How to prove the translation generator commutes with the spinors in SUSY algebra? I was reading Modern Supersymmetry by John Terning, the book starts with SUSY algebra and says
$$
\left[ P_{\mu} , Q_{\alpha} \right] = \left[ P_{\mu} , Q_{\alpha}^{\dagger} \right] = 0
$$
I am wondering how I could prove the above commutation relation given the following anticommutators
\begin{equation}
\left\{ Q_{\alpha} , Q_{\beta} \right\} = \left\{ Q_{\dot{\alpha}}^{\dagger} , Q_{\dot{\beta}}^{\dagger} \right\} = 0 , \quad
\left\{ Q_{\alpha} , Q_{\dot{\alpha}}^{\dagger} \right\} = 2 \sigma _{\alpha \dot{\alpha}}^{\mu} P_{\mu}
\end{equation}
And $\sigma ^{\mu}$ is defined as $\sigma ^{\mu} = \left( 1 , \sigma ^i \right)$.

One way I tried is to use Jacobi identity $\left[ \left\{ Q_{\alpha} , Q_{\dot{\alpha}}^{\dagger} \right\} , Q_{\alpha} \right] + \left[ \left\{ Q_{\dot{\alpha}}^{\dagger} , Q_{\alpha} \right\} , Q_{\alpha} \right] + \left[ \left\{ Q_{\alpha} , Q_{\alpha} \right\} , Q_{\dot{\alpha}}^{\dagger} \right] = 0$ and therefore find $\sigma _{\alpha \dot{\alpha}}^{\mu} \left[ P_{\mu} , Q_{\alpha} \right] = 0$. But this result then leads to the following matrix equation if you substitute back the Pauli matrices
$$
\left( \begin{array}{cc}
\left[ P_0 , Q_1 \right] + \left[ P_3 , Q_1 \right] &  \left[ P_1 , Q_1 \right] - \mathrm{i} \left[ P_2 , Q_1 \right] \\
\left[ P_1 , Q_2 \right] + \mathrm{i} \left[ P_2 , Q_2 \right] & \left[ P_0 , Q_2 \right] - \left[ P_3 , Q_2 \right] \\
\end{array} \right) = 0 
$$
It is clear that for the off diagonal terms, due to the imaginary $\mathrm{i}$, it is clear that $\left[ P_1 , Q_1 \right] = \left[ P_1 , Q_2 \right] = \left[ P_2 , Q_1 \right] = \left[ P_2 , Q_2 \right] = 0$. However, one can not argue the same way for the diagonal terms.
 A: It follows from the SUSY algebra with $[Q_\alpha,~Q^\dagger_{\dot\alpha}]~=~0$. I will drop the spinor matrix index notion for brevity. Then consider the commutator
$$
2\sigma^\mu[P,~Q]~=~[\{Q,~Q^\dagger\}, Q]
$$
$$
=~Q[Q^\dagger,~Q]~+~[Q^\dagger,~Q]Q.
$$
Here obviously $[Q,~Q]~=~0$ and we use the commutator $[Q_\alpha,~Q^\dagger_{\dot\alpha}]~=~0$.
A: Sketched proof:


*

*We can transform the momentum $P_{\mu}\leftrightarrow P_{\alpha\dot{\alpha}}$ using the Pauli sigma matrices. Then$^1$
$$[ Q_{\alpha}, \bar{Q}_{\dot{\alpha}}]~=~2P_{\alpha\dot{\alpha}}, \qquad [P_{\alpha\dot{\alpha}}, P_{\beta\dot{\beta}}] ~=~0.\tag{1}$$

*We will assume that the generators $P_{\alpha\dot{\alpha}}$, $Q_{\alpha}$,   $\bar{Q}_{\dot{\alpha}}$ of translations and super-translations are a linear basis for a super Lie algebra (up to possible central charges).

*We also have 
$$[ Q_{\alpha},Q_{\beta}]~=~0, \qquad [\bar{Q}_{\dot{\alpha}},\bar{Q}_{\dot{\beta}}]~=~0,\tag{2}$$
since the only relevant $SL(2,\mathbb{C})$-covariant tensor structure (the Levi-Civita tensor) for the RHSs of eq. (2) has the wrong symmetry
property under exchange of indices.

*The only $SL(2,\mathbb{C})$-covariant possibility is
$$[ Q_{\alpha}, P_{\beta\dot{\beta}}]~=~c\epsilon_{\alpha\beta} \bar{Q}_{\dot{\beta}}\tag{3} $$
for some constant $c$. By Hermitian conjugation this implies 
$$[ \bar{Q}_{\dot{\alpha}}, P_{\beta\dot{\beta}}]~\stackrel{(3)}{\propto}~\bar{c}\epsilon_{\dot{\alpha}\dot{\beta}} Q_{\beta} \tag{4}$$
so that
$$
|c|^2 \epsilon_{\alpha\beta}\epsilon_{\dot{\beta}\dot{\gamma}} Q_{\gamma}
~\stackrel{(3)+(4)}{\propto}~ 
[[ Q_{\alpha}, P_{\beta\dot{\beta}}], P_{\gamma\dot{\gamma}}]
~\stackrel{\text{Jac.Id.}}{=}~[(\beta\dot{\beta})\leftrightarrow( \gamma\dot{\gamma})].\tag{5}$$
The LHS of eq. (5) only has the symmetry of the RHS of eq. (5) if the constant $c=0$ vanishes. $\Box$.
References:


*

*F. Quevedo & O. Schlotterer,
Supersymmetry and Extra Dimensions, Lecture notes, 2008; p. 22. (Hat tip: knzhou.)


--
$^1$ In this answer the square bracket 
$[A,B]~:=~ AB-(-1)^{|A||B|}BA$ denotes the supercommutator.
