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Solutions to Problems in Quantum Mechanics P. Saltsidis, additions by B. Brinne

1995,1999

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Most of the problems presented here are taken from the book Sakurai, J. J., Modern Quantum Mechanics, Reading, MA: Addison-Wesley, 1985.

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Contents I Problems 1 2 3 4 5

Fundamental Concepts . . . . . . . Quantum Dynamics . . . . . . . . . Theory of Angular Momentum . . . Symmetry in Quantum Mechanics . Approximation Methods . . . . . .

II Solutions 1 2 3 4 5

Fundamental Concepts . . . . . . . Quantum Dynamics . . . . . . . . . Theory of Angular Momentum . . . Symmetry in Quantum Mechanics . Approximation Methods . . . . . .

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CONTENTS

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Part I Problems

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1. FUNDAMENTAL CONCEPTS

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1 Fundamental Concepts

1.1 Consider a ket space spanned by the eigenkets fja0ig of a Hermitian operator A. There is no degeneracy. (a) Prove that Y (A ; a0) a0

is a null operator. (b) What is the signi cance of Y (A ; a00) 0 00 ? a00 6=a0 a ; a

(c) Illustrate (a) and (b) using A set equal to Sz of a spin 21 system. 1.2 A spin 12 system is known to be in an eigenstate of S~
n^ with eigenvalue h=2, where n^ is a unit vector lying in the xz-plane that makes an angle with the positive z-axis. (a) Suppose Sx is measured. What is the probability of getting +h=2? (b) Evaluate the dispersion in Sx, that is, h(Sx ; hSx i)2i: (For your own peace of mind check your answers for the special cases = 0, =2, and .) 1.3 (a) The simplest way to derive the Schwarz inequality goes as follows. First observe (hj + h j)
(ji + j i)  0 for any complex number ; then choose  in such a way that the preceding inequality reduces to the Schwarz inequility.

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(b) Show that the equility sign in the generalized uncertainty relation holds if the state in question satis es
Aji = 
B ji with  purely imaginary. (c) Explicit calculations using the usual rules of wave mechanics show that the wave function for a Gaussian wave packet given by " 0 (x0 ; hxi)2 # i h p i x 0 2 ; 1 = 4 ; hx ji = (2d ) exp h

4d2

satis es the uncertainty relation

q

q

h(
x)2i h(
p)2i = h2 :

Prove that the requirement hx0j
xji = (imaginary number)hx0j
pji is indeed satis ed for such a Gaussian wave packet, in agreement with (b). 1.4 (a) Let x and px be the coordinate and linear momentum in one dimension. Evaluate the classical Poisson bracket [x; F (px)]classical :

(b) Let x and px be the corresponding quantum-mechanical operators this time. Evaluate the commutator   x; exp iphxa :



(c) Using the result obtained in (b), prove that

 ipxa  exp h jx0i; (xjx0i = x0jx0i)

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2. QUANTUM DYNAMICS

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is an eigenstate of the coordinate operator x. What is the corresponding eigenvalue? 1.5 (a) Prove the following: (i) hp0jxji = ih @p@ 0 hp0ji; Z @  (p0); (ii) h jxji = dp0 (p0)ih @p 0  where (p0) = hp0 ji and  (p0) = hp0j i are momentum-space wave functions. (b) What is the physical signi cance of  ix  exp h ;

where x is the position operator and  is some number with the dimension of momentum? Justify your answer.

2 Quantum Dynamics

2.1 Consider the spin-procession problem discussed in section 2.1 in Jackson. It can also be solved in the Heisenberg picture. Using the Hamiltonian  eB  H = ; mc Sz = !Sz ; write the Heisenberg equations of motion for the time-dependent operators Sx(t), Sy (t), and Sz (t). Solve them to obtain Sx;y;z as functions of time. 2.2 Let x(t) be the coordinate operator for a free particle in one dimension in the Heisenberg picture. Evaluate [x(t); x(0)] :

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2.3 Consider a particle in three dimensions whose Hamiltonian is given by 2 H = 2p~m + V (~x):

By calculating [~x
p~; H ] obtain

* + d h~x
~pi = p2 ; h~x
r ~ V i: dt m

To identify the preceding relation with the quantum-mechanical analogue of the virial theorem it is essential that the left-hand side vanish. Under what condition would this happen? 2.4 (a) Write down the wave function (in coordinate space) for the state   exp ;hipa j0i: You may use 2 !23 0 !1=21 0 h  x hx0j0i = ;1=4x;0 1=2 exp 4; 21 x 5 ; @x0  m! A : 0

(b) Obtain a simple expression that the probability that the state is found in the ground state at t = 0. Does this probability change for t > 0? 2.5 Consider a function, known as the correlation function, de ned by C (t) = hx(t)x(0)i; where x(t) is the position operator in the Heisenberg picture. Evaluate the correlation function explicitly for the ground state of a one-dimensional simple harmonic oscillator.

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2. QUANTUM DYNAMICS

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2.6 Consider again a one-dimensional simple harmonic oscillator. Do the following algebraically, that is, without using wave functions. (a) Construct a linear combination of j0i and j1i such that hxi is as large as possible. (b) Suppose the oscillator is in the state constructed in (a) at t = 0. What is the state vector for t > 0 in the Schrodinger picture? Evaluate the expectation value hxi as a function of time for t > 0 using (i) the Schrodinger picture and (ii) the Heisenberg picture. (c) Evaluate h(
x)2i as a function of time using either picture. 2.7 A coherent state of a one-dimensional simple harmonic oscillator is de ned to be an eigenstate of the (non-Hermitian) annihilation operator a: aji = ji; where  is, in general, a complex number. (a) Prove that ji = e;jj =2eay j0i is a normalized coherent state. (b) Prove the minimum uncertainty relation for such a state. (c) Write ji as 1 X ji = f (n)jni: 2

n=0

Show that the distribution of jf (n)j2 with respect to n is of the Poisson form. Find the most probable value of n, hence of E . (d) Show that a coherent state can also be obtained by applying the translation ( nite-displacement) operator e;ipl=h (where p is the momentum operator, and l is the displacement distance) to the ground state.

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(e) Show that the coherent state ji remains coherent under timeevolution and calculate the time-evolved state j(t)i. (Hint: directly apply the time-evolution operator.) 2.8 The quntum mechanical propagator, for a particle with mass m, moving in a potential is given by: K (x; y; E ) =

Z1 0

dteiEt=hK (x; y; t; 0) = A

X sin(nrx) sin(nry) E ; h22mr2 n2 n

where A is a constant. (a) What is the potential? (b) Determine the constant A in terms of the parameters describing the system (such as m, r etc. ). 2.9 Prove the relation

d(x) = (x) dx

where (x) is the (unit) step function, and (x) the Dirac delta function. (Hint: study the eect on testfunctions.) 2.10 Derive the following expression

m! h(x2 + 2x2 ) cos(!T ) ; x x i Scl = 2 sin( 0 T T !T ) 0

for the classical action for a harmonic oscillator moving from the point x0 at t = 0 to the point xT at t = T . 2.11 The Lagrangian of the single harmonic oscillator is L = 21 mx_ 2 ; 12 m!2x2

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2. QUANTUM DYNAMICS

(a) Show that

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  hxbtbjxatai = exp iShcl G(0; tb; 0; ta)

where Scl is the action along the classical path xcl from (xa; ta) to (xb; tb) and G is G(0; tb; 0; ta) =  m  (N2+1) 8 9 Z N m 0):

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(a) Is the energy spectrum continuous or discrete? Write down an approximate expression for the energy eigenfunction speci ed by E. (b) Discuss brie y what changes are needed if V is replaced be V = jxj:

3 Theory of Angular Momentum

3.1 Consider a sequence of Euler rotations represented by !    ;i3  ; i ; i 2 3 (1 = 2) exp exp D (; ; ) = exp 2

=

e;i(+ )=2 cos 2 ei(; )=2 sin 2

2

;e;i(; )=2 sin 2 ei(+ )=2 cos 2

2 !

:

Because of the group properties of rotations, we expect that this sequence of operations is equivalent to a single rotation about some axis by an angle . Find . 3.2 An angular-momentum eigenstate jj; m = mmax = j i is rotated by an in nitesimal angle " about the y-axis. Without using the (j ) explicit form of the dm0m function, obtain an expression for the probability for the new rotated state to be found in the original state up to terms of order "2. 3.3 The wave function of a patricle subjected to a spherically symmetrical potential V (r) is given by (~x) = (x + y + 3z)f (r):

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3. THEORY OF ANGULAR MOMENTUM

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(a) Is an eigenfunction of L~ ? If so, what is the l-value? If not, what are the possible values of l we may obtain when L~ 2 is measured? (b) What are the probabilities for the particle to be found in various ml states? (c) Suppose it is known somehow that (~x) is an energy eigenfunction with eigenvalue E . Indicate how we may nd V (r). 3.4 Consider a particle with an intrinsic angular momentum (or spin) of one unit of h. (One example of such a particle is the %meson). Quantum-mechanically, such a particle is described by a ketvector j%i or in ~x representation a wave function %i(~x) = h~x; ij%i where j~x; ii correspond to a particle at ~x with spin in the i:th direction. (a) Show explicitly that in nitesimal rotations of %i(~x) are obtained by acting with the operator u~" = 1 ; i h~"
(L~ + S~)

(3.1)

where L~ = hi r^  r~ . Determine S~ ! (b) Show that L~ and S~ commute. (c) Show that S~ is a vector operator. (d) Show that r~  %~(~x) = h1 (S~
~p)%~ where p~ is the momentum operator. 2

3.5 We are to add angular momenta j1 = 1 and j2 = 1 to form j = 2; 1; and 0 states. Using the ladder operator method express all

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(nine) j; m eigenkets in terms of jj1j2; m1m2i. Write your answer as jj = 1; m = 1i = p1 j+; 0i ; p1 j0; +i; : : : ;

2 2 where + and 0 stand for m1;2 = 1; 0; respectively.

(3.2)

3.6 (a) Construct a spherical tensor of rank 1 out of two dierent vectors U~ = (Ux; Uy ; Uz ) and V~ = (Vx ; Vy ; Vz ). Explicitly write T(1)1;0 in terms of Ux;y;z and Vx;y;z . (b) Construct a spherical tensor of rank 2 out of two dierent vectors U~ and V~ . Write down explicitly T(2)2;1;0 in terms of Ux;y;z and Vx;y;z . 3.7 (a) Evaluate

j X m=;j

j) jd(mm 0 ( )j2m

for any j (integer or half-integer); then check your answer for j = 21 . (b) Prove, for any j , j X m=;j

m2jdm(j)0 m( )j2 = 12 j (j + 1) sin + m02 + 12 (3 cos2 ; 1):

[Hint: This can be proved in many ways. You may, for instance, examine the rotational properties of Jz2 using the spherical (irreducible) tensor language.] 3.8 (a) Write xy, xz, and (x2 ; y2) as components of a spherical (irreducible) tensor of rank 2.

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4. SYMMETRY IN QUANTUM MECHANICS

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(b) The expectation value Q  eh; j; m = j j(3z2 ; r2)j; j; m = j i is known as the quadrupole moment. Evaluate eh; j; m0j(x2 ; y2)j; j; m = j i; (where m0 = j; j ; 1; j ; 2; : : : )in terms of Q and appropriate ClebschGordan coecients.

4 Symmetry in Quantum Mechanics

4.1 (a) Assuming that the Hamiltonian is invariant under time reversal, prove that the wave function for a spinless nondegenerate system at any given instant of time can always be chosen to be real. (b) The wave function for a plane-wave state at t = 0 is given by a complex function ei~p
~x=h. Why does this not violate time-reversal invariance? 4.2 Let (~p0) be the momentum-space wave function for state ji, that is, (~p0) = h~p0 ji.Is the momentum-space wave function for the time-reversed state ji given by (~p0, (;~p0), (~p0), or (;~p0)? Justify your answer. 4.3 Read section 4.3 in Sakurai to refresh your knowledge of the quantum mechanics of periodic potentials. You know that the energybands in solids are described by the so called Bloch functions n;k full lling, ika n;k (x + a) = e n;k (x)

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where a is the lattice constant, n labels the band, and the lattice momentum k is restricted to the Brillouin zone [;=a; =a]. Prove that any Bloch function can be written as, X n(x ; Ri)eikRi n;k (x) = Ri

where the sum is over all lattice vectors Ri. (In this simble one dimensional problem Ri = ia, but the construction generalizes easily to three dimensions.). The functions n are called Wannier functions, and are important in the tight-binding description of solids. Show that the Wannier functions are corresponding to dierent sites and/or dierent bands are orthogonal, i:e: prove Z dx?m(x ; Ri)n(x ; Rj )  ij mn Hint: Expand the n s in Bloch functions and use their orthonormality properties. 4.4 Suppose a spinless particle is bound to a xed center by a potential V (~x) so assymetrical that no energy level is degenerate. Using the time-reversal invariance prove hL~ i = 0 for any energy eigenstate. (This is known as quenching of orbital angular momemtum.) If the wave function of such a nondegenerate eigenstate is expanded as XX l m

Flm(r)Ylm (; );

what kind of phase restrictions do we obtain on Flm(r)? 4.5 The Hamiltonian for a spin 1 system is given by H = ASz2 + B (Sx2 ; Sy2):

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5. APPROXIMATION METHODS

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Solve this problem exactly to nd the normalized energy eigenstates and eigenvalues. (A spin-dependent Hamiltonian of this kind actually appears in crystal physics.) Is this Hamiltonian invariant under time reversal? How do the normalized eigenstates you obtained transform under time reversal?

5 Approximation Methods

5.1 Consider an isotropic harmonic oscillator in two dimensions. The Hamiltonian is given by 2 p2y m!2 2 2 p x H0 = 2m + 2m + 2 (x + y )

(a) What are the energies of the three lowest-lying states? Is there any degeneracy? (b) We now apply a perturbation V = m!2xy

where  is a dimensionless real number much smaller than unity. Find the zeroth-order energy eigenket and the corresponding energy to rst order [that is the unperturbed energy obtained in (a) plus the rst-order energy shift] for each of the three lowest-lying states. (c) Solve the H0 + V problem exactly. Compare with the perturbation results obtained in q(b). p p [You may use hn0jxjni = h=2m! ( n + 1 0 + n 0 ):] n ;n+1

n ;n;1

5.2 A system that has three unperturbed states can be represented by the perturbed Hamiltonian matrix 0 1 E1 0 a B@ 0 E1 b CA a b E2

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where E2 > E1. The quantities a and b are to be regarded as perturbations that are of the same order and are small compared with E2 ; E1. Use the second-order nondegenerate perturbation theory to calculate the perturbed eigenvalues. (Is this procedure correct?) Then diagonalize the matrix to nd the exact eigenvalues. Finally, use the second-order degenerate perturbation theory. Compare the three results obtained. 5.3 A one-dimensional harmonic oscillator is in its ground state for t < 0. For t  0 it is subjected to a time-dependent but spatially uniform force (not potential!) in the x-direction, F (t) = F0e;t=

(a) Using time-dependent perturbation theory to rst order, obtain the probability of nding the oscillator in its rst excited state for t > 0). Show that the t ! 1 ( nite) limit of your expression is independent of time. Is this reasonable or surprising? (b) Can we nd higher states? q excited p p 0 [You may use hn jxjni = h=2m! ( n + 1n0;n+1 + nn0;n;1 ):] 5.4 Consider a composite system made up of two spin 21 objects. for t < 0, the Hamiltonian does not depend on spin and can be taken to be zero by suitably adjusting the energy scale. For t > 0, the Hamiltonian is given by

 4
 H = 2 S~1
S~2: h Suppose the system is in j + ;i for t  0. Find, as a function of

time, the probability for being found in each of the following states j + +i, j + ;i, j ; +i, j ; ;i: (a) By solving the problem exactly.

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5. APPROXIMATION METHODS

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(b) By solving the problem assuming the validity of rst-order time-dependent perturbation theory with H as a perturbation switched on at t = 0. Under what condition does (b) give the correct results? 5.5 The ground state of a hydrogen atom (n = 1,l = 0) is subjected to a time-dependent potential as follows: V (~x; t) = V0cos(kz ; !t): Using time-dependent perturbation theory, obtain an expression for the transition rate at which the electron is emitted with momentum p~. Show, in particular, how you may compute the angular distribution of the ejected electron (in terms of  and  de ned with respect to the z-axis). Discuss brie y the similarities and the dierences between this problem and the (more realistic) photoelectric eect. (note: For the initial wave function use  Z 2 1 ;Zr=a0 : n=1;l=0(~x) = p e  a0 3

If you have a normalization problem, the nal wave function may be taken to be 1 f (~x) =

L

3 2

ei~p
~x=h

with L very large, but you should be able to show that the observable eects are independent of L.)

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Part II Solutions

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1. FUNDAMENTAL CONCEPTS

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1 Fundamental Concepts 1.1 Consider a ket space spanned by the eigenkets fja0ig of a Hermitian operator A. There is no degeneracy. (a) Prove that Y (A ; a0) a0

is a null operator. (b) What is the signi cance of Y (A ; a00) 0 00 ? a00 6=a0 a ; a

(c) Illustrate (a) and (b) using A set equal to Sz of a spin 21 system. (a) Assume that ji is an arbitrary state ket. Then X Y Y Y X (A ; a0)ji = (A ; a0) ja00i h|a00{zji} = ca00 (A ; a0)ja00i a0 a0 a00 a00 a0 ca00 X Y 00 0 00 a002fa0g = ca00 (a ; a )ja i = 0: (1.1) a00

a0

(b) Again for an arbitrary state ji we will have 2 3 2 3 1 00) 00) zX }| { Y Y ( A ; a ( A ; a 4 5 4 5 ja000iha000 ji a0 ; a00 ji = a0 ; a00 a00 6=a0

a000 X 000 Y (a000 ; a00) 000 ha ji 0 00 ja i = a000 a00 = 6 a0 a ; a X 000 ha jia000a0 ja000i = ha0jija0i ) a000 a00 6=a0

= =

2 3 00) Y ( A ; a 4 5 = ja0iha0j  a0 : 0 ; a00 a 00 0 a 6=a So it projects to the eigenket ja0i.

(1.2)

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26 (c) It is Sz  h=2(j+ih+j;j;ih;j). This operator has eigenkets j+i and j;i with eigenvalues h=2 and -h=2 respectively. So Y Y (Sz ; a0) = (Sz ; a01) a0 "a0 # h  h  = 2 (j+ih+j ; j;ih;j) ; 2 (j+ih+j + j;ih;j) " # h  h   2 (j+ih+j ; j;ih;j) + 2 (j+ih+j + j;ih;j)

0 z }| { = [;hj;ih;j][hj+ih+j] = ;h2j;i h;j+ih+j = 0; (1.3) where we have used that j+ih+j + j;ih;j = 1. For a0 = h=2 we have Y (Sz ; a001) Sz + h2 1 Y (Sz ; a00) 0 00 = 00 00 = h=2 + h=2 a00 6=a0 a ; a a 6=h=2 h=2 ; a " # 1 h  h  = h 2 (j+ih+j ; j;ih;j) + 2 (j+ih+j + j;ih;j) = h1 hj+ih+j = j+ih+j: (1.4)

Similarly for a0 = ;h=2 we have Y (Sz ; a001) Y (Sz ; a00) Sz ; h2 1 = = 0 00 00 ;h=2 ; h=2 a00 6=a0 a ; a a00 6=;h=2 ;h=2 ; a " # 1 h  h  = ; h 2 (j+ih+j ; j;ih;j) ; 2 (j+ih+j + j;ih;j) = ; h1 (;hj;ih;j) = j;ih;j: (1.5)

1.2 A spin 12 system is known to be in an eigenstate of S~
n^ with eigenvalue h=2, where n^ is a unit vector lying in the xz-plane that makes an angle with the positive z-axis. (a) Suppose Sx is measured. What is the probability of getting +h=2?

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1. FUNDAMENTAL CONCEPTS

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(b) Evaluate the dispersion in Sx, that is, h(Sx ; hSx i)2i: (For your own peace of mind check your answers for the special cases = 0, =2, and .) Since the unit vector n^ makes an angle with the positive z-axis and is lying in the xz-plane, it can be written in the following way n^ = e^z cos + e^x sin (1.6) So S~
n^ = S" z cos + Sx sin = # [(S-1.3.36),(S-1.4.18)] " # h  h  = 2 (j+ih+j ; j;ih;j) cos + 2 (j+ih;j + j;ih+j) sin :(1.7)

Since the system is in an eigenstate of S~
n^ with eigenvalue h=2 it has to satisfay the following equation S~
n^ jS~
n^ ; +i = h =2jS~
n^ ; +i: (1.8) From (1.7) we have that ! ~S
n^ =
h cos sin : (1.9) 2 sin ; cos The eigenvalues and eigenfuncions of this operator can be found if one solves the secular equation ! h  = 2 cos

;  h  = 2 sin

det(S~
n^ ; I ) = 0 ) det h=2 sin ;h=2 cos ;  = 0 ) 2 2 2 ; h4 cos2 + 2 ; h4 sin2 = 0 ) 2 ; h4 = 0 )  =  h2 : (1.10) ! a ~ Since the system is in the eigenstate jS
n^; +i  b we will have that ! ! ! ( ) h cos sin a = h a ) a cos + b sin = a ) b a sin ; b cos = b 2 sin ; cos 2 b

= a 2 sin2 2 = a tan : b = a 1 ;sincos (1.11)

2 sin 2 cos 2 2

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28 But we want also the eigenstate jS~
n^ ; +i to be normalized, that is a2 + b2 = 1 ) a2 + a2 tan2 2 = 1 ) a2 cos2 2 a2 sin2 2 = cos2 2 r

2 2 ) a = cos 2 ) a =  cos2 2 = cos 2 ; (1.12) where the real positive convention has been used in the last step. This means that the state in which the system is in, is given in terms of the eigenstates of the Sz operator by (1.13) jS~
n^ ; +i = cos 2 j+i + sin 2 j;i: (a) From (S-1.4.17) we know that jSx; +i = p12 j+i + p12 j;i: (1.14) So the propability of getting +h=2 when Sx is measured is given by !  2 2 1

1

hSx; +jS~
n^ ; +i = p h+j + p h;j cos 2 j+i + sin 2 j;i 2 2 2 1 1

= p cos 2 + p sin 2 2 2 1

1 = 2 cos2 2 + 2 sin2 2 + cos 2 sin 2 = 12 + 12 sin = 12 (1 + sin ): (1.15) For = 0 which means that the system is in the jSz ; +i eigenstate we have

jhSx ; +jSz ; +ij2 = 21 (1) = 12 :

(1.16)

For = =2 which means that the system is in the jSx; +i eigenstate we have

jhSx; +jSx; +ij2 = 1:

(1.17)

For =  which means that the system is in the jSz ; ;i eigenstate we have

jhSx; +jSz ; ;ij2 = 21 (1) = 21 :

(1.18)

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1. FUNDAMENTAL CONCEPTS

29

(b) We have that

h(Sx ; hSxi)2i = hSx2i ; (hSx i)2:

(1.19)

As we know

Sx = h2 (j+ih;j + j;ih+j) ) 2 Sx2 = h4 (j+ih;j + j;ih+j) (j+ih;j + j;ih+j) ) 2 2 h  h  2 Sx = 4 (|j+ih+j + {z j;ih;j)} = 4 : 1

So

hSx i = = (hSxi)2 =

hSx2i = =

(1.20)



 h  

cos 2 h+j + sin 2 h;j 2 (j+ih;j + j;ih+j) cos 2 j+i + sin 2 j;i h cos sin + h sin cos = h sin ) 2 2 2 2 2 2 2 h2 sin2 and  h2   4

cos 2 h+j + sin 2 h;j 4 cos 2 j+i + sin 2 j;i h2 [cos2 + sin2 ] = h2 : (1.21) 4 2 2 4

So substituting in (1.19) we will have 2

2

h(Sx ; hSxi)2i = h4 (1 ; sin2 ) = h4 cos2 :

(1.22)

and nally 2

h(
Sx)2i =0;jSz ;+i = h4 ; h(
Sx)2i ==2;jSx ;+i = 0; 2 h(
Sx)2i =0;jSz ;;i = h4 :

(1.23) (1.24) (1.25)

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30

1.3 (a) The simplest way to derive the Schwarz inequality goes as follows. First observe (hj + h j)
(ji + j i)  0 for any complex number ; then choose  in such a way that the preceding inequality reduces to the Schwarz inequility. (b) Show that the equility sign in the generalized uncertainty relation holds if the state in question satis es
Aji = 
B ji with  purely imaginary. (c) Explicit calculations using the usual rules of wave mechanics show that the wave function for a Gaussian wave packet given by " 0 (x0 ; hxi)2 # i h p i x 0 2 ; 1 = 4 ; hx ji = (2d ) exp satis es the uncertainty relation q

h

4d2

q

h(
x)2i

h(
p)2i = h2 :

Prove that the requirement hx0j
xji = (imaginary number)hx0j
pji is indeed satis ed for such a Gaussian wave packet, in agreement with (b). (a) We know that for an arbitrary state jci the following relation holds hcjci  0: (1.26) This means that if we choose jci = ji + j i where  is a complex number, we will have (hj + h j)
(ji + j i)  0 ) (1.27)  2 hji + hj i +  h ji + jj h j i  0: (1.28)

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1. FUNDAMENTAL CONCEPTS

31

If we now choose  = ;h ji=h j i the previous relation will be

hji ; h jh ihj ij i ; h jh ihj ij i + jhh jj iji  0 ) hjih j i  jh jij2: 2

(1.29)

Notice that the equality sign in the last relation holds when

jci = ji + j i = 0 ) ji = ;j i that is if ji and j i are colinear.

(1.30)

(b) The uncertainty relation is

h(
A)2ih(
B )2i  14 jh[A; B ]ij2 :

(1.31)

h(
A)2ih(
B )2i  jh
A
B ij2:

(1.32)

To prove this relation we use the Schwarz inequality (1.29) for the vectors ji =
Ajai and j i =
B jai which gives The equality sign in this relation holds according to (1.30) when
Ajai = 
B jai:

(1.33)

On the other hand the right-hand side of (1.32) is (1.34) jh
A
B ij2 = 41 jh[A; B ]ij2 + 14 jhf
A;
B gij2 which means that the equality sign in the uncertainty relation (1.31) holds if 1 jhf
A;
B gij2 = 0 ) hf
A;
B gi = 0 4 ) haj
A
B +
B
Ajai = 0 (1):33) haj(
B )2jai + haj(
B )2jai = 0 ) ( + )haj(
B )2jai = 0: (1.35) Thus the equality sign in the uncertainty relation holds when
Ajai = 
B jai with  purely imaginary.

(1.36)

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32 (c) We have

hx0j
xji  hx0j(x ; hxi)ji = x0hx0ji ; hxihx0ji = (x0 ; hxi)hx0ji:

(1.37)

On the other hand

But

hx0j
pji  hx0j(p ; hpi)ji @ hx0ji ; hpihx0 ji = ;ih @x 0

" # @ hx0ji = hx0ji @ ihpix0 ; (x0 ; hxi)2 0 @x0 @x h 4d2 # " = hx0ji ihhpi ; 21d2 (x0 ; hxi)

(1.38)

(1.39)

So substituting in (1.38) we have

hx0j
pji = hpihx0 ji + 2idh2 (x0 ; hxi) hx0ji ; hpihx0 ji

= 2idh2 (x0 ; hxi) hx0ji = 2idh2 hx0j
xji ) 2 ; i 2 d 0 hx j
xji = h hx0j
pji:

(1.40)

1.4 (a) Let x and px be the coordinate and linear momentum in one dimension. Evaluate the classical Poisson bracket [x; F (px)]classical :

(b) Let x and px be the corresponding quantum-mechanical operators this time. Evaluate the commutator  ipxa  x; exp h :



(c) Using the result obtained in (b), prove that

 ipxa  exp h jx0i; (xjx0i = x0jx0i)

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1. FUNDAMENTAL CONCEPTS

33

is an eigenstate of the coordinate operator x. What is the corresponding eigenvalue? (a) We have

@x @F (px) ; @x @F (px) [x; F (px)]classical  @x @px @px @x = @F (px) : @px

(1.41)

(b) When x and px are treated as quantum-mechanical operators we have " X   ipxa  1 (ia)n pn # X 1 1 (ia)n x n] x; exp h = x; = [ x; p n n x n=0 h n! n=0 n! h ;1 1 1 (ia)n nX X k [x; p ] pn;k;1 p = x x x n n=0 n! h k=0 1 1 (ia)n nX ;1 1 n (ia)n;1 X X n;1 (;a) k n ; k ; 1 = p ( i h  ) p p = x x x n n ; 1 n=1 n! h n=1 n! h k=0  n;1  ipxa  1 X = ;a (n ;1 1)! ia p = ; a exp h x h : (1.42) n=1 (c) We have now   ipxa   ipxa   ipxa  (b) 0 0 x exp h jx i = exp h xjx i ; a exp h jx0i     = x0 exp iphxa jx0i ; a exp iphxa jx0i   = (x0 ; a) exp iphxa jx0i: (1.43)   So exp iphxa jx0i is an eigenstate of the operator x with eigenvalue x0 ; a. So we can write  ipxa  0 (1.44) jx ; ai = C exp h jx0i; where C is a constant which due to normalization can be taken to be 1.

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34

1.5 (a) Prove the following: (i) hp0 jxji = ih @ 0 hp0ji; Z @p @  (p0); (ii) h jxji = dp0 (p0)ih @p 0 

where (p0) = hp0 ji and  (p0) = hp0j i are momentum-space wave functions. (b) What is the physical signi cance of  ix  exp h ;

where x is the position operator and  is some number with the dimension of momentum? Justify your answer. (a) We have (i)

hp0 jxji = = = =

hp0 jxji = (ii)

}|1 { Z hp0 jx dx0jx0ihx0j i = dx0hp0 jxjx0ihx0 ji Z Z ip0 x0 dx0x0hp0jx0ihx0ji (S;1=:7:32) dx0x0Ae; h hx0ji   ip0x0  Z Z ip0 x0 A dx0 @p@ 0 e; h (ih)hx0ji = ih @p@ 0 dx0Ae; h hx0ji Z  @ 0 0 0 0 ih @p0 dx hp jx ihx ji = ih @p@ 0 hp0 ji ) ih @p@ 0 hp0ji: (1.45) zZ

Z

Z

h jxji = dp0 h jp0ihp0 jxji = dp0 (p0)ih @p@ 0 (p0);

(1.46)

where we have used (1.45) and that h jp0i =  (p0) and hp0ji = (p0).

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1. FUNDAMENTAL CONCEPTS  ix 

35

(b) The operator exp h gives translation in momentum space. This can be justi ed by calculating the following operator " X   ix  1 1  ix n # X 1 1  i n p; exp h = p; n! h = n! h [p; xn ] n=0 n=0 n 1 1  i n X X xn;k [p; x]xk;1 = n ! h  n=1 k=1   n 1 n 1 1  i n X 1 i X X n ; 1 = (;ih)x = n! h n(;ih)xn;1 n ! h  n=1 n=1 k=1    i  1 n ; 1 1 1  ix n X X 1 i  n ; 1 = x (;ih) h =  n! h n=1 (n ; 1)! h n=0  ix  =  exp h : (1.47) So when this commutator acts on an eigenstate jp0i of the momentum operator we will have   ix    ix     ix  0 p; exp h jp i = p exp h jp0i ; exp h p0jp0i )   ix     ix   ix   exp h = p exp h jp0i ; p0 exp h jp0i )         (1.48) p exp ixh jp0i = (p0 + ) exp ixh jp0i : Thus we have that

  exp ixh jp0i  Ajp0 + i; (1.49) where A is a constant which due to normalization can be taken to be 1.

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36

2 Quantum Dynamics

2.1 Consider the spin-procession problem discussed in section 2.1 in Jackson. It can also be solved in the Heisenberg picture. Using the Hamiltonian  eB  H = ; mc Sz = !Sz ; write the Heisenberg equations of motion for the time-dependent operators Sx(t), Sy (t), and Sz (t). Solve them to obtain Sx;y;z as functions of time. Let us rst prove the following [AS ; BS ] = CS ) [AH ; BH ] = CH : (2.1) Indeed we have h i [AH ; BH ] = U yAS U ; U yBS U = U yAS BS U ; U yBS AS U = U y [AS ; BS ] U = U yCS U = CH : (2.2) The Heisenberg equation of motion gives dSx = 1 [S ; H ] = 1 [S ; !S ] (S;1=:4:20) ! (;ihS ) = ;!S ; (2.3) y y dt ih x ih x z ih dSy = 1 [S ; H ] = 1 [S ; !S ] (S;1=:4:20) ! (ihS ) = !S ; (2.4) x dt ih y ih y z ih x dSz = 1 [S ; H ] = 1 [S ; !S ] (S;1=:4:20) 0 ) S = constant: (2.5) z dt ih z ih z z Dierentiating once more eqs. (2.3) and (2.4) we get d2Sx = ;! dSy (2=:4) ;!2S ) S (t) = A cos !t + B sin !t ) S (0) = A x x x dt2 dt d2Sy = ! dSx (2=:3) ;!2S ) S (t) = C cos !t + D sin !t ) S (0) = C: y y y dt2 dt But on the other hand dSx = ;!S ) y dt ;A! sin !t + B! cos !t = ;C! cos !t ; D! sin !t ) A=D C = ;B: (2.6)

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2. QUANTUM DYNAMICS So, nally

37

Sx(t) = Sx(0) cos !t ; Sy (0) sin !t Sy (t) = Sy (0) cos !t + Sx(0) sin !t Sz (t) = Sz (0):

(2.7) (2.8) (2.9)

2.2 Let x(t) be the coordinate operator for a free particle in one dimension in the Heisenberg picture. Evaluate [x(t); x(0)] :

The Hamiltonian for a free particle in one dimension is given by 2 H = 2pm : (2.10) This means that the Heisenberg equations of motion for the operators x and p will be " # @p(t) = 1 [p(t); H (t)] = 1 p(t); p2(t) = 0 ) @t ih ih 2m p(t) = p(0) (2.11) " # 2 @x(t) = 1 [x; H ] = 1 x(t); p (t) = 1 2p(t)ih = p(t) (2=:11) p(0) ) @t ih ih 2m 2mih m m x(t) = mt p(0) + x(0): (2.12) Thus nally   ht : (2.13) [x(t); x(0)] = mt p(0) + x(0); x(0) = mt [p(0); x(0)] = ; im

2.3 Consider a particle in three dimensions whose Hamiltonian is given by 2 H = 2~pm + V (~x):

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38

By calculating [~x
p~; H ] obtain * + d h~x
~pi = p2 ; h~x
r ~ V i: dt m To identify the preceding relation with the quantum-mechanical analogue of the virial theorem it is essential that the left-hand side vanish. Under what condition would this happen? Let us rst calculate the commutator [~x
~p; H ] 3 " # 2X 3 3 p2 2 X p ~ [~x
~p; H ] = ~x
~p; 2m + V (~x) = 4 xipi; 2mj + V (~x)5 i=1 j =1 # " X X p2 (2.14) = xi; 2mj pi + xi [pi ; V (~x)] : i ij The rst commutator in (2.14) will give " 2# p xi; 2mj = 21m [xi; p2j ] = 21m (pj [xi; pj ] + [xi; pj ]pj ) = 21m (pj ihij + ihij pj ) = 21m 2ihij pj = imh ij pj : (2.15) The second commutator can be calculated if we Taylor expand the function P V (~x) in terms of xi which means that we take V (~x) = n anxni with an independent of xi. So " X # X 1 ;1 X nX [pi; V (~x)] = pi; anxni = an [pi ; xni] = an xki [pi ; xi] xni ;k;1 n=0

n

n

k=0

;1 X nX X @ X a xn = an (;ih)xni ;1 = ;ih annxni ;1 = ;ih @x n i i n n n k=0 @ V (~x): = ;ih @x (2.16) i The right-hand side of (2.14) now becomes X ih X @ V (~x) [~x
~p; H ] =  ij pj pi + (;ih)xi @xi ij m i ~ V (~x): = imh p~2 ; ih~x
r (2.17)

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2. QUANTUM DYNAMICS The Heisenberg equation of motion gives d ~x
~p = 1 [~x
~p; H ] (2=:17) ~p2 ; ~x
r ~ V (~x) ) dt i*h + m d h~x
p~i = p2 ; h~x
r ~ V i; dt m

39

(2.18)

where in the last step we used the fact that the state kets in the Heisenberg picture are independent of time. The left-hand side of the last equation vanishes for a stationary state. Indeed we have d hnj~x
p~jni = 1 hnj [~x
~p; H ] jni = 1 (E hnj~x
p~jni ; E hnj~x
~pjni) = 0: n dt ih ih n So to have the quantum-mechanical analogue of the virial theorem we can take the expectation values with respect to a stationaru state.

2.4 (a) Write down the wave function (in coordinate space) for the state  ;ipa  exp h j0i: You may use 2 !23 0 !1=21 0 x h  hx0j0i = ;1=4x;0 1=2 exp 4; 12 x 5 ; @x0  m! A : 0

(b) Obtain a simple expression that the probability that the state is found in the ground state at t = 0. Does this probability change for t > 0? (a) We have

 ;ipa  j; t = 0i = exp h j0i )  ;ipa  (Pr:1:4:c) 0 0 hx j; t = 0i = hx exp h j0i = hx0 ; aj0i 2 !23 0 x ; a 5: = ;1=4x;0 1=2 exp 4; 12 x 0

(2.19)

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40 (b) This probability is given by the expression  ;ipa  2 (2.20) jh0j; t = 0ij = jhexp h j0ij2: It is  ;ipa   ;ipa  Z 0 0 0 hexp h j0i = dx h0jx ihx j exp h j0i 2 !2 3 Z 0 x = dx0;1=4x;0 1=2 exp 4; 12 x 5 ;1=4x;0 1=2 0 2 3 ! 2 0  exp 4; 12 x x; a 5 0 " Z  02 02 2 # 1 0 ; 1 = 2 ; 1 0 = dx  x0 exp ; 2 x + x + a ; 2ax " 2x0 Z 2 a2 !# 1 2 a a = px dx0 exp ; 2x2 x02 ; 2x0 2 + 4 + 4 0 0 2! 1 p 2 ! a a = exp ; 4x2 px x0 = exp ; 4x2 : (2.21) 0

0

So

jh0j; t = 0ij2 For t > 0

jh0j; tij2

2! a = exp ; 2x2 : 0

0

(2.22)

 iHt  = = jh0j exp ; h j; t = 0ij2 2 = e;iE0t=hh0j; t = 0i = jh0j; t = 0ij2: (2.23)

jh0jU (t)j; t = 0ij2

2.5 Consider a function, known as the correlation function, de ned by C (t) = hx(t)x(0)i; (2.24) where x(t) is the position operator in the Heisenberg picture. Evaluate the correlation function explicitly for the ground state of a one-dimensional simple harmonic oscillator.

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2. QUANTUM DYNAMICS

41

The Hamiltonian for a one-dimensional harmonic oscillator is given by 2 (t) + 12 m!2x2(t): H = p2m (2.25) So the Heisenberg equations of motion will give " # dx(t) = 1 [x(t); H ] = 1 x(t); p2(t) + 1 m!2x2(t) dt ih ih 2m 2 1 hx(t); p2(t)i + 1 m!2 1 hx(t); x2(t)i = 2mi 2 h ih (t) (2.26) = 22ihihm p(t) = pm # " dp(t) = 1 [p(t); H ] = 1 p(t); p2 (t) + 1 m!2x2(t) dt ih ih 2m 2 2h i m!2 m! 2 = 2ih p(t); x (t) = 2ih [;2ihx(t)] = ;m!2x(t): (2.27) Dierentiating once more the equations (2.26) and (2.27) we get d2x(t) = 1 dp(t) (2=:27) ;!2x(t) ) x(t) = A cos !t + B sin !t ) x(0) = A dt2 m dt 2 d p(t) = 1 dx(t) (2=:26) ;!2p(t) ) p(t) = C cos !t + D sin !t ) p(0) = C: dt2 m dt

But on the other hand from (2.26) we have dx(t) = p(t) ) dt m p (0) cos !t + D sin !t ) ;!x(0) sin !t + B! cos !t = m m p (0) B = m! D = ;m!x(0): So (0) sin !t x(t) = x(0) cos !t + pm! and the correlation function will be 1 sin !t: C (t) = hx(t)x(0)i (2=:29) hx2(0)i cos !t + hp(0)x(0)i m!

(2.28) (2.29) (2.30)

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42 Since we are interested in the ground state the expectation values appearing in the last relation will be h (a + ay)(a + ay)j0i = h h0jaayj0i = h (2.31) hx2(0)i = h0j 2m! 2m! 2m! s s h h0j(ay ; a)(a + ay)j0i hp(0)x(0)i = i m2h! 2m! (2.32) = ;i h2 h0jaayj0i = ;i h2 : Thus h cos !t ; i h sin !t = h e;i!t: C (t) = 2m! (2.33) 2m! 2m!

2.6 Consider a one-dimensional simple harmonic oscillator. Do the following algebraically, that is, without using wave functions. (a) Construct a linear combination of j0i and j1i such that hxi is as large as possible. (b) Suppose the oscillator is in the state constructed in (a) at t = 0. What is the state vector for t > 0 in the Schrodinger picture? Evaluate the expectation value hxi as a function of time for t > 0 using (i) the Schrodinger picture and (ii) the Heisenberg picture. (c) Evaluate h(
x)2i as a function of time using either picture. (a) We want to nd a state ji = c0j0i + c1j1i such that hxi is as large as possible. The state ji should be normalized. This means q jc0j2 + jc1j2 = 1 ) jc1j = 1 ; jc0j2: (2.34) We can write the constands c0 and c1 in the following form

c0 = jc0jei0 (2:34) i1 q i 1 c1 = jc1je = e 1 ; jc0j2:

(2.35)

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2. QUANTUM DYNAMICS by

43

The average hxi in a one-dimensional simple harmonic oscillator is given

hxi = hjxji = (c0h0j + c1h1j) x (c0j0i + c1j1i) = jc0j2hs0jxj0i + c0c1h0jxj1i + c1cs0h1jxj0i + jc1j2h1jxj1i h h0ja + ayj0i + cc h h0ja + ayj1i = jc0j2 2m! 0 1 2m!

s h h1ja + ayj0i + jc j2 h h1ja + ayj1i 1 1 0 2m! 2m! s s h (cc + cc ) = 2 h 6 66 > 64 0 0 0 : : : 2 ;1 775 > 40 > : 0 0 0 : : : ;1 2 0 det N0  pN :

59

0 1 0 ... 0 0

0 0 1 ... 0 0

39 ::: 0 0 > > : : : 0 0 777> > = : : : 0 0 77> ... ... 77> = 7> > : : : 1 0 75> > ::: 0 1 ; (2.102)

We de ne j  j matrices j0 that consist of the rst j rows and j columns of N0 . So 2 2 ; "2!2 ;1 : : : 0 0 0 3 66 ;1 2 ; "2!2 : : : 0 0 0 777 66 0 ;1 : : : 0 0 0 77 6 6 . . . . ... 77 : .. .. .. det j0 +1 = det 66 .. 77 66 0 2!2 0 : : : 2 ; " ; 1 0 77 64 0 2 2 0 : : : ;1 2 ; " ! ;1 5 0 0 ::: 0 ;1 2 ; "2!2 From the above it is obvious that det j0 +1 = (2 ; "2!2) det j0 ; det j0 ;1 ) pj+1 = (2 ; "2!2)pj ; pj;1 for j = 2; 3; : : : ; N

(2.103)

with p0 = 1 and p1 = 2 ; "2!2. (e) We have

(t)  (ta + j")  "pj ) (ta + (j + 1)") = "pj+1 = (2 ; "2!2)"pj ; "pj;1 = 2(ta + j") ; "2!2(ta + j") ; (ta + (j ; 1)") ) (t + ") = 2(t) ; "2!2(t) ; (t ; "): (2.104)

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60 So

(t + ") ; (t) = (t) ; (t ; ") ; "2!2(t) )

; (t);"(t;") = ;!2(t) ) " 0(t) ; 0(t ; ") 2  2 (t) ) d  = ;! 2 (t): lim = ; ! (2.105) "!0 " dt2 (t+");(t) "

From (c) we have also that

(ta) = "p0 ! 0

(2.106)

and

d (t ) = (ta + ") ; (ta) = "(p1 ; p0) = p ; p 1 0 dt a " " = 2 ; "2!2 ; 1 ! 1: The general solution to (2.105) is (t) = A sin(!t + ) and from the boundary conditions (2.106) and (2.107) we have (ta) = 0 ) A sin(!ta + ) = 0 )  = ;!ta + n n 2 Z which gives that (t) = A sin !(t ; ta), while d = A! cos(t ; t ) ) 0(t ) = A! (2) :107) a a dt A! = 1 ) A = !1

Thus

(t) = sin !(!t ; ta) :

(f) Gathering all the previous results together we get " (N +1) N #1=2 m p G = Nlim !1 2ih" det

(2.107) (2.108) (2.109)

(2.110) (2.111)

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2. QUANTUM DYNAMICS = (d)

=

(2:111)

=

61

3;1=2 !N  m 1=2 2 2 i h  " 4 lim " 5 N !1 2ih m det  m 1=2  ;1=2 (e)  m 1=2 lim "p = 2ih [(tb)];1=2 N !1 N 2 i h  s m! (2.112) 2ih sin(!T ) :

So from (a)

 iScl  hxbtbjxatai = exp h G(0; tb ; 0; ta) s m!  im! h i (2:88) 2 + x2 ) cos !T ; 2x x : = exp ( x b a 2ih sin(!T ) 2h sin !T b a

2.12 Show the composition property Z

dx1Kf (x2; t2; x1; t1)Kf (x1; t1; x0; t0) = Kf (x2; t2; x0; t0)

where Kf (x1; t1; x0; t0) is the free propagator (Sakurai 2.5.16), by explicitly performing the integral (i.e. do not use completeness). We have Z dx1Kf (x2; t2; x1; t1)Kf (x1; t1; x0; t0) " # Z s m im ( x 2 ; x1)2 = dx1 2ih(t ; t ) exp 2h(t ; t )  2 1 2 " # 1 s 2 m im(x1 ; x0) exp 2ih(t1 ; t0) 2h(t1 ; t0) s m 1 imx22 exp imx20  = 2i exp h (t2 "; t1)(t1 ; t0) 2h(t2 ; t1) 2h(t2 ; t1) # Z im im im im 2 2 dx1 exp 2h(t ; t ) x1 + 2h(t ; t ) x1 ; 2h(t ; t ) 2x1x2 ; 2h(t ; t ) 2x1x0 2 1 2 1 2 1 2 1

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62 =

=

=

=

=

= =

s ( " 2 2 #) 1 m im x x 2 0 exp + 2ih (t2 (; t1)("t1 ; t0) 2h (t2# ; t1) (t1 "; t0)  #) Z im 1 1 im x x 2 0 2 dx1 exp 2h (t ; t ) + (t ; t ) x1 ; h x1 (t ; t ) + (t ; t ) 2 1 2 1 0 s ( 1 "0 2 #)1 2 1 m exp im x2 + x0  2ih (t2 (; t1)("t1 ; t0) 2h #(t2 ; t1) (t1 ; t0) Z t0 dx1 exp ;2imh (t ;tt2 ; 2 1)(t1 ; t0) " " ##) 2 h t im x 2 ; t0 2 (t1 ; t0) + x0 (t2 ; t1) 2 x1 ; im (t ; t )(t ; t ) h x1 s 2 1 1 0 ( " 2 (t2 ; t1)(t1 ;2 t0) #) 1 m im x2(t1 ; t0) + x0(t2 ; t1)  exp 2h (t2 ; t1)(t1 ; t0) 3 22Zih (t2 ;8t1)(t1"; t0) # " #29 = < ;m t2 ; t0 x2(t1 ; t0) + x0(t2 ; t1) 5  4 dx1 exp : x 1; ; 2ih (t2 ; t1)(t1 ; t0) (t2 ; t0) ( 2) im 1 [ x 2(t1 ; t0) + x0(t2 ; t1)] exp ; 2h (t ; t )(t ; t ) (t2 ; t0) 2 1 1 0 v s ( u u 1  2 i h  ( t ; t )( t ; t ) m im 1 2 1 1 0 t exp 2ih (t2 ; t1)(t1 ; t0) m(t2 ; t0) 2h (t2 ; t1)(t1 ; t0)  " 2 x2(t1 ; t0)(t2 ; t0) + x20(t2 ; t1)(t2 ; t0) ; (t2 ; t0) #) 2 2 2 x2(t1 ; t0) ; x0(t2 ; t1)2 ; 2x2x02(t1 ; t0)(t2 ; t1) (t2 ; t0) s m 2ih(t2 ; t1)  ( " 2 x2(t1 ; t0)(t2 ; t0 ; t1 + t0) + x20(t2 ; t1)(t2 ; t0 ; t2 + t1) ; exp im 2h #)(t2 ; t0)(t2 ; t1)(t1 ; t0) 2x2x02(t1 ; t0)(t2 ; t1) (t ; t )(t ; t )(t ; t ) s 2 0m 2 1 1" im0(x ; x )2 # 2 0 exp 2ih(t ; t ) 2h(t ; t ) 2 0 Kf (x2; t2; x0; t0):

2

0

(2.113)

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2. QUANTUM DYNAMICS

63

2.13 (a) Verify the relation

! i h  e [i; j ] = c "ijk Bk

where ~  m dt~x = ~p ; ecA~ and the relation

" !# 2~x ~ d d  1 d~ x d~ x ~ ~ ~ m dt2 = dt = e E + 2c dt  B ; B  dt :

(b) Verify the continuity equation with ~j given by

(a) We have

@ + r ~ 0
~j = 0 @t

! e  h  0 ~ ~ 2 ~j = m =( r ) ; mc Aj j :

 eAi  eA j [i; j ] = pi ; c ; pj ; c = ; ec [pi ; Aj ] ; ec [Ai; pj ] ! i h  e @A j ihe @Ai ihe @Aj @Ai = c @x ; c @x = c @x ; @x j i j !i = ihce "ijk Bk : (2.114) We have also that 2 3 2 3 dxi = 1 [x ; H ] = 1 4x ; ~ 2 + e5 = 1 4x ; ~ 2 5 dt ih i ih i 2m ih i 2m = ih12m f[xi; j ] j + j [xi; j ]g = ih21m f[xi; pj ]j + j [xi; pj ]g = 22ihihm j ij = mi )

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64

d2xi dt2

= = (2:114)

=

= =

m ddtx2i 2

m ddt~x2 2

= =

2 3 " # 1 dxi ; H = 1 4 ; ~ 2 + e5 ih dt ihm i 2m 1 f[ ;  ]  +  [ ;  ]g + e [ ; ] ih2m2 " i j j j i j #ihm i  1 ihe " B  + ihe "  B + e p ; eAi ;  2m2ih c ijk k j c ijk j k ihm i c e (;" B  + "  B ) + e [p ; ] 2m2c  ikj k j ijk j k  ihm i e m " xj B ; " B xj ; e @ ) 2m2c "ijk dt k ! ikj k dt !m# @xi ~x  B~ ; B~  ~x ) eEi + 2ec dt i " !#dt i ~x  B~ ; B~  ~x : e E~ + 21c dt (2.115) dt

(b) The time-dependent Schrodinger equation is

0 12 ~ @ hx0j; t ; ti = hx0jH j; t ; ti = hx0j 1 @~p ; eA A + ej; t ; ti ih @t 0 0 0 2m c 2 3 2 3 0 0 ~ ~ ~ 0 ; eA(~x ) 5
4;ihr ~ 0 ; eA(~x ) 5 hx0j; t0; ti + e(~x0)hx0j; t0; ti = 21m 4;ihr c c " # 2 e e e 1 2 0 0 0 0 0 0 2 0 ~
r ~ + ihr ~
A~ (~x ) + ih A~ (~x )
r ~ + 2 A (~x ) (~x0; t) = 2m ;h r c c c 0 0 +e(~x ) (~x ; t) ~ 0 ~ 0 = 21m ;h2r02 (~x0; t)0 + ec ih r
A (~x ; t) + ec ihA~ (~x0)
r~ 0 (~x0; t) # 2 e e 0 0 0 2 0 0 ~ (~x ; t) + 2 A (~x ) (~x ; t) + e(~x0) (~x0; t) + ih c A~ (~x )
r c " # 2   1 e e e 2 2 0 0 0 2 ~ ~ ~ ~ = 2m ;h r + c ih r
A + 2ih c A
r + c2 A + e : (2.116) Multiplying the last equation by  we get @ = ih  @t

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2. QUANTUM DYNAMICS 65 " # 2  1 ;h2 r02 + e ih r e e 0 2  0 2 2 ~
A~ j j + 2ih A~
r ~ + 2 A j j + ej j2: 2m c c c The complex conjugate of this eqution is @ = ;ih @t " # 2  1 ;h2 r02  ; e ih r e e ~ 0
A~ j j2 ; 2ih A~
r ~ 0  + 2 A2j j2 + ej j2: 2m c c c Thus subtracting the last two equations we get 2 h i ; 2hm r02 ; r02  e   2 e 0 ~ ~ ~0 + r ~ 0 ) + mc ih r
A j j + mc ihA~
( r ! @ @   = ih @t + @t ) 2 h i  e   ~ 0 ~ 2  e  ~ ~ 0 2 ih r
A j j + mc ihA
(r j j ) ; 2hm r~ 0
r~ 0 ; r~ 0  + mc = ih @ j j2 ) @t h i   ~ 0 h ~ 2i @ j j2 = ; h r ~ 0
=(  r ~0 ) + e r
Aj j ) @t m" mc #   @ j j2 + r ~ 0
h =( r ~ 0 ) ; e A~ j j2 = 0 ) @t m mc @ + r ~ 0
~j = 0 (2.117) @t     ~ 0 ) ; mce A~ j j2: and  = j j2 with ~j = mh =( r

2.14 An electron moves in the presence of a uniform magnetic eld in the z-direction (B~ = B z^). (a) Evaluate where

[x; y ];

x  px ; eAc x ;

y  py ; eAc y :

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66

(b) By comparing the Hamiltonian and the commutation relation obtained in (a) with those of the one-dimensional oscillator problem show how we can immediately write the energy eigenvalues as !  2k2 j eB j h  1 h  + n+ ; E = k;n

2m

mc

2 where hk is the continuous eigenvalue of the pz operator and n is a

nonnegative integer including zero.

The magentic eld B~ = B z^ can be derived from a vector petential A~ (~x) of the form Bx ; A = 0: Ax = ; By ; A (2.118) y= z 2 2 Thus we have     eA eA eBy eBx x y (2:118) [x; y ] = px ; c ; py ; c = px + 2c ; py ; 2c eB [y; p ] = iheB + iheB [ p = ; eB x ; x] + 2c 2c y 2c 2c eB = ih c : (2.119) (b) The Hamiltonian for this system is given by 0 1 ~ 2 1 2 1 2 1 2 e A 1 H = 2m @p~ ; c A = 2m x + 2m y + 2m pz = H1 + H2

(2.120)

where H1  21m 2x + 21m 2y and H2  21m p2z . Since " 2  2 # 1 eBy eBx [H1; H2] = 4m2 px + 2c + ; py ; 2c ; p2z = 0 (2.121) there exists a set of simultaneous eigenstates jk; ni of the operators H1 and H2. So if hk is the continious eigegenvalue of the operator pz and jk; ni its eigenstate we will have 2 2 2 H2jk; ni = 2pmz jk; ni = h2mk : (2.122)

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2. QUANTUM DYNAMICS

67

On the other hand H1 is similar to the Hamiltonian of the one-dimensional oscillator problem which is given by H = 21m p2 + 21 m!2x2 (2.123) with [x; p] = ih. In order to use the eigenvalues of the harmonic oscillator En = h ! n + 21 we should have the same commutator between the squared operators in the Hamiltonian. From (a) we have  xc  eB [x; y ] = ih c ) [ eB ; y ] = ih: (2.124) So H1 can be written in the following form  2 j2 H1  21m 2x + 21m 2y = 21m 2y + 21m eBxc jeB c2 !2  2 eB j xc : (2.125) = 21m 2y + 21 m jmc eB eB j to have In this form it is obvious that we can replace ! with jmc !  2k 2 h  j eB j h  1 H jk; ni = H1jk; ni + H2jk; ni = 2m jk; ni + mc n + 2 jk; ni " 2 2 ! # h  k j eB j h  1 = 2m + mc n + 2 jk; ni: (2.126)

2.15 Consider a particle of mass m and charge q in an impenetrable cylinder with radius R and height a. Along the axis of the cylinder runs a thin, impenetrable solenoid carrying a magnetic ux . Calculate the ground state energy and wavefunction. In the case where B~ = 0 the Schrodinger equation of motion in the cylindrical coordinates is h2 [r2 ] = 2E ) ; h @2 1 @ m 1 @2 @2 i 2  h ; m @2 +  @ + 2 @2 + @z2 (~x) = 2E (~x) (2.127)

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68 If we write (; ; z) = ()R()Z (z) and k2 = 2mE h2 we will have 2 2 ()Z (z) d R2 + ()Z (z) 1 dR + R()Z2 (z) d 2 d  d  d 2 +R()() ddzZ2 + k2R()()Z (z) = 0 ) 1 d2R + 1 dR + 1 d2 + 1 d2Z + k2 = 0(2.128) R() d2 R() d 2() d2 Z (z) dz2 with initial conditions (a; ; z) = (R; ; z) = (; ; 0) = (; ; a) = 0. So 1 d2Z = ;l2 ) d2Z + l2Z (z) = 0 ) Z (z) = A eilz + B e;ilz (2.129) 1 1 Z (z) dz2 dz2 with   Z (0) = 0 ) A1 + B1 = 0 ) Z (z) = A1 eilz ; e;ilz = C sin lz Z (a) = 0 ) C sin la = 0 ) la = n ) l = ln = n a n = 1; 2; : : : So

Z (z) = C sin lnz

(2.130)

Now we will have

1 d2R + 1 dR + 1 d2 + k2 ; l2 = 0 R() d2 R() d 2() d2 2 2 1 d2 + 2(k2 ; l2) = 0 ) R() ddR2 + R() dR + d () d2 2 (2.131) ) (1) dd2 = ;m2 ) () = eim:

with ( + 2) = () ) m 2 Z : So the Schrodinger equation is reduced to 2 d2R +  dR ; m2 + 2(k2 ; l2) = 0 R() d2 R() d

(2.132)

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2. QUANTUM DYNAMICS 69 2 R 1 dR " 2# d m 2 2 ) d2 +  d + (k ; l ) ; 2 R() = 0 # " 2 2R d 1 dR m ) d(pk2 ; l2)2 + pk2 ; l2 d(pk2 ; l2) + 1 ; (k2 ; l2)2 R() = 0 p p ) R() = A3Jm( k2 ; l2) + B3Nm( k2 ; l2) (2.133) In the case at hand in which a ! 0 we should take B3 = 0 since Nm ! 1 when  ! 0. From the other boundary condition we get p p R(R) = 0 ) A3Jm(R k2 ; l2) = 0 ) R k2 ; l2 = m (2.134) where m is the  -th zero of the m-th order Bessel function Jm. This means that the energy eigenstates are given by the equation   2 2 2 p m = R k2 ; l2 ) k2 ; l2 = Rm2 ) 2mE ; n a = Rm2 2 h    2# 2 " 2 h  m (2.135) ) E = 2m R2 + n a

while the corresponding eigenfunctions are given by m )eim sin n z  ( x ~ ) = A J ( nm c m R a with n = 1; 2; : : : and m 2 Z . Now suppose that B~ = B z^. We can then write ! 2!  B a A~ = 2 ^ = 2 ^:

(2.136)

(2.137)

The Schrodinger equation in the presence of the magnetic eld B~ can be written as follows 2 3 2 3 ~ ~ 1 4;ihr e A ( x ~ ) e A ( x ~ ) ~; 5 4 ~ 5 (~x) = E (~x) 2m c
;ihr ; c !# 2 " @ h  @ 1 @ ie  ) ; 2m ^@ + z^ @z + ^  @ ; hc 2
" !# @ @ 1 @ ie  ^ ^@ + z^ @z +   @ ; hc 2 (~x) = E (~x): (2.138)

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70 Making now the transformation D  @@ ; hiec 2 we get # " # 2 " @ @ 1 @ @ 1 h  ; 2m ^@ + z^ @z + ^  D
^@ + z^ @z + ^  D (~x) = E (~x) 2 " @2 2# h  1 @ 1 @ 2 ) ; 2m @2 +  @ + 2 D + @z2 (~x) = E (~x); (2.139)  2 where D2 = @@ ; hiec 2 . Leting A = hec 2 we get ! ! 2 2 @ 2 ie  @ @ @ 2 2 2 D = @2 ; hc 2 @ ; A = @2 ; 2iA @ ; A : (2.140) Following the same procedure we used before (i.e. (; ; z) = R()()Z (z)) we will get the same equations with the exception of " 2 # @ ; 2iA @ ; A2  = ;m2 ) d2 ; 2iA d + (m2 ; A2) = 0: @2 @ d2 d The solution to this equation is of the form el. So l + (m2 ; A2)el = 0 ) l2 ; 2iAl + (m2 ; A2) l2el ; 2iAle q 2iA  ;4A2 ; 4(m2 ; A2) 2iA  2im = = i(A  m) ) l= 2 2 which means that () = C2ei(Am): (2.141) But ( + 2) = () ) A  m = m0 m0 2 Z ) m = (m0 ; A) m0 2 Z : (2.142) This means that the energy eigenfunctions will be  z   m im0  (2.143) nm (~x) = Ac Jm ( R )e sin n a but now m is not an integer. As a result the energy of the ground state will be   2# 2 " 2 h  m (2.144) E = 2m R2 + n a

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2. QUANTUM DYNAMICS

71

where now m = m0 ; A is not zero in general but it corresponds to m0 2 Z such that 0  m0 ; A < 1. Notice also that if we require the ground state to be unchanged in the presence of B , we obtain ux quantization 0 m0 ; A = 0 ) hec 2 = m0 )  = 2me hc m0 2 Z : (2.145)

2.16 A particle in one dimension (;1 < x < 1) is subjected to a constant force derivable from V = x; ( > 0):

(a) Is the energy spectrum continuous or discrete? Write down an approximate expression for the energy eigenfunction speci ed by E. (b) Discuss brie y what changes are needed if V is replaced be V = jxj: (a) In the case under construction there is only a continuous spectrum and the eigenfunctions are non degenerate. From the discussion on WKB approximation we had that for E > V (x) i Z q  A ( x ) = exp 2 m [ E ; V ( x )] dx I [E ; V (x)]1=4 h  Zq  B + [E ; V (x)]1=4 exp ; hi 2m[E ; V (x)]dx ! Z x0 q c 1  = [E ; V (x)]1=4 sin h 2m[E ; V (x)]dx ; 4 x p Z x0=E=  ! 1=2 c 2 m E  = [E ; V (x)]1=4 sin h ; x dx ; 4  x 2  3 s  3 = 2 2m ;  5 = [E ; Vc(x)]1=4 sin 4; 32 E ; x h 4   (2.146) = [qc]11=4 sin 32 q3=2 + 4

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72

h i  1=3 where q =  E ; x and  = 2hm2 . On the other hand when E < V (x) ! Zx q c 1 2 II (x) = [x ; E ]1=4 exp " ; h x0=E= 2m(x ; E )dx # Zx q c 1 2 = [x ; E ]1=4 exp ; h2m 0 2m(x ; E )d(2mx) x =E=   = c31=4 exp ; 2 (;q)2=3 : (2.147) [;q] 3 We can nd an exact solution for this problem so we can compare with the approximate solutions we got with the WKB method. We have H ji = E ji ) hpjH ji = hpjE ji 2 ) hpj 2pm + xji = E hpji 2 ) 2pm (p) + ih dpd (p) = E(p) 2! d ; i p ) dp (p) = h E ; 2m (p) 2! d ( p ) ; i p ) (p) = h E ; 2m dp 3! p ; i ) ln (p) = h Ep ; 6m + c1 " !# 3 i p ) E (p) = c exp h 6m ; Ep : (2.148) We also have  (E ; E 0 )

=

Z

Z

hE jE 0i = dphE jpihpjE 0 i = E (p)E0 (p)dp

  Z jcj2 dp exp hi (E ; E 0)p jcj22h(E ; E 0) ) (2.149) c = p1 : 2h (2:148)

=

So

" !# 3 1 i p E (p) = p exp h 6m ; Ep : 2h

(2.150)

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2. QUANTUM DYNAMICS

73

These are the Hamiltonian eigenstates in momentum space. For the eigenfunctions in coordinate space we have Z Z ipx i  p3 ;Ep 1 (2:150) p dpe h e h 6m (x) = dphxjpihpjE i = 2  h  " 3    # Z 1 p i E p dp exp i h6m ; h  ; x p : = (2.151) 2h  Using now the substitution p3 = u3 p ) (2.152) u = (h2m )1=3 h6m 3 we have " 3 # E  1=3 Z +1 iu i ( h 2 m ) p du exp 3 ; h  ; x u(h2m)1=3 (x) = 2h  ;1 " # Z +1 3  iu = p du exp 3 ; iuq ; (2.153) 2  ;1 h  1=3 i and q =  E ; x . So where  = 2hm2 ! ! Z +1 Z +1 3 3 u  u  du cos 3 ; uq = p cos (x) = p 3 ; uq du 2  ;1   0 R +1 sin  u3 ; uq du = 0. In terms of the Airy functions since ;1 3 ! Z +1 3 u 1 cos 3 ; uq du (2.154) Ai(q) = p 0 we will have (x) = p Ai(;q): (2.155)  For large jqj, leading terms in the asymptotic series are as follows  2  1 p Ai(q)  2 q1=4 exp ; 3 q3=2 ; q > 0 (2.156)   (2.157) Ai(q)  p(;1 q)1=4 sin 32 (;q)3=2 + 4 ; q < 0

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74 Using these approximations in (2.155) we get 2   1  3 = 2 (q)  p q1=4 sin 3 q + 4 ; for E > V (x)    2   1 3 = 2 p exp ; 3 (;q) ; for E < V (x) (2.158) (q)  2  (;q)1=4 as expected from the WKB approximation. (b) When V = jxj we have bound states and therefore the energy spectrum is discrete. So in this case the energy eigenstates heve to satisfy the consistency relation Z x1 q   dx 2m[E ; jxj] = n + 21 h; n = 0; 1; 2; : : : (2.159) x2

The turning points are x1 = ; E and x2 = E . So Z E= q Z E= q  1 n + 2 h = 2m[E ; x]dx dx 2m[E ; jxj] = 2 0 ;E= p Z E=  E 1=2 = ;2 2m  ; x d(;x) 0 p 2  E 3=2 E= p 2  E 3=2 = ;2 2m 3  ; x = 2 2m 3  ) 0      E 3=2 3 n + 12 h  E  [3 n + 21 h]2=3 p = )  = 42=3(2m)1=3 )  4 2m 2  1  32=3 3 n + 2 h 5 p En = 4 : (2.160) 4 2m

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3. THEORY OF ANGULAR MOMENTUM

75

3 Theory of Angular Momentum

3.1 Consider a sequence of Euler rotations represented by !   ;i3   ; i ; i 2 3 (1 = 2) D (; ; ) = exp exp exp 2

=

e;i(+ )=2 cos 2 ei(; )=2 sin 2

2

;e;i(; )=2 sin 2 ei(+ )=2 cos 2

2!

:

Because of the group properties of rotations, we expect that this sequence of operations is equivalent to a single rotation about some axis by an angle . Find . In the case of Euler angles we have ;i(+ )=2 cos ;e;i(; )=2 sin ! e (1 = 2) D (; ; ) = ei(; )=2 sin 2 ei(+ )=2 cos 2 (3.1) 2 2 while the same rotation will be represented by   1 0  ; n ) sin (S ;3:2:45) @ cos 2 ; inz sin 2 (;in x y (1 = 2)  2  A : (3.2) D (; n^ ) =  (;inx + ny ) sin 2 cos 2 + inz sin 2 Since these two operators must have the same eect, each matrix element should be the same. That is ! !   ; i (  +

) = 2 e cos 2 = cos 2 ; inz sin 2 !  ) cos 2 = cos ( +2 ) cos 2 ) cos  = 2 cos2 2 cos2 ( +2 ) ; 1 " # (  +

) 2 2 )  = arccos 2 cos 2 cos 2 ; 1 : (3.3)

3.2 An angular-momentum eigenstate jj; m = mmax = j i is rotated by an in nitesimal angle " about the y-axis. Without using the

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76

explicit form of the d(mj)0m function, obtain an expression for the probability for the new rotated state to be found in the original state up to terms of order "2. The rotated state is given by

jj; j iR =

R("; y^)jj; j i = d(j)(")jj; j i =

"



 iJy "  exp ; h jj; j i

# iJ y " (;i)2"2 2 = 1; + J jj; j i (3.4) h 2h2 y up to terms of order "2. We can write Jy in terms of the ladder operators ) J+ = Jx + iJy ) J = J+ ; J; : (3.5) y J; = Jx ; iJy 2i Subtitution of this in (3.4), gives " # 2 " " 2 jj; j iR = 1 ; 2h (J+ ; J;) + 8h2 (J+ ; J; ) jj; j i (3.6) We know that for the ladder operators the following relations hold q J+jj; mi = h (j ; m)(j + m + 1)jj; m + 1i (3.7) q J;jj; mi = h (j + m)(j ; m + 1)jj; m ; 1i (3.8) So q (J+ ; J;)jj; j i = ;J; jj; j i = ;h 2j jj; j ; 1i (3.9) q (J+ ; J; )2jj; j i = ;h 2j (J+ ; J;)jj; j ; 1i q = ;h 2j (J+jj; j ; 1i ; J;jj; j ; 1i)  q q q = ;h 2j 2j jj; j i ; 2(2j ; 1)jj; j ; 2i and from (3.6)

2 q 2 q " " " jj; j iR = jj; j i + 2 2j jj; j ; 1i ; 8 2j jj; j i + 8 2 j (2j ; 1)jj; j ; 2i 2 ! 2q q " = 1 ; 4 j jj; j i + 2" 2j jj; j ; 1i + "4 j (2j ; 1)jj; j ; 2i:

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3. THEORY OF ANGULAR MOMENTUM

77

Thus the probability for the rotated state to be found in the original state will be 2 ! 2 2 " 2 jhj; j jj; j iRj = 1 ; 4 j = 1 ; "2 j + O("4): (3.10)

3.3 The wave function of a particle subjected to a spherically symmetrical potential V (r) is given by (~x) = (x + y + 3z)f (r):

(a) Is an eigenfunction of L~ ? If so, what is the l-value? If not, what are the possible values of l we may obtain when L~ 2 is measured? (b) What are the probabilities for the particle to be found in various ml states? (c) Suppose it is known somehow that (~x) is an energy eigenfunction with eigenvalue E . Indicate how we may nd V (r). (a) We have (~x)  h~xj i = (x + y + 3z)f (r): So

(3.11)

" !# 2 1 @ 1 @ @ (S ;3:6:15) 2 2 ~ h~xjL j i = ;h sin2  @2 + sin  @ sin  @ (~x): (3.12) If we write (~x) in terms of spherical coordinates (x = r sin  cos ; y = r sin  sin ; z = r cos ) we will have (~x) = rf (r) (sin  cos  + sin  sin  + 3 cos ) : (3.13) Then 1 @ 2 (~x) = rf (r) sin  @ (cos  ; sin ) = ; rf (r) (cos  + sin ()3.14) sin  sin2  @2 sin2  @

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78 and ! 1 @ sin  @ (~x) = rf (r) @ h;3 sin2  + (cos  + sin ) sin  cos i = sin  @ @ sin  @ i rf (r) h;6 sin  cos  + (cos  + sin )(cos2  ; sin2 )(3.15) : sin  Substitution of (3.14) and (3.15) in (3.11) gives  1  2 2 2 2 ~ h~xjL j i = ;h rf (r) ; sin  (cos  + sin )(1 ; cos  + sin ) ; 6 cos    = h2rf (r) sin1  2 sin2 (cos  + sin ) + 6 cos  = 2h2rf (r) [sin  cos  + sin  sin  + 3 cos ] = 2h2 (~x) ) L2 (~x) = 2h2 (~x) = 1(1 + 1)h2 (~x) = l(l + l)h2 (~x) (3.16) which means that (~x) is en eigenfunction of L~ 2 with eigenvalue l = 1. (b) Since we already know that l = 1 we can try to write (~x) in terms of the spherical harmonics Y1m (; ). We know that s s s 3 3 z Y10 = 4 cos  = 4 r ) z = r 43 Y10 9 8 q   q Y1+1 = ;q 83 (x+riy) = < x = r q23 Y1;1 ; Y1+1  ) Y1;1 = 83 (x;riy) ; : y = ir 23 Y1;1 + Y1+1 So we can write s i hp (~x) = r 23 f (r) 3 2Y10 + Y1;1 ; Y1+1 + iY1+1 + iY1;1 s hp i = 23 rf (r) 3 2Y10 + (1 + i)Y1;1 + (i ; 1)Y1+1 : (3.17) But this means that the part of the state that depends on the values of m can be written in the following way hp i j im = N 3 2jl = 1; m = 0i + (1 + i)jl = 1; m = ;1i + (1 ; i)jl = 1; m = 1i and if we want it normalized we will have jN j2(18 + 2 + 2) = 1 ) N = p122 : (3.18)

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3. THEORY OF ANGULAR MOMENTUM So

(c) If

9; P (m = 0) = jhl = 1; m = 0j ij2 = 922
2 = 11 1; P (m = +1) = jhl = 1; m = +1j ij2 = 222 = 11 1: P (m = ;1) = jhl = 1; m == 1j ij2 = 222 = 11

79 (3.19) (3.20) (3.21)

E (~x) is an energy eigenfunction then it solves the Schrodinger equation ;h2 " @ 2 (~x) + 2 @ (~x) ; L2 (~x)# + V (r) (~x) = E 2m @r2 E r @r E h2r2 E

E E (~x) " # 2 2 d 2 d 2 ; h  m ) 2m Yl dr2 [rf (r)] + r dr [rf (r)] ; r2 [rf (r)] + V (r)rf (r)Ylm = Erf (r)Ylm ) " # 2 d 1 h  2 2 0 0 V (r) = E + rf (r) 2m dr [f (r) + rf (r)] + r [f (r) + rf (r)] ; r f (r) ) 2 V (r) = E + rf1(r) 2hm [f 0(r) + f 0(r) + rf 00(r) + 2f 0(r)]] ) 2 rf 00 (r) + 4f 0 (r) h  V (r) = E + 2m : rf (r)

(3.22)

3.4 Consider a particle with an intrinsic angular momentum (or spin) of one unit of h. (One example of such a particle is the %meson). Quantum-mechanically, such a particle is described by a ketvector j%i or in ~x representation a wave function %i(~x) = h~x; ij%i where j~x; ii correspond to a particle at ~x with spin in the i:th direction. (a) Show explicitly that in nitesimal rotations of %i(~x) are obtained by acting with the operator (3.23) u~" = 1 ; i h~"
(L~ + S~)

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80

where L~ = hi r^  r~ . Determine S~ ! (b) Show that L~ and S~ commute. (c) Show that S~ is a vector operator. (d) Show that r~  %~(~x) = h1 (S~
~p)%~ where ~p is the momentum operator. 2

(a) We have

j%i =

3 Z X

i=1

j~x; iih~x; ij%i =

3 Z X

i=1

j~x; ii%i(~x)d3x:

(3.24)

Under a rotation R we will have 3 Z X j%0i = U (R)j%i = U (R) [j~xi jii] %i(~x)d3x i=1 Z 3 3 Z X X = jR~xi jiiDil(1)(R)%l (~x)d3x det=R=1 j~x; iiDil(1)(R)%l (R;1~x)d3x i=1 i=1 Z 3 X = j~x; ii%i0~x)d3x )

%i0(~x)

=

i=1 Dil(1)(R)%l(R;1~x) ) %~0(~x) = R~%(R;1~x):

(3.25)

Under an in nitesimal rotation we will have

R(; n^ )~r = ~r + ~r = ~r + (^n  ~r) = ~r + ~"  ~r:

(3.26)

So

%~0(~x) = R()%~(R;1~x) = R()%~(~x ; ~"  ~x) = ~%(~x ; ~"  ~x) + ~"  %~(~x ; ~"  ~x): (3.27) On the other hand ~ %~(~x) = ~%(~x) ; ~"
(~x  r ~ )%~(~x) ~%(~x ; ~"  ~x) = ~%(~x) ; (~"  ~x)
r (3.28) = ~%(~x) ; hi ~"
L~ %~(~x)

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3. THEORY OF ANGULAR MOMENTUM ~ %~(~x)  hr ~ %i (~x)i jii. Using this in (3.27) we get where r   i i 0 ~ ~ %~ (~x) = ~%(~x) ; h ~"
L%~(~x) + ~"  %~(~x) ; h ~"
L%~(~x) = ~%(~x) ; hi ~"
L~ %~(~x) + ~"  ~%(~x): But

81

(3.29)

      "~  %~ = "y %3 ; "z %2 e^x + "z %1 ; "x%1 e^y + "x%2 ; "y %1 e^z (3.30) or in0matrix 1 form 0 10 1 1 1 0 % 0 ;"z "y B@ %20 CA = B@ "z 0 ;"x CA B@ %%2 CA = %30 ;"y "x 0 %3 2 0 1 0 1 0 13 0 1 1 0 0 0 0 0 1 0 ;1 0 % 64"x B C B C B C 7 B @ 0 0 ;1 A + "y @ 0 0 0 A + "z @ 1 0 0 A5 @ %2 CA = 0 1 0 ;1 0 0 0 0 0 %3 2 0 1 0 1 0 13 0 1 1 0 0 0 0 0 i h  0 ; i h  0 % i 6 B C B C B C 7 B ; h 4"x @ 0 0 ;ih A + "y @ 0 0 0 A + "z @ ih 0 0 A5 @ %2 CA 0 ih 0 ;ih 0 0 0 0 0 %3 which means that ~"  %~ = ; hi ~"
S~~%(~x) with (S)kl = ;ihkl. Thus we will have that  i 0 ~ ~ %~ (~x) = U~"~%(~x) = 1 ; h ~"
(L + S ) %~(~x) ) U~" = 1 ; hi ~"
(L~ + S~): (3.31) (b) From their de nition it is obvious that L~ and S~ commute since L~ acts only on the j~xi basis and S~ only on jii. (c) S~ is a vector operator since X [Si; Sj ]km = [SiSj ; Sj Si]km = [(;ih)ikl(;ih)jlm ; (;ih)jkl(;ih)ilm] i Xh 2 = h ikljml ; h2jkliml X = h2 (ij km ; imjk ; ij km + jmki) X = h2 (jmki ; imjk ) X X X = h2 ijlkml = ihijl(;ihkml) = ihijl(Sl)km : (3.32)

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82 (d) It is

r~  %~(~x) = hi ~p  %~(~x) = hi ilp %l(~x)jii = h12 (S)lm p%m jii = 12 (S~
~p)%~: (3.33) h

3.5 We are to add angular momenta j1 = 1 and j2 = 1 to form j = 2; 1; and 0 states. Using the ladder operator method express all (nine) j; m eigenkets in terms of jj1j2; m1m2i. Write your answer as jj = 1; m = 1i = p12 j+; 0i ; p12 j0; +i; : : : ;

(3.34)

where + and 0 stand for m1;2 = 1; 0; respectively. We want to add the angular momenta j1 = 1 and j2 = 1 to form j = jj1 ; j2j; : : :; j1 + j2 = 0; 1; 2 states. Let us take rst the state j = 2, m = 2. This state is related to jj1m1; j2m2i through the following equation X jj; mi = hj1j2; m1m2jj1j2; jmijj1j2; m1m2i (3.35) m=m1 +m2

So setting j = 2, m = 2 in (3.35) we get

jj = 2; m = 2i = hj1j2; + + jj1j2; jmij + +i norm. = j + +i (3.36) If we apply the J; operator on this statet we will get

Jq ; jj = 2; m = 2i = (J1; + J2; )j + +i ) h (j + m)(j ; m + 1)jj = 2; m = 1i = q q h (j + m )(j ; m + 1)j0+i + h (j2 + m2)(j2 ; m2 + 1)j + 0i p 1 1 1 1p p ) 4jj = 2; m = 1i = 2j0+i + 2j + 0i ) jj = 2; m = 1i = p12 j0+i + p12 j + 0i: (3.37)

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3. THEORY OF ANGULAR MOMENTUM

83

In the same way we have J;jj = 2; m = 1i = p1 (J1; + J2;)j0+i + p1 (J1; + J2;)j + 0i ) 2 2 i i h hp p p p p 6jj = 2; m = 0i = p1 2j ; +i + 2j00i + p1 2j00i + 2j + ;i ) 2 2 p 6jj = 2; m = 0i = 2sj00i + j ; +i + j + ;i ) (3.38) jj = 2; m = 0i = 23 j00i + p16 j + ;i + p16 j ; +i

s

2 (J + J )j00i 3 1; 2; + p1 (J1; + J2;)j + ;i + p1 (J1; + J2;)j ; +i ) 6 s 6h i p p p p p 6jj = 2; m = ;1i = 23 2j ; 0i + 2j0;i + p1 2j0;i + p1 2j ; 0i ) 6 6 p p p p 2 2 1 1 jj = 2; m = ;1i = 6 2j0;i + 6 2j ; 0i + 6 2j0;i + 6 2j ; 0i ) (3.39) jj = 2; m = ;1i = p12 j0;i + p12 j ; 0i

J;jj = 2; m = 0i =

J;jj = 2; m = ;1i = p1 (J1; + J2;)j0;i + p1 (J1; + J2; )j ; 0i ) 2 2 p p p 1 1 4jj = 2; m = ;2i = p 2j ; ;i + p 2j ; ;i ) 2 2 jj = 2; m = ;2i = j ; ;i: (3.40) Now let us return to equation (3.35). If j = 1, m = 1 we will have jj = 1; m = 1i = aj + 0i + bj0+i (3.41) This state should be orthogonal to all jj; mi states and in particular to jj = 2; m = 1i. So hj = 2; m = 1jj = 1; m = 1i = 0 ) p12 a + p12 b = 0 ) a + b = 0 ) a = ;b : (3.42)

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84 In addition the state jj = 1; m = 1i should be normalized so

hj = 1; m = 1jj = 1; m = 1i = 1 ) jaj2 + jbj2 = 1 (3):42) 2jaj2 = 1 ) jaj = p12 :

By convention we take a to be real and positive so a = p12 and b = ; p12 . That is (3.43) jj = 1; m = 1i = p12 j + 0i ; p12 j0+i: Using the same procedure we used before J;jj = 1; m = 1i = p1 (J1; + J2;)j + 0i ; p1 (J1; + J2;)j0+i ) 2 2 i 1 hp hp p p p i 1 2jj = 1; m = 0i = p 2j00i + 2j + ;i ; p 2j ; +i + 2j00i ) 2 2 1 1 (3.44) jj = 1; m = 0i = p2 j + ;i ; p2 j ; +i

J;jj = 1; m = 0i = p1 (J1; + J2;)j + ;i ; p1 (J1; + J2;)j ; +i ) 2 2 p p p 2jj = 1; m = ;1i = p1 2j0;i ; p1 2j ; 0i ) 2 2 1 1 jj = 1; m = ;1i = p2 j0;i ; p2 j ; 0i: (3.45) Returning back to (3.35) we see that the state jj = 0; m = 0i can be written as jj = 0; m = 0i = c1j00i + c2j + ;i + c3j ; +i: (3.46) This state should be orthogonal to all states jj; mi and in particulat to jj = 2; m = 0i and to j = 1; m = 0i. So s hj = 2; m = 0jj = 0; m = 0i = 0 ) 23 c1 + p16 c2 + p16 c3 ) 2c1 + c2 + c3 = 0 (3.47) hj = 1; m = 0jj = 0; m = 0i = 0 ) p12 c2 ; p12 c3 ) c2 = c3: (3.48)

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3. THEORY OF ANGULAR MOMENTUM

85

Using the last relation in (3.47), we get 2c1 + 2c2 = 0 ) c1 + c2 = 0 ) c1 = ;c2:

(3.49)

The state jj = 0; m = 0i should be normalized so

hj = 0; m = 0jj = 0; m = 0i = 1 ) jc1j2 + jc2j2 + jc3j2 = 1 ) 3jc2j2 = 1 ) jc2j = p13 : (3.50) By convention we take c2 to be real and positive so c2 = c3 = c1 = ; p13 . Thus

jj = 0; m = 0i = p13 j + ;i + p13 j ; +i ; p13 j00i:

p1

3

and

(3.51)

So gathering all the previous results together

jj = 2; m = 2i jj = 2; m = 1i jj = 2; m = 0i jj = 2; m = ;1i jj = 2; m = ;2i jj = 1; m = 1i jj = 1; m = 0i jj = 1; m = ;1i jj = 0; m = 0i

= j + +i p1 j0+i + p1 j + 0i = q 2 2 2 1 p = 3 j00i + 6 j + ;i + p16 j ; +i = p12 j0;i + p12 j ; 0i = j ; ;i = p12 j + 0i ; p12 j0+i = p12 j + ;i ; p12 j ; +i = p12 j0;i ; p12 j ; 0i = p13 j + ;i + p13 j ; +i ; p13 j00i:

(3.52)

3.6 (a) Construct a spherical tensor of rank 1 out of two dierent vectors U~ = (Ux; Uy ; Uz ) and V~ = (Vx ; Vy ; Vz ). Explicitly write T(1)1;0 in terms of Ux;y;z and Vx;y;z . (b) Construct a spherical tensor of rank 2 out of two dierent vectors U~ and V~ . Write down explicitly T(2)2;1;0 in terms of Ux;y;z and Vx;y;z .

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86 (a) Since U~ and V~ are vector operators they will satisfy the following commutation relations [Ui; Jj ] = ih"ijk Uk [Vi; Jj ] = ih"ijk Vk :

(3.53)

From the components of a vector operator we can construct a spherical tensor of rank 1 in the following way. The de ning properties of a spherical tensor of rank 1 are the following q [Jz ; Uq(1)] = h qUq(1); [J; Uq(1)] = h (1  q)(2  q)Uq(1)1 : (3.54) It is :54) [Jz ; Uz ] (3=:53) 0hUz (3) Uz = U0 p [J+; U0] (3=:54) 2hU+1 = [J+; Uz ] = [Jx + iJy ; Uz ] (3:53) = ;ihUy + i(ih)Ux = ;h(Ux + iUy ) ) U+1 = ; p1 (Ux + iUy ) 2 (3:54) p 2hU;1 = [J;; Uz ] = [Jx ; iJy ; Uz ] [J;; U0] = (3:53) = ;ihUy ; i(ih)Ux = h(Ux ; iUy ) ) U;1 = p1 (Ux ; iUy ) 2 So from the vector operators U~ and V~ we can construct spherical with components

U0 = Uz U+1 = ; p12 (Ux + iUy ) U;1 = p12 (Ux ; iUy )

V0 = Vz V+1 = ; p12 (Vx + IVy ) V;1 = p12 (Vx ; iVy )

(3.55)

(3.56)

(3.57) tensors (3.58)

It is known (S-3.10.27) that if Xq(1k1 ) and Zq(2k2) are irreducible spherical tensors of rank k1 and k2 respectively then we can construct a spherical tensor of rank k X Tq(k) = hk1 k2; q1q2jk1k2 ; kqiXq(1k1)Zq(2k2) (3.59) q1 q 2

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3. THEORY OF ANGULAR MOMENTUM In this case we have (1) T+1 = h11; +10j11; 11iU+1 V0 + h11; 0 + 1j11; 11iU0 V+1 (3:52) 1 = p U+1V0 ; p1 U0V+1 2 2 1 = ; 2 (Ux + iUy )Vz + 21 Uz (Vx + iVy )

87

(3.60)

T0(1) = h11; 00j11; 10iU0 V0 + h11; ;1 + 1j11; 10iU;1 V+1 +h11; +1 ; 1j11; 10iU+1 V;1 (3:52) = ; p1 U;1V+1 + p1 U+1V;1 2 2 1 = p 12 (Ux ; iUy )(Vx + iVy ) ; p1 21 (Ux + iUy )(Vx ; iVy ) 2 2 1 = p [UxVx + iUxVy ; iUy Vx + Uy Vy ; UxVx + iUxVy ; iUy Vx ; Uy Vy ] 2 2 = pi (UxVy ; Uy Vx) (3.61) 2 T;(1)1 = h11; ;10j11; 11iU;1 V0 + h11; 0 ; 1j11; 11iU0 V;1 (3:52) = ; p1 U;1V0 + p1 U0V;1 2 2 = ; 21 (Ux ; iUy )Vz + 21 Uz (Vx ; iVy ): (3.62) (b) In the same manner we will have (2) T+2 = h11; +1 + 1j11; 2 + 2iU+1 V+1 (3=:52) U+1 V+1 = 12 (Ux + iUy )(Vx + iVy ) = ; 21 (UxVx ; Uy Vy + iUxVy + iUy Vx ) (3.63) (2) T+1 = (3:52) = =

h11; 0 + 1j11; 2 + 1iU0 V+1 + h11; +10j11; 2 + 1iU+1 V0 p1 U0V+1 + p1 U+1V0 2 2 ; 12 (Uz Vx + UxVz + iUz Vy + iUy Vz )

T0(2) = h11; 00j11; 20iU0 V0 + h11; ;1 + 1j11; 20iU;1 V+1 +h11; +1 ; 1j11; 20iU+1 V;1

(3.64)

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88 (3:52)

=

= = =

T;(2)1 = (3:52) = =

s

s s 2U V + 1U V + 1U V 0 0 ;1 +1 +1 ;1 3 6 6 s s s 2 U V ; 1 1 (U ; iU )(V + iV ) ; 1 p1 1 (U + iU )(V ; iV ) y x y y x y z z 62 x 6 22 x s3  1 2U V ; 1 U V ; i U V + i U V 6 z z 2 x x 2 x y 2 y x  i i 1 1 1 ; 2 Uy Vy ; 2 UxVx + 2 UxVy ; 2 Uy Vx ; 2 Uy Vy s 1 (2U V ; U V ; U V ) (3.65) 6 z z x x y y

h11; 0 ; 1j11; 2 ; 1iU0 V;1 + h11; ;10j11; 2 + 1iU;1 V0 p1 U0V;1 + p1 U;1 V0 2 2 1 (U V + U V ; iU V ; iU V ) x z z y y z 2 z x

(3.66)

T;(2)2 = h11; ;1 ; 1j11; 2 ; 2iU;1 V;1 (3=:52) U;1V;1 = 21 (Ux ; iUy )(Vx ; iVy ) = 12 (UxVx ; Uy Vy ; iUxVy ; iUy Vx): (3.67)

3.7 (a) Evaluate

j X m=;j

j) 2 jd(mm 0 ( )j m

for any j (integer or half-integer); then check your answer for j = 21 . (b) Prove, for any j , j X m=;j

m2jdm(j)0 m( )j2 = 12 j (j + 1) sin + m02 + 12 (3 cos2 ; 1):

[Hint: This can be proved in many ways. You may, for instance, examine the rotational properties of Jz2 using the spherical (irreducible) tensor language.]

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3. THEORY OF ANGULAR MOMENTUM (a) We have

j X

= = = =

m=;j j X m=;j j X m=;j j X m=;j j X m=;j

89

j ) ( )j2m jd(mm 0

mjhjmje;iJy =hjjm0ij2

  mhjmje;iJy =hjjm0i hjmje;iJy =hjjm0i mhjmje;iJy =hjjm0ihjm0jeiJy =hjjmi

hjm0jeiJy =hmjjmihjmje;iJy =hjjm0i

2 j 3 X 1 0 iJ =  h = h hjm je y Jz 4 jjmihjmj5 e;iJy =hjjm0i m=;j 1 = h hjm0jeiJy =hJz e;iJy =hjjm0i (3.68) = h1 hjm0jD( ; e^y )Jz D( ; e^y )jjm0i: But the momentum J~ is a vector operator so from (S-3.10.3) we will have that X (3.69) D ( ; e^y)Jz D( ; e^y ) = Rzj ( ; e^y )Jj : j

On the other hand we know (S-3.1.5b) that 0 1 cos 0 sin R( ; e^y ) = B (3.70) @ 0 1 0 CA : ; sin 0 cos So j X j ) ( )j2m = 1 [; sin hjm0jJ jjm0 i + cos hjm0jJ jjm0i] jd(mm 0 x z h m=;j   1 J + + J; 0 0 0 = h ; sin hjm j 2 jjm i + hm cos = m0 cos : (3.71)

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90 For j = 1=2 we know from (S-3.2.44) that ; sin ! cos (1=2) dmm0 ( ) = sin 2 cos 2 : 2 2 0 So for m = 1=2 1=2 X jd(mj)1=2( )j2m = ; 12 sin2 2 + 12 cos2 2 m=;1=2 = 21 cos = m0 cos while for m0 = ;1=2 1=2 X jd(mj)1=2( )j2m = ; 21 cos2 2 + 21 sin2 2 m=;1=2 = ; 12 cos = m0 cos :

(3.72)

(3.73)

(3.74)

(b) We have j X

= = = =

m=;j j X m=;j j X m=;j j X m=;j j X m=;j

m2jd(mj)0 m( )j2 m2jhjm0je;iJy =hjjmij2

  m2hjm0je;iJy =hjjmi hjm0je;iJy =hjjmi m2hjm0je;iJy =hjjmihjmjeiJy =hjjm0i

hjm0je;iJy =hm2jjmihjmjeiJy =hjjm0i

2 j 3 X 1 = 2 hjm0je;iJy =hJz2 4 jjmihjmj5 eiJy =hjjm0i h m=;j = 12 hjm0je;iJy =hJz2eiJy =hjjm0i h = h1 hjm0jD( ; e^y )Jz2Dy( ; e^y)jjm0i:

(3.75)

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3. THEORY OF ANGULAR MOMENTUM From (3.65) we know that

T0(2) =

s

91

1 (3J 2 ; J 2) 6 z

(3.76)

where T0(2) is the 0-component of a second rank tensor. So

p

Jz2 = 36 T0(2) + 31 J 2 (3.77) and since D(R)J 2Dy(R) = J 2D(R)Dy (R) = J 2 we will have Pj 2 (j ) 2 m=;j m jdmm0 ( )j = s 1 1 hjm0jJ 2jjm0i + 2 1 hjm0jD( ; e^ )J 2Dy( ; e^ )jjm0(3.78) i: y z y 3 h2 h3 3 We know that for a spherical tensor (S-3.10.22b) k X ( k ) y D(R)Tq D (R) = Dq(k0q)(R)Tq(0k) (3.79) q0 =;k

which means in our case that

hjm0jD( ; e^y )Jz2Dy( ; e^y )jjm0i = hjm0j =

2 X

2 X

q0 =;2

(2) Tq(2) 0 Dq0 0 ( ; e^y )jjm0i

q0 =;2

Dq(2)00 ( ; e^y)hjm0jTq(2)0 jjm0i:

(3.80)

But we know from the Wigner-Eckart theorem that hjm0jTq(2) 06=0 jjm0i = 0. So j X m=;j

j) 2 m2jd(mm 0 ( )j

s 1 1 2 = 2 h j (j + 1) + 2 23 D00(2)( ; e^y)hjm0jT0(2)jjm0i 3h h 1 1 (2) = 3 j (j + 1) + 2 d00 ( )hjm0jJz2 ; 13 J 2jjm0i  2 1  1 1 (2) 0 = 3 j (j + 1) + 2 d00 ( ) m ; 3 j (j + 1)

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92

 2 1  1 1 2 0 = 3 j (j + 1) + 2 (3 cos ; 1) m ; 3 j (j + 1) 02 1 1 1 m 2 = ; j (j + 1) cos + j (j + 1) + j (j + 1) + (3 cos2 ; 1) 2 6 3 2 2 2 0 2 1 1 (3.81) = 2 j (j + 1) sin + m 2 (3 cos ; 1) 1 2 where we have used d(2) 00 ( ) = P2 (cos ) = 2 (3 cos ; 1).

3.8 (a) Write xy, xz, and (x2 ; y2) as components of a spherical (irreducible) tensor of rank 2. (b) The expectation value Q  eh; j; m = j j(3z2 ; r2)j; j; m = j i is known as the quadrupole moment. Evaluate eh; j; m0j(x2 ; y2)j; j; m = j i; (where m0 = j; j ; 1; j ; 2; : : : )in terms of Q and appropriate ClebschGordan coecients. (a) Using the relations (3.63-3.67) we can nd that in the case where U~ = V~ = ~x the components of a spherical tensor of rank 2 will be (2) T+2 = 12 (x2 ; y2) + ixy T;(2)2 = 21 (x2 ; y2) ; ixy (2) = ;(xz + izy ) T+1 T;(2)1 = xz ; izy (3.82) q q (2) 1 1 2 2 2 2 2 T0 = 6 (2z ; x ; y ) = 6 (3z ; r )

So from the above we have (2) ; T (2) (2) ; T (2)  2 2 (2) (2) T T +2 ; 2 ; 1 +1 : (3.83) x ; y = T+2 + T;2 ; xy = ; xz = 2i 2 (b) We have

Q = eh; j; m = j j(3z2 ; r2)j; j; m = j i

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3. THEORY OF ANGULAR MOMENTUM (3:82)

=

)

93

p

(2) kj i p h j k T (W:;E:) (2) p 6eh; j; m = j jT0 j; j; m = j i = hj 2; j 0jj 2; jj i 2j + 1 6e

hj kT (2)kj i =

p

pQ hj 2; j20jjj+2;1jj i : 6e

(3.84)

So

e h; j; m0j(x2 ; y2)j; j; m = j i (3:83) (2) = eh; j; m0jT+2 j; j; m = j i + eh; j; m0jT;(2)2 j; j; m = j i z }|0 { (2) (2) = e hj 2; j 2jj 2; jm0i hjpk2Tj +k1j i + em0;j;2hj 2; j ; 2jj 2; jj ; 2i hjpk2Tj +k1j i (3:84) Q hj 2; j; ;2jj 2; j; j ; 2i = p hj 2; j; 0jj 2; j; j i m0;j;2: (3.85) 6

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94

4 Symmetry in Quantum Mechanics

4.1 (a) Assuming that the Hamiltonian is invariant under time reversal, prove that the wave function for a spinless nondegenerate system at any given instant of time can always be chosen to be real. (b) The wave function for a plane-wave state at t = 0 is given by a complex function ei~p
~x=h. Why does this not violate time-reversal invariance? (a) Suppose that jni in a nondegenerate energy eigenstate. Then H jni = H jni = En jni ) jni = ei jni ) jn; t0 = 0; ti = e;itH=hjni = e;itEn=hjni = 2E t E t eitEZn=hjni = ei( hn +) jni = ei( hn +)jn;Zt0 = 0; ti  )  d3xj~xih~xj jn; t0 = 0; ti = ei( 2Ehn t +) d3xj~xih~xj jn; t0 = 0; ti Z Z 3  ) d xh~xjn; t0 = 0; ti j~xi = d3xei( 2Ehn t +)h~xjn; t0 = 0; tij~xi

) n(~x; t) = ei( Ehn t +) n(~x; t): 2



(4.1) the wave function will be

So if we choose at any instant of time  = ; 2Ehnt real. (b) In the case of a free particle the Schrodinger equation is p2 jni = E jni ) ; h2 r ~ 2m 2m n (x) = En(x) ) n (x) = Aei~p
~x=h + Be;i~p
~x=h (4.2) The wave functions n(2x) = e;i~p
~x=h and 0n (x) = ei~p
~x=h correspond to the same eigenvalue E = 2pm and so there is degeneracy since these correspond to dierent state kets j~pi and j ; p~i. So we cannot apply the previous result.

4.2 Let (~p0) be the momentum-space wave function for state ji, that is, (~p0) = h~p0ji.Is the momentum-space wave function for the

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4. SYMMETRY IN QUANTUM MECHANICS

95

time-reversed state ji given by (~p0), (;~p0), (~p0 ), or (;~p0)? Justify your answer. In the momentum space we have Z Z ji = d3p0h~p0 jij~p0i ) ji = d3p0(~p0)j~p0i Z Z 3 0 0 0 ) ji = d p  [h~p jij~p i] = d3p0h~p0 jij~p0i:

(4.3)

For the momentum it is natural to require

hj~pji = ;h~j~pj~ i ) h~ j~p;1j~i ) ~p;1 = ;~p

(4.4)

So ~pj~p0i (4=:4) ;~pj~p0i ) j~p0i = j ; ~p0i up to a phase factor. So nally Z Z ji = d3p0h~p0ji j ; p~0i = d3p0h;~p0 jij~p0i ) h~p0jji = ~(~p0) = h;~p0ji = (;~p0):

(4.5)

(4.6)

4.3 Read section 4.3 in Sakurai to refresh your knowledge of the quantum mechanics of periodic potentials. You know that the energybands in solids are described by the so called Bloch functions n;k full lling, ika n;k (x + a) = e n;k (x)

where a is the lattice constant, n labels the band, and the lattice momentum k is restricted to the Brillouin zone [;=a; =a]. Prove that any Bloch function can be written as, X n(x ; Ri)eikRi n;k (x) = Ri

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96

where the sum is over all lattice vectors Ri. (In this simble one dimensional problem Ri = ia, but the construction generalizes easily to three dimensions.). The functions n are called Wannier functions, and are important in the tight-binding description of solids. Show that the Wannier functions are corresponding to dierent sites and/or dierent bands are orthogonal, i:e: prove Z

dx?m(x ; Ri)n(x ; Rj )  ij mn

Hint: Expand the n s in Bloch functions and use their orthonormality properties. The de ning property of a Bloch function n;k (x + a) = e

ika

n;k (x)

is

n;k (x):

(4.7)

We can show that the functions PRi n(x ; Ri)eikRi satisfy the same relation X X n(x + a ; Ri)eikRi = n [x ; (Ri ; a)]eik(Ri;a)eika Ri Ri Ri ;a=Rj ika X = e n(x ; Rj )eikRj (4.8) Rj

which means that it is a Bloch function X n(x ; Ri )eikRi : n;k (x) =

(4.9)

Ri

The last relation gives the Bloch functions in terms of Wannier functions. To nd the expansion of a Wannier function in terms of Bloch functions we multiply this relation by e;ikRj and integrate over k. X n(x ; Ri)eikRi n;k (x) = Ri Z =a Z =a X ) dke;ikRj n;k (x) = n (x ; Ri) eik(Ri;Rj )dk (4.10) ;=a

Ri

;=a

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4. SYMMETRY IN QUANTUM MECHANICS But

97

ik(Ri;Rj ) =a (Ri ; Rj )] e eik(Ri;Rj )dk = i(R ; R ) = 2 sin [=a Ri ; Rj ;=a i j ;=a = ij 2a (4.11) where in the last step we used that Ri ; Rj = na, with n 2 Z . So Z =a X dke;ikRj n;k (x) = n(x ; Ri)ij 2a ;=a R Z =a i a ) n(x ; Ri) = 2 ;=a e;ikRi n;k (x)dk (4.12) So using the orthonormality properties of the Bloch functions Z dxm(x ; Ri )n(x ; Rj ) Z Z Z a2 ikRi  (x)e;ik0 Rj 0 = e n;k0 (x)dkdk dx m;k 2 (2) Z Z a2 Z 0 Rj ikR ; ik  0 i = m;k (x) n;k0 (x)dxdkdk (2)2 e Z Z a2 ikRi ;ik0 Rj   (k ; k 0 )dkdk 0 e = mn 2 (2) Z =a 2 ik(Ri ;Rj ) dk = a   : = a 2 mn e (4.13) (2) 2 mn ij

Z =a

;=a

4.4 Suppose a spinless particle is bound to a xed center by a potential V (~x) so assymetrical that no energy level is degenerate. Using the time-reversal invariance prove hL~ i = 0 for any energy eigenstate. (This is known as quenching of orbital angular momemtum.) If the wave function of such a nondegenerate eigenstate is expanded as XX l m

Flm(r)Ylm(; );

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98

what kind of phase restrictions do we obtain on Flm(r)? Since the Hamiltonian is invariant under time reversal

H  = H: (4.14) So if jni is an energy eigenstate with eigenvalue En we will have H jni = H jni = Enjni: (4.15) If there is no degeneracy jni and jni can dier at most by a phase factor. Hence jn~ i  jni = eijni: (4.16) For the angular-momentum operator we have from (S-4.4.53)

hnjL~ jni = ;hn~ jL~ jn~ i (4=:16) ;hnjL~ jni ) hnjL~ jni = 0 : We have

Z

Z

jni =  d3xh~xjni j~xi Z = d3xh~xjnij~xi (4=:16) eijni ) hx~0jjni = hx~0jni = ei hx~0jni: (4.18) So if we use h~xjni = Pl Pm Flm(r)Ylm(; ) X  X Flm(r)Ylm(; ) = ei Flm(r)Ylm(; ) ml ml X (S ;4:4:57) X  m ; m ) Flm(r)(;1) Yl (; ) = ei Flm(r)Ylm(; ) ml Zml 0 X Z X m   m ; m ) Yl0 Flm(r)(;1) Yl (; )d = ei Ylm0 0  Flm(r)Ylm (; )d

ml X  ml m X ) Flm(r)(;1) m0;;m l0l = ei Flm(r)m0;m l0l

)

ml Fl0;;m0 (r)(;1);m0

d3xj~xih~xjni =

(4.17)

ml i 0 0 = e Fl m (r) ) Fl0;;m0 (r) = (;1)m0 Fl0m0 (r)ei :

(4.19)

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4. SYMMETRY IN QUANTUM MECHANICS

99

4.5 The Hamiltonian for a spin 1 system is given by H = ASz2 + B (Sx2 ; Sy2): Solve this problem exactly to nd the normalized energy eigenstates and eigenvalues. (A spin-dependent Hamiltonian of this kind actually appears in crystal physics.) Is this Hamiltonian invariant under time reversal? How do the normalized eigenstates you obtained transform under time reversal? For a spin 1 system l = 1 and m = ;1; 0; +1. For the operator Sz we have Sz jl; mi = hmjl; mi ) hlnjSz jl; mi = hmhnjmi ) (Sz )nm = hmnm (4.20) So 0 1 0 1 1 0 0 1 0 0 Sz =
h B @ 0 0 0 CA ) Sz2 =
h2 B@ 0 0 0 CA 0 0 ;1 0 0 1 For the operator Sx we have S+ + S; j1; mi = 1 S j1; mi + 1 S j1; mi ! Sxjl; mi = 2 + 2 ; 2 h1; njSxj1; mi = 12 h1; njS+ j1; mi + 12 h1; njS; j1; mi q (S ;3:5:39) 1 q 1 h (1 + m)(2 ; m) = h  (1 ; m )(2 + m )  + n;m+1 2 n;m;1 : 2 So 1 0 0 0 01 0 p 0 2 0 p p Sx =
h2 B @ 0 0 2 CA + h2 B@ 2 p0 0 CA 0 0 0 0 2 0 p 0 1 p0 2 p0 = h2 B @ 2 p0 2 CA ) 0 2 0 0 1 01 1 0 12 2 2 0 2 2 h  (4.21) Sx2 = 4 B @ 0 4 0 CA = h2 B@ 0 1 0 CA : 1 1 2 0 2 2 0 2

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100 In the same manner for the operator Sy = S+2;iS; we nd 1 0 0 p 2 0 p p Sx =
2hi B 0 2C A) @; 2 p 0 ; 2 0 0 1 0 1 1 1 0 ; ; 2 0 2 2 2 2 Sx2 =
; h4 B @ 0 ;4 0 CA = h 2 B@ 0 1 0 CA : ; 21 0 12 2 0 ;2 Thus the Hamiltonian can be represented by the matrix 0 1 A 0 B H =
h2 B @ 0 0 0 CA : B 0 A

(4.22)

(4.23)

To nd the energy eigenvalues we have to solve the secular equation 0 2 2 1 A h  ;  0 B h  det(H ; I ) = 0 ) det B ; 0 CA = 0 @ 0 B h2 0 Ah2 ;  h i ) (Ah2 ; )2(;) + (B h2)2 = 0 )  (Ah2 ; )2 ; (B h2)2 = 0 ) (Ah2 ;  ; B h2)(Ah2 ;  + B h2) = 0 ) 1 = 0; 2 = h2(A + B ); 3 = h2(A ; B ): (4.24) To nd the eigenstate jnc i that corresponds to the eigenvalue c we have to solve the following equation 0 10 1 0 1 A 0 B a aC B C B C B 2 h @ 0 0 0 A @ b A = c @ b A : (4.25) B 0 A c c For 1 = 0

0 10 1 ( A 0 B a + cB = 0 C B C B h 2 @ 0 0 0 A @ b A = 0 ) aA aB + cA = 0 c B 0 A ( ( B a = ; c A ) ;c B2 + cA = 0 ) ac == 00 A

(4.26)

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4. SYMMETRY IN QUANTUM MECHANICS So

0 1 0 1 0 0
B b C norm: B 1 C jn0i = @ A = @ A ) 0 0 jn0i = j10i:

In the same way for  = h 2(A + B ) 0 10 1 0 A 0 B a B@ 0 0 0 CA B@ b C A = (A + B ) B@ B 0 A c ( ) ab == 0c So

So

(4.27)

1 8 aC > < aA + cB = a(A + B ) b A)> 0 = b(A + B ) : aB + cA = c(A + B ) c (4.28)

0 1 0 1 c 1 1
norm: p B 0 C B C 0 jnA+B i = @ A = 2 @ A ) c 1 jnA+B i = p12 j1; +1i + p12 j1; ;1i:

For  = h2(A ; B ) we have 0 10 1 0 A 0 B a B@ 0 0 0 CA B@ b CA = (A ; B ) B@ B 0 A c ( ) a =b =;0c

101

(4.29)

1 8 > a < aA + cB = a(A ; B ) C b A)> 0 = b(A ; B ) : aB + cA = c(A ; B ) c

0 1 0 1 c 1
norm: p1 B B C jnA+B i = @ 0 A = @ 0 CA ) 2 ;1 ;c jnA;B i = p12 j1; +1i ; p12 j1; ;1i:

(4.30)

(4.31)

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102 Now we are going to check if the Hamiltonian is invariant under time reversal H ;1 = ASz2;1 + B (Sx2;1 ; Sy2;1) = ASz ;1Sz ;1 + B (Sx;1Sx;1 ; Sy ;1Sy ;1) = ASz2 + B (Sx2 ; Sy2) = H: (4.32) To nd the transformation of the eigenstates under time reversal we use the relation (S-4.4.58) jl; mi = (;1)mjl; ;mi:

(4.33)

So jn0i = j10i (4=:33) j10i = jn0i jnA+B i =

p1 j1; +1i + p1 j1; ;1i 2 2

= ; p1 j1; ;1i ; p1 j1; +1i 2 2 = ;jnA+B i

(4:33)

jnA;B i =

p1 j1; +1i ; p1 j1; ;1i 2 2

= ; p1 j1; ;1i + p1 j1; +1i 2 2 = jnA;B i:

(4:33)

(4.34) (4.35)

(4.36) (4.37)

(4.38)

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5. APPROXIMATION METHODS

103

5 Approximation Methods

5.1 Consider an isotropic harmonic oscillator in two dimensions. The Hamiltonian is given by 2 p2y m!2 2 2 p x H0 = 2m + 2m + 2 (x + y )

(a) What are the energies of the three lowest-lying states? Is there any degeneracy? (b) We now apply a perturbation V = m!2xy

where  is a dimensionless real number much smaller than unity. Find the zeroth-order energy eigenket and the corresponding energy to rst order [that is the unperturbed energy obtained in (a) plus the rst-order energy shift] for each of the three lowest-lying states. (c) Solve the H0 + V problem exactly. Compare with the perturbation results obtained in q(b). p p [You may use hn0jxjni = h=2m! ( n + 1 0 + n 0 ):] n ;n+1

n ;n;1

De ne step operators:

r m! ipx ); ax  2h (x + m! r m! ipx ); y ax  2h (x ; m! r m! ipy ); ay  2h (y + m! r (y ; ipy ): ayy  m! 2h m! From the fundamental commutation relations we can see that [ax; ayx] = [ay ; ayy ] = 1:

(5.1)

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104 De ning the number operators

Nx  ayxax;

Ny  ayy ay

we nd

N  Nx + Ny = hH!0 ; 1 ) H0 = h !(N + 1):

(5.2)

I.e. energy eigenkets are also eigenkets of N :

Nx j m; n i = m j m; n i; Ny j m; n i = n j m; n i ) N j m; n i = (m + n) j m; n i so that

(5.3)

H0 j m; n i = Em;n j m; n i = h!(m + n + 1) j m; n i:

(a) The lowest lying states are state degeneracy E0;0 = h ! 1 E1;0 = E0;1 = 2h! 2 E2;0 = E0;2 = E1;1 = 3h! 3 (b) Apply the perturbation V = m!2xy. Full problem: (H0 + V ) j l i = E j l i Unperturbed problem: H0 j l0 i = E 0 j l0 i Expand the energy levels and the eigenkets as

E = E 0 +
1 +
2 + : : : j l i = j l0 i + j l1 i + : : :

(5.4)

so that the full problem becomes h i h i (E 0 ; H0 ) j l0 i + j l1 i + : : : = (V ;
1 ;
2 : : :) j l0 i + j l1 i + : : : :

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5. APPROXIMATION METHODS

105

To 1'st order: (E 0 ; H0) j l1 i = (V ;
1) j l0 i:

(5.5)

Multiply with h l0 j to nd

h l0 j E 0 ; H0 j l1 i = 0 = h l0 j V ;
1 j l0 i )
1h l0 j l0 i =
1 = h l0 j V j l0 i

(5.6)

In the degenerate case this does not work since we're not using the right basis kets. Wind back to (5.5) and multiply it with another degenerate basis ket

h m0 j E 0 ; H0 j l1 i = 0 = h m0 j V ;
1 j l0 i )
1h m0 j l0 i = h m0 j V j l0 i:

(5.7)

Now, h m0 j l0 i is not necessarily kl since only states corresponding to dierent eigenvalues have to be orthogonal! Insert a 1: X 0 h m j V j k0 ih k0 j l0 i =
1h m0 j l0 i: k 2D

This is the eigenvalue equation which gives the correct zeroth order eigenvectors! Let us use all this: 1. The ground state is non-degenerate )
100 = h 0; 0 j V j 0; 0 i = m!2h 0; 0 j xy j 0; 0 i  h 0; 0 j (ax +ayx)(ay +ayy ) j 0; 0 i = 0 2. First excited state is degenerate j 1; 0 i, j 0; 1 i. We need the matrix elements h 1; 0 j V j 1; 0 i, h 1; 0 j V j 0; 1 i, h 0; 1 j V j 1; 0 i, h 0; 1 j V j 0; 1 i. h (a +ay )(a +ay ) = h! (a a +ay a +a ay +ay ay ) V = m!2xy = m!2 2m! x x y y 2 xy xy x y x y and

p

ax j m; n i = m j m ; 1; n i

p

ayx j m; n i = m + 1 j m + 1; n i etc:

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106 Together this gives V10;10 = V01;01 = 0; V10;01 = h2! h 1; 0 j ax ayy j 0; 1 i = h2! ; V01;10 = h2! h 1; 0 j ayx ay j 0; 1 i = h2! : The V -matrix becomes

(5.8)

! h! 0 1 2 1 0 and so the eigenvalues (=
1) are
1 =  h! : 2 To get the eigenvectors we solve ! ! ! 0 1 x = x 1 0 y y and get j


i+ = p12 ( j 0; 1 i + j 1; 0 i); E+ = h!(2 + 2 ); j


i; = p12 ( j 0; 1 i ; j 1; 0 i); E; = h !(2 ; 2 ): (5.9) 3. The second excited state is also degenerate j 2; 0 i, j 1; 1 i, j 0; 2 i, so we need the corresponding 9 matrix elements. However the only nonvanishing ones are: V11;20 = V20;11 = V11;02 = V02;11 = ph! (5.10) 2 p (where the 2 came from going from level 1 to 2 in either of the oscillators) and thus to get the eigenvalues we evaluate 0 1 ;  1 0 0 = det B @ 1 ; 1 CA = ;(2 ; 1) +  = (2 ; 2) 0 1 ;

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5. APPROXIMATION METHODS

107

which means that the eigenvalues are f0; h!g. By the same method as above we get the eigenvectors

p j


i+ = 21 ( j 2; 0 i + 2 j 1; 1 i + j 0; 2 i); j


i0 = p12 (; j 2; 0 i + j 0; 2 i); p j


i; = 21 ( j 2; 0 i ; 2 j 1; 1 i + j 0; 2 i);

E+ = h !(3 + ); E0 = 3h!; E; = h !(3 ; ):

(c) To solve the problem exactly we will make a variable change. The potential is h i m!2 21 (x2 + y2) + xy = # " 1  2 2 2 2 2 = m! 4 ((x + y) + (x ; y) ) + 4 (x + y) ; (x ; y) ) : (5.11) Now it is natural to introduce

p0x  p1 (p0x + p0y ); x0  p1 (x + y); 2 2 1 1 y0  p (x ; y); p0y  p (p0x ; p0y ): (5.12) 2 2 Note: [x0; p0x] = [y0; p0y ] = ih, so that (x0, p0x ) and (y0, p0y ) are canonically conjugate. In these new variables the problem takes the form 2 02 02 H = 21m (p0x2 + p0y2) + m! 2 [(1 + )x + (1 ; )y ]: p p So we get one oscillator with !x0 = ! 1 +  and another with !y0 = ! 1 ; . The energy levels are:

E0;0 = h!; p E1;0 = h ! + h!x0 = h !(1 + 1 + ) = = h!(1 + 1 + 12  + : : :) = h !(2 + 21 ) + O(2);

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108

E0;1 E2;0 E1;1 E0;2

= = = =

h ! + h!y0 = : : : = h !(2 ; 21 ) + O(2); h! + 2h!x0 = : : : = h!(3 + ) + O(2); h! + h!x0 + h!y0 = : : : = 3h! + O(2); h! + 2h!y0 = : : : = h!(3 ; ) + O(2):

(5.13)

So rst order perturbation theory worked!

5.2 A system that has three unperturbed states can be represented by the perturbed Hamiltonian matrix 0 1 E 1 0 a B@ 0 E1 b CA a b E2 where E2 > E1. The quantities a and b are to be regarded as per-

turbations that are of the same order and are small compared with E2 ; E1. Use the second-order nondegenerate perturbation theory to calculate the perturbed eigenvalues. (Is this procedure correct?) Then diagonalize the matrix to nd the exact eigenvalues. Finally, use the second-order degenerate perturbation theory. Compare the three results obtained. (a) First, nd the exact result by diagonalizing the Hamiltonian: 0 a E1 ; E 0 = 0 E1 ; E b =  a E2 ; E h b i = (E1 ; E ) (E1 ; E )(E2 ; E ) ; jbj2 + a [0 ; a(E1 ; E )] = = (E1 ; E )2(E2 ; E ) ; (E1 ; E )(jbj2 + jaj2): (5.14) So, E = E1 or (E1 ; E )(E2 ; E ) ; (jbj2 + jaj2) = 0 i.e. E 2 ; (E1 + E2)Es+ E1E2 ; (jaj2 + jbj2) = 0 ) 2 E = E1 +2 E2  E1 +2 E2 ; E1E2 + jaj2 + jbj2 =

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5. APPROXIMATION METHODS s 2 E + E = 1 2  E1 ; E2 + jaj2 + jbj2:

109

(5.15) 2 2 Since jaj2 + jbj2 is small we can expand the square root and write the three energy levels as:

E = E1 ;   2 E 1 + E2 E1 ; E2 2 2 2 1 E = 1 + 2 (jaj + jbj )( E ; E ) + : : : = 2 + 2 1 2 2 2 + jbj ; = E1 + jaEj ; 1 E2 2 + jbj2 E = E1 +2 E2 ; E1 ;2 E2 (: : :) = E2 ; jaEj ; : 1 E2 (5.16) (b) Non degenerate perturbation theory to 2'nd order. The basis we use is 0 1 0 1 0 1 1 0 0 j 1 i = B@ 0 CA ; j 2 i = B@ 1 CA ; j 3 i = B@ 0 CA : 0 0 1 0 1 0 0 a The matrix elements of the perturbation V = B @ 0 0 b CA are a b 0

h 1 j V j 3 i = a; h 2 j V j 3 i = b; h 1 j V j 2 i = h k j V j k i = 0: Since
(1) k = h k j V j k i = 0 1'st order gives nothing. But the 2'nd order shifts are

X jVk1 j2 jh 3 j V j 1 ij2 = jaj2 ; = 0 0 E1 ; E2 E1 ; E2 k6=1 E1 ; Ek 2 2 2 X jVk2 j jh 3 j V j 2 ij j b j = 0 0 = E1 ; E2 = E1 ; E2 ; k6=2 E2 ; Ek X jVk3 j2 jaj2 + jbj2 = ; jaj2 + jbj2 : = = 0 0 E2 ; E1 E2 ; E1 E1 ; E2 k6=3 E3 ; Ek


(2) 1 =
(2) 2
(2) 3

(5.17)

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110 The unperturbed problem has two (degenerate) states j 1 i and j 2 i with energy E1, and one (non-degenerate) state j 3 i with energy E2. Using nondegenerate perturbation theory we expect only the correction to E2 (i.e.
(2) 3 ) to give the correct result, and indeed this turns out to be the case. (c) To nd the correct energy shifts for the two degenerate states we have to use degenerate perturbation theory. The V -matrix for the degenerate ! 0 0 subspace is 0 0 , so 1'st order pert.thy. will again give nothing. We have to go to 2'nd order. The problem we want to solve is (H0 + V ) j l i = E j l i using the expansion j l i = j l0 i + j l1 i + : : : E = E 0 +
(1) +
(2) + : : : (5.18) where H0 j l0 i = E 0 j l0 i. Note that the superscript index in a bra or ket denotes which order it has in the perturbation expansion. Dierent solutions to the full problem are denoted by dierent l's. Since the (sub-) problem we are now solving is 2-dimensional we expect to nd two solutions corresponding to l = 1; 2. Inserting the expansions in (5.18) leaves us with h i (E 0 ; H0 ) j l0 i + j l1 i + : : : = h i (V ;
(1) ;
(2) : : :) j l0 i + j l1 i + : : : : (5.19) At rst order in the perturbation this says: (E 0 ; H0) j l1 i = (V ;
(1)) j l0 i; where of course
(1) = 0 as noted above. Multiply this from the left with a bra h k0 j from outside the deg. subspace h k0 j E 0 ; H0 j l1 i = h k0 j V j l0 i 0 X 0 0 (5.20) ) j l1 i = j k Eih0k;jEV j l i : k6=D

k

This expression for j l1 i we will use in the 2'nd order equation from (5.19) (E 0 ; H0) j l2 i = V j l1 i ;
(2) j l0 i: To get rid of the left hand side, multiply with a degenerate bra h m0 j (H0 j m0 i = E 0 j m0 i) h m0 j E 0 ; H0 j l2 i = 0 = h m0 j V j l1 i ;
(2)h m0 j l0 i:

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5. APPROXIMATION METHODS

111

Inserting the expression (5.20) for j l1 i we get X h m0 j V j k0 ih k0 j V j l0 i =
(2)h m0 j l0 i: 0 ; Ek E k6=D To make this look like an eigenvalue equation we have to insert a 1: X X h m0 j V j k0 ih k0 j V j n0 i 0 0 h n j l i =
(2)h m0 j l0 i: 0 ; Ek E n2D k6=D Maybe it looks more familiar in matrix form X Mmn xn =
(2)xm n2D

where

X h m0 j V j k0 ih k0 j V j n0 i ; E 0 ; Ek k6=D = 0 h m j l0 i

Mmn = xm

are expressed in the basis de ned by j l0 i. Evaluate M in the degenerate subspace basis D = f j 1 i; j 2 ig 2  M11 = EV13;V31E 0 = E ja;j E ; M12 = EV13;V32E 0 = E ab; E ; 1 1 2 1 1 2 3 3  2 2 M21 = EV23;V31E 0 = E a;bE ; M22 = EjV;23jE 0 = E jb;j E : 1

3

1

2

1

3

1

2

With this explicit expression for M , solve the eigenvalue equation (de ne  =
(2)(E1 ; E2), and take out a common factor E1;1 E2 ) 2 ;  ! j a j ab 0 = det ab jbj2 ;  = = (jaj2 ; )(jbj2 ; ) ; jaj2jbj2 = = 2 ; (jaj2 + jbj2) )  = 0; jaj2 + jbj2 2 + jbj2 j a j (2) (2) )
1 = 0
2 = E ; E : (5.21) 1

2

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112 From before we knew the non-degenerate energy shift, and now we see that degenerate perturbation theory leads to the correct shifts for the other two levels. Everything is as we would have expected.

5.3 A one-dimensional harmonic oscillator is in its ground state for t < 0. For t  0 it is subjected to a time-dependent but spatially uniform force (not potential!) in the x-direction, F (t) = F0e;t=

(a) Using time-dependent perturbation theory to rst order, obtain the probability of nding the oscillator in its rst excited state for t > 0. Show that the t ! 1 ( nite) limit of your expression is independent of time. Is this reasonable or surprising? (b) Can we nd higher q excited p states?0 0 [You may use hn jxjni = h=2m! ( n + 1n ;n+1 + pnn0;n;1 ):] (a) The problem is de ned by 2 2 H0 = 2pm + m!22x V (t) = ;F0xe;t= (F = ; @V @x ) At t = 0 the system is in its ground state j ; 0 i = j 0 i. We want to calculate X j ; t i = cn (t)e;Ent=h j n i

En0

n

= h!(n + 12 ) where we get cn (t) from its di. eqn. (S. 5.5.15): @ c (t) = X V ei!nmtc (t) ih @t n nm m m Vnm = h n j V j m i !nm = En ;h Em = !(n ; m) We need the matrix elements Vnm Vnm = h n j ; F0sxe;t= j m i = ;F0e;t= h n j x j m i = p h (pm = ;F0e;t= 2m! n;m;1 + m + 1n;m+1 ):

(5.22)

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5. APPROXIMATION METHODS

113

Put it back into (5.22)

s @ h pn + 1e;i!tc (t) + pnei!tc (t) : ; t= ih @t cn(t) = ;F0e n+1 n;1 2m! (1) Perturbation theory means expanding cn(t) = c(0) n + cn + : : :, and to zeroth order this is @ c(0)(t) = 0 ) c(0) =  n0 n @t n To rst order we get 1 Z t dt0 X V (t0)ei!nmt0 c(0) = c(1) ( t ) = n m ih 0 m nm s h Z t dt0e;t= pn + 1e;i!t0 c(0) (t) + pnei!t0 c(0) (t) = ; Fih0 2m! n+1 n;1 0 We get one non-vanishing term for n = 1, i.e. at rst order in perturbation theory with the H.O. in the ground state at t = 0 there is just one non-zero expansion coecient s h Z t dt0ei!t0;t0 = p1 F 0 c(1) ( t ) = ; 1;1;0 = 1 ih s 2m! 0 #t " F 1 )t0 h  1 0 ( i! ; = ; ih 2m! i! ; 1 e   0 s  h 1 1 ; e(i!; 1 )t = Fih0 2m! i! ; 1 and X ;iE1 t ;iEn t (1) j ; t i = c(1) n (t)e h j n i = c1 (t)e h j 1 i: n

The probability of nding the H.O. in j 1 i is As t ! 1

2 jh 1 j ; t ij2 = jc(1) 1 (t)j :

s F h 1 0 c(1) 1 ! ih 2m! i! ; 1 = const:  This is of course reasonable since applying a static force means that the system asymptotically nds a new equilibrium.

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114 (b) As remarked earlier there are no other non-vanishing cn's at rst order, so no higher excited states can be found. However, going to higher order in perturbation theory such states will be excited.

5.4 Consider a composite system made up of two spin 21 objects. for t < 0, the Hamiltonian does not depend on spin and can be taken to be zero by suitably adjusting the energy scale. For t > 0, the Hamiltonian is given by

 4
 H = 2 S~1
S~2: h Suppose the system is in j + ;i for t  0. Find, as a function of

time, the probability for being found in each of the following states j + +i, j + ;i, j ; +i, j ; ;i: (a) By solving the problem exactly. (b) By solving the problem assuming the validity of rst-order time-dependent perturbation theory with H as a perturbation switched on at t = 0. Under what condition does (b) give the correct results? (a) The basis we are using is of course j S1z ; S2z i. Expand the interaction potential in this basis: S~1
S~2 = S1x"S2x + S1y S2y + S1z S2z = fin this basisg 2 = h4 ( j + ih ; j + j ; ih + j )1 ( j + ih ; j + j ; ih + j )2 + + i2(; j + ih ; j + j ; ih + j )1 (; j + ih ; j + j ; ih + # j )2 + + ( j + ih + j ; j ; ih ; j )1 ( j + ih + j ; j ; ih ; j )2 = 2" h  = 4 j + + ih ; ; j + j + ; ih ; + j + + j ; + ih + ; j + j ; ; ih + + j + 2 + i ( j + + ih ; ; j ; j + ; ih ; + j + ; j ; + ih + ; j + j ; ; ih + + j ) +

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115

+ j + + ih + + j ; j + ; ih + ; j + # ; j ; + ih ; + j + j ; ; ih ; ; j = In matrix form this is (using j 1 i = j + + i j 2 i = j + ; i j 3 i = j ; + i j 4 i = j ; ; i) 01 0 0 01 B 0 ;1 2 0 CC H =
B B@ 0 2 ;1 0 CA : (5.23) 0 0 0 1 This basis is nice to use, since even though the problem is 4-dimensional we get a 2-dimensional matrix to diagonalize. Lucky us! (Of course this luck is due to the rotational invariance of the problem.) Now diagonalize the 2  2 matrix to nd the eigenvalues and eigenkets ! ; 1 ;  2 2 2 0 = det 2 ;1 ;  = (;1 ; ) ; 4 =  + 2 ; 3

)  = 1; ;3

=1:

! ! x = x y y ) ;x + 2y = x ) x = y = p12

;1 2 2 ;1

 = ;3 :

!

! ! x = ;3 x y y ) ;x + 2y = ;3x ) x = ;y = p12 So, the complete spectrum is: 8 1 > < j + + i; j ; ; i; p2 ( j + ; i + j ; + i with energy
> : p1 ( j + ; i ; j ; + i with energy ; 3


;1 2 2 ;1

2

!

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116 This was a cumbersome but straightforward way to calculate the spectrum. A smarter way would have been to use S~ = S~1 + S~2 to nd   S~ 2 = S 2 = S~12 + S~22 + 2S~1
S~2 ) S~1
S~2 = 21 S~ 2 ; S~12 ; S~22   We know that S~12 = S~22 = h2 21 21 + 1 = 3h42 so 2! 3 h 2 1 S~1
S~2 = 2 S ; 2 Also, we know that two spin 21 systems add up to one triplet (spin 1) and one singlet (spin 0), i.e. S = 1 (3 states) ) S~1
S~2 = 21 (h21(1 + 1) ; 3h22 ) = 14 h2 : (5.24) 2 2 3 h 3 1 ~ ~ S = 0 (1 state) ) S1
S2 = 2 (; 2 ) = ; 4 h ~ ~ Since H = 4
h2 S1
S2 we get 2 E(spin=1) = 4
2 14h =
; h 4
h2 = ;3
: E(spin=0) = 2 ;3 h 4

n

(5.25)

From Clebsch-Gordan odecomposition we know that j + + i; j ; ; i; p1 ( j + ; i + j ; + i) are spin 1, and p1 ( j + ; i ; j ; + i) is spin 0! 2 2 Let's get back on track and nd the dynamics. In the new basis H is diagonal and time-independent, so we can use the simple form of tthe time-evolution operator:  i  U (t; t0) = exp ; h H (t ; t0) : The initial state was j + ; i. In the new basis n j 1 i = j + + i; j 2 i = j ; ; i; j 3 i = p12 ( j + ; i + j ; + i); o j 4 i = p12 ( j + ; i ; j ; + i)

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5. APPROXIMATION METHODS the initial state is

117

j + ; i = p12 ( j 3 i + j 4 i):

Acting with U (t; 0) on that we get   j ; t i = p1 exp ; hi Ht ( j 3 i + j 4 i) = 2    3i   1 i = p exp ;
t j 3 i + exp
t j 4 i = h h "2   = exp ;ih
t p1 ( j + ; i + j ; + i)+ 2 #  3i
t  1 +exp h p ( j + ; i ; j ; + i) = 2 h ;i!t 3i!t i 1 = 2 (e + e ) j + ; i + (e;i!t + e3i!t) j ; + i where (5.26) ! 
h : The probability to nd the system in the state j i is as usual jh j ; t ij2 8 > h + + j ; t i = h ; ; j ; t i = 0 > > < jh + ; j ; t ij2 = 14 (2 + e4i!t + e;4i!t) = 21 (1 + cos4!t) ' 1 ; 4(!t)2 : : : > > > : jh ; + j ; t ij2 = 1 (2 ; e4i!t ; e;4i!t) = 1 (1 ; cos4!t) ' 4(!t)2 : : : 4 2 (b) First order perturbation theory (use S. 5.6.17):

c(0) n = ni ; Z ;i t dt0ei!nit0 V (t0): c(1) ( t ) = ni n h t0 Here we have (using the original basis) H0 = 0, V given by (5.23)

j i i = j + ; i;

(5.27)

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118

j f i = j ; + i; Ei = fE = 0g = 0; !ni = En ; n h

Vfi = 2
; Vni = 0; n 6= f: Inserting this into (5.27) yields (0) c(0) i = c j +; i = 1; Zt i (1) (1) cf = c j ;+ i = ; h dt2
= ;2i!t: (5.28) 0 as the only non-vanishing coecients up to rst order. The probability of nding the system in j ; ; i or j + + i is thus obviously zero, whereas for the other two states P ( j + ; i) = 1 (2) 2 2 2 P ( j ; + i) = jc(1) f (t) + cf (t) + : : : j = j2i!tj = 4(!t) to rst order, in correspondence with the exact result. The approximation breaks down when !t  1 is no longer valid, so for a given t: !t  1 )
 ht :

5.5 The ground state of a hydrogen atom (n = 1,l = 0) is subjected to a time-dependent potential as follows: V (~x; t) = V0 cos(kz ; !t): Using time-dependent perturbation theory, obtain an expression for the transition rate at which the electron is emitted with momentum ~p. Show, in particular, how you may compute the angular distribution of the ejected electron (in terms of  and  de ned with respect to the z-axis). Discuss brie y the similarities and the dierences between this problem and the (more realistic) photoelectric eect. (note: For the initial wave function use  Z  32 1 n=1;l=0(~x) = p a e;Zr=a0 : 0

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If you have a normalization problem, the nal wave function may be taken to be 1 f (~x) =

L 32

ei~p
~x=h

with L very large, but you should be able to show that the observable eects are independent of L.) To begin with the atom is in the n = 1; l = 0 state. At t = 0 the perturbation V = V0 cos(kz ; !t) is turned on. We want to nd the transition rate at which the electron is emitted with momentum ~pf . The initial wave-function is  1 3=2 1 i(~x) = p a e;r=a0 0 and the nal wave-function is  1  f (~x) = L3=2 ei~p
~x=h: The perturbation is h i V = V0 ei(kz;!t) + e;i(kz;!t) = V ei!t + V ye;i!t: (5.29) Time-dependent perturbation theory (S.5.6.44) gives us the transition rate 2 wi!n = 2h Vniy (En ; (Ei + h!)) because the atom absorbs a photon h!. The matrix element is y 2 V02  ikz  2 Vni = 4 e ni and Z  ikz  ikz ~ e ni = h kf j e j n = 1; l = 0 i = d3xh ~kf j eikz j x ih x j n = 1; l = 0 i =  1 3=2 Z e;i~kf
~x 1 3 ikx 3 = d x L3=2 e p a e;r=a0 = 0 Z = 3=2p1 3=2 d3xe;i(~kf
~x;kx3 );r=a0 : (5.30) L a0

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120

  So eikz ni is the 3D Fourier transform of the initial wave-function (and some constant) with ~q = ~kf ; k~ez . That can be extracted from (Sakurai problem 5.39)  ikz  642 1 e ni = L3a5 h 1 i4 0 a2 + (~kf ; k~ez )2 0 The transition rate is understood to be integrated over the density of states. We need to get that as a function of ~pf = h~kf . As in (S.5.7.31), the volume element is dn dp : n2dnd = n2d dp f f Using 2  )2 p2 kf2 = f2 = n (2 L2 h we get dn = 1 2L2pf = 2h L2pf = L dp 2n (2h)2 Lp (2h)2 2h which leaves

f

f

L3kf2 L3p2f = (2)3h d dpf = (2h)3 d dpf and this is the sought density. Finally, 2 2 L3p2f 1 d dpf : wi!~pf = 2h V40 L643a5 h 1 i 4 0 a2 + (~kf ; k~ez )2 (2 h)3 0 n2dnd

Note that the L's cancel. The angular dependence is in the denominator: ~ 2 kf ; k~ez = [(jkf jcos ; k) ~ez + jkf jsin (cos'~ex + sin'~ey )]2 = = jkf j2cos2 + k2 ; 2kjkf jcos + jkf j2sin2 = = kf2 + k2 ; 2kjkf jcos: (5.31) In a comparison between this problem and the photoelectric eect as discussed in (S. 5.7) we note that since there is no polarization vector involved, w has no dependence on the azimuthal angle . On the other hand we did not make any dipole approximation but performed the x-integral exactly.