On SUSY irreps: eigenvalues of Pauli-Lubanski operator?

[lennyleonard]

Hi everyone,

Just an easy question that came to my mind while studying basics of SUSY.
Consider in N=1, D=4 a massive clifford vacuum $|m,s,s_3\rangle$, and for cconcreteness take its spin to be s=1/2.
Now, acting with the four supercharges on both the $|m,1/2,1/2\rangle$ and $|m,1/2,-1/2\rangle$ states of the cliffor vacuum i get a total of 8 states, namely
$$\begin{align} &|m,1/2,1/2\rangle\\ &\overline Q_{\dot 1}|m,1/2,1/2\rangle \text{ with }s_3=0\\[2mm] &\overline Q_{\dot 2}|m,1/2,1/2\rangle \text{ with }s_3=1\\[2mm] &\overline Q_{\dot 1}\overline Q_{\dot 2}|m,1/2,1/2\rangle \text{ with }s_3=1/2\\[2mm] \end{align}$$
and
$$\begin{align} &|m,1/2,-1/2\rangle\\ &\overline Q_{\dot 1}|m,1/2,-1/2\rangle \text{ with }s_3=-1\\[2mm] &\overline Q_{\dot 2}|m,1/2,-1/2\rangle \text{ with }s_3=0\\[2mm] &\overline Q_{\dot 1}\overline Q_{\dot 2}|m,1/2,1/2\rangle \text{ with }s_3=-1/2\\[2mm] &\end{align}$$

Now, as far as I understood, these should form a massive vector, which I'd say then is set up by ($\overline Q_{\dot 2}|m,1/2,1/2\rangle\,,\, \overline Q_{\dot 1}|m,1/2,-1/2\rangle\,,\,\frac{1}{\sqrt{2}}(\overline Q_{\dot 2}|m,1/2,-1/2\rangle+\overline Q_{\dot 1}|m,1/2,1/2\rangle)$), two weyl fermions of opposite chirality (thus a dirac spinor) ($|m,1/2,1/2\rangle\,,|m,1/2,-1/2\rangle$) and ($\overline Q_{\dot 1}\overline Q_{\dot 2}|m,1/2,1/2\rangle\,,\overline Q_{\dot 1}\overline Q_{\dot 2}|m,1/2,-1/2\rangle$) and at last one real scalar $\frac{1}{\sqrt{2}}(\overline Q_{\dot 2}|m,1/2,-1/2\rangle-\overline Q_{\dot 1}|m,1/2,1/2\rangle)$.

Now, my questions:

1) How do I know that the state $\frac{1}{\sqrt{2}}(\overline Q_{\dot 2}|m,1/2,-1/2\rangle-\overline Q_{\dot 1}|m,1/2,1/2\rangle)$ is indeed a scalar? I mean, I only know that its $s_3$ component is zero, but I don't know whether ist spin is! Equivalently, how do I know that the state $\frac{1}{\sqrt{2}}(\overline Q_{\dot 2}|m,1/2,-1/2\rangle+\overline Q_{\dot 1}|m,1/2,1/2\rangle)$ is part of a vector and not just another scalar (or maybe that the plus sign linear combination above is a scalar and the minus sign combination is part of the vector field)?
To answer this question I thought of letting the square of the Pauli-Lubanski operator W act on these states: even if this is no longer a casimir it should nevertheless give me the spin of the state, right? Anyway, to this end i tried to compute the commutator $[W^2,Q]$ (where Q here is just any of the supercharges), but I got a non-sense answer, ($[W^2,Q] =\frac{-3m^2}{4}Q$, which seems reasonable but actually isn't - i probably messed up somewhere) so that I'really missing something here...could you please help me on this matter?2) Also, if you still have time you could also explain me why in the multiplet above the scalar is real (without referring to the fact that bosonic modes must be equal in number to fermionic ones) and why the two weyl fermions have opposite chirality?
Cheers!

 

[AndyW]

Hello,

Thank you for your question. The states you have listed do indeed form a massive vector multiplet in N=1, D=4 SUSY. To answer your first question, the state \frac{1}{\sqrt{2}}(\overline Q_{\dot 2}|m,1/2,-1/2\rangle-\overline Q_{\dot 1}|m,1/2,1/2\rangle) is indeed a scalar because it transforms as a singlet under the Lorentz group, meaning its spin is zero. This can also be seen by noting that it is a linear combination of two states with opposite spin projections, so the overall spin projection must be zero.

To understand why the two Weyl fermions have opposite chirality, we can look at the transformation properties of the supercharges. The supercharges \overline Q_{\dot 1} and \overline Q_{\dot 2} have opposite chirality, meaning they transform under different representations of the Lorentz group. This is why their actions on the states |m,1/2,1/2\rangle and |m,1/2,-1/2\rangle result in opposite spin projections. As for why the scalar is real, this is a consequence of SUSY. In SUSY, the number of bosonic and fermionic degrees of freedom must be equal, and since we have two Weyl fermions in this multiplet, we must also have two real scalars.

To answer your second question, we can consider the transformation properties of the supercharges again. The supercharges \overline Q_{\dot 1} and \overline Q_{\dot 2} transform under different representations of the Lorentz group, which means that they act on states with different chirality in opposite ways. This is why the two Weyl fermions in the multiplet have opposite chirality.

I hope this helps clarify your questions. If you have any further questions, please don't hesitate to ask.