7.5 Double Groups
179
and there is another twofold axis in G perpendicular to this ˆ
C 2 , both this ˆ
C 2 and
ℵ ˆ
C 2 belong to the same class.
The theorem can easily be demonstrated in an algebraic way. If a symmetry element ˆ
A is conjugate to ˆ
B, then element ℵ ˆ
A is conjugate to ℵ ˆ
B, since the Bethe
operation commutes with all symmetry elements:
ˆ
A = ˆ
X ˆ
B ˆ
X
−1 →ℵ ˆ
A = ˆ
Xℵ ˆ
B ˆ
X
−1
(7.45)
Since multiplication by ℵ corresponds to multiplication by −I, a symmetry element
ˆ
A and its double-group partner ℵ ˆ
A have characters that differ by sign. Unless their
characters are zero, they cannot belong to the same class since symmetry elements
in the same class must have the same character. This explains the first rule of the
theorem. 2
Exceptions can exist when the character is zero. The character of the spinor irrep
is given by
χ
spin = a +¯ a =±2 cos
α
2
(7.46)
This character can be zero only for α =±π and, hence, for binary rotations with
n = 2. To examine whether or not the matrix for a binary rotation can be classconjugated to minus itself, we may limit ourselves to the study of one orientation
of the rotation axis, say ˆ
C
z
2 . Indeed, in SU(2) any orientation can always be transformed backward to this standard choice by a unitary transformation. The problem
thus reduces to finding a spinor operation ˆ
X represented by a matrix X with Cayley–
Klein parameters a x ,b x , which transforms D(C
z
2 ) into minus itself:
ˆ
X ˆ
C
z
2
ˆ
X
−1 =ℵ ˆ
C
z
2
(7.47)
or, in terms of the spinor matrices,
a x b x
− ¯
b x ¯
a x
−i 0
0 i
¯
a x −b x
¯
b x a x
=
i 0
0 −i
(7.48)
This expression reduces to a set of two equations:
|a x |
2 −|b x |
2 =−1
a x b x = 0
(7.49)
These equations can be solved only when a x is equal to zero. This implies that both
the real and complex parts of this parameter must be equal to zero; hence,
α = π mod 2π
n z = 0
(7.50)
2 Note that not all elements in the same class will be oriented in the positive hemisphere, as is clear
from Tables 7.4 and 7.6.
179
and there is another twofold axis in G perpendicular to this ˆ
C 2 , both this ˆ
C 2 and
ℵ ˆ
C 2 belong to the same class.
The theorem can easily be demonstrated in an algebraic way. If a symmetry element ˆ
A is conjugate to ˆ
B, then element ℵ ˆ
A is conjugate to ℵ ˆ
B, since the Bethe
operation commutes with all symmetry elements:
ˆ
A = ˆ
X ˆ
B ˆ
X
−1 →ℵ ˆ
A = ˆ
Xℵ ˆ
B ˆ
X
−1
(7.45)
Since multiplication by ℵ corresponds to multiplication by −I, a symmetry element
ˆ
A and its double-group partner ℵ ˆ
A have characters that differ by sign. Unless their
characters are zero, they cannot belong to the same class since symmetry elements
in the same class must have the same character. This explains the first rule of the
theorem. 2
Exceptions can exist when the character is zero. The character of the spinor irrep
is given by
χ
spin = a +¯ a =±2 cos
α
2
(7.46)
This character can be zero only for α =±π and, hence, for binary rotations with
n = 2. To examine whether or not the matrix for a binary rotation can be classconjugated to minus itself, we may limit ourselves to the study of one orientation
of the rotation axis, say ˆ
C
z
2 . Indeed, in SU(2) any orientation can always be transformed backward to this standard choice by a unitary transformation. The problem
thus reduces to finding a spinor operation ˆ
X represented by a matrix X with Cayley–
Klein parameters a x ,b x , which transforms D(C
z
2 ) into minus itself:
ˆ
X ˆ
C
z
2
ˆ
X
−1 =ℵ ˆ
C
z
2
(7.47)
or, in terms of the spinor matrices,
a x b x
− ¯
b x ¯
a x
−i 0
0 i
¯
a x −b x
¯
b x a x
=
i 0
0 −i
(7.48)
This expression reduces to a set of two equations:
|a x |
2 −|b x |
2 =−1
a x b x = 0
(7.49)
These equations can be solved only when a x is equal to zero. This implies that both
the real and complex parts of this parameter must be equal to zero; hence,
α = π mod 2π
n z = 0
(7.50)
2 Note that not all elements in the same class will be oriented in the positive hemisphere, as is clear
from Tables 7.4 and 7.6.