Problem Sheet 3 Solutions

8/9/2019 Problem Sheet 3 Solutions

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2010/11

DEPARTMENT OF PHYSICS & ASTRONOMY

PHAS3226 QUANTUM MECHANICS

Problem Paper 3 - solutions

Solutions to be handed in on Tuesday 16 November 2010

1. (a) The component of spinSin the direction of the unit vector nisSn = S nand has two normalizedeigenfunctions

j+ni= cos (=2) ji+ sin (=2) eiji and j

ni=sin(=2) eiji+ cos (=2) ji

satisfying Snj+n i = h2 j+n i and Snjn i =h2 jn i, where (; ) are the polar angles of theunit vector n. If ji= 1(r) ji+2(r) ji is a normalized state, show that the probability ofobtaining the value h=2 in a measurement of Sn is [10]

P(Sn = h=2) =j1(r)j2 cos2 (=2) +j2(r)j2 sin2 (=2) + sin Re12e

i :Expand the stateji in terms of the eigenstates ofSn as

ji= aj+ni+ bj

ni= 1(r) ji+ 2(r) ji: (1)

The probability of obtaining state with Sn = 1

2hi.e. to be in statej+

niis given byjaj2. But

a = h+

n ji= 1(r

) h+

n ji+ 2(r

) h+

n ji (2)= 1(r)cos(=2) + 2(r)sin(=2) e

i: (3)

Hence

jaj2 = 1(r)cos(=2) + 2(r)sin(=2) ei 1(r)cos(=2) + 2(r)sin(=2) ei ;= j1(r) j2 cos2 (=2) +j2(r) j2 sin2 (=2)

+2(r)sin(=2) ei1(r)cos(=2) +

1(r)cos(=2)2(r)sin(=2) e

i;

= j1(r) j2 cos2 (=2) +j2(r) j2 sin2 (=2) + 2Re1(r)2(r)cos(=2)sin(=2) e

i ;= j1(r) j2 cos2 (=2) +j2(r) j2 sin2 (=2) + Re

1(r)2(r)sin() e

i ;= j1(r) j2 cos2 (=2) +j2(r) j2 sin2 (=2) + sin Re

1(r)2(r)sin() e

i

:(b) Obtain, using the raising and/or lowering operators, the six wave functionsjj;mifor thep-states

of an electron in terms of the spherical harmonics Ym1 ( j1;mi, m=1; 0; 1) and the spinorsji ( j1

2; 12i) andji ( j1

2;1

2i). [10]

For the p-state electron the orbital quantum number ` = 1, has three z -projections m = 1; 0;1,i.e. statesj1; 1i,j1; 0i,j1;1i ( Ym` = Y11; Y01; Y11 ). The two spin statesji ( j 12 ; 12 i) andji ( j1

2;1

2i) have ms = 12 and 12 respectively. The maximum total angular momentum has

J=` + 12

, the minimum is J=` 12

. The highest (top) state has J= 32

, MJ= 3

2and is clearly

given by

3

2;3

2

=Y11; j

3

2;3

2i=j1; 1ij1

2;1

2i (4)

asMJ =m`+ ms. The state J= 3

2, MJ=

3

2is clearly given by

3

2;3

2

=Y11 ; j

3

2;3

2i=j1;1ij1

2;1

2i: (5)

The state J = 32

, MJ = 1

2 can be obtained from the J = 3

2, MJ =

3

2one by application of the

lowering operator J. Recall that in general JjJ;Mi =pJ(J+ 1)M(M1)hjJ;M1i,

thus

Jj32;3

2i=

s3

2

3

2+ 1

3

2

3

21

hj3

2;1

2i=

p3hj3

2;1

2i: (6)

In terms of the individual orbital and spin angular momenta, J = L +S so using Lj`;m`i=

p` (` + 1)m`(m`1)hj`;m` 1iandSjs;msi= ps (s + 1)ms(ms1)hjs;ms 1i, with

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`= 1, m` = 1, s = 1

2, ms=

1

2,

Jj32;3

2i = L+

S j1; 1ij1

2;1

2i; (7)

=Lj1; 1i

j12;1

2i+j1; 1iSj1

2;1

2i;

p3hj3

2;1

2i =

p2hj1; 0ij1

2;1

2i+ hj1; 1ij1

2;1

2i

p2hY01 + hY

11; (8)

j32;1

2i =

r2

3hj1; 0ij1

2;1

2i+ 1p

3j1; 1ij1

2;1

2i

r2

3hY01 +

1p3

hY11: (9)

The state J= 32

, MJ =12 can be obtained from the J= 32 , MJ = 12 state by application ofJor by applying J+ to the state J=

3

2, MJ =32 ,

J+j32;3

2i =

L++ S+

j1;1ij12;1

2i; (10)

=L+j1;1i

j12;1

2i+j1;1iS+j1

2;1

2i;

p3hj3

2;1

2i =

p2hj1; 0ij1

2;1

2i+ hj1;1ij1

2;1

2i

p2hY01+ hY

11 ; (11)

j32;1

2i =

r2

3hj1; 0ij1

2;1

2i+ 1p

3j1;1ij1

2;1

2i

r2

3hY01+

1p3

hY11 : (12)

The state J = 12

, MJ = 1

2 must be a linear combination of Y01 =j1; 0ij12 ; 12 i and Y11 =j1; 1ij1

2;1

2isince MJ=m`+ ms, thus

j12;1

2i= aj1; 0ij1

2;1

2i+ bj1; 1ij1

2;1

2i: (13)

But

J+j12;1

2i = 0 =

L++ S+

aj1; 0ij1

2;1

2i+ bj1; 1ij1

2;1

2i

(14)

=

aL+j1; 0i

j12;1

2i+ bj1; 1iS+j1

2;1

2i

0 = ap

2hj1; 1ij12;1

2i+ bhj1; 1ij1

2;1

2i=

ap

2 + b

hj1; 1ij12;1

2i (15)

givingap

2 + b= 0. Normalization requiresjaj2 + jbj2 = 1, sojaj2 + 2jaj2 = 1and a = 1p3

. Hence

j1

2

;1

2 i=

1

p3 j1; 0

ij1

2

;1

2 i r2

3 j1; 1

ij1

2

;

1

2 i 1

p3Y01

r2

3

Y11: (16)

Applying J gives

Jj12;1

2i =

L+ S

1p3j1; 0ij1

2;1

2i

r2

3j1; 1ij1

2;1

2i!; (17)

hj12;1

2i =

r2

3hj1;1ij1

2;1

2i+ 1p

3j1; 0ihj1

2;1

2i

r2

3

p2hj1; 0ij1

2;1

2i;

j12;1

2i =

r2

3j1;1ij1

2;1

2i 1p

3j1; 0ij1

2;1

2i

r2

3Y11

r1

3Y01: (18)

Note the state J = 12

, MJ

= 12

could also be found by requiring it to be orthogonal to the state

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J= 32

, MJ= 1

2 i.e.

h32;1

2j12;1

2i = 0;

0 =r

23

hh1; 0jh12;1

2j+ 1p

3h1; 1jh1

2;1

2j!

aj1; 0ij12;1

2i+ bj1; 1ij1

2;1

2i;

0 =

r2

3ha +

1p3b;

as before.

2. (a) Show that the operatorL Scan be expressed in terms of the raising and lowering operators L+,L, S+, S, and the componentsLz, Sz by L S= 12

L+S+LS+

+LzSz .} [4]

The operator L S= LxSx +LySy+LzS. But the the raising/lowering operators L = Lx iLygive Lx =

1

2 L++ L and Ly = 1

2i L+ L and similar relations for Sx and Sy. ThusL S = 1

2

L++ L

12

S++ S

+

1

2i

L+L

12i

S+S

+LzS (19)

= 1

2

L+S+ LS+

+LzS: (20)

(b) If a particle has spin 12

and is in a state with orbital angular momentum `, there are two ba-sis states,j`;s;`z; szi which can be expressed in terms of the individual states asj`;s;`z; szi =j`; `zijs; szi with total z-component of angular momentummh, namely

jai j ;m 12; 12i=j ;m 1

2ij1

2; 12i

and

jb

i j`; 1

2;m + 1

2;

1

2

i=

j`;m + 1

2

ij1

2;

1

2

i:

With these two statesjai,jbi as a basis show that the matrix representation of the operator L Sis (Hint: expressLS in terms of the raising and lowering operatorsL+, L, S+, S, and thecomponentsLz, Sz.) [16]

L S=

0B@ 12

m 1

2

1

2

h` + 1

2

2 m2i1=21

2

h` + 1

2

2 m2i1=2 12

m + 1

2

1CA h2:

The matrix representation ofL S with the basis states,jai,jbi is

L S=hajL Sjai hajL Sjbi

hb

jL

S

ja

i hb

jL

S

jb

i : (21)

In general Lj`;mi =p

(` + 1)m (m1)hj`;m1i with a similar relation for S. Thusexplicitly

L+j`;m 12 i =q

(` + 1) m 12

m + 1

2

hj`;m + 1

2i; (22)

=

r`2 + ` +

1

4m2hj`;m + 1

2i;

=

q` + 1

2

2 m2hj`;m + 12i: (23)

Similarly for

Lj`;m + 1

2 i = q(` + 1) m + 12 m 12hj`;m 12 i; (24)

=

q` + 1

2

2 m2hj`;m 12i: (25)

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For S, S+j12 ;12 i =q

1

2

1

2+ 1

12

1

2

hj1

2; 12i = hj1

2; 12i, S+j12 ; 12 i = Sj12 ;12 i = 0 and

Sj12 ; 12 i= hj12 ;12 i.To evaluate the matrix elements, L

S

jai

= 12L+S+ LS++LzS j`;m 12 ij12 ; 12 i note that

L operators only operate on thej`;m 12i states and Soperators only operate on thej1

2;1

2i states.

Hence using the relations above for the action ofLand Soperators on the states gives

L Sjai =

1

2

L+S+ LS+

+LzS

j`;m 1

2ij1

2; 12i (26)

L Sj`;m 12ij1

2; 12i = 1

2

q` + 1

2

2 m2hj`;m + 12ihj1

2;1

2i+ m 1

2

h

1

2hj`;m 1

2ij12; 12i(27)

and

L Sjbi =

1

2 L+S+ LS+

+LzSz

j`;m + 1

2ij12;1

2i (28)

= 12

q` + 1

2

2 m2hj`;m 1

2ihj1

2;1

2i+ m + 1

2

h

12

h

j`;m 12ij12; 12i (29)

Hence using the orthogonality of the statesj`;m + 12i,j`;m 1

2i,j1

2; 12iandj1

2;1

2i gives

hajL Sjai = 12

m 1

2

h2;

hajL Sjbi = 12

` + 1

2

2 m2 h2;hbjL Sjai = 1

2

` + 1

2

2 m2 h2;hbjL Sjbi = 1

2 m + 1

2 h2:

3. (a) Write down, without proof, the wave functions for two spin- 12

particles which as eigenstatesjS;Sziof denite total angular momentumS and z-component Sz. [4]

The singlet state, S= 0, is

j0; 0i= 1p2

(1212) : (30)

The triplet states , S= 1, are

j1; 1i = 12; (31)j1; 0i = 1p

2(12+ 12) ; (32)

j1;

1i

= 12: (33)

(b) Suppose that particles a and b interact through a magnetic dipole-dipole potential

V =A(a b) r2 3 (a r) (b r)

r5 ;

wherer is the inter-particle separation anda andb are the Pauli matrices referring to particlesa and b respectively. If the two particles are a xed distance d apart along the z-axis, showthat (Hint; express V in terms of the total spin operator S2 and z-componentSz and recall,2 =2x+

2y+

2z = 3)

i. Vdoes not mix the statesjS;Szi, i.e. the o-diagonal elements are zero,ii. and that the diagonal elements are

h1; 1jVj1; 1i=h1;1jVj1;1i=2Ad3

; h1; 0jVj1; 0i= 4Ad3

; h0; 0; jVj0; 0i= 0:

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As the two particles are separated by a xed distance d along the z-axis, the dipole-dipoleinteraction reduces to

V =A(a b) d2 3 (az) (bz) d2

d5 =

A

d3[(a

b)

3 (az) (bz)] (34)

since r= dz so a r= a zd= az, and similarly for b r.The total spin

S= Sa+ Sb= h

2(a+ b) (35)

and

S2 =

h2

4

2a+

2b+ 2ab

(36)

and so

ab = 12

4

h2S2 2a2b

: (37)

But 2

=2

x+ 2

y+ 2

z = 3 soa b = 2

h2S2 3: (38)

Similarly sinceSz = h2

(az+ bz), then

azbz = 1

2

4

h2S2z 2az 2bz

;

= 2

h2S2z1: (39)

Using eq(??) and eq(??) in eq(34) gives

V = A

d3 2

h2S2 33 2h2 S2z1 ;

2A

d3h2

hS2 3S2z

i: (40)

Thus since S2jS;Szi= S(S+ 1) h2jS;Sziand SzjS;Szi= SzhjS;Szithen S2z jS;Szi= S2zh2jS;Sziand

VjS;Szi= 2Ad3h2

hS2 3S2z

ijS;Szi= 2A

d3h2S(S+ 1) h2 3S2zh2

jS;Szi: (41)Hence the matrix elements are

hS0; S0zjVjS;Szi= 2A

d3h2 S(S+ 1) h2 3S2zh2

hS0; S0zjS;Szi: (42)

Since the statesjS;Szi are orthonormal, i.e. hS0; S0zjS;Szi = S0SS0zSz eq(??) shows that theo-diagonal matrix elements are all zero. [2]

The diagonal elements are [8]

h1; 1jVj1; 1i = 2Ad3h2

2h2 3h2 =2A

d3; (43)

h1; 0jV1; 0i = 2Ad3h2

2h2

=

4A

d3; (44)

h1;1jVj1;1i = 2Ad3h2

2h2 3h2 =2A

d3; (45)

h0; 0jVj0; 0i = 0: (46)

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(c) At time t= 0 the two spins are pointing towards each other. By expressing this state in terms of

a linear combination of thejS;Szistates show that after a time t= hd3

4Athe spins will be pointing

away from each other. [6]

Recall from the second year course, that the general solution of the time-dependent Schrdingerequation is j (t)i =Pn cnjnieiEnt=h where jni satises the time-independent Schrdingerequation Hjni= Enjni with energy eigenvalueEn.At timet = 0if the two spins are pointing towards each other the spin statej (0)imust be either12 or12 depending on which particles are labelled a and b. Taking the rst case12, thenstate can be expressed in terms of the total spin statesjS;Szias

j (0)i= 11 = 1p

2[j1; 0i+j0; 0i] : (47)

Thus the general time-dependent wavefunction is

j (t)i= aj1; 0ieiE10t=h + bj0; 0ieiE00t=h (48)

whereE10and E00are the energies of thej1; 0iandj0; 0istates respectively. These are found fromthe time-independent Schrdinger equation VjS;Szi= ESS

zjS;Szi. Hence ESS

z=hS;SzjVjS;Szi

and so E10 =h1; 0jVj1; 0i= 4Ad3 and E00 =h0; 0jVj0; 0i= 0, and thus

j (t)i= aj1; 0iei4At=hd3 + bj0; 0i: (49)

Thus evaluating eq(??) att = 0and comparing with equ(??) gives a = b = 1p2

, and hence

j (t)i= 1p2j1; 0iei4At=hd3 + 1p

2j0; 0i: (50)

Express this in terms of the individual particle spin states gives

j (t)i = 1p2

1p

2(12+ 12)

ei4At=hd

3

+ 1p

2

1p

2(12 12)

;

= 1

2

ei4At=hd

3

+ 112+

1

2

ei4At=hd

3 112

= ei2At=hd31

2

hei2At=hd

3

+ ei2At=hd312+

ei2At=hd

3 ei2At=hd3i

12

= ei2At=hd3

12cos

2At

hd3

i12sin

2At

hd3

: (51)

Thus initially at t = 0 the system is in a pure 12 state. It is in a pure 12 state whencos

2Athd3

= 0, i.e. when 2Athd3 = (2n + 1) =2, with n a positive integer. Thus the spins are rst

pointing away from each other at a time t = hd3=4A.

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Problem Sheet 3 Solutions