Quantum Mechanics for Mathematicians

Irreducible representations will be characterized by a highest weight vector,
as follows

Theorem(Highest weight theorem).Finite dimensional irreducible represen-
tations ofSU(2)have weights of the form

−n,−n+ 2,···,n− 2 ,n

forna non-negative integer, each with multiplicity 1 , withna highest weight.

Proof.Finite dimensionality implies there is a highest weightn, and we can
choose any highest weight vectorvn∈Vn. Repeatedly applyingπ′(S−) tovn
will give new vectors
vn− 2 j=π′(S−)jvn∈Vn− 2 j

with weightsn− 2 j.
Consider the span of thevn− 2 j,j≥0. To show that this is a representation
one needs to show that theπ′(S 3 ) andπ′(S+) leave it invariant. Forπ′(S 3 ) this
is obvious, forπ′(S+) one can show that

π′(S+)vn− 2 j=j(n−j+ 1)vn−2(j−1) (8.1)

by an induction argument. Forj= 0 this is the highest weight condition onvn.
Assuming validity forj, validity forj+ 1 can be checked by

π′(S+)vn−2(j+1)=π′(S+)π′(S−)vn− 2 j

=(π′([S+,S−]) +π′(S−)π′(S+))vn− 2 j

=(π′(2S 3 ) +π′(S−)π′(S+))vn− 2 j

=((n− 2 j)vn− 2 j+π′(S−)j(n−j+ 1)vn−2(j−1)

=((n− 2 j) +j(n−j+ 1))vn− 2 j

=(j+ 1)(n−(j+ 1) + 1)vn−2((j+1)−1)

where we have used the commutation relation

[S+,S−] = 2S 3

By finite dimensionality, there must be some integerksuch thatvn− 2 j 6 = 0
forj≤k, andvn− 2 j= 0 forj=k+ 1. But then, forj=k+ 1, we must have

π′(S+)vn− 2 j= 0

By equation 8.1 this will happen only fork+ 1 =n+ 1 (and we need to taken
positive). We thus see thatv−nwill be a “lowest weight vector”, annihilated by
π′(S−). As expected, the pattern of weights is invariant under change of sign,
with non-zero weight spaces for

−n,−n+ 2,···,n− 2 ,n