bounce back and travel to the left But with with choice there are no particles with z gt 0 which are traveling to the left particles are only incoming from the left side This means that G 0 in our solution but the other coef cients are not zero Then the matching conditions give i3 1 17i A and F717i A Now a further bit of interpretation for this result The incoming wave represents the motion of a single particle Though experimental setups send a beam of particles toward a target the incoming wave here does not represent more than one particle Certainly in the elementary sense of the word A is the amplitude of the incoming wave B is the amplitude of the re ected wave and F is the amplitude of the transmitted wave If we take these values to also be quantum mechanical amplitudes for the different processes then squaring them should give probabilities and we can take ratios to get the relative probability of re ection or transmission Thus the relative probability for re ection is B 1B1 62 F 11412 7 13 and the relative probability for transmission is 1F 12 1 W T62 also called the re ection and transmission coef cients These give R T 1 as they should Substituting for 3 gives the results 1 1 2 T 2 1 2 Emozz 1 maZQ E These results can be made for rigorous if one wants the point is that we must be very careful about how we handle non normalizable states and what the solutions mean One can also solve the problem of the positive delta function potential ie a delta function barrier There are no bound states for that potential but an incident wave will have a probability 27 FHVITE SQUARE WELL 21 of passing through this barrier in fact the same as for the well since R and T just depend on 042 So the particle has a probability of passing through an in nitely high though very thin barrier In classical motion a particle cannot get through a barrier if its energy is less than the maximum potential heigh 7 of that barrier The phenomenon of quantum motion through barriers where it is forbidden classically is known as tunneling 27 Finite Square Well Now consider the nite square well potential V f V 0 or altmlta 23918 0 for lmlgta where V0 is a positive constant Again we expect to get both bound and scattering states This problem requires more algebra to solve it than the delta function potential but the math is not hard First look for bound states for which E lt 0 As before de ne 72mE h Outside the well where V 0 the TlSE can be written as 121 2 W 1quot which has exponential solutions Choosing the ones with the proper behavior as x a ioo gives Bem xlt 7a m 1M Frfm xgta In the region 7a lt z lt a with V 7V0 the TlSE is 52 12 V01 E1 Then if we de ne 7 2mE V0 7 h it can be written 2 d 11 2 7 5 dm2 w and this is a simple DE Here we do want to use trig fuctions and the solution is 11m Csin m D cos z for 7 a lt z lt a and now we have 5 unknowns to solve for B F C D and the energy E buried inside H We will nd them though the boundary conditions normalization and symmetry 22 CHAPTER 2 THE THVIEJNDEPENDENT SCHRODHV GER EQUATION First symmetry an important tool in quantum mechanics Certainly the potential is sym metric for z lt gt ix Are the solutions symmetric in some way One can show that for a symmetric potential one can at least choose the solutions to be either symmetric or anti symmetric that is 1ampm symmetric M790 i 71m anti symmetric In one could have gotten this fact from the boundary conditions but this is useful to understand and it does save some work The condition of symmetry gives B iF We will now explicitly look for symmetric solutions for the nite square well The well will have such a solution It may have an anti symmetric solution that case is dealt with in one of the Problems So then we have B F and it also gives C 0 since sin m is an odd function Our solution is now Fem m lt a 1ampm Dcos z 7a lt z lt a Fe m m gt a Applying continuity of the wave function and its derivative at z a gives Fe m D cos a 7 hFe m 6D sin a Note For this problem the potential Vm makes a nite jump but otherwise is not pathological at the boundary We are thus also required to make 1ampm smooth there so the derivative is continuous One can then show that with the de nitions 226a 202 2mV0 we have the condition 2 tanz lt30 71 219 2 And from this transcendental1 equation we can get 2 and then the value of E A useful strategy is to plot both sides of 219 together and note where the curves intersect see text Of course computers can give a numerical solution We see that even a simple problem can require some messy math for a solution Such is life We can consider limiting cases of the potential to check these results 0 A wide deep wellZ In this case 20 is big One can see graphically that there are many crossings for the left and right sides of 219 always occuring just below 2 E mr2 For these solutions one can show 2 2 2 ii 7139 h E V x 7 where is odd 0 2m2a2 n that is to say these are the energies of the states measured from the bottom of the well This agrees with our previous box result But here we miss half the solutions because we ve only considered symmetric states See Prob 229 for the rest 1Meaning that it s mathematically messy not that you re going to meditate on it 2Deep and wide deeeeep and wide 27 FHVITE SQUARE WELL 23 0 Shallow narrow well Then 20 is small There is still one intersection of the curves plots of left and right sides of 219 so there is always one symmetric bound state As for solving for all the coef cients we can get them from normalization of the wave function That s uninteresting so we move on to do the scattering states Now consider E gt O For the regions where V O we have with k de ned as k E v 2mE Aeikz Beiikm for z lt 7a since we we want a solution with both an incident wave and a re ected wave as when we did the delta function potential In the well with Z 7 2mE V0 h the solution is 1ampm Csin m D cosh and for z gt a we want a solution with only a right going transmitted wave Solving We an take E gt 0 as given and nd relations between the coef cients from the boundary conditions The conditions are the 11 and are continuous at both z 7a and at z a Here the algebra starts to be a little challenging and it involves complex numbers but one can show 72H LBW 2 7 k2F and F 5 k 22 2 cos2 a 7 i 2 sin2 a and we will just calculate the transmission coef cient T FFAF In fact due to the messy algebra it s easier to present T l and for that we nd T 1 V OZ sin2 lt 2mE W 220 4EE V0 Of course T 1 if V0 0 but also T 1 if the sin in 220 gives zero That occurs when 2 gal 2mEn 1 V0 n7r for any integer n En being the energies of these perfectlyitransmitted waves And this occurs when 2 nzwz 2m2a2 which in fact are the energies of the in nite square well Interestingly this resonancektype effect can actually be observed in forward electron scattering on atoms EnVO CHAPTER 2 THE THVIEJNDEPENDENT SCHRODHV GER EQUATION Chapter 3 Formalism 31 Introduction We ve now covered the basics of the meaning of the wave functions and gotten tons 07 practice by solving the 1 D Schrodinger equation for a good assortment of potentials What s left Well the obvious things We need to solve problems in three dimensions we need to deal with multi particle systems and also learn about spin a dynamical quantity which occurs in the quantum world And more mathematical methods But we also need a rmer grounding in the fundamentals of quantum theory This is because we will encounter other important dynamical quantities possibly abstract like spin and we need to know how to work with their operators and associated quantum states We need to know how to deal with any observable in QM beyond the ones we ve seen z and p 32 Hilbert Space Quantum Mechanics is as we ve seen based on wave functions and operators The wave functions are treated much like abstract vectors as we saw from considering linear combinations ofthem and the operators are also linear operators which makes of think of matrices This leads us to the branch of mathematics known as Linear Algebra The natural language of quantum mechanics as Grif ths sez Alas most of us don t haven t learned linear algebra so well when we get to QM Were it not for QM it might be totally neglected in our mathiphys education In any case it is necessary to learn some of it Grif ths has a big important appendix giving a refresher course in linear algebra First we have vectors in an Nidimensional space represented by an N tuple of complex components Vector la gt a Two vectors have an inner product also called a scalar product 043 aibl agbg a7VbN 25 26 CHAPTER 3 FORMALISM Note the complex conjugation of the components for the rst vector The inner product is not commutative switch the order and the result is complexiconjugated Vectors are transformed by linear operators ie they are linear transformations These transformations are represented by matrices When we transform vector la to vector 3 by means of the transformation T represented by matrix T we have 7511 7512 tiN a1 7521 7522 tzN a2 l Tla gt bTa tNl tNN aN As written here la and 3 denote the abstract vectors and T the abstract operator while a and b and T give their representations for a particular basis In linear algebra one studies the types of matrices which can arise in linear transformations and the operations we perform with and on these matrices In some of this work we study how we can change between different bases giving different representations The vectors in QM are the wave functions The ones we ve seen are functions of z could it be otherwise so these are vectors with an in nite number of components that is the values at each point of which there are an in nite number So things get very abstract here What we need to do to adapt the normal language of linear algebra to that of the new vectors is change from the discrete index of regular vectors to the continuous index of functions In doing this sums go over to integrals Actually later on when we encounter spin more generally angular momentum we will go back to the familiar nitedimensional vector space A vector wave function that we encounter in QM must be normalizeable so that we can demand f 1 Idm 1 So we will deal with a special class of functions those for which vamltw We very loosely refer to these functions as our vectors in a Hilbert space and hope that no mathematicians are listening to our sloppy de nitions These vectors functions will have an inner product de ned as b Umfmwmw where a and b are limits appropriate for the vector space which must exist for functions in this space Note ltfl9gt lt9lfgt so the inner product is not commutative A function f is normalized if f 1 A set of basis functions in the space is orthonormal if A set of set of functions fn in the space is complete if any other function fz in the space can be expressed as a linear combination of them 1 23 an n1 3 3 OBSERVABLES 27 If the set of functions here is orthonormal then the Gas can be found by Fourier s trick 0n WW 3 3 Observables We use a similar notation for the expectation value of an observable Qzp ltQgt3Wdmltmww Q will always be a linear operator Since Q is the average of many measurements it must be real Q Q This implies MQ IO ltl ilgt 1410 for any function 1 in the space So operators corresponding to observables must have the property lt Q Q for all Such an operator is said to be Hermitian So a Hermitian operator can be applied to the second or rst member of an inner product with the same result The operator i z clearly is Hermitian since it simply multiplies One can show that the momentum operator 13 aim is Hermitian but note the derivative operator by itself is not Hermitian 34 Determinate States Measuring the value of the observable Q on an ensemble of particles all prepared in the same state in general does not give the same value every time generally we can only get the expectation value Q Quantum mechanics is this sense indeterminate But could we prepare a state such that it did give the same value for Q each time Such a state would be a determinate state for the observable Q Actually we ve seen this already as stationary states are determinate states for the Hamiltonian For such a state 7 if the measurement of Q each time gives the value qi the standard deviation of Q would be zero One can then show that for this state 11 q This is an eigenvalue equation for the operator Q 1 is an eigenfunction of Q with corresponding eigenvalue q We note that an eigenvalue is a number An eigenfunction can be multiplied by a number and it will still be an eigenfunction with the same eigenvalue An eigenfunction can t be zero otherwise it wouldn t be a legal vector for our states but an eigenualue can certainly be zero Sometimes several independent eigenfunctions will have the same eigenvalue In such a case we say the eigenfunctions are degenerate In the TlSE we had flip EiJ E is the eigenvalue of the H operator And the full wave function Ilz t iJze mh is still an eigenfunction of 28 CHAPTER 3 FORMALISM 35 Eigenfunctions of Hermitian Operators We thus have reason to care about the general problem of eigenfunctions of Hermitian operators That is because the QM operators corresponding to observables must be Hermitian and because the determinate state for an observable must be an eigenstate of the corresponding operator Eigenfunctions and eigenvalues fall into two classes 0 The spectrum is discrete Eigenvalues are separated from one another though there may be an in nite number of them In this case the eigenfunctions are in our Hilbert space as they are normalizeable o The spectrum is continuous Eigenvalues ll out an entire range of the real numbers In this case the eigenfunctions are not normalizeable though a linear combination of them could be but then this linear combination is not itself an eigen function We saw this in the case of the free particle In some problems there is both a discrete spectrum and a continuous spectrum as in the case of the nite square well Now some more depth on these two cases and some theorems 351 Discrete Spectrum Here we get normalizeable eigenfunctions They satisfy two conditions one can show that 1 The eigenvalues are real 2 Eigenfunctions belonging to di erent eigenvalues are orthongonal We saw that the eigenfunctions of the in nite square well and the harmonic oscillator were orthogonal But it is a general property Alas if the there are degenerate that is they have the same eigenvalue the second theorem does not apply But in this case we can construct orthogonal eigenfunctions from the ones that we nd If necessary one can use the Gramischmidt process from Linear Algebra in practice we won t need to do this So actually in all cases we will have a set of mutually orthogonal eigenfunctions The eigenfunctions of an observable operator are also complete that is any other function in the Hilbert space can be expressed as a linear combination of them We also say that they span the space This property will later on prove to be extremely useful when we have to construct approximate solutions to the Schrodinger equation 352 Continuous Spectrum Here the eigenfunctions are not normalizeable There are problems with this case because the inner product of two functions may not even exist in the conventional sense But we can get the above properties for the discrete spectrum to hold in a more abstract sense The theory can be made more rigorous but we will deal with a few examples The eigenfunctions of the momentum operator with eigenvalue p satisfy h d ppr This gives I W Aerh 35 EIGENFUNCTIONS OF HERMITIAN OPERATORS 29 which is not squarekintegrable for any value ofp even a complex one So there are no eigenfunctions of the momentum operator in the proper Hilbert space However if we restrict ourselves to real eigen values p which is what we would expect then we nd that the inner product of two such functions is 00 00 I I cum fp ltzgtfpltzgtdz W emith OO 00 But what is this One can show i 00 eiqmh dm 6q 27139 700 this is an exercise elsewhere in the text and then ltfp lfpgt lAlZQWWp p If we choose 1 1 A m then mas so ltfplfpgt6pip which looks a lot like real orthogonality where we get a Kronecker delta for the inner product Here it s a delta function Grif ths calls this state of affairs Dirac orthonormality The momentum eigenfunctions are also complete as long as we use an integral in place of the sum to forma linear combinations of basis functions Consider an arbitrary squareiintegrable function It can be written in the form Cp fpz dp Cpeipzhdp We can get the coef cient Cp with the continuum version of Fourier s trick 00 00 mm cltpgtltfp vplt Cp6pp dp cltp gt OO 00 This is essentially what occurred in a previous discussion where we inverted a Fourier transform via Plancheral s theorem The eigenfunctions of momentum are complex sinusoidal functions with wavelength 27r p which is the old De Broglie formula using h 271 But the use of momentum here is more subtle than what De Broglie hypothesized since there is no possible quantum motion with a de nite value of the momentum Thus we have dealt with the eigenfunctions of p They are not in the true Hilbert space of quantum states but our ability to use them puts them in the suburbs of the Hilbert space And thus we have given the eigenfunctions of momentum and energy later we ll deal with eigenfunctions of other dynamical quantities such as angular momentuym and spin But we ve overlooked something what about the most basic operator of all the coordinate x As an operator it is simply multiplicative but what are its eigenfunctions and eigenvalues The answer is so simple it s actually a bit confusing The eigenfunction of the operator z having eigenvalue y is gyz 6z 7 y which satis es gy lgy 6y 7 3 which again is a Dirac delta rather than a Kronecker delta so that these eigenfunctions use Dirac orthonormality and so are not proper As with the momentum eigenfunctions these are complete 30 CHAPTER 3 FORMALISM 36 General Statistical Interpretation The wave function m t was introduced in Chapter 1 as something we would use to nd the probabilty of a measurement of coordinate m We will now discuss the general theory of probabilities of measurement of any dynamical quantity Actually these are the postulates of QM we haven t derived them But up to now we have gotten some experience in the mathematics needed to know what they are talking about If you measure the quantity Qxp of a particle in the state z t you will get one of the eigenvalues of the hermitian operator Qz j If the spectrum of Q is discrete the probability of getting eigenvalue qn is lcnlz where If the spectrum of Q is continuous the probability of the eigenvalue 2 to be in the range dz is lcz 2 12 where 02 ltle 1 gt Upon measurement of Q the wave function collapses to the corresponding eigenstate 37 The Uncertainty Principle We ve already seen examined but not proven a special case we found that h Two39p Z 5 for any state satisfying the Schrodinger equation For any two observables A and B with the commutator of their operators given by A B E A 7 BA one can show that for any quantum state 1 2 2 1 A A 2 gt UAUB 7 lt2 ltlAiBlgtgt 31 This is the generalized uncertainty principle With A m and B 13 77 using 3213 in Eq 31 gives 22 h h amagt 7 gt amapZE 371 Minimum Uncertainty Wavepacket In a couple of the problems we have already encountered a quantum state which gave the minimum uncertainty product They were gaussian wavepackets but can we show that this must be the case Grif ths shows that if we demand that uncertainty product give then we can produce a differential equation for 11 h d W 7 lt2 i we 7 ltzgtgti idm 38 ENERG YiTIME UN CERTAHV TY 31 for some constant a This has the solution Aeiamiltzgt22heiltpgtmh which as in our examples is a gaussian wavepacket 38 EnergyiTime Uncertainty gm GHQ 32 Rant on the abuse of this relation too long to be included in the notes Perhaps next year One can show 39 Dirac Notation We can now express the general quantum theory taking the things far beyond our initial simplistic View of the quantum state We work in analogy with nding the components of the vector A To get them take the dot product of A with the different unit vectors AmiA AyjA But if we consider a coordinate system where the axes are rotated call it the m 3 system then the components are given by A mi A Aj A The idea here is that the vector A exists independent of our choice of coordinates but to express that vectors we must nd its components using a particular choice of coordinates and then those components are found by taking a dot product with the basis vectors CHAPTER 3 FORMALISM Chapter 4 The Schrodinger Equation in Three Dimensions There is lots left to do We need to know how solve for quantum motion in three dimension adding two more dimensions brings in more features to our quantum states 41 Schrodinger Equation in Three Dimensions First7 we need to be clear about how the make our Schrodinger equation applicable to three dimen sions As the classical energy is now 1 imvz V0quot 09 1932 193 V1quot7 we make the replacements h 8 h 8 h 8 gt if gt if gt if Pm i 69 7 Py i 8y 7 P2 482 which can be summarized as h P H V We note that V is a function of all three coordinates Making these replacements in our old equation i H I7 with now 1 being I1quott7 we nd 81 h E7 if 2 1 V 2 3t QmV w where V2 is the laplacian7 V2 82 82 82 w t 872 Q The meaning of Ilr7 t is just as it was before the probability of nding the particle at r within dgr dz dy dz at time t is l llr tl2dz dy dz If the potential is independent of time there will be a complete set of of stationary eigenstates n t mamh 33 34 CHAPTER 4 THE SCHRODHVGER EQUATION IN THREE DIMENSIONS where the 11r s satisfy the TlSE 52 fivzw V18 E18 2m Then the general solution is rv t Z Cn nreiiEnth n All of this is ne if the problem can be easily expressed in Cartesian coordinates a few problems have easy solution in terms of z y An example is the isotropic harmonic oscillator which has the potential Vr kr2 but which can be expressed as Vr kz2 y2 22 But more often the potential depends on the distance r from the origin in a way such that Cartesian coordinates are not easy to work with Then one is forced to use spherical coordinates 42 Spherical Coordinates Separation of Variables We will assume Vr is function of r only We recall from our other ne coursework that the laplacian can be expressed in spherical coordinates as Vziig 23 13 sing 1872 r28r T 8r rzsin080 88 rzsin208 gt2 so that the Schrodinger equation is 8218 28 18 a 818 18211 7 725 a t 72 slum 10 t 72 smzoWl VW E1 Gakl How can we ever solve this The same way as always Assume a solution with separated variables 1W 07 lt8 RUM07 lt8 Recall what we do with this Substitute into the differential equation then diVide through by RY We get an equation where one side depends only on r and the other only on 0 lt8 One can then show that each side of that equation must be equal to a constant Now that constant could be anything we have a good reason to write it as 66 1 We will nd that 6 must be an integer but for now with no loss in generality we use that form and then the separated equations are mTZ ii g 7 212 Vr 7 E 6 1 1 1 8 8Y 1 821 7 7i 07 7 76 Z 1 Ysin080 lt81 80gtsin208 gt2 l which still looks pretty grim but the separation has made things easier and the rst is an ordinary differential equation Eventually we want to solve this for a particular choice of Vr but we note that the angular equation doesn t depend on that choice 43 THE ANGULAR EQUATION 35 43 The Angular Equation Rearranging we have 2 sin0 110887 71 46 1 sin20Y This equation might look familiar It also occurs in E amp M when we solve the Laplace equation in spherical coordinates except that we now include a j dependence which is not considered in our E amp M book The problems didn t need it Here in QM we must consider the j dependence of the wave function As this is still a differential equation in two coordinates we do a further separation of variables and write 3097 gtgt 90 1 gt Again we substitute split up the equation and set both sides equal to the same constant which this time we call m2 again for good reasons to be seen later The resulting separated equations are sin0 611103 6 1 sin20 m2 1 121 7 2 6 7 7m The rst one is still pretty messy but the second one is easy The solution for I is 121 7 2 I 7 imq W 7 m gt 7 5 where we absorb the constant in all the other factors and we deal with the possibilities e by letting m be positive or negative We chose the complex form here because it is easier to work with wave functions don t have to be real We do require that the wave function have one value at each point in space so I must have the same value if we incrase j by 27139 One can show that this implies iimq m0 i1i2 Using this we go back to the 9 equation which we now write as i d 1 d9 1 2 2 7 sin 0 lts1n0wgt 1 66 1 sin 9 7 m 9 7 0 41 We won t even attempt to derive the solution to this It s enough to present them and give their properties The solution to 41 is 90 Alecos 0 where Pl is the associated Legendre function de ned by d lml PZW lt17 WW2 7 PM 13 and where m can take on the 26 1 values m441z 36 CHAPTER 4 THE SCHRODHV GER EQUATION IN THREE DIMENSIONS and the Pzx are the familiar Legendre polynomials which in fact are given by the Rodrigues formula 1 d l 2 l P195 7 W 95 1 of which the rst few are P0z1 P1mm P2z3m271 Note that lez is in general not a polynomial eg for Z 2 132935 3x2 71 1321a 3m17 m2 132 317 2 and by the de nitions we are using Pfquot Pgquot and P Pl but conventions differ amongst the reference books While the le s look messy if we substitute z cos0 and v1 7 z sin0 then we have P31 P10cos0 P11sin0 P29 3cos207 1 P21 3sin0 cos0 P22 3sin20 Polar plots of these functions remind one of the orbitals from basic chemistry for good reason as we ll see eventually We should note that since the 9 equation was second order there should be two solutions here The others called QAm blow up at 9 0 and t9 7139 so they re not permitted Now we put 90 and ltIgt gt together to get the full angular wave function which we will call Y0 gt We need to choose how the product 91 is normalized Of course we still have to nd RU but when we do the normalization condition on the whole wave function will be 00 27r 7r lwl2d3r lRlzrzdr lYlZsin0d0d gt1 0 0 0 and for later convenience we will choose things so that the separate factors are normalized co 27r 7r lRlzrzdr1 and lYlZsin0d0d gt1 0 0 0 The following formula will do the job 25 1 67 l 71 gt0 elm PZquotcos0 where e1 mgo 42 Y M E 47139 lmll m 7 The nm s are called the spherical harmonics and they are of enormous importance in theoretical physics Note different references may use different phases from Eq 42 The spherical harmonics are mutually orthonormal 27r 7r Y e an w sin0d0 d gt 616mm O O 44 THE RADIAL EQUATION 37 The rst few are 1 3 3 of i o 7 i1 iA 11 Y0 7 4 Y1 4W cos0 Y1 8 sin 05 Y0 6 cos2 9 7 1 Yi1 175 sin0 cos 0eii etc 2 V 167r 2 V 87r 39 Note7 YOO is not equal to 1 We will have more to say about the angular function Yt97 gt and its relation to the angular momentum of a particle later on For now7 we return to the equation for the radial function RU 44 The Radial Equation The Schrodinger equation for a particular radial potential Vr gives the radial equation d ZdR 2m 7 7 7 V 7 E R Z Z 1 R QC thr l ltgt And how do we solve this A useful trick is to de ne a sort of reduced radial wave function ur 717 E 7137 We then get 52 d2 h z z 1 777 V g UEu 43 2m 172 2m 72 We can note that 43 has the same form as a one dimensional Schrodinger equation except that the potential is replaced by the effective potential 52 z a 1 Veff V 1 772 2m 7 and where the extra term is called the centrifugal term We have seen some similar in classical mechanics for the central force problem The terms acts like a potential which pushes the particle outward7 just like the corresponding pseudo force in classical mechanics Note that with this de nition of uh the normalization condition on the radial function is f0 dr 1 That all we can do with the radial function until we make a speci c choice for the potential Vr7 ie the physical problem we want to solve 45 Example Spherical Box As an example leading to more complicated H atom7 Grif ths considers the case of the potential VTO rlta oo rgta which might call a spherical in nite well7 or something 38 CHAPTER 4 THE SCHRODHV GER EQUATION IN THREE DIMENSIONS Outside the well we must have 1 0 Inside the well the radial equation is dzu 7 1 xQmE h W 2 7 k2 u where k 7 We want to solve this DE one of the boundary conditions is that ua 0 in order for the wave function 1 to be continuous Being a pathological potential as usual we do not demand W 0 The Z 0 case is easy dzu 7 drz 7 Recalling that the actual radial function is R and that the cosine term would blow up at r O we must have B 0 then the boundary condition at r a gives sinka 0 or ka mr and then ikzu gt ur Asinkr B coskr 71271252 Qmaz Enl0 71123 and the complete solution after normalizing is 1 sinmrra V 27m T But what about arbitrary Z for this problem That solution introduces some new functions often seen in mathematical physics which we will just give here For the general case the general solution for u is a linear combination of spherical Bessel functions W0quot 717 ArjAkr Brn kr where jlz is the spherical Bessel function of order 6 and ngz is the spherical Neumann function of order 6 de ned as Mmkkwc dgtzsim nl7illt1dgtlcosm mdm m mdm m The rst few are cos x 7 W 7 7 See the text or standard reference books We nd that the nzm s blow up at the origin so they are not permitted So the B coef cient must be zero and the radial wave function is RU Aszw The permitted values of the energy E come from the boundary condition at r a The radial wave function must be continuous at r a and since it is zero outside the well we must have Ra AjAka 0 so that ka is a zero root of the function jzz This function has many roots so that if we let BM be the nth root of jzm then ki which ives E ii 52 7 a g M7 Qmaz M 46 THE H ATOM 39 as the nth energy level of the states of angular index 6 With this the complete wave function is me AszWMH WWQ gt where AM will be determined by normalization We note that each energy level actually has 26 1 states because for each 6 there are 26 1 possible values of m and the energy is the same for each 46 The H Atom 461 Introduction At last we are ready to solve the rst and last problem which is an exact rendering of something in the physical world With a positive charge 5 nailed down to the origin a point charge 75 and mass m me moves around in three dimensions with the potential energy given by 1 52 4713960 7 WT It should be stated that what we will do here for the H atom is not exact in that there is lots that is not put into our solution Motion of the proton special relativity the dependence of the interaction on the spins of the particles and the correct quantum treatment of the electromagnetic interaction But apart from that we will have put in most of the physics for a real physical problem We should recall that Bohr solved this system using a simple model which is instmctiue but which we now know is basically incorrect He assumed that the electron executes a circular orbit constrained so that its angular momentum is an integral multiple of h He was able to show that the permitted energies of the system are given by 7136 eV 2 71 En n123 which with relations for photon energy and wavelength E w hf and f 0 gave the observed spectra for hydrogen atoms in particular the visible Balmer lines In the years from 1912 to 1925 people tried to extend Bohr s simple theory to more complex system largely without success With the advent of quantum mechanics it was realized that the approach to atomic motion using speci c trajectories is fundamentally wrong 462 The Radial Equation and Asymptotic Behavior As the angular part of the wave function has already been found it remains only to solve the radial equation for the Coulomb potential With ur 5Rr it is h dzu 52 1 52 MM 2 747 g T u Eu 44 Our problem is to nd the allowed functions ur which will give the allowed energies the eigen functions and eigenvalues Clearly we will require that ur a 0 as r a 00 This is certainly the hardest DE to be solved analytically in this entire course But as it is also the most important we will want to work out its solution using tricks similar to those used in the analytic solution for the harmonic oscillator 40 CHAPTER 4 THE SCHRODHV GER EQUATION IN THREE DIMENSIONS Though there are indeed scattering solutions for the Coulomb problem the solution of which is a notorious sticky point in the theory of scattering we are de nitely dealing with bound state solutions here7 so we can assume E lt 0 We will make the de nitions 72mE mez Is 7 p m and p0 7 27r60lm The reduced radii p and p0 are dimensionless allowing us to focus on the math with physical constants ying around Substituting into 44 gives d2 P0 7 17 102 0 w u 45 which is super cially simpler than 44 but is still a nasty differential equation At this point we recall our strategy for solving the HO differential equation We looked at the behavior of the DE and thus its solutions at large distance As we might have expected7 it turned out to be some kind of exponential decay which we then factored out of the solution we then solved for the remaining part For a radial function we actually have two extremes one at large distance and another at r a 0 that is7 very small r For large p large r7 the radial equation and its solution becomes 7 u gt up Ae p Bep for which only the e P term is possible As p a 0 we get dzu ll 1 172 p2 u ultpgt0p 1Dp for which we can keep only the pl1 for proper behavior at small p Now the idea is to pull the factors for these limiting solutions and solve for what s left De ne upp 16 vp 46 and now solve for vp One might hope that the equation for it is simpler but that isn t really the case Looking ahead7 there is a restriction on vp in that it can t be such that up blows up at large p From 467 vp can t overpower the factor e P Substitution gives d2 d p722l17plP072l1v0 47 dp dp The only thing we can do now is to write a power series expansion for 12p7 00 up chp j0 and try to solve for the 0739s Substitution of the series solution gives the recurrence relation 61 7 2jl1700 C 7 j1j 2l 2 7 46 THE H ATOM 41 which will give all of the 0739s if we have the rst one But p0 remains unknown because we don t have a value for the energy E What xes this value It is the condition mentioned above that vp can t overpower the factor 5 One can show that unless the series for vp stops truncates at some term the series will give a function like 52 which we can t allow So there has to be a maximal index jmax for which we get ijax1 0 2jmax 17po0 If we de ne n E jmax Z 1 then p0 2n and we can derive a condition on the energy E zaz 7 me4 7 2m 7 87126371ng which then gives m 52 2 1 E1 E77 727 f 123 48 77 22 n2 72 or n 7 7 7 lt gt In 48 n has those values because jmax and 6 can possibly be zero We can scrape up everything and write down the wave functions remembering that p was de ned using re 2 2 a me 2 l i where a E w 0529 X10710 m 47reoh 71 an mez and a is the Bohr radius namely the radius of the smallest orbit of the simple Bohr model The wave function for the state with the three quantum numbers n 6 m is wnzmr EMMYquot07 gtgt where 1 EMT Pme p W 7 and vp is a polynomial of degree jmax n 7 Z 7 1 in p with the recursion relation C i 2j 17n C 11 j1j 26 2 7 463 The States of the H atom The ground state has n 1 and gives E1 7136 eV so the binding energy of H in the ground state is 136 eV The ground state wave function is 1 7m 11100T707 3107 300097 W 3540 49 For n 2 the energy is 7136 eV E2 f 734 eV ie the energy of the rst excited state and we can have 6 0 Z 1 for this case the latter having the possibilities m 71 0 1 All 4 states have n 2 42 CHAPTER 4 THE SCHRODHV GER EQUATION IN THREE DIMENSIONS The Z 0 and Z 1 cases give two different truncates series hence two different radial functions When we nd and normalize them such that f6 172 dr 1 they are 1 17 7 R20 70732 lt17 SirZa R21 a732geir2a 1 xE x For arbitrary n 6 can have the values 012n71 and for each 6 there are 26 1 possible values for m Thus the degeneracy of energy level En is n71 1a 226 1 n2 0 The polynomials we get for the radial function for each 716 are in fact known to olditimey mathematicians They even have a name they are called the associated Laguerre polynomials mm d I7 up Li ilap where L211 E 71 a Lqx and the Lqm are the regular Laguerre polynomials LN em y am The radial function still needs to be normalized though When we do this and put everything together the big result for the H atom wave function for the state given by 711771 is which is probably the most messiest thing we ll ever produce in this course but we note again that this is an exact answer for a realistic system and those are rare in quantum mechanics We note again that the full wave function depends on the numbers 711771 while the radial function depends on both 71 and Z The energy E depends only on n The H atom wave functions are orthonormal pgyew 1pm 73 sin0 dr d0 d gt 6xn6ll6mm For states with different energy eigenvalues this has to be the case but the wave functions were chosen so that all are mutually orthonormal There are various ways to visualize these solutions one can make shaded density plots of 11112 and plot surfaces of constant W12 Note that if we are ignoring the factor of the angular function and considering a radial probability distribution we need to plot rlerlZ because that is the function which is a normalized probability distribution 47 ANGULAR MOMENTUM 43 464 The H atom spectrum It s worth pointing out that using the only physics of basic QM if a particle is in one of stationary states of the H atom found above it stays in that state Of course from basic modern physics we know that the electron can make transitions to other states and in doing so emits or absorbs a photon of a characteristic energy but that process involves some ideas that for the time being we have to clumsily attach to the present picture The missing physics is the interaction of the electron with the electromagnetic eld and such a treatment must include the possibly of the creation and destruction of photons if not also electrons With such physics one could calculate the times involved in these transitions We ll have to come back to that later With that in mind having derived the possible energies En we know that when the electron makes a transition from state m to the state 71f there is emitted a photon whose energy given by E he 11 with f c is E7AE427 y7BseV474 711 71f This can expressed in a form which gives the photon wavelength 1 1 1 m amp 2 7 A 713 72772 where 1373 7 1097x10 m n1 iif 47rc 4713960 With 71f 1 we get the Lyman series of spectral lines for 71f 2 we have the Balmer series for 71f 3 the Paschen series and so on1 One can literally see the photons from the Balmer series by electrically exciting a tube of hydrogen gas In such a tube the hydrogen molecules have been dissociated with the resulting photons coming from the electron motion in the isolated atoms We also note that these series the rst few at least are of enormous importance in astronomy 47 Angular Momentum With motion in three dimensions we can consider a new dynamical quantity that had no meaning in 1 D and that is angular momentum We recall the importance of angular momentum in classical mechanics It has the de nition L r x p and the total angular momentum of a system is conserved if there is no net external torque Angular momentum is important in quantum systems which have rotational symmetry of some sort and that s a lot of them Lrgtltp gt Lmypzizpy We also de ne L where we make the replacement pm a 1In order the rst six series are named Lyman Balmer Paschen Brackett Pfund Humphreys Few people are impressed by this knowledge 44 CHAPTER 4 THE SCHRODHV GER EQUATION IN THREE DIMENSIONS We would like to nd the eigenfunctions and eigenvalues of the various L operators The rst order of business is to nd the commutation relations between them to see how many of them are compatible as discussed in Chap 3 One can show L1 Ly i Lz Ly L1 MLm L1 Lm ihLy This shows that the three components of the L operator are fit not compatible and we can only get eigenfunctions of one of them for the stationary states However7 Lin 0 LiLy 0 Win 0 so that L2 is compatible with any one component so we can possibly nd simultaneous eigen functions of both L2 and L1 Now we recall that for the solution of the QM harmonic oscillator there were two approaches one using just the operator language algebraic and the other using a normal differential equation analytic We will use the same kinds of approaches here 471 Algebraic Approach to Angular Momentum De ne what we later justify as ladder operators Li E Lm i iLy 411 First off7 one can show the following Suppose f is an eigenfunction of of L2 with eigenvalue and is also an eigenfunction of L1 with eigenvalue ILL Then 0 Lif is also an eigenfunction of L2 with eigenvalue o Lif is also an eigenfunction of L1 with eigenvalue ILL i h This property justi es the names ladder operator we note that its application changes Lz eigenvalues within the same L2 eigenvalue One can show that the eigenvalues of L2 are 52M 1 and the eigenvalues of L1 are m where m goes from 76 to Z in integer steps Thus we denote the eigenfunctions by ff with L217quot 52M 1 szlm hm quot The number 6 can be any integer of half integer that is7 012 And we will see that even though the integers apply to the spatial motion of the electron in the H atom7 the half integer values also have a role in physics 4 7 ANG ULAR MOMENTUM 472 Analytic Approach to Angular Momentum By making operators out of the classical expression L r x p7 we can show that the operator for is h A 8 A 1 8 L 7 i 7 97 i lt 819 sin 0 13 and the the Lz and ladder operators are 7 h 8 7 int 8 4 8 L1 7 Li 7 i e izcot0ggt and the L2 operator is 1 8 8 1 82 L2 in if of 7 111980 lt81 80 1 190812 As this is the same as the operator in the angular azimuthal equation7 we conclude that the spherical harmonics are the eigenfunctions of L2 and L1 473 Spin 1 5y msz 52 51 msm 5151 155 412 Szlsmgt zss1lsmgt Sz smgt m smgt SHsm xss17mmi1 s m XltZgtaXbx where XltC1gt and XltC1gt 2732 10 in 1 0 8 1h 01 81 0 71 h 01 h 0 72 h 1 0 Sm lt1 0 8y ltz 0 SZ lt0 71 7 01 0 72 7 1 0 W 10 7 1 0 01 0 71 ltzgt1 1 z 1 1 1 X 1 474 Electron in a Magnetic Field 1 I iiEthh B Bok H VBOSZ X05 aX5 Eth bxieilE th lt ZZJEJh gt CHAPTER 4 THE SCHRODHV GER EQUATION IN THREE DIMENSIONS

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.