# Solution Set 22 ( ) ```Physics 745 - Group Theory
Solution Set 22
1.  It is rare we will actually use the representation matrices Γ( j ) ( R ) , but
occasionally it is useful. We want to work out explicitly Γ( 2 ) ( R ( x ) ) , generated
1
by the generators
Ta = 12 σ a ,
⎛0 1⎞
⎛ 0 −i ⎞
⎛1 0 ⎞
where σ 1 = ⎜
⎟, σ2 = ⎜
⎟, σ3 = ⎜
⎟.
⎝1 0⎠
⎝i 0 ⎠
⎝ 0 −1 ⎠
We will write x in the form x = xrˆ , where r̂ is a unit vector, and x is the
magnitude of x.
2
(a)  Show that ( rˆ ⋅ σ ) = 1 , where 1 is the unit matrix, for any unit vector r̂ .
Furthermore, find a simplification for ( rˆ ⋅ σ ) for arbitrary positive integer n
n
when n is even or odd.
We simply write out the expression explicitly, writing rˆ = ( r1 , r2 , r3 ) , so that
⎛ r
2
( rˆ ⋅ σ ) = ⎜ 3
⎝ r1 + ir2
r1 − ir2 ⎞⎛ r3
⎟⎜
− r3 ⎠⎝ r1 + ir2
r1 − ir2 ⎞ ⎛ r32 + r12 + r22
⎟=⎜
− r3 ⎠ ⎝
0
⎞
= 1 ( rˆ 2 ) = 1
2
2
2⎟
r3 + r1 + r2 ⎠
0
Now, if you have it raised to an even power, you will simply get ( rˆ ⋅ σ ) = 1n = 1 , while
2n
for an odd power, you will have ( rˆ ⋅ σ )
2 n +1
= 1n ( rˆ ⋅ σ ) = rˆ ⋅ σ . That’s pretty simple.
(b)  Expand out the representation
Γ
( 12 )
∞
( R ( x ) ) = exp ( −iT ⋅ x ) = ∑
n=0
( −i T ⋅ x )
n
n!
dividing it into even and odd terms. Simplify as much as possible
1
(c)  Show that Γ( 2 ) ( R ( x ) ) = 1 cos ( 12 x ) − i ( rˆ ⋅ σ ) sin ( 12 x )
We simply follow the instructions, and then divide it into even and odd terms:
Γ
( 12 )
∞
( R ( x)) = ∑
( −iT ⋅ x )
n
∞
1
n
( −i 12 rˆ ⋅ σx )
n!
n =0
n =0 n !
∞
∞
1
1
n
n
n
n
= ∑ ( −i 12 x ) ( rˆ ⋅ σ ) + ∑ ( −i 12 x ) ( rˆ ⋅ σ )
n even n !
n odd n !
=∑
2
4
3
5
= 1 ⎡1 − 2!1 ( 12 x ) + 4!1 ( 12 x ) − + &quot;⎤ + ( rˆ ⋅ σ ) ⎡ −i ( 12 x ) + i 3!1 ( 12 x ) − i 5!1 ( 12 x ) + −&quot;⎤
⎣
⎦
⎣
⎦
1
1
= 1 cos ( 2 x ) − i ( rˆ ⋅ σ ) sin ( 2 x )
```