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

1995,1999

0

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

ContentsI Problems1 2 3 4 5 Fundamental Concepts . . . . . . . Quantum Dynamics . . . . . . . . . Theory of Angular Momentum . . . Symmetry in Quantum Mechanics . Approximation Methods . . . . . . Fundamental Concepts . . . . . . . Quantum Dynamics . . . . . . . . . Theory of Angular Momentum . . . Symmetry in Quantum Mechanics . Approximation Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 7 14 17 19

3

II Solutions1 2 3 4 5

. 25 . 36 . 75 . 94 . 103

23

1

2

CONTENTS

Part I Problems

3

1. FUNDAMENTAL CONCEPTS

5

1 Fundamental Conceptsa0

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

1 (c) Illustrate (a) and (b) using A set equal to Sz of a spin 2 system.

~ ^ 1.2 A spin 1 system is known to be in an eigenstate of S n with 2 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 (h j + h j) (j i + j i) 0 for any complex number then choose in such a way that the preceding inequality reduces to the Schwarz inequility.

6

(b) Show that the equility sign in the generalized uncertainty relation holds if the state in question satis es Aj i = B j i 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 " 2# 0 0 0j i = (2 d2 );1=4 exp ihpix ; (x ; hxi) hxh4d2

satis es the uncertainty relationq

Prove that the requirement hx0j xj i = (imaginary number)hx0j pj i 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 bracketx F (px)]classical :

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

q

(b) Let x and px be the corresponding quantum-mechanical operators this time. Evaluate the commutator (c) Using the result obtained in (b), prove thatx exp iph a jx0i x x exp iph a :

(xjx0i = x0jx0i)

2. QUANTUM DYNAMICS

7

is an eigenstate of the coordinate operator x. What is the corresponding eigenvalue? 1.5 (a) Prove the following: @ (i) hp0jxj i = ih @p0 hp0j i Z @ (ii) h jxj i = dp0 (p0)ih @p0 (p0) where (p0) = hp0 j i and (p0) = hp0j i are momentum-space wave functions. (b) What is the physical signi cance of where x is the position operator and is some number with the dimension of momentum? Justify your answer.exp ix h

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. Evaluatex(t) x(0)] :

8

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

* + d h~ ~i = p2 ; h~ rV i: x ~ dt x p 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 ipa exp ;h j0i: You may use2 !3 x0 25 1 hx0j0i = ;1=4x;1=2 exp 4; 2 x 0 0 0 @x0h m!

!1=21 A:

(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.

2. QUANTUM DYNAMICS

9

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: aj i = j i where is, in general, a complex number. (a) Prove that j i = e;j j =2e ay j0i is a normalized coherent state. (b) Prove the minimum uncertainty relation for such a state. (c) Write j i as 1 X j i = f (n)jni:2

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.

n=0

10

(e) Show that the coherent state j i 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 ) =

Z10

dteiEt=hK (x y t 0) = A

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 relationd (x) = (x) dx

X sin(nrx) sin(nry) r E ; h22m2 n2 n

where (x) is the (unit) step function, and (x) the Dirac delta function. (Hint: study the e ect on testfunctions.) 2.10 Derive the following expressioni m! h Scl = 2 sin(!T ) (x2 + 2x2 ) cos(!T ) ; x0xT 0 T

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 1 _ L = 2 mx2 ; 1 m!2x2 2

2. QUANTUM DYNAMICS

11

(a) Show that where Scl is the action along the classical path xcl from (xa ta) to (xb tb) and G isG(0 tb 0 ta) = Z lim dy1 : : : dyN 2 mh" N !1 i(N +1) 2

hxbtbjxatai = exp iScl G(0 tb 0 ta) h

8 N 9 0?(a) We have

ipa j t = 0i = exp ;h j0i ) 1 hx0j t = 0i = hx0 exp ;ipa j0i (Pr:=:4:c) hx0 ; aj0i=

h 2 !3 x0 ; a 25 : ;1=4x;1=2 exp 4; 1 0 2 x0

(2.19)

40 (b) This probability is given by the expression (2.20) jh0j t = 0ij2 = jhexp ;ipa j0ij2: h It is ;ipa j0i = Z dx0h0jx0ihx0 j exp ;ipa j0i hexp h 2 h 0 !2 3 Z x = dx0 ;1=4x;1=2 exp 4; 1 x 5 ;1=4x;1=2 0 0 2 0 2 !3 x0 ; a 25 exp 4; 1 x 2 0 " # Z 0 ;1=2x;1 exp ; 1 x02 + x02 + a2 ; 2ax0 = dx 0 2 " 2x0 !# 1 Z dx0 exp ; 2 x02 ; 2x0 a + a2 + a2 = p x 2x2 2 4 4 0 0 ! ! 2 a p 1 p x = exp ; a2 : (2.21) = exp ; 4x2 0 x 4x2 So0 0 0

jh0jFor t > 0

t = 0ij2

! a2 : = exp ; 2x2 0

(2.22)

jh0j tij2 = jh0jU (t)j t = 0ij2 = jh0j exp ; iHt j t = 0ij2 h 2 = e;iE t=hh0j t = 0i = jh0j t = 0ij2: (2.23)0

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.

2. QUANTUM DYNAMICS

41

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 x(t) = x(0) cos !t + p(0) sin !t m! and the correlation function will be 1 :29) C (t) = hx(t)x(0)i (2= hx2(0)i cos !t + hp(0)x(0)i m! sin !t:

The Hamiltonian for a one-dimensional harmonic oscillator is given by 2 (t H = p2m) + 1 m!2x2(t): (2.25) 2 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 i 1 h i 1 h = 2mih x(t) p2(t) + 2 m!2 i1 x(t) x2(t) h (t ih (2.26) = 22hm p(t) = pm) i # " dp(t) = 1 p(t) H ] = 1 p(t) p2 (t) + 1 m!2x2(t) dt ih ih 2m 2 m!2 hp(t) x2(t)i = m!2 ;2ihx(t)] = ;m!2x(t): (2.27) = 2ih 2 ih Di erentiating once more the equations (2.26) and (2.27) we get d2x(t) = 1 dp(t) (2= ;!2x(t) ) x(t) = A cos !t + B sin !t ) x(0) = A :27) 2 dt m dt d2p(t) = 1 dx(t) (2= ;!2p(t) ) p(t) = C cos !t + D sin !t ) p(0) = C: :26) 2 dt m dt

(2.28) (2.29) (2.30)

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

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 j i = c0j0i + c1j1i such that hxi is as large as possible. The state j i 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 = jc0jei 0 q :34) i 1 (2= ei 1 1 ; jc j2 : c1 = jc1je 0

(2.35)

2. QUANTUM DYNAMICSby

43

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

s h h1ja + ayj0i + jc j2 h h1ja + ayj1i +c1c0 2m! 1 2m! s s h h = 2m! (c0c1 + c1c0) = 2 2m! 6 . . . 6. . .7 . .7 6 .. .. .. 6. . .7 >6 6 >6 0 0 0 : : : 2 ;1 7 60 >4 5 4 > : 0 0 0 : : : ;1 2 0 0 det N pN :

59

0 1 0 . . . 0 0

0 0 1 . . . 0 0

39 ::: 0 0 > : : : 0 0 7> 7> > 7= : : : 0 0 7> 7 = . . 7> . . 7> . . 7> : : : 1 0 7> 5> ::: 0 1 (2.102)

We de ne j j matrices j0 that consist of the rst j rows and j columns of 0 N . So 2 2 ; "2!2 ;1 : : : 0 0 0 3 6 ;1 2 ; "2!2 : : : 0 0 0 7 6 7 6 0 6 7 ;1 : : : 0 0 0 7 6 . 6 .. . . . . 7: . . . . 7 det j0 +1 = det 6 . . . . 7 6 7 6 0 2!2 0 ::: 2 ; " ;1 0 7 6 7 6 0 2!2 4 5 0 : : : ;1 2 ; " ;1 7 2! 2 0 0 ::: 0 ;1 2 ; " From the above it is obvious that det0 j +1 pj+1

= (2 ; "2!2) det j0 ; det j0 ;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") ) (ta + (j + 1)") = = ) (t + ") =

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

60 So (t + ") ; (t) = (t) ; (t ; ") ; "2!2 (t) ) = ;!2 (t) ) 0 0 2 lim (t) ; " (t ; ") = ;!2 (t) ) d 2 = ;!2 (t): "!0 dt From (c) we have also that(t+"); (t) "

;"

(t); (t;") "

(2.105) (2.106)

(ta) = "p0 ! 0

Thus

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 1 A! = 1 ) A = ! t (t) = sin !(! ; ta) :

and

(2.107) (2.108) (2.109)

(2.110) (2.111)

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

2. QUANTUM DYNAMICS=(d)

61

=

(2:111)

=

2 3;1=2 ! m 1=2 4 lim " 2ih" N det 5 N !1 2 ih m m 1=2 lim "p ;1=2 (=) m 1=2 (t )];1=2 e N b 2 ih s2 ih m! N !1 (2.112) 2 ih sin(!T ) :

So from (a)

hxbtbjxatai = exp iScl G(0 tb 0 ta) s h(2:88)

=

m! exp im! h(x2 + x2 ) cos !T ; 2x x i : b a 2 ih sin(!T ) 2h sin !T b a

2.12 Show the composition propertyZdx1Kf (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(x2 ; x1)2 = dx1 2 ih(t ; t ) exp 2h(t ; t ) 2 1 2 1 " s 2# m (x ) exp imh(t1 ; x0) 2 ih(t1 ; t0) 2 1 ; t0 s 2 2 1 = 2 mh (t ; t )(t ; t ) exp 2h(imx2 t ) exp 2h(imx0 t ) i t2 ; 1 t2 ; 1 2" 1 1 0 # Z im x2 + im x2 ; im 2x x ; im 2x x dx1 exp 2h(t ; t ) 1 2h(t ; t ) 1 2h(t ; t ) 1 2 2h(t ; t ) 1 02 1 2 1 2 1 2 1

62 =

=

=

=

=

= =

s ( " 2 #) 1 m im x2 + x2 0 2 ih (t2 ( t1)(t1 ; t0) exp 2h (t2 ; t1) (t1 " t0) ; " ; # #) Z im 1 + 1 x2 + x0 2 ; im x dx1 exp 2h (t ; t ) (t ; t ) x1 h 1 (t ; t ) (t ; t ) 2 1 2 1 1 0 s ( 1 "0 2 2 #) 1 m exp im x2 + x0 2 ih (t2 ( t1)("1 ; t0) ; t 2h # t2 ; t1) (t1 ; t0) ( Z dx1 exp ;im (t ;tt2 ;tt0; t ) 2 h 2 1)( 1 0" " ##) t2 ; t0 im x x2(t1 ; t0) + x0(t2 ; t1) 2 ; 2h x1 im (t ; t )(t ; t ) h 1 s 2 1 1 0 ( " 2 (t2 ; t1)(t1 ; t0) #) 2( 1 m t exp im x2(t(1t;;0t) +tx0;t2 ; t1) 2 2h 2 1)( 1 t0 ) 8 2Z ih (t2 ; t1)(t1"; t0) #" #293 = < ;m t2 ; t0 + 4 dx1 exp : x1 ; x2(t1 ; t0) ; x0)(t2 ; t1) 5 2ih (t2 ; t1)(t1 ; t0) (t2 t0 ( ) im 1 x2(t1 ; t0) + x0(t2 ; t1)]2 exp ; 2h (t ; t )(t ; t ) (t2 ; t0) 2 1 1 v0 s u 2ih(t ; t )(t ; t ) ( im u 1 m 1 2 1 1 0 t exp 2h (t ; t )(t ; t ) 2 ih (t2 ; t1)(t1 ; t0) m(t2 ; t0) 2 1 1 0 " 2 2(t2 ; t1)(t2 ; t0) x2(t1 ; t0)(t2 ; t0) + x0 ; (t2 ; t0) #) x2(t1 ; t0)2 ; x2(t2 ; t1)2 ; 2x2x02(t1 ; t0)(t2 ; t1) 2 0 (t2 ; t0) s m 2 ih(t2 ; t1) ( " 2 t t + 2( exp im x2(t1 ; t0)(t2 ; t0(t;;1 t+)(0) ; x0)(tt2 ; tt1)(t2 ; t0 ; t2 + t1) ; 2h #) 2 0 t2 t1 1 ; 0) 2x2x02(t1 ; t0)(t2 ; t1) (t ; t )(t ; t )(t ; t ) s 2 0m 2 1 1" im0(x ; x )2 # exp 2h(t2 ; t 0) 2 ih(t ; t )2 0 Kf (x2 t2 x0 t0): 2 0

(2.113)

2. QUANTUM DYNAMICS

63

2.13 (a) Verify the relationi

! ihe " B j] = c ijk k

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

" !# d2~ = d~ = e E + 1 d~ B ; B d~ : x x ~ ~ x ~ m dt2 dt 2c dt dt@ + r0 ~ = 0 ~ j @t

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

! h =( r0 ) ; e Aj j2: ~ ~ ~= j m mc

(a) We havei j]

= pi ; eAi pj ; eAj = ; e pi Aj ] ; e Ai pj ] c c c c! = ihe @Aj ; ihe @Ai = ihe @Aj ; @Ai c @xi c @xj c @xi @xj ! ihe " B : = c ijk k (2.114)

We have also that 2 3 2 3 dxi = 1 x H ] = 1 4x ~ 2 + e 5 = 1 4x ~ 2 5 dt ih i ih i 2m ih i 2m 1 = ih1m f xi j ] j + j xi j ]g = ih2m f xi pj ] 2 2ih = 2ihm j ij = mi )

j+ j

xi pj ]g

64

d2xi dt2

= =(2:114)

=

= =

m ddtxi 22 2

= =

x md ~ dt2

2 3 " # 1 dxi H = 1 4 ~ 2 + e 5 ih dt ihm i 2m 1 f e i j ] j + j i j ]g + i ] 2 ih2m " i #hm 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 # e ~ B ; B ~ ) x ~ x ~ eEi + 2c dt dt i " !# i ~ ~ ~ ~ x x ~ e E + 21c dt B ; B dt : (2.115)

(b) The time-dependent Schrodinger equation is

0 1 ~ A2 @ p A ih @t hx0j t0 ti = hx0jH j t0 ti = hx0j 21 @~ ; ec + e j t0 ti m 2 3 2 3 ~ (~ 0) 5 4 ~ 0 eA(~ 0) 5 0 ~x ~ = 21 4;ihr0 ; eAcx ;ihr ; c hx j t0 ti + e (~ 0)hx0j t0 ti x m " # 1 ;h2r0 r0 + e ihr0 A(~ 0) + ih e A(~ 0) r0 + e2 A2(~ 0) (~ 0 t) ~ ~ ~ ~x ~ ~ x = 2m c c x c2 x +e (~ 0) (~ 0 t) x x ~ ~ x ~ ~ x = 21 ;h2r02 (~ 0 t)0 + e ih r0 A (~ 0 t) + e ihA(~ 0) r0 (~ 0 t) x m c c x # e A(~ 0) r0 (~ 0 t) + e2 A2(~ 0) (~ 0 t) + e (~ 0) (~ 0 t) + ih c ~ x ~ x x x c2 x x " # 1 ;h2r02 + e ih r0 A + 2ih e A r0 + e2 A2 + e : (2.116) ~ ~ ~ ~ = 2m c c c2 Multiplying the last equation by we get @ ih @t =

2. QUANTUM DYNAMICS " 1 ;h2 r02 + e ih r0 A j j2 + 2ih e A ~ ~ ~ 2m c c

65

~ r0

# e2 A2j j2 + e j j2: + c2

The complex conjugate of this eqution is @ ;ih @t = " # 1 ;h2 r02 ; e ih r0 A j j2 ; 2ih e A r0 + e2 A2j j2 + e j j2: ~ ~ ~ ~ 2m c c c2 Thus subtracting the last two equations we get 2 h i ; 2hm r02 ; r02 e e ~ ~ ~ ~ ~ + mc ih r0 A j j2 + mc ihA ( r0 + r0 ) ! @ + @ = ih @t @t ) 2 e e ~ h ~ ~ i ~ ~ ~ ~ ; 2hm r0 r0 ; r0 + mc ih r0 A j j2 + mc ihA (r0j j2) = ih @ j j2 ) @t @ j j2 = ; h r0 h=( r0 )i + e r0 hAj j2i ) ~ ~ ~ ~ @t m" mc # @ j j2 + r0 h =( r0 ) ; e Aj j2 = 0 ) ~ ~ ~ @t m mc @ + r0 ~ = 0 ~ j (2.117) @t e ~ ~ with ~ = m =( r0 ) ; mc Aj j2: and = j j2 j h

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

px ; eAx c

y

py ; eAy : c

66

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

(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 ! h2k2 + jeB jh n + 1 E =kn

2m

mc

nonnegative integer including zero.

~ ~x The magentic eld B = B z can be derived from a vector petential A(~ ) ^ of the form Ax = ; By Ay = Bx Az = 0: (2.118) 2 2 Thus we have eAx p ; eAy (2:= p + eBy p ; eBx 118) x y ] = px ; y x c c 2c y 2c = ; eB px x] + eB y py ] = iheB + iheB 2c 2c 2c 2c eB : = ih c (2.119)(b) The Hamiltonian for this system is given by 0 1 ~ 2 1 @p ; eA A = 1 2 + 1 2 + 1 p2 = H + H H = 2m ~ c 1 2 2m x 2m y 2m z where H11 2 1 2 2m x + 2m y

(2.120)

and H2 21 p2. Since m z " # 1 p + eBy 2 + p ; eBx 2 p2 = 0 (2.121) H1 H2] = 4m2 x 2c y z 2c 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 2k2 2 z H2jk ni = 2pm jk ni = h m : (2.122) 2

2. QUANTUM DYNAMICS

67

On the other hand H1 is similar to the Hamiltonian of the one-dimensional oscillator problem which is given by 1 H = 21 p2 + 2 m!2x2 (2.123) m with x p] = ih. In order to use the eigenvalues of the harmonic oscillator 1 En = h! n + 2 we should have the same commutator between the squared operators in the Hamiltonian. From (a) we have eB ) xc (2.124) x y ] = ih y ] = ih: c eB So H1 can be written in the following form 2 1 2+ 1 2= 1 2+ 1 x c jeB j2 H1 2m x 2m y 2m y 2m eB c2 !2 2 1 2 + 1 m jeB j xc (2.125) = 2m y 2 mc eB :eB In this form it is obvious that we can replace ! with jmcj to have ! h2k2 jk ni + jeB jh n + 1 jk ni H jk ni = H1jk ni + H2jk ni = 2m mc 2 " 2 2 ! # k 1 = h m + jeB jh n + 2 jk ni: (2.126) 2 mc

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 i ) m h h2 @ 2 + 1 @ + 1 @ 2 + @ 2 ; m @ 2 @ 2 @ 2 @z2 (~ ) = 2E (~ ) x x (2.127)

68

z) = ( )R( )Z (z) and k2 = 2mE we will have h2 2 2 ( )Z (z) d R + ( )Z (z) 1 dR + R( )Z (z) d 2 2 d2 d d 2Z +R( ) ( ) d 2 + k2R( ) ( )Z (z) = 0 ) dz 2R 1 d + 1 dR + 1 d2 + 1 d2Z + k2 = 0(2.128) 2 ( )d 2 R( ) d 2 R( ) d 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 : : :If we write ( So

Z (z) = C sin lnzNow we will have

(2.130)

1 d2R + 1 dR + 1 d2 + k2 ; l2 = 0 2 ( )d 2 R( ) d 2 R( ) d 2 2 1 2 ) R( ) d R + R( ) dR + ( ) d 2 + 2(k2 ; l2) = 0 d2 d d 2 (2.131) ) (1 ) d 2 = ;m2 ) ( ) = e im : d (2.132)

with ( + 2 ) = ( ) ) m 2 Z: So the Schrodinger equation is reduced to 2 d2 R dR 2 2 2 2 2 + R( ) d ; m + (k ; l ) = 0 R( ) d

2. QUANTUM DYNAMICS 69 " # d2R + 1 dR + (k2 ; l2) ; m2 R( ) = 0 ) d2 2 d # " d2R 1 dR m2 ) 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 ! 1when ! 0. From the other boundary condition we get

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 m = R k2 ; l2 ) k2 ; l2 = R2 ) 2mE ; n a = R2 m h2 " # h2 2 + n 2 m (2.135) ) E = 2m R2 a while the corresponding eigenfunctions are given by m x )eim sin n az (2.136) nm (~ ) = Ac Jm ( R with n = 1 2 : : : and m 2 Z . ~ Now suppose that B = B z. We can then write ^ ! ! B 2 ^= a ^: ~ (2.137) A= 2 2 ~ The Schrodinger equation in the presence of the magnetic eld B can be written as follows 2 3 2 3 ~x ~x 1 4;ihr ; eA(~ ) 5 4;ihr ; eA(~ ) 5 (~ ) = E (~ ) ~ ~ x x 2m c c " !# h2 ^ @ + z @ + ^ 1 @ ; ie ) ; 2m @ ^ @z @ !# 2 hc " ie ^@@ + z @z + ^ 1 @@ ; hc 2 ^@ (~ ) = E (~ ): x x (2.138)

p

p

70ie Making now the transformation D @@ ; hc 2 we get " # " # @ + z @ + ^ 1 D (~ ) = E (~ ) h2 ^ @ + z @ + ^ 1 D ^@ ^ @z x x ; 2m @ ^ @z " # h2 @ 2 + 1 @ + 1 D2 + @ 2 (~ ) = E (~ ) ) ; 2m @ 2 @ x (2.139) 2 @z2 x

where D2 =

! ! @ 2 ; 2ie @ ; A2 = @ 2 ; 2iA @ ; A2 : (2.140) = @ 2 hc 2 @ @ 2 @ 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 @ d 2 d The solution to this equation is of the form el . So l2el ; 2iAlel + (m2 ; A2)el = 0 ) l2 ; 2iAl + (m2 ; A2) q 2iA ;4A2 ; 4(m2 ; A2) 2iA 2im = = i(A m) ) l= 2 2 which means that ( ) = C2ei(A m) : (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 m x )eim0 sin n az (2.143) nm (~ ) = Ac Jm ( R but now m is not an integer. As a result the energy of the ground state will be # " h2 2 + n 2 m (2.144) E = 2m R2 aD2

@ @

ie e ; hc 2 2. Leting A = hc 2 we get

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 e m0 ; A = 0 ) hc 2 = m0 ) = 2 m hc m0 2 Z : (2.145) e

2.16 A particle in one dimension (;1 < x < 1) is subjected to a constant force derivable fromV= 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) A iZ q (x) = E ; V (x)]1=4 exp h 2m E ; V (x)]dx I iZ q B + E ; V (x)]1=4 exp ; h 2m E ; V (x)]dx ! c 1 Z x0 q2m E ; V (x)]dx ; = E ; V (x)]1=4 sin h 4 p x Z x0=E= ! 1=2 c 2m E ; x dx ; = E ; V (x)]1=4 sin h x 2 3 4 s 3=2 2m c 4; 2 E ; x 5 = E ; V (x)]1=4 sin 3 h ;4 2 1 = qc1=4 sin 3 q3=2 + 4 ] (2.146)

72

h i m 1=3 where q = E ; x and = 2h2 . On the other hand when E < V (x) ! c2 1 Z x q2m( x ; E )dx II (x) = x ; E ]1=4 exp " ; h x0=E= # c2 1 Z x q2m( x ; E )d(2m x) = x ; E ]1=4 exp ; h2m 0 x =E= c3 exp ; 2 (;q)2=3 : = (2.147) ;q]1=4 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 j i = E j i ) hpjH j i = hpjE j i 2 ) hpj 2pm + xj i = E hpj i 2 d ) 2pm (p) + ih dp (p) = E (p) ! d (p) = ;i E ; p2 (p) ) dp h 2m ! d (p) = ;i E ; p2 dp ) (p) h 2m ! 3 ; ) ln (p) = h i Ep ; 6pm + c1 " !# i p3 ; Ep : ) E (p) = c exp h 6m (2.148)We also have (E ; E 0 )

So

Z jcj2 dp exp hi (E ; E 0)p jcj22 h (E ; E 0) ) (2.149) c = p1 : 2 h(2:148)

= =

hE jE 0i = dphE jpihpjE 0 i =

Z

Z

E (p) E 0 (p)dp

" 1 exp i E (p) = p h 2 h

!# p3 ; Ep : 6m

(2.150)

2. QUANTUM DYNAMICS

73

These are the Hamiltonian eigenstates in momentum space. For the eigenfunctions in coordinate space we have 3 Z 1p Z dpe ipx e hi 6pm ;Ep (2:150) h (x) = dphxjpihpjE i = " 2 h # Z 1p dp exp i p3 ; i E ; x p : = (2.151) h 6m h 2 h Using now the substitution 3 3 p (2.152) u = (h2m )1=3 ) h p6m = u 3 we have " # (h2mp 1=3 Z +1 du exp iu3 ; i E ; x u(h2m )1=3 ) (x) = ;1 2 h " 3 3 #h Z +1 = p du exp iu ; iuq (2.153) 3 ;1 2 hE i m 1=3 and q = where = 2h2 ; x . So ! ! Z +1 u3 ; uq = p Z +1 cos u3 ; uq du du cos 3 (x) = p 3 ;1 0 2 R +1 since ;1 sin u33 ; uq du = 0. In terms of the Airy functions ! 1 Z +1 cos u3 ; uq du (2.154) Ai(q) = p 3 0 we will have (x) = p Ai(;q): (2.155) For large jqj, leading terms in the asymptotic series are as follows 2 p 1q1=4 exp ; 3 q3=2 q > 0 Ai(q) 2 2 1 p (;q)1=4 sin 3 (;q)3=2 + 4 q < 0 Ai(q) (2.156) (2.157)

74 Using these approximations in (2.155) we get p q11=4 sin 2 q3=2 + 4 for E > V (x) (q) 3 2 p (;q1)1=4 exp ; 3 (;q)3=2 for E < V (x) (2.158) (q) 2 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 1 dx 2m E ; jxj] = n + 2 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 h = n+ 2 2m E ; x]dx dx 2m E ; jxj] = 2 0 ;E= 1=2 p Z E= E = ;2 2m ; x d(;x)

2 E ; x 3=2 E= = 2p2m 2 E 3=2 ) = ;2 2m 3 3 0 1 +2 E 3=2 = 3 np 1 h ) E = 3 n + 2 h]2=3 ) 42=3(2m )1=3 4 2m 2 32=3 1 43 n + 2 h 5 : p En = (2.160) 4 2m

p

0

3. THEORY OF ANGULAR MOMENTUM

75

3 Theory of Angular Momentum=

3.1 Consider a sequence of Euler rotations represented by ;i 3 exp ;i 2 ! exp ;i 3 D(1=2)( ) = expe;i( + )=2 cos 2 ei( ; )=2 sin2

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 ! e;i( + )=2 cos 2 ;e;i( ; )=2 sin 2 (1=2) ( D ) = ei( ; )=2 sin (3.1) ei( + )=2 cos 2 2 while the same rotation will be represented by 1 0 (;inx ; ny ) sin 2 A (1=2) ( n) (S ;3:2:45) @ cos 2 ; inz sin 2 : (3.2) D ^ = (;inx + ny ) sin 2 cos 2 + inz sin 2 Since these two operators must have the same e ect, each matrix element should be the same. That is ! ! ;i( + )=2 cos = cos e 2 2 ; inz sin 2 ! ) cos 2 = cos ( + ) cos 2 2 ) cos = 2 cos2 2 cos2 ( + ) ; 1 2 " # 2 cos2 ( + ) ; 1 : ) = arccos 2 cos 2 (3.3) 2

;e;i( ; )=2 sin 2 ei( + )=2 cos2

2

2 !

:

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

76j explicit form of the d(m)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

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 " # " (J ; J ) + "2 (J ; J )2 jj j i jj j iR = 1 ; 2h + ; 8h2 + ; (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)

jj j iR = R(" y)jj j i = d(j)(")jj j i = exp ; iJhy " jj j i ^ " # iJy " + (;i)2"2 J 2 jj j i = 1; (3.4) h 2h2 y

" q2j jj j ; 1i ; "2 2j jj j i + "2 2qj (2j ; 1)jj j ; 2i jj j iR = jj j i + 2 8 8 ! "2 j jj j i + " q2j jj j ; 1i + "2 qj (2j ; 1)jj j ; 2i: = 1; 4 2 4

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 = 1; " j jhj j jj j iRj = 1 ; " j + O("4): (3.10) 4 2

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

~ (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 L2 is measured? (b) What are the probabilities for the particle to be found in various ml states? (c) Suppose it is known somehow that (~ ) is an energy eigenfuncx tion with eigenvalue E . Indicate how we may nd V (r).(a) We have (~ ) h~ j i = (x + y + 3z)f (r): x x (3.11) (3.12)

" !# 2 ~ 2j i (S;3:6:15) ;h2 12 @ 2 + 1 @ sin @ h~ jL x = (~ ): x sin @ sin @ @ If we write (~ ) in terms of spherical coordinates (x = r sin cos x r sin sin z = r cos ) we will have (~ ) = rf (r) (sin cos + sin sin + 3 cos ) : x Then 1 @ 2 (~ ) = rf (r) sin @ (cos ; sin ) = ; rf (r) (cos + sin sin sin2 @ 2 x sin2 @

So

y=(3.13) (3.14) )

78 and ! 1 @ sin @ (~ ) = rf (r) @ h;3 sin2 + (cos + sin ) sin cos i = x 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 h~ jL2j i = ;h2rf (r) ; sin (cos + sin )(1 ; cos2 + sin2 ) ; 6 cos x~ 1 = h2rf (r) sin 2 sin2 (cos + sin ) + 6 cos = 2h2rf (r) sin cos + sin sin + 3 cos ] = 2h2 (~ ) ) x 2 (~ ) = 2h2 (~ ) = 1(1 + 1)h2 (~ ) = l(l + l)h2 (~ ) L x x x x (3.16) ~ which means that (~ ) is en eigenfunction of L2 with eigenvalue l = 1. x (b) Since we already know that l = 1 we can try to write (~ ) in terms of x m ( ). We know that the spherical harmonics Y1 s s s 3 cos = 3 z ) z = r 4 Y 0 Y10 = 4 1 r 3 q q 3 (x+iy4) 9 8 Y1+1 = ; 8 r = < x = r q23 Y1;1 ; Y1+1 q ): Y1;1 = 83 (x;iy) y = ir 23 Y1;1 + Y1+1 r So we can write s i hp (~ ) = r 23 f (r) 3 2Y10 + Y1;1 ; Y1+1 + iY1+1 + iY1;1 x 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 = p1 : (3.18) 22

3. THEORY OF ANGULAR MOMENTUMSo 9 P (m = 0) = jhl = 1 m = 0j ij2 = 9222 = 11 2 1 P (m = +1) = jhl = 1 m = +1j ij2 = 22 = 11 2 1 P (m = ;1) = jhl = 1 m == 1j ij2 = 22 = 11 :

79 (3.19) (3.20) (3.21)

(c) If

E E (~ ) " x 2 ;h Y m d2 rf (r)] + 2 d rf (r)] ; 2 rf (r)]# + V (r)rf (r)Y m = ) 2m l dr2 l r dr r2 Erf (r)Ylm ) " # 1 h2 d f (r) + rf 0 (r)] + 2 f (r) + rf 0 (r)] ; 2 f (r) ) V (r) = E + rf (r) 2m dr r r2 V (r) = E + rf1r) 2hm f 0(r) + f 0(r) + rf 00(r) + 2f 0(r)]] ) ( 2 rf 00 (r) + 4f 0 (r) V (r) = E + 2hm : rf (r)

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

(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 ~ representation a wave function x %i(~ ) = h~ ij%i x x where j~ ii correspond to a particle at ~ with spin in the i:th dix x rection. (a) Show explicitly that in nitesimal rotations of %i(~ ) are obtained x by acting with the operator " ~ ~ ~ (3.23) u~ = 1 ; i h (L + S ) "

80

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

(a) We have

j%i =

3 XZ

i=1

j~ iih~ ij%i = x x

3 XZ

i=1

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

(3.24)

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

%i0(~ ) x

=

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

(3.25) (3.26)

Under an in nitesimal rotation we will have

R(So

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

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

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

81

(3.29)

" ~ ~ % = "y %3 ; "z %2 ex + "z %1 ; "x%1 ey + "x%2 ; "y %1 ez (3.30) ^ ^ ^ or in matrix form 0 10 1 0 10 1 1 % 0 ;"z "y B %20 C = B "z 0 ;"x C B %2 C = @ A @ A@ % A 30 % ;"y "x 0 %3 2 0 1 0 1 0 13 0 1 1 0 0 0 0 0 1 0 ;1 0 % 6"x B 0 0 ;1 C + "y B 0 0 0 C + "z B 1 0 0 C7 B %2 C = 4 @ A @ A @ A5 @ A 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 ih 0 ;ih 0 i 6" B 0 0 ;ih C + " B 0 0 0 C + " B ih 0 0 C7 B %2 C ;h 4 x @ A y@ A z@ A5 @ % A 0 ih 0 ;ih 0 0 0 0 0 %3 which means that i " ~% x " ~ ~ % = ; h ~ S ~(~ ) with (S )kl = ;ih kl. Thus we will have that i" ~ ~ i" ~ ~ ~ x %0(~ ) = U~~(~ ) = 1 ; h ~ (L + S ) %(~ ) ) U~ = 1 ; h ~ (L + S ): (3.31) ~ x " "% x ~ ~ ~ (b) From their de nition it is obvious that L and S commute since L acts ~ only on jii. only on the j~ i basis and S x ~ (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 ikl jml ; h2 jkl iml X = h2 ( ij km ; im jk ; ij km + jm ki) X = h2 ( jm ki ; im jk ) X X X = h2 ijl kml = ih ijl(;ih kml) = ih ijl(Sl)km : (3.32)

82 (d) It is

ip ~ x i 1 ~ ~x r %(~ ) = h ~ %(~ ) = h i lp %l(~ )jii = h2 (S )lm p %m jii x ~ p~ = 12 (S ~)%: h

(3.33)

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 as1 1 jj = 1 m = 1i = p2 j+ 0i ; p2 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 j j 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

Jqjj = 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 1 p p ) 4jj = 2 m = 1i = 2j0+i + 2j + 0i 1 1 ) jj = 2 m = 1i = p2 j0+i + p2 j + 0i: (3.37)

3. THEORY OF ANGULAR MOMENTUM

83

In the same way we have 1 1 J;jj = 2 m = 1i = p (J1; + J2;)j0+i + p (J1; + J2;)j + 0i ) 2 2 i hp p i 1 hp p p 1 6jj = 2 m = 0i = p 2j ; +i + 2j00i + p 2j00i + 2j + ;i ) 2 2 p 6jj = 2 m = 0i = 2j00i + j ; +i + j + ;i ) s 1 1 (3.38) jj = 2 m = 0i = 2 j00i + p6 j + ;i + p6 j ; +i 3 2 (J + J )j00i 3 1; 2; 1 1 + p (J1; + J2;)j + ;i + p (J1; + J2;)j ; +i ) 6 s 6 i 1p hp p p 1p 6jj = 2 m = ;1i = 2 2j ; 0i + 2j0;i + p 2j0;i + p 2j ; 0i ) 3 6 6 2 p2j0;i + 2 p2j ; 0i + 1 p2j0;i + 1 p2j ; 0i ) jj = 2 m = ;1i = 6 6 6 6 1 1 j0;i + p j ; 0i (3.39) jj = 2 m = ;1i = p2 2

J;jj = 2 m = 0i =

s

1 1 J;jj = 2 m = ;1i = p (J1; + J2;)j0;i + p (J1; + J2; )j ; 0i ) 2 2 p 1 p2j ; ;i + p p2j ; ;i ) 1 4jj = 2 m = ;2i = p 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)

84 In addition the state jj = 1 m = 1i should be normalized so 1 :42) hj = 1 m = 1jj = 1 m = 1i = 1 ) jaj2 + jbj2 = 1 (3) 2jaj2 = 1 ) jaj = p2 :

1 By convention we take a to be real and positive so a = p12 and b = ; p2 . That is 1 1 (3.43) jj = 1 m = 1i = p2 j + 0i ; p2 j0+i: Using the same procedure we used before 1 1 J;jj = 1 m = 1i = p (J1; + J2;)j + 0i ; p (J1; + J2;)j0+i ) 2 2 p 1 hp2j00i + p2j + ;ii ; p hp2j ; +i + p2j00ii ) 1 2jj = 1 m = 0i = p 2 2 1 1 j + ;i ; p j ; +i (3.44) jj = 1 m = 0i = p2 2

1 1 J;jj = 1 m = 0i = p (J1; + J2;)j + ;i ; p (J1; + J2;)j ; +i ) 2 2 p p p 1 1 2jj = 1 m = ;1i = p 2j0;i ; p 2j ; 0i ) 2 2 1 j0;i ; p j ; 0i: 1 jj = 1 m = ;1i = p2 (3.45) 2 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 1 1 hj = 2 m = 0jj = 0 m = 0i = 0 ) 2 c1 + p6 c2 + p6 c3 3 ) 2c1 + c2 + c3 = 0 (3.47) 1 1 hj = 1 m = 0jj = 0 m = 0i = 0 ) p2 c2 ; p2 c3 ) c2 = c3: (3.48)

3. THEORY OF ANGULAR MOMENTUMUsing the last relation in (3.47), we get 2c1 + 2c2 = 0 ) c1 + c2 = 0 ) c1 = ;c2: The state jj = 0 m = 0i should be normalized so

85

(3.49)

hj = 0 m = 0jj = 0 m = 0i = 1 ) jc1j2 + jc2j2 + jc3j2 = 1 ) 3jc2j2 = 1 1 ) jc2j = p3 : (3.50)By convention we take c2 to be real and positive so c2 = c3 = 1 c1 = ; p3 . Thus 1 1 1 jj = 0 m = 0i = p3 j + ;i + p3 j ; +i ; p3 j00i: So gathering all the previous results together1 p 3

and

(3.51)

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 1 p = q2 j0+i + p12 j + 0i 1 1 = 2 j00i + p6 j + ;i + p6 j ; +i 3 1 = p2 j0;i + p12 j ; 0i = j ; ;i 1 = p2 j + 0i ; p12 j0+i 1 1 = p2 j + ;i ; p2 j ; +i 1 = p2 j0;i ; p12 j ; 0i 1 1 1 = p3 j + ;i + p3 j ; +i ; p3 j00i:

(3.52)

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

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)] = hqUq(1) J Uq(1)] = h (1 q)(2 q)Uq(1)1 : (3.54) It is:53) :54) Jz Uz ] (3= 0hUz (3) Uz = U0 (3.55) p :54) J+ U0] (3= 2hU+1 = J+ Uz ] = Jx + iJy Uz ] (3:53) = ;ihUy + i(ih)Ux = ;h(Ux + iUy ) ) 1 U+1 = ; p (Ux + iUy ) (3.56) 2 :54) p 2hU;1 = J; Uz ] = Jx ; iJy Uz ] J; U0] (3= (3:53) = ;ihUy ; i(ih)Ux = h(Ux ; iUy ) ) 1 U;1 = p (Ux ; iUy ) (3.57) 2 ~ ~ So from the vector operators U and V we can construct spherical tensors with components U0 = Uz V0 = Vz 1 1 p (Ux + iUy ) V+1 = ; p2 (Vx + IVy ) (3.58) U+1 = ; 2 1 1 U;1 = p2 (Ux ; iUy ) V;1 = p2 (Vx ; iVy )

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

3. THEORY OF ANGULAR MOMENTUMIn this case we have (1) T+1 = h11 +10j11 11iU+1 V0 + h11 0 + 1j11 11iU0 V+1 1 (3:52) 1 = p U+1V0 ; p U0V+1 2 2 1 (U + iU )V + 1 U (V + iV ) = ;2 x y z 2 z x y

87

(3.60)

T0(1) = h11 00j11 10iU0 V0 + h11 ;1 + 1j11 10iU;1 V+1 +h11 +1 ; 1j11 10iU+1 V;1 1 1 (3:52) = ; p U;1V+1 + p U+1V;1 2 2 1 1 (U ; iU )(V + iV ) ; p 1 (U + iU )(V ; iV ) 1 = p 2 x y x y y x y 2 22 x 1 = p UxVx + iUxVy ; iUy Vx + Uy Vy ; UxVx + iUxVy ; iUy Vx ; Uy Vy ] 2 2 i = p (UxVy ; Uy Vx) (3.61) 2 (1) T;1 = h11 ;10j11 11iU;1 V0 + h11 0 ; 1j11 11iU0 V;1 1 1 (3:52) = ; p U;1V0 + p U0V;1 2 2 1 1 = ; 2 (Ux ; iUy )Vz + 2 Uz (Vx ; iVy ): (3.62)(b) In the same manner we will have:52) (2) T+2 = h11 +1 + 1j11 2 + 2iU+1 V+1 (3= U+1 V+1 = 1 (Ux + iUy )(Vx + iVy ) 2 1 (U V ; U V + iU V + iU V ) = ;2 x x y y (3.63) x y y x(2) T+1 = (3:52) = =

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

(3.64)

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

88(3:52)

=

s

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

s s 2U V + 1U V + 1U V 3 0 0 s 6 ;1 +1 6 +1 ;1 s s 1 2 U V ; 1 1 (U ; iU )(V + iV ) ; 1 p 1 (U + iU )(V ; iV ) y x y y x y z z 2 x 6 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 ; 2 Uy Vy ; 2 UxVx + 2 UxVy ; 2 Uy Vx ; 1 Uy Vy 2 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 1 1 p U0V;1 + p 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)

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

3.7 (a) Evaluate

j X m=;j

j) jd(mm0 ( )j2m

1 for any j (integer or half-integer) then check your answer for j = 2 . (b) Prove, for any j ,

j X m=;j

(j m2jdm)0 m( )j2 = 1 j (j + 1) sin + m02 + 1 (3 cos2 ; 1): 2 2

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. THEORY OF ANGULAR MOMENTUM(a) We havej X j) jd(mm0 ( )j2m

89

= = = =

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

mjhjmje;iJy =hjjm0ij2 mhjmje;iJy =hjjm0i hjmje;iJy =hjjm0i mhjmje;iJy =hjjm0ihjm0jeiJy =hjjmi

2 j 3 1 hjm0jeiJy =hJ 4 X jjmihjmj5 e;iJy =hjjm0i = h z m=;j 1 hjm0jeiJy =hJ e;iJy =hjjm0i = h z 1 ^ ^ (3.68) = h hjm0jD ( ey )Jz D( ey )jjm0i: ~ But the momentum J is a vector operator so from (S-3.10.3) we will have that X ^ (3.69) D ( ey)Jz D( ey ) = Rzj ( ey )Jj : ^ ^On the other hand we know (S-3.1.5b) that 0 1 cos 0 sin R( ey ) = B 0 1 0 C : ^ @ (3.70) A ; sin 0 cos So j X (j) 1 jdmm0 ( )j2m = h ; sin hjm0jJxjjm0i + cos hjm0jJz jjm0i] m=;j 1 = h ; sin hjm0j J+ + J; jjm0i + hm0 cos 2 0 cos : = m (3.71)j

hjm0jeiJy =hmjjmihjmje;iJy =hjjm0i

90 For j = 1=2 we know from (S-3.2.44) that ! cos 2 ; sin 2 : (1=2) dmm0 ( ) = sin cos 2 2 0 = 1=2 So for m 1=2 X (j) jdm1=2( )j2m = ; 1 sin2 2 + 1 cos2 2 2 2m=;1=2

(3.72)

while for m0 = ;1=2 1=2 Xm=;1=2

=

1 2 cos

= m0 cos

(3.73)

j 1 1 jd(m)1=2( )j2m = ; 2 cos2 2 + 2 sin2 2

= ; 1 cos = m0 cos : 2

(3.74)

(b) We havej X m=;j j X m=;j j X m=;j j X m=;j j X m=;j j m2jd(m)0 m( )j2

= = = =

m2jhjm0je;iJy =hjjmij2 m2hjm0je;iJy =hjjmi hjm0je;iJy =hjjmi m2hjm0je;iJy =hjjmihjmjeiJy =hjjm0i

2 j 3 1 hjm0je;iJy =hJ 2 4 X jjmihjmj5 eiJy =hjjm0i = 2 z h m=;j = 12 hjm0je;iJy =hJz2eiJy =hjjm0i h 1 = h hjm0jD( ey )Jz2Dy( ey)jjm0i: ^ ^

hjm0je;iJy =hm2jjmihjmjeiJy =hjjm0i

(3.75)

3. THEORY OF ANGULAR MOMENTUMFrom (3.65) we know that

91

T0(2) =

s

1 (3J 2 ; J 2) 6 z

(3.76)

where T0(2) is the 0-component of a second rank tensor. So 1 Jz2 = 36 T0(2) + 3 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) ^y z ^y i: 3 h2 h3 3 We know that for a spherical tensor (S-3.10.22b) k (k) Dy(R) = X D(k) (R)T (0k) D(R)Tq (3.79) q0 q qq0 =;k

p

which means in our case that

hjm0jD( ey )Jz2Dy( ey )jjm0i = hjm0j ^ ^=2 X

2 X

q0 =;2

(2) Tq(2)Dq0 0( ey )jjm0i ^ 0

q0 =;2

Dq(2)( ey)hjm0jTq(2)jjm0i: ^ 00 0

(3.80)

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

s 1 h2j (j + 1) + 1 2 D(2)( e )hjm0jT (2)jjm0i = 2 ^y 0 3h h2 3 00 = 1 j (j + 1) + 1 d(2)( )hjm0jJz2 ; 1 J 2jjm0i 3 2 00 3 1 j (j + 1) + 1 d(2)( ) m02 ; 1 j (j + 1) = 3 2 00 3

j) m2jd(mm0 ( )j2

92 1 1 = 3 j (j + 1) + 2 (3 cos2 ; 1) m02 ; 1 j (j + 1) 3 1 j (j + 1) cos2 + 1 j (j + 1) + 1 j (j + 1) + m02 (3 cos2 ; 1) = ; 2 6 3 2 1 j (j + 1) sin2 + m02 1 (3 cos2 ; 1) (3.81) = 2 2 where we have used d(2)( ) = P2(cos ) = 1 (3 cos2 ; 1). 00 2

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 coe cients.~ (a) Using the relations (3.63-3.67) we can nd that in the case where U = ~ x V = ~ the components of a spherical tensor of rank 2 will be(2) (2) 1 T+2 = 1 (x2 ; y2) + ixy T;2 = 2 (x2 ; y2) ; ixy 2 (2) (2) T+1 = ;(xz + izy) (3.82) q 1 2 2 2 q 1 2 2 T;1 = xz ; izy (2) = T0 6 (2z ; x ; y ) = 6 (3z ; r )

So from the above we have

x2 ; y2(b) We have

(2) (2) = T+2 + T;2

(2) (2) T+2 ; T;2 xy = 2i

(2) (2) T;1 ; T+1 : (3.83) xz = 2

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

3. THEORY OF ANGULAR MOMENTUM(3:82)

93

=

p

6eh j

m = j jT0(2)j

Q ) h j kT (2)k j i = p6e hj 2 j20jjj+ 1jj i : 2So

h jpT (2) k j i p6e j m = j i = hj 2 j 0jj 2 jj i k2j + 1 p(W:;E:)

(3.84)

e h j m0j(x2 ; y2)j j m = j i (3:83) (2) (2) = eh j m0jT+2 j j m = j i + eh j m0jT;2 j j m = j i 0 z }| 0 { h j kT (2)k j i (2) pT = e hj 2 j 2jj 2 jm i p2j + 1 + e m0 j;2hj 2 j ; 2jj 2 jj ; 2i h j k2j +k1 j 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

94

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 px a complex function ei~ ~=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 En t n eitEZ =h jni = ei( h + ) jni = ei( hn + )jn tZ = 0 ti 0 2 En t ) d3xj~ ih~ j jn t0 = 0 ti = ei( h + ) d3xj~ ih~ j jn t0 = 0 ti x x x x Z Z x x ) d3xh~ jn t0 = 0 ti j~ i = d3xei( 2Ehn t + )h~ jn t0 = 0 tij~ i x x

4 Symmetry in Quantum Mechanics

2E t t) = ei( hn + ) n(~ t): x (4.1) So if we choose at any instant of time = ; 2Ent the wave function will be h real. (b) In the case of a free particle the Schrodinger equation is p2 jni = E jni ) ; h2 r (x) = E (x) ~ n 2m 2m n px px ) n (x) = Aei~ ~=h + Be;i~ ~=h (4.2) px px The wave functions n(2x) = e;i~ ~=h and 0n (x) = ei~ ~=h correspond to the p and so there is degeneracy since these correspond same eigenvalue E = 2m to di erent state kets j~i and j ; pi. So we cannot apply the previous result. p ~

)

x n (~

4.2 Let (~0) be the momentum-space wave function for state j i, p that is, (~0) = h~0j i.Is the momentum-space wave function for the p p

4. SYMMETRY IN QUANTUM MECHANICS

95

time-reversed state j i given by (~0), (;~0), (~0 ), or (;~0)? p p p p Justify your answer.In the momentum space we have Z Z j i = d3p0h~0 j ij~0i ) j i = d3p0 (~0)j~0i p p p p Z Z ) j i = d3p0 h~0j ij~0 i] = d3p0h~0 j i j~0i: p p p p For the momentum it is natural to require

(4.3)

h j~j i = ;h ~j~j ~ i ) p p ;1 j ~ i ) ~ ;1 = ;~ h ~j ~ p p pSo

(4.4) (4.5)

pp = p p ~j~0i (4:4) ;~ j~0i ) j~0i = j ; ~0i p pup to a phase factor. So nally Z Z j i = d3p0h~0j i j ; p0i = d3p0h;~0 j i j~0i p ~ p p ) h~0j j i = ~(~0) = h;~0j i = (;~0): p p p p

(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, 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 ikR n k (x) = n (x ; Ri )e iRi ika n k (x + a) = e n k (x)

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 di erent sites and/or di erent bands are orthogonal, i:e: proveZdx ? (x ; Ri) n(x ; Rj ) mij mn

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

is (4.7)

We can show that the functions PRi n(x ; Ri)eikRi satisfy the same relation X X ikR ik(R ;a) ika n (x + a ; Ri )e i = n x ; (Ri ; a)]e i e Ri Ri Ri ;a=Rj ika X ikR = e (4.8) n (x ; Rj )e jRj

n k (x):

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

(4.9)

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 ikR n k (x) = n (x ; Ri )e i Ri Z =a Z =a X ) dke;ikRj n k (x) = (x ; Ri) eik(Ri;Rj )dk (4.10) n; =aRi

; =a

4. SYMMETRY IN QUANTUM MECHANICSBut

97

eik(Ri;Rj ) =a = 2 sin =a(Ri ; Rj )] = i(R ; R ) Ri ; Rj 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 2 dke;ikRj n k (x) = n (x ; Ri ) ij a ; =a R Z =a i ) n(x ; Ri) = 2a ; =a e;ikRi n k (x)dk (4.12) So using the orthonormality properties of the Bloch functions Z dx m(x ; Ri ) n(x ; Rj ) Z Z Z a2 ikRi ;ik0 Rj n k0 (x)dkdk 0 dx = m k (x)e 2e (2 ) Z Z a2 Z ikRi ;ik0 Rj 0 = m k (x) n k0 (x)dxdkdk (2 )2 e Z Z a2 ikRi ;ik0 Rj 0 0 = mn (k ; k )dkdk 2e (2 ) Z =a 2 = a 2 mn eik(Ri;Rj )dk = 2a mn ij : (4.13) (2 )=a ik(R ;R ) e i j dk ; =a

Z

; =a

4.4 Suppose a spinless particle is bound to a xed center by a potential V (~ ) so assymetrical that no energy level is degenerate. x Using the time-reversal invariance prove ~ hLi = 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 asXXl m

Flm(r)Ylm(

)

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 = En jni: (4.15) If there is no degeneracy jni and jni can di er at most by a phase factor. Hence jni jni = ei jni: ~ (4.16) For the angular-momentum operator we have from (S-4.4.53) ~ ~ hnjLjni = ;hnjLjni (4= ;hnjLjni ) ~ ~ ~ :16) ~ hnjLjni = 0 :We have (4.17)

Z = d3xh~ jni j~ i (4= ei jni ) x x :16) ~ ~ ~ hx0j jni = hx0jni = ei hx0jni: (4.18) So if we use h~ jni = Pl Pm Flm(r)Ylm( ) x X X Flm(r)Ylm ( ) = ei Flm(r)Ylm( ) ml ml (S ;4:4:57) X m Y ;m ( ) = ei X F (r)Y m ( ) ) Flm(r)(;1) l lm l ml ml Z 0 X Z X ) Ylm Flm(r)(;1)m Yl;m ( )d = ei Ylm0 Flm(r)Ylm ( )d 0 0 ml X ml m X ) Flm(r)(;1) m0 ;m l0l = ei Flm(r) m0 m l0l

jni =

Z

d3xj~ ih~ jni = x x

Z

d3xh~ jni j~ i x x

)

Fl0 ;m

ml

0 0 (r)(;1);m

= ei

Fl0m0 (r)

) Fl0 ;m0 (r) = (;1)m0 Fl0m0 (r)ei : (4.19)

ml

4. SYMMETRY IN QUANTUM MECHANICS

99

4.5 The Hamiltonian for a spin 1 system is given by 2 H = ASz2 + B (Sx ; 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 = hm nm (4.20) So 0 1 0 1 1 0 0 1 0 0 Sz = h B 0 0 0 C ) Sz2 = h2 B 0 0 0 C @ A @ A 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 = 1 h1 njS+ j1 mi + 1 h1 njS; j1 mi 2 2 q (S ;3:5:39) 1 q 1 = h (1 ; m)(2 + m) n m+1 + 2 h (1 + m)(2 ; m) n m;1: 2 So 1 0 0 0 p 1 0 2 p C h p 0 0 0 Sx = h B 0 0 2 A + 2 B 2 p 0 C 0 @ A 2@ 0 0 0 0 2 0 p 0 1 0 0 p 2 p C = hB 2 p 0 2 A) @ 2 0 2 0 0 1 01 1 1 2 0 2 2 2 h2 B 0 4 0 C = h2 B 0 0 0 C : 2 (4.21) Sx = 4 @ A @ 1 A 1 0 1 2 0 2 2 2

100; In the same manner for the operator Sy = S+2iS; we 1 0 0 p 2 p 0 p h Sx = 2i B ; 2 p 2 C ) 0 A @ 0 ; 2 0 0 1 0 1 2 2 ;2 0 2 2 Sx = ; h B 0 ;4 0 C = h2 B 0 @ A @ 4 2 0 ;2 1 ;2

nd

1 0 ;1 2 C 1 0 A: 0 1 2

(4.22)

Thus the Hamiltonian can be represented by the matrix 0 1 A 0 B H = h2 B 0 0 0 C : @ A B 0 A

(4.23)

To nd the energy eigenvalues we have to solve the secular equation 0 2 1 Ah ; 0 B h2 C det(H ; I ) = 0 ) det B 0 ; 0 A=0 @ 2 2; Bh h 0 2Ah 2 i ) (Ah2 ; )2(; ) + (B h2)2 = 0 ) (Ah ; ) ; (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 jn c i that corresponds to the eigenvalue solve the following equation 0 10 1 0 1 A 0 B CB a C a 2 B 0 0 0 A@ b A = B b C: h @ c@ A B 0 A c c For1

c

we have to (4.25)

=0

0 10 1 ( A 0 B B 0 0 0 C B a C = 0 ) aA + cB = 0 h2 @ A@ b A aB + cA = 0 c B 0 A ( ( a = ;c B ) a = 0 A ) ;c B2 + cA = 0 c=0 A

(4.26)

4. SYMMETRY IN QUANTUM MECHANICSSo

101

0 1 0 1 0 norm: 0 jn0i = B b C = B 1 C ) @ A @ A 0 0 jn0i = j10i:= h2(A + B ) 1 0 aC b A = (A + B ) B @ c

(4.27)

In the same way for 0 10 A 0 B CB B 0 0 0 A@ @ B 0 A ( =c ) a= 0 b So

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

0 1 0 1 c norm: 1 1 jnA+B i = B 0 C = p2 B 0 C ) @ A @ A c 1 1 1 jnA+B i = p2 j1 +1i + p2 j1 ;1i:

(4.29)

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

1 8 > aA + cB = a(A ; B ) a C)< bA > 0 = b(A ; B ) : aB + cA = c(A ; B ) c(4.30)

0 1 0 1 c norm: 1 B 0 C = p B 1 C) jnA+B i = @ A @ 0 A 2 ;1 ;c 1 1 jnA;B i = p2 j1 +1i ; p2 j1 ;1i:

(4.31)

102 Now we are going to check if the Hamiltonian is invariant under time reversal

H

;1

2 = A Sz2 ;1 + B ( Sx ;1 ; Sy2 ;1) = A Sz ;1 Sz ;1 + B ( Sx ;1 Sx 2 = ASz2 + B (Sx ; Sy2) = H:

;1 ;

Sy

;1

Sy

(4.32)

;1 )

To nd the transformation of the eigenstates under time reversal we use the relation (S-4.4.58)

jl mi = (;1)mjl ;mi:So

(4.33)

jn0i =

:33) j10i (4= j10i = jn0i

1 1 jnA+B i = p2 j1 +1i + p2 j1 ;1i 1 1 (4:33) = ; p j1 ;1i ; p j1 +1i 2 2 = ;jnA+B i 1 1 jnA;B i = p2 j1 +1i ; p2 j1 ;1i(4:33)

(4.34) (4.35)

(4.36) (4.37)

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

(4.38)

5. APPROXIMATION METHODS

103

5 Approximation Methods

5.1 Consider an isotropic harmonic oscillator in two dimensions. The Hamiltonian is given by (a) What are the energies of the three lowest-lying states? Is there any degeneracy? (b) We now apply a perturbationV = m!2xy p2 + p2 + m!2 (x2 + y2) y x H0 = 2m 2m 2

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 0 jxjni = h=2m! (pn + 1 0 You may use hn + n 0 ):]n n+1 n n;1

De ne step operators:

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

(5.1)

104 De ning the number operators

Nx ay ax xwe nd

Ny ay ay y

H0 N Nx + Ny = h! ; 1 ) H0 = h!(N + 1):I.e. energy eigenkets are also eigenkets of N :

(5.2)

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 iso 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 = E0 + 1 + 2 + : : : j l i = j l0 i + j l1 i + : : :so that the full problem becomes h i (E 0 ; H0 ) j l0 i + j l1 i + : : : = (V ;1; 2 : : :)

(5.4)

h

j l0 i + j l1 i + : : : :

i

5. APPROXIMATION METHODSTo 1'st order: (E 0 ; H0) j l1 i = (V ; Multiply with h l0 j to nd1 ) j l0 i:

105

(5.5)

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 ; 1h m0 j l0 i = h m0 j V j l0 i:

1 j l0 i )

(5.7)

Now, h m0 j l0 i is not necessarily kl since only states corresponding to di erent 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: This is the eigenvalue equation which gives the correct zeroth order eigenvectors! Let us use all this: 1. The ground state is non-degenerate )1 = h0 00

k 2D

0 j V j 0 0 i = m!2h 0 0 j xy j 0 0 i h 0 0 j (ax +ay )(ay +ay ) j 0 0 i = 0 x y

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 V = m!2xy = m!2 2m! (ax+ay )(ay +ay ) = h! (axay +ay ay +axay +ay ay ) x y x y x y 2 and ax j m n i = m j m ; 1 n i

p

ay j m n i = m + 1 j m + 1 n i etc: x

p

106 Together this gives V10 10 = V01 01 = 0 V10 01 = h! h 1 0 j ax ay j 0 1 i = y 2 V01 10 = h! h 1 0 j ay ay j 0 1 i = x 2 The V -matrix becomes

h! 2 h! : 2

(5.8)

! h! 0 1 2 1 0 and so the eigenvalues (= 1) are h! : 1= 2 To get the eigenvectors we solve ! ! ! 0 1 x = x 1 0 y y and get 1 E+ = h!(2 + 2 ) j i+ = p2 ( j 0 1 i + j 1 0 i) 1 j i; = p2 ( 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: h! V11 20 = V20 11 = V11 02 = V02 11 = p (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 0 1 0 = det B 1 ; 1 C = ; ( 2 ; 1) + = (2 ; 2) @ A 0 1 ;

5. APPROXIMATION METHODSwhich means that the eigenvalues are f0 as above we get the eigenvectors

107

h!g. By the same method E+ = h!(3 + ) E0 = 3h! E; = h!(3 ; ):

p 1 j i+ = 2 ( j 2 0 i + 2 j 1 1 i + j 0 2 i) 1 j i0 = p2 (; j 2 0 i + j 0 2 i) p 1 j i; = 2 ( j 2 0 i ; 2 j 1 1 i + j 0 2 i )

(c) To solve the problem exactly we will make a variable change. The potential is h1 i m!2 2 (x2 + y2) + xy = # " 2 1 ((x + y )2 + (x ; y )2) + (x + y )2 ; (x ; y )2) : (5.11) = m! 4 4 Now it is natural to introduce 2 2 1 (x ; y) 1 (p0 ; p0 ): y0 p p0y p x y (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 H = 21 (p0x2 + p0y2) + m! (1 + )x02 + (1 ; )y02]: m 2 p p 0 0 So we get one oscillator with !x = ! 1 + and another with !y = ! 1 ; . The energy levels are: 1 p (x + y) 1 p (p0x + p0y )

x0

p0x

E0 0 = h! p 0 E1 0 = h! + h!x = h!(1 + 1 + ) = 1 = h!(1 + 1 + 1 + : : :) = h!(2 + 2 ) + O( 2) 2

108

E0 1 E2 0 E1 1 E0 2

= = = =

0 1 h! + h!y = : : : = h!(2 ; 2 ) + O( 2) 0 h! + 2h!x = : : : = h!(3 + ) + O( 2) 0 0 h! + h!x + h!y = : : : = 3h! + O( 2) 0 h! + 2h!y = : : : = 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 matrix0 1 E1 0 a C B 0 E1 b A @ 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: E1 ; E 0 a 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 ) + E1 + E2 2 ; E E + jaj2 + jbj2 = E = E1 + E2 1 2 2 2

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

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 1 E = E1 + E2 + E1 ; E2 1 + 2 (jaj2 + jbj2)( E ; E )2 + : : : = 2 2 1 2 2 + jbj2 = E1 + jaj ; E E1 2 2 + jb 2 E = E1 + E2 ; E1 ; E2 (: : :) = E2 ; jaj ; Ej : 2 2 E1 2 (5.16)(b) Non degenerate perturbation theory to 2'nd order. The basis we use is 0 1 0 1 0 1 1C 0C 0 j1i = B 0 A j2i = B 1 A j3i = B 0 C: @ @ @ A 0 0 1 0 1 0 0 aC The matrix elements of the perturbation V = B 0 0 b A 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) = h k j V j k i = 0 1'st order gives nothing. But the 2'nd order kshifts are(2) 1 (2) 2 (2) 3 2 X jVk1 j2 3 1 2 = jh Ej V jE ij = E jaj E 0 0 1; 2 1; 2 k6=1 E1 ; Ek 2 2 X jVk2 j jh 3 j V j 2 ij = jbj2 = 0 0 = E1 ; E2 E1 ; E2 k6=2 E2 ; Ek 2 2 2 + jb 2 X jVk3 j2 = = E jaj E + E jbj E = ; jaj ; Ej : 0 0 E1 2 2; 1 2; 1 k6=3 E3 ; Ek

=

(5.17)

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 , so 1'st order pert.thy. will again give nothing. We have subspace is 0 0 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. Di erent solutions to the full problem are denoted by di erent 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 X 0 ih 0 V 0 (5.20) ) j l1 i = j k E 0k;jE j l i : 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:k6=D k

5. APPROXIMATION METHODSInserting the expression (5.20) for j l1 i we get X h m0 j V j k0 ih k0 j V j l0 i = E 0 ; Ek k6=D

111

(2)h m0 j l0 i:

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: E 0 ; Ek n2D k6=D Maybe it looks more familiar in matrix form X Mmn xn = (2)xmn2D

where

Mmn = xm

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

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 M12 = EV13V32 0 = E ab E M11 = EV13V31 0 = E jaj E 1 ; E3 1; 2 1 ; E3 1; 2 2 V23V31 = a b jV23j = jbj2 : M21 = E ; E 0 E ; E M22 = E ; E 0 E ; E1 3 1 2 1 3 1 2

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

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, (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 excited states? q You may use hn0jxjni = h=2m! (pn + 1 n0 n+1 + pn n0 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 = h!(n + 1 ) 2 where we get cn (t) from its di . eqn. (S. 5.5.15): X @ ih @t cn (t) = Vnm ei!nmtcm(t) m Vnm = h n j V j m i !nm = En ; Em = !(n ; m) h We need the matrix elements Vnm Vnm = h n j ; F0s ;t= j m i = ;F0e;t= h n j x j m i = xe p h p = ;F0e;t= 2m! ( m n m;1 + m + 1 n m+1):0 En

F (t) = F0e;t=

n

(5.22)

5. APPROXIMATION METHODSPut it back into (5.22)

113

s @ c (t) = ;F e;t= h pn + 1e;i!tc (t) + pnei!tc (t) : ih @t n 0 n+1 n;1 2m! Perturbation theory means expanding cn(t) = c(0) + c(1) + : : :, and to zeroth n n order this is @ c(0)(t) = 0 ) c(0) = n0 n @t n To rst order we get Zt X c(1)(t) = i1 dt0 Vnm (t0)ei!nmt0 c(0) = n m h 0 m s p p h Zt 0 = ; Fh 2m! 0 dt0e;t= n + 1e;i!t0 c(0) (t) + nei!t0 c(0)1 (t) n+1 n; i 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 coe cient s F0 h Z t dt0ei!t0;t0 = p1 c(1)(t) = ; ih 2m! 1;1 0 = 1 0 s # " F0 h 1 e(i!; 1 )t0 t = ; ih 2m! i! ; 1 0 s h 0 = Fh 2m! 1 1 1 ; e(i!; 1 )t i i! ; and X h h j t i = c(1)(t)e ;iEnt j n i = c(1)(t)e ;iE1t j 1 i: 1 nThe probability of nding the H.O. in j 1 i is As t ! 1n

jh 1 j t ij2 = jc(1)(t)j2: 1

s F0 h 1 = const: c(1) ! ih 2m! i! ; 1 1 This is of course reasonable since applying a static force means that the system asymptotically nds a new equilibrium.

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.1 5.4 Consider a composite system made up of two spin 2 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

~ ~ H = 4 2 S1 S2: 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: ~ ~ S1 S2 = S1x" 2x + S1y S2y + S1z S2z = fin this basisg S 2 = h ( j + ih ; j + j ; ih + j )1 ( j + ih ; j + j ; ih + j )2 + 4 + 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 = " h2 j + + ih ; ; j + j + ; ih ; + j + = 4 + j ; + ih + ; j + j ; ; ih + + j + 2 ( j + + ih ; ; j ; j + ; ih ; + j + + i ; j ; + ih + ; j + j ; ; ih + + j ) +

5. APPROXIMATION METHODS+ j + + ih + + j ; j + ; ih + ; j + # ; j ; + ih ; + j + j ; ; ih ; ; j =

115

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 C H = B 0 ;1 ;1 0 C : B0 2 2 0C (5.23) @ A 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:

) = 1 ;3! ! x = x y y 1 ) ;x + 2y = x ) x = y = p2

;1 2 2 ;1

!

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

= ;3 :

;1 2 2 ;1

!

2

116 This was a cumbersome but straightforward way to calculate the spectrum. ~ ~ ~ A smarter way would have been to use S = S1 + S2 to nd ~ ~2 ~2 ~ ~ ~ ~ 1 ~ ~2 ~2 S 2 = S 2 = S1 + S2 + 2S1 S2 ) S1 S2 = 2 S 2 ; S1 ; S21 ~2 ~2 We know that S1 = S2 = h2 2 1 2 +1 1 2

= 3h2 so 4

~ ~ S1 S2 =

S2 ;

3h2 2

!

1 Also, we know that two spin 2 systems add up to one triplet (spin 1) and one singlet (spin 0), i.e. ~ ~ 1 S = 1 (3 states) ) S1 S2 = 2 (h21(1 + 1) ; 3h2 ) = 1 h2 2 4 : (5.24) 3 ~1 S2 = 1 (; 3h2 ) = ; 4 h2 ~ 2 2 S = 0 (1 state) ) S ~ ~ Since H = 4h2 S1 S2 we get

h2 E(spin=1) = 4 2 14 = h 4 ;3h2 = ;3 : E(spin=0) = 2 4 h

From Clebsch-Gordan o decomposition we know that j + + i j ; ; i 1 1 p ( j + ; i + j ; + i) are spin 1, and p ( 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 1 j 1 i = j + + i j 2 i = j ; ; i j 3 i = p2 ( j + ; i + j ; + i) o 1 j 4 i = p2 ( j + ; i ; j ; + i)

n

(5.25)

5. APPROXIMATION METHODSthe initial state is 1 j + ; i = p2 ( j 3 i + j 4 i):

117

Acting with U (t 0) on that we get 1 i j t i = p exp ; h Ht ( j 3 i + j 4 i) = 2 1 exp ; i t j 3 i + exp 3i t j 4 i = = p h h "2 1 = exp ;ih t p ( j + ; i + j ; + i)+ 2 # 3 i t p ( j + ; i ; j ; + i) = 1 +exp h 2 h i 1 (e;i!t + e3i!t) j + ; i + (e;i!t + e3i!t) j ; + i = 2 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 ti = h; ; j ti = 0 > > < 1 2 1 4i!t ;4i!t 2 > jh + ; j t ij = 4 (2 + e + e ) = 2 (1 + cos4!t) ' 1 ; 4(!t) : : : > > : 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) = ni n Zt (1) (t) = ;i dt0ei!nit0 Vni (t0): cn h t0 Here we have (using the original basis) H0 = 0, V given by (5.23)

(5.27)

jii = j + ;i

118

Vfi = 2 Vni = 0 n 6= f: Inserting this into (5.27) yields c(0) = c(0) ; i = 1 i j+ Z (1) = c(1) = ; i t cf (5.28) j ;+ i h 0 dt2 = ;2i!t: as the only non-vanishing coe cients 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 P ( j ; + i) = jc(1)(t) + c(2)(t) + : : : j2 = j2i!tj2 = 4(!t)2 f f 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: h: !t 1 ) t

jf i = j ; +i !ni = En ; Ei = fEn = 0g = 0 h

5.5 The ground state of a hydrogen atom (n = 1,l = 0) is subjected to a time-dependent potential as follows: V (~ t) = V0 cos(kz ; !t): x Using time-dependent perturbation theory, obtain an expression for the transition rate at which the electron is emitted with momentum ~. Show, in particular, how you may compute the angular p distribution of the ejected electron (in terms of and de ned with respect to the z-axis). Discuss brie y the similarities and the di erences between this problem and the (more realistic) photoelectric e ect. (note: For the initial wave function use2 1 Z 3e;Zr=a0 : x n=1 l=0 (~ ) = p a0

5. APPROXIMATION METHODS

119

If you have a normalization problem, the nal wave function may be taken to be 1 with L very large, but you should be able to show that the observable e ects 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 ~f . The initial wave-function is p 1 1 3=2e;r=a0 x i (~ ) = p a0 and the nal wave-function is 1 ei~ ~=h: px x f (~ ) = L3=2 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 y 2 wi!n = 2h Vni (En ; (Ei + h!)) because the atom absorbs a photon h!. The matrix element is 2 y Vni 2 = V40 eikz ni 2 and Z ikz ~ f j eikz j n = 1 l = 0 i = d3xh ~ f j eikz j x ih x j n = 1 l = 0 i = e ni = h k k Z e;i~f ~ k x 1 1 3=2 = d3x L3=2 eikx3 p a e;r=a0 = 0 Z 1 k x = 3=2p 3=2 d3xe;i(~f ~;kx3 );r=a0 : (5.30) L a0

x f (~ ) =

2 L3

px ei~ ~=h

120 So eikz ni is the 3D Fourier transform of the initial wave-function (and some constant) with ~ = ~ f ; k~z . That can be extracted from (Sakurai problem q k e 5.39) 1 64 2 eikz ni = L3a5 h 1 i4 0 a2 + (~ f ; k~z )2 k e 0 The transition rate is understood to be integrated over the density of states. We need to get that as a function of ~f = h~ f . As in (S.5.7.31), the volume p k element is dn n2dnd = n2d dp dpf : f Using 2 2 2 2 p kf = f2 = n (22 ) L h we get dn = 1 2L2pf = 2 h L2pf = L dp 2n (2 h)2 Lp (2 h)2 2 h which leavesf f2 L3kf L3p2 = (2 )3h d dpf = (2 hf)3 d dpf and this is the sought density. Finally, 2 64 2 L3 p2 1 wi!~f = 2h V40 L3a5 h 1 i4 (2 hf)3 d dpf : p 0 a2 + (~ f ; k~z )2 k e 0

n2dnd

Note that the L's cancel. The angular dependence is in the denominator: ~ f ; k~z 2 = (jkf jcos ; k) ~z + jkf jsin (cos'~x + sin'~y )]2 = k e e e e = jkf j2cos2 + k2 ; 2kjkf jcos + jkf j2sin2 = 2 = kf + k2 ; 2kjkf jcos : (5.31) In a comparison between this problem and the photoelectric e ect 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.