The $\mathcal N = 1$ algebra \begin{equation} \{Q_\alpha,\overline Q_\beta\} = -2P_\mu\Gamma^\mu_{\alpha\beta} \end{equation} is in a frame where $k_1 = k_0$ \begin{equation} \{Q_\alpha, Q_\beta^\dagger\} = 2k^0(1 + 2S_0)_{\alpha\beta} \end{equation} or in $\bf s$ basis, \begin{equation} \{Q_{s_0',s_1'},Q^\dagger_{s_0,s_1}\} = 4k^0\delta_{s_0,1/2}\delta_{\bf s'\bf s}. \end{equation}

The matrix elements of $Q_{-1/2,s_1}$ vanish. The remaining operator gives the raising and lowering operator: \begin{equation} b = (4k^0)^{-1/2}Q_{1/2,-1/2},\;\; b^\dagger = (4k^0)^{-1/2}Q_{1/2,1/2},\;\; \{b,b^\dagger\} = 1,\;\; b^2 = (b^\dagger)^2 = 0 \end{equation}

We construct the representation by starting from $|\lambda\rangle$ {\it s.t.}, \begin{equation} S_1|\lambda\rangle = \lambda|\lambda\rangle,\;\; b|\lambda\rangle = 0. \end{equation} By operating $b^\dagger$, \begin{equation} b^\dagger|\lambda\rangle = \Big|\lambda + \frac12\Big\rangle,\;\; S_1\Big|\lambda + \frac12\Big\rangle = \Big(\lambda + \frac12\Big) \Big|\lambda + \frac12\Big\rangle. \end{equation}

My question is why the state $b^\dagger|\lambda\rangle$ is $|\lambda + 1/2\rangle$?

P.S. I think that in order to confirm it we need to calculate the commutator $[b^\dagger,S_1]$ and it is equal to $1/2$.


1 Answer 1


You need to use the super Poncaré algebra, that besides the commutator $[q,q]\sim p$ there is also the commutators $[M,p]\sim p$, $[M,q]\sim q$, and $[M,M]\sim M$. Here $q$ are the supercharges, $p$ are the momentum components and $M$ the Lorentz generators.

Turns out that using Jacobi identity

$$ [M,[q,q]]=[[M,q],q]+[q,[M,q]] $$

and that $[q,q]\sim p$ we learn that $q$ must be a spinor, so it has charge $\pm \frac{1}{2}$ under $S_0, \dots, S_{\frac{d}{2}}$ Cartan subalgebra of the Lorentz generators

$$ [S,q]\sim \frac{1}{2} q $$

I did not put indices up to here since all I said is valid to any dimension and notation. For your case where we have $d=1+3$, we can work with the $SL(2,\mathbb{C})$ notation

$$ \alpha=(1\equiv ++,\,2\equiv --)\qquad \dot\alpha=(\dot 1\equiv +-,\,\dot 2\equiv -+) $$

where the first sign is the charge under $S_0$ and the second is the charge under $S_1$. The momentum is given by $p_{\alpha\dot\alpha}$ and the supercharges are $q_{\alpha}$ and $\bar q_{\dot\alpha}$ such that

$$ [q_{\alpha},\bar q_{\dot\alpha}]=2 p_{\alpha\dot\alpha} $$

and we have

$$ [S_0,q_{\pm \,\star}] = \pm\frac{1}{2} q_{\pm \,\star},\qquad [S_1,q_{\star\, \pm}] = \pm\frac{1}{2} q_{\star \,\pm} $$

that follows from the Jacobi identity of

$$ [S_i,2p_{\alpha\dot\alpha}]=[S_{i}, [q_{\alpha},\bar q_{\dot\alpha}]]=[[S_i,q_{\alpha}],\bar q_{\dot\alpha}]+[q_{\alpha},[S_i,\bar q_{\dot\alpha}]] $$


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