PHY6426/Fall 2007: CLASSICAL MECHANICSHOMEWORK ASSIGNMENT #7: Solutions

due by 9:35 a.m. Mon 10/15Instructor: D. L. Maslov

maslov@phys.ufl.edu 392-0513 Rm. 2114

Please help your instructor by doing your work neatly.

1. Prove the following two properties of the Pauli matrices

a) [10 points]

σiσj = δij I + iεijkσk,

where εijk is the Levi-Civita antisymmetric tensor defined as equal to +1, if ijk is a cyclic permutation of 123,equal −1, if ijk is an anti-cyclic permutation of 123, and equal to zero, if at least two of the three indicescoincide.

Solution:

1) i = j

(σi)2 = I

εiik = 0, so the formula works

2) i 6= j → δij = 0. Choose i = 1, j = 2

σ1σ2 =

(

0 11 0

) (

0 −ii 0

)

=

(

i 00 −i

)

= i

(

1 00 −1

)

= iσ3

σ2σ1 =

(

0 −ii 0

) (

0 11 0

)

= −

(

i 00 −i

)

= −i

(

1 00 −1

)

= −iσ3

ε123 = 1. ε213 = −1, so the formula works again. Similar for other matrices.

b) [20 points]

(a · σ) (b · σ) = (a · b) I + i (a× b) · σ (1)

Solution

a · σ =a1σ1 + a2σ2 + a3σ3

b · σ =b1σ1 + b2σ2 + b3σ3

(a · σ) (b · σ) = (a1σ1 + a2σ2 + a3σ3) (b1σ1 + b2σ2 + b3σ3)

= a1b1σ21 + a2b2σ

22 + a3b3σ

22 + a1b2σ1σ2 + a2b1σ2σ1

+a1b3σ1σ3 + a3b1σ3σ1 + a2b3σ2σ3 + a3b2σ3σ2

Using σ2i = I , σ1σ2 = −σ2σ1 = iσ3, σ2σ3 = −σ2σ3 = iσ1, σ3σ1 = −σ1σ3 = iσ2, one proves the required

relation.

2. a) [40 points] Prove that for any function f,

f(

aI + b · σ)

=f (a+ |b|) + f (a− |b|)

2I +

f (a+ |b|) − f (a− |b|)

2

(

b

|b|· σ

)

, (2)

where, by convention, lima→0 f(

aI)

= f (0) I .

Solution

2

Expand f(

aI + b · σ)

into formal Taylor series

f(

aI + b · σ)

=

∞∑

n=0

f (n) (0)

n!

(

aI + b · σ)n

,

where f (n) (0) = dnf (x) /dxn|x=0. Do a couple of terms to get the patter, using Eq.(1) for (b · σ)2

= b2I

(

aI + b · σ)2

= a2I + 2ab · σ+ (b · σ) (b · σ)

= a2I + 2ab · σ + b2I =(

a2 + b2)

I + 2ab · σ(

aI + b · σ)3

= a3I + 3a2 (b · σ) + 3a (b · σ)2+(b · σ)

3

= a3I + 3a2 (b · σ) + 3ab2I + b2 (b · σ)

=(

a3 + 3ab2)

I +(

3a2 + b2)

(b · σ)

Groupping the terms containing I and b · σ, we obtain

f(

aI + b · σ)

=

{

f (0) + f (1) (0)a+f (2) (0)

2!

(

a2 + b2)

+f (3) (0)

3!

(

a3 + 3ab2)

+ . . .

}

I

+

{

f (1) (0) |b| +f (2) (0)

2!2a |b| +

f (3) (0)

3!

(

3a2 + b2)

|b| + . . .

}

b

|b|·σ.

A direct check shows that the coefficients of the I and b

|b| ·σ coincide with those in the required relation.

b) [10 points] Using the result derived in part a), prove that the general form of an SU (2) matrix, describinga rotation by angle θ about axis n, is given by

U = exp

(

iθ

2n · σ

)

.

Solution

As it was shown in the class,

U = cosθ

2I + i sin

θ

2n · σ.

Applying formula (2) for a = 0 and b = (θ/2)n, we obtain the required result.

3. [30 points] Find the Cayley-Klein parameters of a general rotation in terms of its Euler angles.

The Euler’s rotations by angles φ and ψ are the z−axes. Therefore,

U (γ) = cosγ

2I − i sin

γ

2σ3 =

(

e−iγ/2 0

0 eiγ/2

)

(Notice that the angle enters with the negative sign. This is because the Euler rotation is a transformationapplied to the coordinate system and the SU(2) rotation is a transformation applied to the vector itself.). Therotation by angle θ about the intermediate position of the x-axis (nodal line) is given by

U (θ) = cosθ

2I − i sin

θ

2σ1 =

(

cos θ/2 −i sin θ/2−i sin θ/2 cos θ/2

)

The net result is the product of the three rotations

U = U (φ) U (θ) U (ψ) =

(

e−i(φ+ψ)/2 cos θ/2 −ie−i(φ−ψ)/2 sin θ/2

−iei(φ−ψ)/2 sin θ/2 ei(φ+ψ)/2 cos θ/2

)

3

(Also notice that the order of rotations is opposite to the order of A matrices in SO(3)). Comparing the lastresult with the definition of the Cayley-Klein parameters

U =

(

α0 + iα3 α2 + iα1

−α2 + iα1 α0 − iα3

)

,

we obtain

α0 = cosθ

2cos

1

2(φ+ ψ)

α1 = − sinθ

2cos

1

2(φ− ψ)

α2 = sinθ

2sin

1

2(φ− ψ)

α3 = − cosθ

2sin

1

2(φ+ ψ)