I have been struggling with these Majorana fermion identities for quite sometime now. I would be grateful if someone can help me with them.

Let $\lambda$,$\theta$ and $\psi$ be $4$-component Majorana fermions. Then apparently the following are true,

  • $(\bar{\theta}\gamma_5 \theta)(\bar{\psi}_L\theta)(\bar{\theta}\gamma^\mu \partial_\mu \psi_L) = \frac{1}{4}(\bar{\theta}\gamma_5 \theta)^2 (\bar{\psi}_L\gamma^\mu \partial_\mu \psi_L)$

  • $(\bar{\theta}\gamma_5 \theta)(\partial _ \mu \bar{\psi}_L \gamma^\mu \theta)(\bar{\theta}\psi_L) = -\frac{1}{4}(\bar{\theta}\gamma_5 \theta)^2 (\partial_\mu \bar{\psi}_L\gamma^\mu \psi_L)$

  • $(\bar{\theta}\gamma _5 \gamma _\mu \theta)(\bar{\psi}_L\theta)(\bar{\theta}\psi_L) = \frac{1}{4} (\bar{\theta}\gamma _5 \theta)^2 (\bar{\psi}\gamma ^\mu \psi)$

  • $(\bar{\theta}\gamma_5 \theta)(\bar{\psi}_L\theta)(\bar{\theta}\lambda) = \frac{1}{4} (\bar{\theta}\gamma _5 \theta)^2 (\bar{\psi}_L\lambda)$

I guess looking at the above that all the 4 have some generic pattern and hence probably require some same key idea which I am missing. Its not clear to me as to how to "pull out" the $\theta$s between the other fermions to outside and then again repack then into a $(\bar{\theta}\gamma_5 \theta)$. I will be happy to get some help regarding the above.

  • $\begingroup$ On another question and in the comment thread below, I saw that you were using Weinberg III as your primary textbook for Susy. His eq(26.A.8) shows that the right hand side of your 3rd equation vanishes. All your equations have 4 $\theta$'s and so can be put in a standard form and proved using his (26.A.15). This formula (basically a completeness relation) answers your question on how to "pull out" and "repack" the $\theta$'s. $\endgroup$
    – Simon
    Mar 6 '11 at 1:40

They're more complicated cousins of the Fierz identities,


The article above also recommends you Okun's book for the general recipe to prove similar identities. Note that all the identities you wrote except for the third one are just normal Fierz identities because the first factor may be cancelled as it appears (once) both on left-hand side and right-hand side.

The fact that it's not trivial to prove those identities doesn't mean that they're not true. If you rewrote them correctly, they are true. You may trust that they're true. In principle, you may verify them by writing the most general values of the spinors $\theta$ and $\psi$ (and $\lambda$, in the last case) - in terms of four complex components each (reduced to two complex by the Majorana condition) - and by calculating the explicit values of the products of the inner products. The identities above will hold. It's kind of inevitable that some identities of a similar form hold because there are just four components in each variable and the number of monomials of the right degree in those components is limited and may be therefore written in different ways.

Spinor identities may be annoyingly technical, especially if one deals with higher dimensions or extended supersymmetry.


If you're going to do a lot of work in 4 dimensions, it might be worth learning two-component notation. This is used in the text books, e.g., of Srednicki, Buchbinder & Kuzenko or the comprehensive article by Dreiner, Haber and Martin. Two component notation has the advantage that the 4 dimensional objects decompose into the direct sum (or something similar) of 2 dimensional objects, which have fewer invariants and identities.

I'll use the conventions of B&K (which I think slightly differ from those in the question).

The four component spinors takes the form $$ \psi_a = \begin{pmatrix}\psi_\alpha \\ \bar\psi^{\dot\alpha}\end{pmatrix} \,, \quad \bar\psi^a = \begin{pmatrix}\psi^\alpha & \bar\psi_{\dot\alpha}\end{pmatrix} $$ and the gamma and projection matrices $$ \gamma_a = \begin{pmatrix} 0 & \sigma_a \\ \tilde\sigma_a & 0 \end{pmatrix}\,,\quad \gamma_5 = -i \gamma_0\gamma_1\gamma_2\gamma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \,,\quad $$ $$ P_L = \frac12 (1 + \gamma_5) \,, \quad P_R = \frac12 (1 - \gamma_5) $$ the left and right spinors are then $$ \psi_L = P_L\psi = \begin{pmatrix}\psi \\ 0 \end{pmatrix} \,, \quad \psi_R = P_R\psi = \begin{pmatrix} 0 \\ \bar\psi \end{pmatrix} \,, \quad $$ where there should never be any reason to confuse a 2-component and 4-component spinor.

The spinor bilinears are $$ \bar\theta \theta = \theta^2 + \bar\theta^2 \,,\quad \bar\theta \gamma_5 \theta = \theta^2 - \bar\theta^2 $$ etc..., where the contraction convention is $\theta^2 = \theta^\alpha \theta_\alpha$ and $\bar\theta^2 = \bar\theta_{\dot\alpha} \bar\theta^{\dot\alpha}$.

Then, to check your final identity (for example), the left hand side is $$ (\bar\theta\gamma_5\theta)(\bar\psi_L\theta)(\bar\theta\lambda) = (\theta^2-\bar\theta^2)(\bar\psi\bar\theta)(\theta\lambda+\bar\theta\bar\lambda) = (\theta^2-0)(-\frac12\bar\theta^2)(\bar\psi\bar\lambda) = -\frac12\theta^2\,\bar\theta^2\,\bar\psi\bar\lambda $$ where we used the fact that $\theta^3=\theta_{\alpha_1}\theta_{\alpha_2}\theta_{\alpha_3}=0$ and similarly for $\bar\theta^3=0$. We also used the important identities $$ \begin{align} \theta_\alpha \theta^\beta &= -\frac12\theta^2\delta_\alpha^\beta \,,& \bar\theta^{\dot\alpha}\bar\theta_{\dot\beta} &=-\frac12\bar\theta^2\delta_{\dot\beta}^{\dot\alpha} \ . \end{align}$$ The right hand side is $$ \frac14(\bar\theta\gamma_5\theta)^2(\bar\psi_L\lambda) = \frac14(\theta^2-\bar\theta^2)(\theta^2-\bar\theta^2)(\bar\psi\bar\lambda) = -\frac12\theta^2\bar\theta^2\bar\psi\bar\lambda $$ so the last identity checks out.

The rest of the identities can be similarly checked.


Since there seems to be a bit of confusion for the other identities, here's their proofs. Because I'm lazy, I've suppressed all indices. For example $\bar\theta\partial\psi =\bar\theta_{\dot\alpha}\tilde\sigma^{\dot\alpha\alpha}_a\partial^a\psi_\alpha =-(\partial^a\psi_\alpha)\tilde\sigma^{\dot\alpha\alpha}_a\bar\theta_{\dot\alpha} =-(\partial_a\psi^\alpha)\sigma_{\alpha\dot\alpha}^a\bar\theta^{\dot\alpha} =-\psi\overleftarrow{\partial}\bar\theta\ .$

Identity 1: $$ LHS = (\theta^2-\bar\theta^2)(\bar\psi\bar\theta)(\bar\theta\partial\psi) = (\theta^2-0)(-\frac12\bar\theta^2)(\bar\psi\partial\psi) = -\frac12\theta^2\bar\theta^2(\bar\psi\partial\psi) $$ $$ RHS = \frac14(\theta^2-\bar\theta^2)^2(\bar\psi\partial\psi) = LHS $$

Identity 2 is basically just the complex conjugate result.

Identity 3: I think you have a mistake in this one (ignoring the obvious problem of the non-matching $\mu$ index). This is because $(\bar{\psi}\gamma ^\mu \psi)=0$ if we insert a $\gamma_5$ so it's like the term on the left it's nonvanishing $(\bar{\psi}\gamma_5\gamma_\mu \psi)=2\psi\sigma_\mu\bar\psi$. So $$ LHS = (\theta\sigma_\mu\bar\theta-\bar\theta\tilde\sigma_\mu\theta)(\bar\psi\bar\theta)(\psi\theta) = 2(\psi\theta)(\theta\sigma_\mu\bar\theta)(\bar\theta\bar\psi) = \frac12\theta^2\bar\theta^2 (\psi\sigma_\mu\bar\psi) $$ $$ RHS = \frac{1}{4} (\bar{\theta}\gamma _5 \theta)^2 (\bar{\psi}\gamma_5\gamma_\mu\psi) = -\theta^2\bar\theta^2(\psi\sigma_\mu\bar\psi) $$ so it seems to be out by a factor of $-1/2$. (Although, I might have miscounted the factors of 1/2).

  • $\begingroup$ Nice answer. However I would not recommend a 311 page physics reports article as a reference for learning about two-spinor notation ;] $\endgroup$
    – user346
    Mar 3 '11 at 6:49
  • 1
    $\begingroup$ @DeepakVaid: I did call it comprehensive! :) $\endgroup$
    – Simon
    Mar 3 '11 at 7:12
  • $\begingroup$ @Simon I wonder what you are calling as "2-component" notation. Yours seem to be the kind of notation that is used in Wess and Bagger. But would you call that 2-component? I am using the convention in Weinberg's books of having all fermions as 4-component. $\endgroup$
    – user6818
    Mar 3 '11 at 17:32
  • $\begingroup$ @Anirbit: Yep, this is the notation that Wess and Bagger use. It's either called "2-component" or Van de Waerden notation. $\endgroup$
    – Simon
    Mar 3 '11 at 21:46
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    $\begingroup$ @Simon; I was using Burgess & Moore as a source. But I need to understand this better. What I want is not to be trained to calculate like a computer program / monkey, but instead to be able to understand it from the bottom up. I'm thinking that for me, the right thing to do is to rewrite Majorana spinors into geometric algebra / Pauli spin matrices using my usual density matrix form. But before I do this, I'll do the regular spinors. $\endgroup$ Mar 4 '11 at 3:54

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