Physics 745 - Group Theory
Solution Set 13
The original basis is:
a = axˆ
b = b ( cos γ xˆ + sin γ yˆ )
c = czˆ
An arbitrary element of the translation group is then n1a + n2b + n3c , where ( n1 , n2 , n3 ) are
all integers, or instead we can choose n1 and n3 to be half-integers, but n2 is still an
integer.
Imagine switching basis to
T1 =
1
2
(a + c)
T2 = b
T3 =
1
2
(a − c)
Then it is pretty easy to see that T1 ± T3 yields the two vectors a and c. It follows that
n1a + n2b + n3c = n1 ( T1 + T3 ) + n2 T2 + n3 ( T1 − T2 ) = ( n1 + n3 ) T1 + n2 T2 + ( n1 − n3 ) T3
Given the restrictions on ( n1 , n2 , n3 ) , it is obvious that all three of ( n1 + n3 , n2 , n1 − n3 ) will
be integers, and we have accomplished our goal.
The area of the parallelogram bounded by a and b is ab sin γ , which is the same
as the magnitude of a × b . This must then be multiplied by the amount that c sticks out
of the plane of a and b, so the total volume is V = a × b c cos θ , where θ is the angle
between the vector c and the perpendicular to a and b. A little thought will convince you
that this implies V = ( a × b ) ⋅ c , with the absolute value taking into account the fact that
the cross product a × b might point the opposite direction from c. We therefore have
V = ( a × b ) ⋅ c = [ ab sin γ zˆ ] ⋅ czˆ = abc sin γ
V ′ = ( T1 × T2 ) ⋅ T3 =
1
4
⎡⎣( axˆ + czˆ ) × b ( cos γ xˆ + sin γ yˆ ) ⎤⎦ ⋅ ( axˆ − czˆ )
= 14 b ( a sin γ zˆ + c cos γ yˆ − c sin γ xˆ ) ⋅ ( axˆ − czˆ ) = 14 abc − sin γ − sin γ
= 12 abc sin γ = 12 V
The minus sign inside the absolute value just indicates that the basis set T1 , T2 , T3 is lefthanded. If you want to, this can be fixed in numerous ways, such as changing the sign of
T3. It is also easy to find the volume by taking the determinant of a matrix consisting of
the three vectors.