PHY–396 K. Solutions for problem set #3.

PHY?396 K. Solutions for problem set #3.

Problem 1(a): Let's begin by working out the indexology of eq. (2). Note that for all 9 combinations of i, j = x, y, z, there is only one non-vanishing term on the RHS of eq. (2), namely unit matrix I for i = j or ?ik for i = j and k = i, j. Specifically,

xx = I, yx = -iz , zx = +iy ,

xy = +iz , yy = I, zy = -ix

xz = -iy , yz = +ix , zz = I.

(S.1)

Actually multiplying the Pauli matrices (1) in all possible combinations verifies all nine of these equations and hence eq. (2).

Problem 1(b): Take any 3?vector v whose components vi are either ordinary numbers or else operators which have nothing to do with spin and therefore commute with the Pauli matrices. For all such vectors v,

v ? 2 = (vii)(vjj) = vivj ij = ij I + iijkk = (vivi)I + iijkvivjk , (S.2)

or in 3?vector notations

v ? 2 = (v2)I + i(v ? v) ? .

(S.3)

Moreover, if the components of v commute with each other, vivj = vjvi, then v ? v = 0. Indeed,

ijkvivj = -jikvivj

by antisymmetry of ijk

= -ijkvjvi

=

1 2

ijk

[vi

,

vj

]

re-labeling the summation indices i j averaging the LHS and the previous line

0 for [vi, vj] = 0.

(S.4)

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Consequently, eq. (S.3) reduces to its first term, v ? 2 = (v2)I.

(S.5)

Problem 1(c): Thanks to eq. (S.5), for any even n = 2m = 0, 2, 4, 6, . . .

v ? 2m = (v2)m I = v2m I

while for any odd n = 2m + 1 = 1, 3, 5, . . .

v ? 2m+1 = (v2)m v ?

=

v2m+1

v? v

where v = v2. Consequently,

exp v ?

=

1 n=0 n!

v?

n

=

1 even n n!

v?

n

+

1 n!

odd n

v?

n

=

even

n

1 n!

vn

I

+

1 n!

vn

v

? v

odd n

=

vn even n n!

I+

vn n!

odd n

=

cosh(v) I

+

sinh(v)

v

? v

.

v? v

(S.6) (S.7)

(S.8)

Problem 1(d): Classically, a magnetic moment m misaligned with the external magnetic field B precesses with the Larmor frequency = GB where G is the gyromagnetic ratio of the magnetic moment to the angular momentum, G = m/J. For the atomic magnetic moment due to

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orbital motion of the electrons G = e/2cMe (in Gauss units) while for the magnetic moment due to electrons' spins G = 2 ? e/2cMe, thus precession frequency

L

=

eB cMe

.

(S.9)

Now consider the quantum magnetic moment due to electron's spin. For the spin Hamiltonian (3), the time evolution operators is

U^ (t, t0) = exp -i(t - t0)H^ /h?

=

exp

-i(t

-

t0

)

mB h?

(B

?

)

.

(S.10)

Or in terms of the magnitude B and the direction n of the magnetic field,

U^ (t, t0) = exp

-i(t

-

t0

)

B

mB h?

(n

?

)

(S.11)

where hence

BmB h?

=

B h?

?

eh? 2cMe

=

Be 2cMe

=

L 2

,

U^ (t, t0) = exp

-i

(t

- 2

t0

)

(n

?

)

.

(S.12) (S.13)

To actually evaluate this matrix exponential, we use eq. (S.8) for v having direction n and imaginary magnitude v = -i(t - t0)/2, thus

U^ (t, t0) = cosh

-i

(t

- 2

t0)

I + sinh

-i

(t

- 2

t0)

n?

(4)

= cos

(t - t0) 2

I - i sin

(t - t0) 2

n? ,

quod erat demonstrandum.

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Problem 1(e): Let

| = |Z+ + |Z-

(S.14)

for some complex coefficients , such that ||2 + ||2 = 1 (so that | = 1); or in

column-vector notations

| =

.

(S.15)

In this state and therefore

| ^x | | ^y | | ^z |

= + = 2 Re(), = -i + i = 2 Im(), = - = ||2 - ||2,

(S.16)

mx = -mB ? 2 Re(),

my = -mB ? 2 Im(),

mz = -mB(||2 - ||2). (S.17)

Now consider the time evolution of the state -- and hence of the magnetic moment (S.17) in a constant uniform magnetic field N . As we saw in part (d), the evolution operator U^ (t, t0) is spelled out in eq. (4). In particular, for the magnetic field in z direction, the matrix (4) becomes diagonal

U^ (t, t0) = cos = cos

(t - t0) 2

(t - t0) 2

I - i sin

(t - t0) 2

z

10 01

- i sin

(t - t0) 2

10 0 -1

=

cos

(t-t0) 2

- i sin

(t-t0) 2

0

cos

(t-t0) 2

0

+ i sin

(t-t0) 2

=

exp

-i

(t-t0) 2

0

0

exp

+i

(t-t0) 2

.

4

(S.18)

Consequently, if at the time t0

|(t0) =

0 , 0

(S.19)

then at a later time t > t0

|(t) = U^ (t, t0) |(t0) =

(t) (t)

(S.20)

for

(t) = exp

-i

(t

- 2

t0)

? 0 ,

(t) = exp

+i

(t

- 2

t0)

? 0 .

(S.21)

For the expectation value (S.17) of the electron's magnetic moment, this means that

mz (t) = -mB ||2-||2) = -mB |0|2-|0|2) = const while

i.e., time-independent , (S.22)

mx (t) + i my (t) = -mB ? 2(t)(t) = -mB ? 200 ? exp +i(t - t0) = -mB ? 2 |0| |0| ? exp i(t - t0) + i0

(S.23)

for some constant phase 0 = arg(00). Consequently,

mx (t) = -mB ? 2 |0| |0| ? cos (t - t0) + 0 , my (y) = -mB ? 2 |0| |0| ? sin (t - t0) + 0 ,

(S.24)

which in 2d terms describes a rotation of the vector ( mx , my ) with frequency . In 3d terms, this rotation of the x and y components while the z components stays constant means precession around the z axis of the magnetic field.

The bottom line is, in quantum mechanics the expectation value m of the electron's magnetic moment behaves exactly like a classical magnetic moment m: it precesses around the magnetic field's direction with the classical Larmor frequency .

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