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.

  • 2
    $\begingroup$ It's probably best to get this standard material from books or lecture notes like these (page 22). $\endgroup$ – knzhou Aug 9 '18 at 20:26
  • $\begingroup$ @knzhou That is a good proof, but did they specify why they considered $\left[ Q_{\alpha} , P_{\mu} \right]$ as $c \sigma _{\alpha \dot{\alpha}}^{\mu} Q_{\dot{\alpha}}^{\dagger}$? $\endgroup$ – zyy Aug 9 '18 at 22:00
  • $\begingroup$ 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 $. $\endgroup$ – Prahar Aug 10 '18 at 3:57
  • $\begingroup$ @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. $\endgroup$ – zyy Aug 10 '18 at 13:48
  • 1
    $\begingroup$ 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. $\endgroup$ – 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 $[Q_\alpha,~Q^\dagger_{\dot\alpha}]~=~0$.

  • $\begingroup$ 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. $\endgroup$ – zyy Aug 9 '18 at 21:34
  • 1
    $\begingroup$ 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. $\endgroup$ – 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. $\Box$.


  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.

  • $\begingroup$ Thanks so much! There is one point in your answer that I would like to ask, how did you conclude equation (3)? $\endgroup$ – zyy Aug 10 '18 at 14:20
  • $\begingroup$ Before answering that, let me turn your question around: Which other terms would you like to add to the RHS of eq. (3)? $\endgroup$ – Qmechanic Aug 10 '18 at 14:51
  • $\begingroup$ 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 $Q_{\dot{\alpha}}^{\dagger}$. $\endgroup$ – zyy Aug 10 '18 at 15:39
  • $\begingroup$ 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. $\endgroup$ – Qmechanic Aug 10 '18 at 15:56
  • $\begingroup$ I am sorry that I could not understand point 2 immediately, I will study on that and try, thanks! $\endgroup$ – zyy Aug 10 '18 at 22:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.