# How to count states in SUSY multiplets?

There is an easy proof of the structure of multiplet that I don't reproduce here (it can be found in Bertolini, Lecture on Supersymmetry, pp.40-41 for the massless case and p.47 for the massive one). Here, I am referring to the massless case, to avoid useless complication.

In this proof the author constructs a Clifford algebra starting with the susy charges $Q^I_1$ and $\bar{Q}^I_{\dot{1}}$, so he has $\mathcal{N}$ creation operators and $\mathcal{N}$ destruction ones.

He argues that the states made by the application of $k$ destruction operators are $\binom{\mathcal{N}}{k}$, because of complete antisymmetry due to the algebra of anticommutation of creation operators, and here is my doubt:

• I understood perfectly well why there are no more than $\binom{\mathcal{N}}{k}$: because I can create new states only with the application of creation operators and they anticommute.
• I didn't understand why they can't be fewer: maybe I can find that two different (inequivalent under permutation) strings of creation operators (whose number is fixed by the helicity of the state) generate the same state (e.g.: two susy charges are represented as exactly the same operator)

The author doesn't consider this case, and I'm inclined to think that this cannot happen, but I don't understand why.

## 1 Answer

It's not clear in your question what you expect to conceivably happen; so what, exactly, you failed to understand.

From the N -oscillator algebra (3.10) it is evident that the state of k creation operators $a^\dagger_a ...a^\dagger_b | \lambda_0\rangle$ is unique for each k-tuple (a,...,b) of distinct integer labels. All such states have helicity $\lambda_0 + k/2$, but they cannot be the same.

Take N=5, k=2 for specificity. If you are asking why no two of the 10 states of helicity $\lambda_0 +1$ can be the same, imagine two such, say $a^\dagger_1 a^\dagger_2 | \lambda_0\rangle$ and $a^\dagger_3 a^\dagger_4 | \lambda_0\rangle$ for the sake of argument. The first state is annihilated by $a_3$, but the second is not, since this operation yields $a^\dagger_4| \lambda_0\rangle$, a state of helicity $\lambda_0 + 1/2$.

• Thank you for your answer, maybe my question wasn’t very well expressed but you have understood what I meant. What I was not so confident about is if a_3 annihilates the first state, speaking of your example. I didn’t find the way to prove it using only the algebra (the reason is exactly that I didn’t succeed/tried to define correctly the question, I didn’t think enough honestly), but now is clear: I can employ the algebra (the only thing that I have) to distinguish states verifying what operators annihilate the state. – Annibale Apr 8 '18 at 22:31
• Yes, $a_3$ will commute with $a_1^\dagger a_2^\dagger$ and destroy the Clifford vacuum. – Cosmas Zachos Apr 8 '18 at 22:42