# Infinite dimensional representation of su(2)

1. Apr 3, 2009

### matematikawan

I'm trying to understand this paper which the author claimed that he had constructed an infinite dimensional representation of the su(2) algebra. The hermitian generators are given by

$$J_x=\frac{i}{2}(\sqrt{N+1}a-a^\dagger\sqrt{N+1} )$$
$$J_y=-\frac{i}{2}(\sqrt{N+1}a+a^\dagger\sqrt{N+1} )$$
$$J_z=N+\frac{1}{2}$$

where the creation and the annihilation operators $$a^\dagger$$ and a satisfy the commutator relation $$[a,a^\dagger]=-1$$ and $$N\equiv-a^\dagger a$$.

My question is, what is the meaning of the expression $$\sqrt{N+1}$$? So that I can proceed with the manipulation.
Does it means the binomial expression
$$\sqrt{N+1}= 1 + \frac{1}{2}N + ...$$?

I have been thinking along that line. How do we show that the representation is infinite dimensional?

ps How do you create the superscript operator dagger in tex?

Last edited: Apr 3, 2009
2. Apr 3, 2009

### Ben Niehoff

The dagger symbol is just \dagger. I don't know the answer to any of your other questions. :D

3. Apr 3, 2009

### matematikawan

Thanks I have done the necessary changes. They don't have the template for dagger.

4. Apr 5, 2009

Hi,

I think you are a little confused. The N in J_z represents the Number Operator which is defined as a^dagger*a
Where as the N in the square root is just a number and is equal to the eigen value of the Number operator.

Conventionally, that's what I would think of, when I see the equations you have written
But of course, I haven't seen the paper you are referring to and I might be wrong.

5. Apr 5, 2009

### matematikawan

If N is just a number why the author bother to write the expression to the right of $$a^\dagger$$ .

Yes I agree usually the number operator $$N=a^\dagger a$$. But that is in the case of finite dimensional representation.

For your information, the papers that I trying to understand are:

1. Andre van Tonder, Ghosts as Negative Spinors, Nuc. Phys. B 645(2002) pp 371-386.
2. Andre van Tonder, On the representation theory of negative spin, Nuc. Phys. B 645(2002) pp 387-402.

6. Apr 5, 2009

### vkroom

What is attempted in your equations is known in physics as Schwinger's Boson representation of angular momentum algebra ( SU(2) in your case. ) The$$J_i$$ are the generators of this algebra.
Therefore, one would conclude that the 'N' in the first two eqns are actually numbers:eigenvalue of the operator $$\hat{N} = a ^\dagger a$$, whose spectrum is infact the set of all non-negative integers, denoted by $$N$$.
In the third eqn its actually the operator $$\hat{N}$$. As u can see it doesn't matter if it is $$N$$ or, $$\hat{N}$$. Both yields the same behavior. If it were a simple number, that eqn denotes (N+1/2)Id, where Id-> identity operator.

7. Apr 6, 2009

### matematikawan

OK I make a mistake. But it is in the expression Jy. There shouldn't be the imaginary number i. Other than that I think I have copied correctly.

$$J_y=-\frac{1}{2}(\sqrt{N+1}a+a^\dagger\sqrt{N+1} )$$

If it true that $$\sqrt{N+1}$$ is just a number then

$$[J_x , J_y]=-\frac{i(N+1)}{4}[a , a^\dagger] = \frac{i(N+1)}{4}.$$

which is not equivalent to the su(2) algebra $$[J_x , J_y]=iJ_z$$.

However if I assume $$\sqrt{N+1}$$ to be an operator I'm able to verify the relation $$[J_x , J_y]=iJ_z$$.