HOMEWORK ASSIGNMENT 13: Solutions

[Pages:13]PHYS851 Quantum Mechanics I, Fall 2009

HOMEWORK ASSIGNMENT 13: Solutions

1. In this problem you will derive the 2?2 matrix representations of the three spin observables from first principles:

(a) In the basis {| z , | z }, the matrix representation of Sz is of course

Sz =

z |Sz| z z |Sz| z

z |Sz| z z |Sz| z

.

(1)

Use Eqs. (1) and (2) to find the four matrix elements of Sz in the basis of its own eigenstates.

From Sz| z = 2 | z and the orthonormality of the basis, it follows that z |Sz| z = 2 and z |Sz| z = 0.

From Sz| z = - 2 | z and the orthonormality of the basis, it follows that z |Sz| z = 0 and z |Sz| z = - 2 . This gives us

Sz =

10 0 -1

(2)

(b) Invert the definitions S+ = Sx + iSy and S- = Sx - iSy, to express Sx and Sy in terms of S+ and S-.

Inverting these equations gives

Sx

=

1 2

(S+

+

Sz

)

(3)

Sy

=

1 2i

(S+

-

S-)

(4)

(c) Use the equation

S?|s, ms = s(s+1) - ms(ms?1)|s, ms?1

(5)

to find the matrix elements of S+ and S- in the basis {| z , | z .

This formula gives S+| z = 0, S+| z = | z , so that orthonormality gives

S+ =

01 00

(6)

Likewise, S-| z = | z and S-| z = 0, so that

S- =

00 10

(7)

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(d) From your answers to 13.1.b and 13.1.c, derive the matrix representations of Sx and Sy for spin-1/2.

Sx

=

1 2

(S+

+

S-)

=

2

01 10

(8)

Sy

=

1 2i

(S+

-

S-)

=

2i

01 -1 0

=2

0 -i i0

(9)

(e) Explicitly verify that these operators satisfy the angular momentum commutation relations.

2

[Sx, Sy] = 4

01 10

0 -i i0

-

0 -i i0

01 10

2

=4

i0 0 -i

-

-i 0 0i

2

= i2

10 0 -1

= i Sz

(10)

2

[Sy, Sz] = 4

0 -i i0

10 0 -1

-

10 0 -1

0 -i i0

2

=4

0i i0

-

0 -i -i 0

2

= i2

01 10

= i Sx

(11)

2

[Sz, Sx] = 4

10 0 -1

01 10

-

01 10

10 0 -1

2

=4

01 -1 0

-

0 -1 10

2

= i2

0 -i i0

= i Sy

(12)

2

(f) Show explicitly that S2 = 2s(s+1)I.

S2 = Sx2 + Sy2 + Sz2

(13)

2

=4

01 10

01 10

2

+4

0 -i i0

0 -i i0

2

+4

10 2 0 -1

2

=4

10 01

2

+4

10 01

2

+4

10 01

=

32 4

10 01

= 2s(s + 1)I

(14)

(g) Based on symmetry, write the 2?2 matrix representations of Sx, Sy, and Sz in the basis of eigenstates of Sy.

A cyclic permutation (relabeling) of the indices leaves the commutation relations unchanged. Thus we can relabel the indices according to x = y, y = zm and z = x. Thus the eigenstates of y are just the eigenstates of z, in terms of the primed indices we have

Sx = Sy =

0 -i i0

(15)

Sy = Sz =

10 0 -1

(16)

Sz = Sx =

01 10

(17)

Note that the eigenstates of Sy are only defined up to a phase-factor, different phase factor choices will lead to different sets of matrices. The above result, which reflects the permutation symmetry of the angular momentum group, will be generated by setting the phase factors so that

| y

=

ei 2

(|

z

+ i| z)

(18)

| y

=

ei 2

(|

z

- i| z )

(19)

where is an arbitrary phase.

3

2. Pauli spin matrices: The Pauli spin matrices, x, y, and z are defined via

S = s

(20)

(a) Use this definition and your answers to problem 13.1 to derive the 2?2 matrix representations of the three Pauli matrices in the basis of eigenstates of Sz.

With s = 1/2, this gives

x =

01 10

(21)

y =

0 -i i0

(22)

z =

10 0 -1

(23)

(b) For each Pauli matrix, find its eigenvalues, and the components of its normalized eigenvectors in the basis of eigenstates of Sz.

Each Pauli matrix has eigenvalues 1 and -1. The eigenvectors are

| z

1 0

(24)

| z

0 1

(25)

| x

1 2

1 1

(26)

| x

1 2

1 -1

(27)

| y

1 2

1 i

(28)

| y

1 2

1 -i

(29)

(c) Use your answer to 13.2.b to obtain the eigenvalues of Sx, Sy, and Sz, as well as the components of the corresponding normalized eigenvectors in the basis of eigenstates of Sz.

Each component of S has eigenvalues /2 and - /2. The eigenvectors are the same as in 13.2(b).

4

3. Repeat problems 13.1.(a-d) and 13.2.a for the case of a spin-1 particle.

For s = 1, the eigenvalues of Sz are 1, 0, and -1. Thus we can introduce the basis {|1 , |0 , | - 1 },

defined by Sz|m = m|m .

In this basis we must have

1 0 0

Sz = 0 0 0

(30)

0 0 -1

We

still

have

Sx

=

1 2

(S+

+

S-)

and

Sy

=

1 2i

(S-

-

S+).

From S?|m = s(s+1)-m(m?1)|m?1 we find

S+|m =

2-m(m+1)|m+1

(31)

S-|m =

2 - m(m-1)|m-1

(32)

this leads to

1|S+|1

1|S+|0

1|S+| - 1 0 2 0

S+ = 0|S+|1

0|S+|0

0|S+| - 1 = 0 0 2

(33)

-1|S+|1 -1|S+|0 -1|S+| - 1

00 0

and

1|S-|1

1|S-|0

1|S-| - 1 0 0 0

S- = 0|S-|1

0|S-|0

0|S-| - 1 = 2 0 0

(34)

-1|S-|1 -1|S-|0 -1|S-| - 1

0 20

which gives

0

Sx = 2

2 0

2 0

2

0 2 = 0

1

0 1

20

1 0 1

0 1 0

(35)

and

0

Sy

=

2i

- 0

2

2 0 -2

0 2 = 0

1

0 i

20

-i 0 i

0 -i 0

(36)

The generalized Pauli matrices for spin-1 would then be

x

=

1

0 1

20

1 0 1

0 1 0

(37)

y

=

1

0 i

20

-i 0 i

0 -i 0

(38)

1 0 0

z = 0 0 0

(39)

0 0 -1

5

4. Consider an electron whose position is held fixed, so that it can be described by a simple twocomponent spinor (i.e. no r dependence). Let the initial state of the electron be spin up relative to the z-axis, | z . At time t = 0, a uniform magnetic field is applied along the y-axis. What is the state-vector of the electron at time t > 0?

Hint: Start by writing the Hamiltonian, which should contain only the spin-contribution to the

magnetic dipole energy. Then propagate the state using the energy eigenvalue representation of the propagator, U (t) = n |n e-int n|.

The initial state of the system is |(0) = | z

The Hamiltonian is

H

=

-?

?

B

=

-

gq 2M

S

?

B

=

|e| B0 2me

y

=

?B B0y

(40)

With 0 =

|e|B0 me

,

the

eigenvalues

of

H

are then ?

0.

The state at time t is then

|(t) = e-iHt/ |(0)

(41)

= (cos(Ht/ ) - i sin(Ht/ )) |0

(42)

= (cos(0t)I - i sin(0t)y) | z

(43)

= cos(0t)| z + sin(0t)| z

(44)

6

5. Consider the most general normalized spin-1/2 state | = c| z + c| z . (a) Compute Sx , Sy , and Sz , with respect to this state.

Sx = 2 c+, c-

01 10

Sy = 2 c+, c-

0 -i i0

Sz = 2 c+, c-

10 0 -1

c+ c-

= 2 c+, c-

c+ c-

= 2 c+, c-

c+ c-

= 2 c+, c-

c- c+

= 2 (c+c- + c-c+)

-ic- ic+

= 2i (c+c- - c-c+)

c+ -c-

= 2 (c+c+ - c-c-)

(b) Compute the variances Sx, Sy, and Sz.

We know that So that

2

Sx2 = Sy2 = Sz2 = 4 .

Sx = Sy = Sz =

Sx2 - Sx 2 = 2 1 - (c+c- + c-c+)2 Sy2 - Sy 2 = 2 1 + (c+c- - c-c+)2 Sz2 - Sz 2 = 2 1 - (c+c+ - c-c-)2

(c) Prove that Sx = 2 c2 - c2 , Sy = 2 c2 + c2 , and Sz = |c||c|.

With we can write Sx as

(c+c+ + c-c-)2 = 12 = 1

Sx = 2 (c+c+ + c-c-)2 - (c+c- + c-c+)2 = 2 |c+|4 + 2|c+|2|c-|2 + |c-|4 - (c+ )2c2- - 2|c+|2|c-|2 - (c-)2c2+ = 2 (c+)2c2+ - (c+)2c2- - (c- )2c2+ + (c-)2c2- = 2 ((c+)2 - (c-)2)(c2+ - c2-) = 2 |c2+ - c2-|

Similarly, we obtain

Sy = 2 (c+)2c2+ + 2c+c-c+c- + (c-)2c2- + (c+ )2c2- - 2c+c-c+c- + (c-)2c2+ = 2 (c+)2c2+ + (c-)2c2- + (c+ )2c2- + (c- )2c2+ = 2 |c2+ + c2-|

7

and Sz = 2 (|c+|2 + |c-|2)2 - (|c+|2 - |c-|2)2 = 2 4|c+|2|c-|2 = |c+||c-|

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