4Representations of Lie algebras

III Symmetries, Fields and Particles

4.3 Representations of su(2)

We are now going to study the complex representations of

su

(2), which are

equivalently the representations of the complexification

su

C

(2). In this section,

we will write

su

(2) when we actually mean

su

C

(2) for brevity. There are a

number of reasons why we care about this.

(i)

We’ve already done this before, in the guise of “spin”/“angular momentum”

in quantum mechanics.

(ii)

The representations are pretty easy to classify, and form a motivation for

our later general classification of representations of complex simple Lie

algebras.

(iii)

Our later work on the study of simple Lie algebras will be based on our

knowledge of representations of su(2).

Recall that the uncomplexified su(2) has a basis

su(2) = span

R

T

a

= −

1

2

iσ

a

: a = 1, 2, 3

,

where

σ

a

are the Pauli matrices. Since the Pauli matrices are independent over

C, they give us a basis of the complexification of su(2):

su

C

(2) = span

C

{σ

a

: a = 1, 2, 3}.

However, it is more convenient to use the following basis:

H = σ

3

=

1 0

0 −1

, E

±

=

1

2

(σ

1

± iσ

2

) =

0 1

0 0

,

0 0

1 0

.

We then have

[H, E

±

] = ±2E

±

, [E

+

, E

−

] = H.

We can rewrite the first relation as saying

ad

H

(E

±

) = ±2E

±

.

Together with the trivial result that

ad

H

(

H

) = 0, we know

ad

H

has eigenvalues

±2 and 0, with eigenvectors E

±

, H. These are known as roots of su(2).

Now let’s look at irreducible representations of

su

(2). Suppose (

V, ρ

) is a

representation. By linear algebra, we know ρ(H) has an eigenvector v

λ

, say

ρ(H)v

λ

= λv

λ

.

The eigenvalues of

ρ

(

H

) are known as the weights of the representation

ρ

. The

operators E

±

are known as step operators. We have

ρ(H)ρ(E

±

)v

λ

= (ρ(E

±

)ρ(H) + [ρ(H), ρ(E

±

)])v

λ

= (ρ(E

±

)ρ(H) + ρ([H, E

±

])v

λ

= (λ ± 2)ρ(E

±

)v

λ

.

So

ρ

(

E

±

)

v

λ

are also eigenvectors of

ρ

(

H

) with eigenvalues

λ ±

2, provided

ρ

(

E

±

)

v

λ

are non-zero. This constrains what can happen in a finite-dimensional

representation. By finite-dimensionality, we cannot have infinitely many eigen-

values. So at some point, we have to stop. In other words, a finite-dimensional

representation must have a highest weight Λ ∈ C, with

ρ(H)v

Λ

= Λv

Λ

, ρ(E

+

)v

Λ

= 0.

Now if we start off with

v

Λ

we can keep applying

E

−

to get the lower weights.

Indeed, we get

v

Λ−2n

= (ρ(E

−

))

n

v

Λ

for each

n

. Again, this sequence must terminate somewhere, as we only have

finitely many eigenvectors. We might think that the irrep would consist of the

basis vectors

{v

Λ

, v

Λ−2

, v

Λ−4

, ··· , v

Λ−2n

}.

However, we need to make sure we don’t create something new when we act on

these by

ρ

(

E

+

) again. We would certainly get that

ρ

(

E

+

)

v

Λ−2n

is an eigenvector

of eigenvalue Λ

−

2

n

+ 2, but we might get some other eigenvector that is not

v

Λ−2n+2

. So we have to check.

We can compute

ρ(E

+

)v

Λ−2n

= ρ(E

+

)ρ(E

−

)v

Λ−2n+2

= (ρ(E

−

)ρ(E

+

) + [ρ(E

+

), ρ(E

−

)])v

Λ−2n+2

= ρ(E

−

)ρ(E

+

)v

Λ−2n+2

+ (Λ − 2n + 2)v

Λ−2n+2

.

using the fact that [

E

+

, E

−

] =

H

. We now get a recursive relation. We can

analyze this by looking at small cases. When n = 1, the first term is

ρ(E

−

)ρ(E

+

)v

Λ−2n+2

= ρ(E

−

)ρ(E

+

)v

Λ

= 0,

by definition of v

Λ

. So we have

ρ(E

+

)v

Λ−2

= Λv

Λ

.

When n = 2, we have

ρ(E

+

)v

Λ−4

= ρ(E

−

)ρ(E

+

)v

Λ−2

+ (Λ − 2)v

Λ−2

= (2Λ − 2)v

Λ−2

.

In general, ρ(E

+

)v

Λ−2n

is some multiple of v

Λ−2n+2

, say

ρ(E

+

)v

Λ−2n

= r

n

V

Λ−2n+2

.

Plugging this into our equation gives

r

n

= r

n−1

+ Λ − 2n + 2,

with the boundary condition

ρ

(

E

+

)

v

Λ

= 0, i.e.

r

1

= Λ. If we solve this recurrence

relation, we determine an explicit relation for r

n

given by

r

n

= (Λ + 1 − n)n.

Now returning to the problem that the sequence must terminate, we figure that

we also need to have a lowest weight of the form Λ

−

2

n

. Hence we must have

some non-zero vector v

Λ−2n

6= 0 such that

ρ(E

−

)v

Λ−2n

= 0.

For this to be true, we need

r

N+1

= 0 for some non-negative integer

N

. So we

have

(Λ + 1 − (N + 1))(N + 1) = 0.

This is equivalent to saying

(Λ − N)(N + 1) = 0.

Since

N

+ 1 is a non-negative integer, we must have Λ

− N

= 0, i.e. Λ =

N

. So

in fact the highest weight is an integer!

In summary, we’ve got

Proposition.

The finite-dimensional irreducible representations of

su

(2) are

labelled by Λ ∈ Z

≥0

, which we call ρ

Λ

, with weights given by

{−Λ, −Λ + 2, ··· , Λ − 2, Λ}.

The weights are all non-degenerate, i.e. each only has one eigenvector. We have

dim(ρ

Λ

) = Λ + 1.

Proof.

We’ve done most of the work. Given any irrep, we can pick any eigenvector

of

ρ

(

H

) and keep applying

E

+

to get a highest weight vector

v

Λ

, then the above

computations show that

{v

Λ

, v

Λ−2

, ··· , v

−Λ

}

is a subspace of the irrep closed under the action of

ρ

. By irreducibility, this

must be the whole of the representation space.

The representation

ρ

0

is the trivial representation;

ρ

1

is the fundamental one,

and ρ

2

is the adjoint representation.

Now what do these tell us about representations of the Lie group

SU

(2)? We

know

SU

(2) is simply connected, so by our discussion previously, we know the

complex representations of

SU

(2) are exactly the representations of

su

(2). This

is not too exciting.

We know there are other Lie groups with Lie algebra

su

(2), namely

SO

(3).

Now that

SO

(3) is not simply connected, when does a representation of

su

(2)

give us a representation of SO(3)? Recall that we had

SO(3)

∼

=

SU(2)

Z/2Z

.

So an element in

SO

(3) is a pair

{A, −A}

of elements in

SU

(2). Given a

representation

ρ

Λ

of

su

(2), we obtain a corresponding representation

D

Λ

of

SU

(2). Now we get a representation of

SO

(3) if and only if

D

Λ

respects the

identification between A and −A. In particular, we need

D

Λ

(−I) = D

Λ

(I). (∗)

On the other hand, if this were true, then multiplying both sides by

D

Λ

(

A

) we

get

D

Λ

(−A) = D

Λ

(A).

So (

∗

) is a necessary and sufficient condition for

D

Λ

to descend to a representation

of SO(3).

We know that

−I = exp(iπH).

So we have

D

Λ

(−I) = exp(iπρ

Λ

(H)).

We know that ρ

Λ

has eigenvalues Λ ∈ S

Λ

. So D

Λ

(−I) has eigenvalues

exp(iπλ) = (−1)

λ

= (−1)

Λ

,

since

λ

and Λ differ by an even integer. So we know

D

Λ

(

−I

) =

D

Λ

(

I

) if and

only if Λ is even. In other words, we get a representation of

SO

(3) iff Λ is even.

In physics, we have already seen this as integer and half-integer spin. We

have integer spin exactly when we have a representation of

SO

(3). The fact that

half-integer spin particles exist means that “spin” is really about

SU

(2), rather

than SO(3).