PHY6426/Fall 2007: CLASSICAL MECHANICS
HOMEWORK ASSIGNMENT #7: Solutions
due by 9:35 a.m. Mon 10/15
Instructor: 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]
σ̂i σ̂j = δij Iˆ + iijk σ̂k ,
where ijk 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 indices
coincide.
Solution:
1) i = j
(σ̂i )2 = Iˆ
iik = 0, so the formula works
2) i 6= j → δij = 0. Choose i = 1, j = 2
0 1
0 −i
i
σ̂1 σ̂2 =
=
1 0
i 0
0
0 −i
0 1
σ̂2 σ̂1 =
=−
i 0
1 0
1 0
=i
= iσ̂3
0 −1
i 0
1 0
= −i
= −iσ̂3
0 −i
0 −1
0
−i
123 = 1. 213 = −1, so the formula works again. Similar for other matrices.
b) [20 points]
(a · σ̂) (b · σ̂) = (a · b) Iˆ + i (a × b) · σ̂
(1)
Solution
a · σ̂ =a1 σ̂1 + a2 σ̂2 + a3 σ̂3
b · σ̂ =b1 σ̂1 + b2 σ̂2 + b3 σ̂3
(a · σ̂) (b · σ̂) = (a1 σ̂1 + a2 σ̂2 + a3 σ̂3 ) (b1 σ̂1 + b2 σ̂2 + b3 σ̂3 )
= a1 b1 σ̂12 + a2 b2 σ̂22 + a3 b3 σ̂22 + a1 b2 σ̂1 σ̂2 + a2 b1 σ̂2 σ̂1
+a1 b3 σ̂1 σ̂3 + a3 b1 σ̂3 σ̂1 + a2 b3 σ̂2 σ̂3 + a3 b2 σ̂3 σ̂2
Using σ̂i2 = Iˆ , σ̂1 σ̂2 = −σ̂2 σ̂1 = iσ̂3 , σ̂2 σ̂3 = −σ̂2 σ̂3 = iσ̂1 , σ̂3 σ̂1 = −σ̂1 σ̂3 = iσ̂2 , one proves the required
relation.
2. a) [40 points] Prove that for any function f,
f (a + |b|) + f (a − |b|)
f (a + |b|) − f (a − |b|) b
ˆ
ˆ
I+
· σ̂ ,
f aI + b · σ̂ =
2
2
|b|
ˆ
where, by convention, lima→0 f aIˆ = f (0) I.
Solution
(2)
2
Expand f aIˆ + b · σ̂ into formal Taylor series
∞
X
n
f (n) (0) ˆ
aI + b · σ̂ ,
f aIˆ + b · σ̂ =
n!
n=0
2
where f (n) (0) = dn f (x) /dxn |x=0 . Do a couple of terms to get the patter, using Eq.(1) for (b · σ̂) = b2 Iˆ
2
= a2 Iˆ + 2ab · σ̂+ (b · σ̂) (b · σ̂)
aIˆ + b · σ̂
= a2 Iˆ + 2ab · σ̂ + b2 Iˆ = a2 + b2 Iˆ + 2ab · σ̂
3
2
3
= a3 Iˆ + 3a2 (b · σ̂) + 3a (b · σ̂) + (b · σ̂)
aIˆ + b · σ̂
= a3 Iˆ + 3a2 (b · σ̂) + 3ab2 Iˆ + b2 (b · σ̂)
= a3 + 3ab2 Iˆ + 3a2 + b2 (b · σ̂)
Groupping the terms containing Iˆ and b · σ̂, we obtain
f (3) (0) 3
f (2) (0) 2
(1)
2
2
ˆ
f aI + b · σ̂ = f (0) + f (0) a +
a +b +
a + 3ab + . . . Iˆ
2!
3!
b
f (3) (0)
f (2) (0)
3a2 + b2 |b| + . . .
2a |b| +
·σ̂.
+ f (1) (0) |b| +
2!
3!
|b|
A direct check shows that the coefficients of the Iˆ and
b
|b| ·σ̂
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, describing
a rotation by angle θ about axis n̂, is given by
θ
Û = exp i n̂ · σ̂ .
2
Solution
As it was shown in the class,
θ
θ
Û = cos Iˆ + i sin n̂ · σ̂.
2
2
Applying formula (2) for a = 0 and b = (θ/2)n̂, 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,
−iγ/2
γ
γˆ
e
0
Û (γ) = cos I − i sin σ̂3 =
0
eiγ/2
2
2
(Notice that the angle enters with the negative sign. This is because the Euler rotation is a transformation
applied to the coordinate system and the SU(2) rotation is a transformation applied to the vector itself.). The
rotation by angle θ about the intermediate position of the x-axis (nodal line) is given by
θˆ
θ
cos θ/2 −i sin θ/2
Û (θ) = cos I − i sin σ̂1 =
−i sin θ/2 cos θ/2
2
2
The net result is the product of the three rotations
−i(φ+ψ)/2
e
cos θ/2 −ie−i(φ−ψ)/2 sin θ/2
Û = Û (φ) Û (θ) Û (ψ) =
−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 last
result with the definition of the Cayley-Klein parameters
α0 + iα3 α2 + iα1
,
Û =
−α2 + iα1 α0 − iα3
we obtain
θ
1
cos (φ + ψ)
2
2
θ
1
= − sin cos (φ − ψ)
2
2
θ
1
= sin sin (φ − ψ)
2
2
1
θ
= − cos sin (φ + ψ)
2
2
α0 = cos
α1
α2
α3