246
H Solutions to Problems
The angular momentum operator is given by
L z = xp y − yp x
=
i
x
∂
∂y
− y
∂
∂x
=−
i
lim
α→0
ˆ
O(α) − ˆ
E
α
The angular momentum operator thus is proportional to an infinitesimal rotation
in the neighborhood of the unit element.
2.1 The condition that C be unitary gives rise to six equations:
1 =|a|
2 +|b|
2
1 =|a|
2 +|c|
2
1 =|b|
2 +|d|
2
1 =|c|
2 +|d|
2
0 =|ac|e
i(α−γ) +|bd|e
i(β−δ)
0 =|ab|e
i(α−β) +|cd|e
i(γ−δ)
From these equations it is clear that |a|=|d| and |b|=|c|. The phase relationships may be reduced to
e
i(β+γ) =−e
i(α+δ)
With the help of these results the four matrix entries can be rewritten as
|a|e
iα =|a|e
i(α+δ)/2 e
i(α−δ)/2
|d|e
iδ =|a|e
i(α+δ)/2 e
−i(α−δ)/2
|b|e
iβ =|b|e
i(α+δ)/2 e
i[β−
α+δ
2 ]
|c|e
iγ =−|b|e
i(α+δ)/2 e
i[−β+
α+δ
2 ]
The general U(2) matrix may thus be rewritten as
U = e
i(α+δ)/2
|a|e i(α−δ)/2
|b|e
i[β−
α+δ
2 ]
−|b|e
i[−β+
α+δ
2 ] |a|e −i(α−δ)/2
with |a| 2 +|b| 2 = 1. Note that a general phase factor has been taken out. The
remaining matrix has determinant +1 and is called a special unitary matrix (see
further in Chap. 7).
2.2 The relevant integrals are given by
2π
0
e
−ikφ e
ikφ dφ =[φ]
2π
0 = 2π
2π
0
e
±2ikφ dφ =
1
±2ik
e
±2ikφ 2π
0
= 0