# 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

$$\left\{ Q_{\alpha} , Q_{\beta} \right\} = \left\{ Q_{\dot{\alpha}}^{\dagger} , Q_{\dot{\beta}}^{\dagger} \right\} = 0 , \quad \left\{ Q_{\alpha} , Q_{\dot{\alpha}}^{\dagger} \right\} = 2 \sigma _{\alpha \dot{\alpha}}^{\mu} P_{\mu}$$

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.

• It's probably best to get this standard material from books or lecture notes like these (page 22). – knzhou Aug 9 '18 at 20:26
• @knzhou That is a good proof, but did they specify why they considered $\left[ Q_{\alpha} , P_{\mu} \right]$ as $c \sigma _{\alpha \dot{\alpha}}^{\mu} Q_{\dot{\alpha}}^{\dagger}$? – zyy Aug 9 '18 at 22:00
• Are you summing over $\alpha$ in the Jacobi identity that you have considered? You should consider the more general Jacobi identity $[ \{ Q_\alpha , Q^\dagger_{\dot \alpha} \} , Q_\beta ] + \cdots = 0$. – Prahar Aug 10 '18 at 3:57
• @Prahar I was not summing over index $\alpha$. As for general Jacobi identity, I did that before, which would end up giving you $2 \sigma _{\alpha \dot{\alpha}}^{\mu} \left[ P_{\mu} , Q_{\beta} \right] + 2 \sigma _{\beta \dot{\alpha}}^{\mu} \left[ P_{\mu} , Q_{\alpha} \right] = 0$ and it is hard to proceed from that. – zyy Aug 10 '18 at 13:48
• The way I would proceed is to note that the commutator is a tensor product of a spin 1 and spin 1/2 representation of the Lorentz algebra. Thus on the RHS you get either a spin 1/2 charge or a spin 3/2 one. The latter is forbidden by Coleman Mandela so only the spin 1/2 could possibly appear in that commutator which is either Q or Q-bar. Next, write down the most general possible such term matching all the indices. Then you have a general ansatz for the commutator up to two constants (maybe one?). Fix the constants using the Jacobi identity. – Prahar Aug 10 '18 at 13:52

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 $[Q_\alpha,~Q^\dagger_{\dot\alpha}]~=~0$.

• Thank you so much! I was able to show $\sigma _{\alpha \dot{\alpha}}^{\mu} \left[ P_{\mu} , Q_{\alpha} \right] = 0$ from Jacobi identity involving anticommutators. But I was not sure how to proceed after that. I will make edits to my question to reflect that. – zyy Aug 9 '18 at 21:34
• Comments to the answer (v1): 1. Spinor indices can not be dropped consistently. 2. In a super Lie algebra, only supercommutators should be relevant. E.g. the anticommutator (but not the commutator) of two supercharges should be relevant. – Qmechanic Aug 10 '18 at 14:47

Sketched proof:

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

2. 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).

3. 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.

4. 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:

1. 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.

• Thanks so much! There is one point in your answer that I would like to ask, how did you conclude equation (3)? – zyy Aug 10 '18 at 14:20
• Before answering that, let me turn your question around: Which other terms would you like to add to the RHS of eq. (3)? – Qmechanic Aug 10 '18 at 14:51
• Sicne $P_{\mu}$ is essentially some anticommutator of $Q_{\alpha}$ and $Q_{\dot{\alpha}}^{\dagger}$, I would assume generally the right side of equation (3) should be a cubic polynomial of $Q_{\alpha}$ and $Q_{\dot{\alpha}}^{\dagger}$. – zyy Aug 10 '18 at 15:39
• In the operator language, the LHS of eq. (3) is indeed a cubic polynomial of generators. The RHS of eq. (3) must be a linear polynomial of generators because the point 2. – Qmechanic Aug 10 '18 at 15:56
• I am sorry that I could not understand point 2 immediately, I will study on that and try, thanks! – zyy Aug 10 '18 at 22:30