A Question About $4$-Spinor Contractions

Question

Let $f_{abc}$ be a constant which is totally anti-symmetric with respect to indices $a$, $b$ and $c$. Let $\psi^{a}$, $\psi^{b}$, $\psi^{c}$ and $\epsilon$ be Grassmann-valued Majorana fermions. How to prove the following famous identity?

$$f_{abc}\left(\bar{\epsilon}\gamma_{\mu}\psi^{a}\right)\left(\bar{\psi}^{b}\gamma^{\mu}\psi^{c}\right)=0 \tag{1}$$

I am trying to use the Fierz identity show the following equation:

$$f_{abc}\left(\bar{\epsilon}\gamma_{\mu}\psi^{a}\right)\left(\bar{\psi}^{b}\gamma^{\mu}\psi^{c}\right)=f_{abc}\left(\bar{\epsilon}\gamma_{\mu}\psi^{b}\right)\left(\bar{\psi}^{a}\gamma^{\mu}\psi^{c}\right). \tag{2}$$

Then, using cyclic permutation symmetry of $f_{abc}$, one has (1).

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My calculation goes as follows, but I cannot find where I made mistakes.

$$f_{abc}\left(\bar{\epsilon}\gamma_{\mu}\psi^{a}\right)\left(\bar{\psi}^{b}\gamma^{\mu}\psi^{c}\right)=f_{abc}\bar{\epsilon}\gamma_{\mu}\left[(\psi^{a}\bar{\psi}^{b})(\gamma^{\mu}\psi^{c})\right] \tag{3}$$

The Fierz identity is

$$(\lambda\bar{\rho})\chi=-\frac{1}{4}(\bar{\lambda}\chi)\rho-\frac{1}{4}(\bar{\lambda}\gamma_{\mu}\chi)(\gamma^{\mu}\rho)-\frac{1}{4}(\bar{\lambda}\gamma_{5}\chi)(\gamma_{5}\rho)+\frac{1}{4}(\bar{\lambda}\gamma_{\mu}\gamma_{5}\chi)(\gamma^{\mu}\gamma_{5}\rho)+\frac{1}{8}(\bar{\lambda}\gamma_{\mu\nu}\chi)(\gamma^{\mu\nu}\rho)$$

where $\lambda$, $\rho$, $\chi$ are three arbitrary Grassmann-valued Majorana $4$-spinors.

Using the above identity, one has

\begin{gather} f_{abc}\bar{\epsilon}\gamma_{\mu}\left[(\psi^{a}\bar{\psi}^{b})(\gamma^{\mu}\psi^{c})\right] \\ =f_{abc}\bar{\epsilon}\gamma_{\mu}\left[-\frac{1}{4}(\bar{\psi}^{a}\gamma^{\mu}\psi^{c})\psi^{b}-\frac{1}{4}(\bar{\psi}^{a}\gamma^{\rho}\gamma^{\mu}\psi^{c})(\gamma_{\rho}\psi^{b})-\frac{1}{4}(\psi^{a}\gamma_{5}\gamma^{\mu}\psi^{c})(\gamma_{5}\psi^{b})+\frac{1}{4}(\bar{\psi}^{a}\gamma^{\rho}\gamma_{5}\gamma^{\mu}\psi^{c})(\gamma_{\rho}\gamma_{5}\psi^{b})+\frac{1}{8}(\bar{\psi}^{a}\gamma^{\rho\sigma}\gamma^{\mu}\psi^{c})(\gamma_{\rho\sigma}\psi^{b})\right] \tag{4.1} \end{gather}

Since

\begin{align} &(\bar{\psi}^{a}\gamma^{\rho}\gamma^{\mu}\psi^{c})=(\bar{\psi}^{c}\gamma^{\rho}\gamma^{\mu}\psi^{a}), \\ &(\bar{\psi}^{a}\gamma_{5}\gamma^{\mu}\psi^{c})=(\bar{\psi}^{c}\gamma_{5}\gamma^{\mu}\psi^{a}), \end{align}

The second and third terms in (4.1) vanish. One ends up with

$$f_{abc}\bar{\epsilon}\gamma_{\mu}\left[(\psi^{a}\bar{\psi}^{b})(\gamma^{\mu}\psi^{c})\right]=f_{abc}\bar{\epsilon}\gamma_{\mu}\left[-\frac{1}{4}(\bar{\psi}^{a}\gamma^{\mu}\psi^{c})\psi^{b}+\frac{1}{4}(\bar{\psi}^{a}\gamma^{\rho}\gamma_{5}\gamma^{\mu}\psi^{c})(\gamma_{\rho}\gamma_{5}\psi^{b})+\frac{1}{8}(\bar{\psi}^{a}\gamma^{\rho\sigma}\gamma^{\mu}\psi^{c})(\gamma_{\rho\sigma}\psi^{b})\right] \tag{4.2} $$

The last term in the above expression can be simplified by using the identity.

$$\gamma^{\rho\sigma}\gamma^{\mu}=\gamma^{\rho\sigma\mu}+\eta^{\sigma\mu}\gamma^{\rho}-\eta^{\rho\mu}\gamma^{\sigma}=\epsilon^{\rho\sigma\mu\lambda}\gamma_{\lambda}\gamma_{5}++\eta^{\sigma\mu}\gamma^{\rho}-\eta^{\rho\mu}\gamma^{\sigma} \tag{5}$$

Thus,

$$\frac{1}{8}(\bar{\psi}^{a}\gamma^{\rho\sigma}\gamma^{\mu}\psi^{c})(\gamma_{\rho\sigma}\psi^{b})=\frac{1}{8}\left[\bar{\psi}^{a}\left(\epsilon^{\rho\sigma\mu\lambda}\gamma_{\lambda}\gamma_{5}+\eta^{\sigma\mu}\gamma^{\rho}-\eta^{\rho\mu}\gamma^{\sigma}\right)\psi^{c}\right]\gamma_{\rho\sigma}\psi^{b}=\frac{1}{8}\epsilon^{\rho\sigma\mu\lambda}\left(\bar{\psi}^{a}\gamma_{\lambda}\gamma_{5}\psi^{c}\right)\gamma_{\rho\sigma}\psi^{b}+\frac{1}{8}\left(\bar{\psi}^{a}\gamma^{\rho}\psi^{c}\right)\gamma_{\rho\sigma}\eta^{\sigma\mu}\psi^{b}-\frac{1}{8}\left(\bar{\psi}^{a}\gamma^{\sigma}\psi^{c}\right)\gamma_{\rho\sigma}\eta^{\rho\mu}\psi^{b}.$$

Thus,

\begin{gather} \frac{1}{8}\bar{\epsilon}\gamma_{\mu}\left(\bar{\psi}^{a}\gamma^{\rho\sigma}\gamma^{\mu}\psi^{c}\right)\gamma_{\rho\sigma}\psi^{b}=\frac{1}{8}\left(\bar{\epsilon}\gamma_{\mu}\gamma_{\rho\sigma}\psi^{b}\right)\left(\bar{\psi}^{a}\gamma^{\rho\sigma}\gamma^{\mu}\psi^{c}\right) \\ =\frac{1}{8}\epsilon^{\rho\sigma\mu\lambda}\left(\bar{\epsilon}\gamma_{\mu}\gamma_{\rho\sigma}\psi^{b}\right)\left(\bar{\psi}^{a}\gamma_{\lambda}\gamma_{5}\psi^{c}\right)+\frac{1}{8}\left(\bar{\epsilon}\gamma^{\sigma}\gamma_{\rho\sigma}\psi^{b}\right)\left(\bar{\psi}^{a}\gamma^{\rho}\psi^{c}\right)-\frac{1}{8}\left(\bar{\epsilon}\gamma^{\rho}\gamma_{\rho\sigma}\psi^{b}\right)\left(\bar{\psi}^{a}\gamma^{\sigma}\psi^{c}\right) \end{gather}

Since $(\bar{\psi}^{a}\gamma_{\lambda}\gamma_{5}\psi^{c})=(\bar{\psi}^{c}\gamma_{\lambda}\gamma_{5}\psi^{a})$, the first term in the above line does not contribute. The last two terms are equal due to the anti-symmetry of $\gamma_{\rho\sigma}$.

From (5), one finds

$$\gamma^{\sigma}\gamma_{\rho\sigma}=\eta^{\sigma\alpha}\gamma_{\alpha}\gamma_{\rho\sigma}=-3\gamma_{\rho}$$

Thus,

$$\frac{1}{8}\bar{\epsilon}\gamma_{\mu}\left(\bar{\psi}^{a}\gamma^{\rho\sigma}\gamma^{\mu}\psi^{c}\right)\gamma_{\rho\sigma}\psi^{b}=\frac{1}{8}\epsilon^{\rho\sigma\mu\lambda}\left(\bar{\epsilon}\gamma_{\mu}\gamma_{\rho\sigma}\psi^{b}\right)\left(\bar{\psi}^{a}\gamma_{\lambda}\gamma_{5}\psi^{c}\right)-\frac{3}{4}\left(\bar{\psi}^{a}\gamma^{\rho}\psi^{c}\right)\left(\bar{\epsilon}\gamma_{\rho}\psi^{b}\right)$$

Plugging the above result into (4.2), one finds

\begin{gather} f_{abc}\bar{\epsilon}\gamma_{\mu}\left[(\psi^{a}\bar{\psi}^{b})(\gamma^{\mu}\psi^{c})\right] \\ =f_{abc}\bar{\epsilon}\gamma_{\mu}\left[-\frac{1}{4}(\bar{\psi}^{a}\gamma^{\mu}\psi^{c})\psi^{b}+\frac{1}{4}(\bar{\psi}^{a}\gamma^{\rho}\gamma_{5}\gamma^{\mu}\psi^{c})(\gamma_{\rho}\gamma_{5}\psi^{b})+\frac{1}{8}(\bar{\psi}^{a}\gamma^{\rho\sigma}\gamma^{\mu}\psi^{c})(\gamma_{\rho\sigma}\psi^{b})\right] \\ =f_{abc}\bar{\epsilon}\gamma_{\mu}\left[-\frac{1}{4}(\bar{\psi}^{a}\gamma^{\mu}\psi^{c})\psi^{b}+\frac{1}{4}(\bar{\psi}^{a}\gamma^{\rho}\gamma_{5}\gamma^{\mu}\psi^{c})(\gamma_{\rho}\gamma_{5}\psi^{b})\right]-\frac{3}{4}f_{abc}\left(\bar{\psi}^{a}\gamma^{\rho}\psi^{c}\right)\left(\bar{\epsilon}\gamma_{\rho}\psi^{b}\right)+\require{cancel}\bcancel{\frac{1}{8}f_{abc}\epsilon^{\rho\sigma\mu\lambda}\left(\bar{\epsilon}\gamma_{\mu}\gamma_{\rho\sigma}\psi^{b}\right)\left(\bar{\psi}^{a}\gamma_{\lambda}\gamma_{5}\psi^{c}\right)} \\ =-f_{abc}\left(\bar{\psi}^{a}\gamma^{\mu}\psi^{c}\right)\left(\bar{\epsilon}\gamma_{\mu}\psi^{b}\right)-\frac{1}{4}f_{abc}\left(\bar{\psi}^{a}\gamma^{\rho}\gamma^{\mu}\gamma_{5}\psi^{c}\right)\left(\bar{\epsilon}\gamma_{\mu}\gamma_{\rho}\gamma_{5}\psi^{b}\right). \tag{6} \end{gather}

For the last term, one can use the identity

$$\bar{\psi}^{a}\gamma^{\rho}\gamma^{\mu}\gamma_{5}\psi^{c}=-\bar{\psi}^{c}\gamma_{5}\gamma^{\mu}\gamma^{\rho}\psi^{a}=-\bar{\psi}^{c}\left(2\eta^{\mu\rho}\gamma_{5}-\gamma^{\rho}\gamma^{\mu}\gamma_{5}\right)\psi^{a},$$

which implies

$$\bar{\psi}^{a}\gamma^{\rho}\gamma^{\mu}\gamma_{5}\psi^{c}-\bar{\psi}^{c}\gamma^{\rho}\gamma^{\mu}\gamma_{5}\psi^{a}=-2\eta^{\mu\rho}\bar{\psi}^{c}\gamma_{5}\psi^{a}. \tag{7}$$

Thus,

\begin{gather} f_{abc}\bar{\epsilon}\gamma_{\mu}\left[(\psi^{a}\bar{\psi}^{b})(\gamma^{\mu}\psi^{c})\right] \\ =-f_{abc}\left(\bar{\psi}^{a}\gamma^{\mu}\psi^{c}\right)\left(\bar{\epsilon}\gamma_{\mu}\psi^{b}\right)-\frac{1}{4}f_{abc}\left(\bar{\psi}^{a}\gamma^{\rho}\gamma^{\mu}\gamma_{5}\psi^{c}\right)\left(\bar{\epsilon}\gamma_{\mu}\gamma_{\rho}\gamma_{5}\psi^{b}\right) \\ =-f_{abc}\left(\bar{\psi}^{a}\gamma^{\mu}\psi^{c}\right)\left(\bar{\epsilon}\gamma_{\mu}\psi^{b}\right)-\frac{1}{8}\left[f_{abc}\left(\bar{\psi}^{a}\gamma^{\rho}\gamma^{\mu}\gamma_{5}\psi^{c}\right)+f_{cba}\left(\bar{\psi}^{c}\gamma^{\rho}\gamma^{\mu}\gamma_{5}\psi^{a}\right)\right]\left(\bar{\epsilon}\gamma_{\mu}\gamma_{\rho}\gamma_{5}\psi^{b}\right) \\ =-f_{abc}\left(\bar{\psi}^{a}\gamma^{\mu}\psi^{c}\right)\left(\bar{\epsilon}\gamma_{\mu}\psi^{b}\right)-\frac{1}{8}\left[f_{abc}\left(\bar{\psi}^{a}\gamma^{\rho}\gamma^{\mu}\gamma_{5}\psi^{c}\right)-f_{abc}\left(\bar{\psi}^{c}\gamma^{\rho}\gamma^{\mu}\gamma_{5}\psi^{a}\right)\right]\left(\bar{\epsilon}\gamma_{\mu}\gamma_{\rho}\gamma_{5}\psi^{b}\right) \\ =-f_{abc}\left(\bar{\psi}^{a}\gamma^{\mu}\psi^{c}\right)\left(\bar{\epsilon}\gamma_{\mu}\psi^{b}\right)-\frac{1}{8}f_{abc}\left[-2\eta^{\mu\rho}\bar{\psi}^{c}\gamma_{5}\psi^{a}\right]\left(\bar{\epsilon}\gamma_{\mu}\gamma_{\rho}\gamma_{5}\psi^{b}\right) \end{gather}

But since $\bar{\psi}^{c}\gamma_{5}\psi^{a}=\bar{\psi}^{a}\gamma_{5}\psi^{c}$, the last term in the above expression vanish. Thus, one ends up with

$$f_{abc}\left(\bar{\epsilon}\gamma_{\mu}\psi^{a}\right)\left(\bar{\psi}^{b}\gamma^{\mu}\psi^{c}\right)=f_{abc}\bar{\epsilon}\gamma_{\mu}\left[(\psi^{a}\bar{\psi}^{b})(\gamma^{\mu}\psi^{c})\right]=-f_{abc}\left(\bar{\psi}^{a}\gamma^{\mu}\psi^{c}\right)\left(\bar{\epsilon}\gamma_{\mu}\psi^{b}\right)=f_{abc}\left(\bar{\psi}^{c}\gamma^{\mu}\psi^{a}\right)\left(\bar{\epsilon}\gamma_{\mu}\psi^{b}\right),$$

which is trivially correct due to the cyclic property of $f_{abc}$.

Other than using the Fierz identity to rearrange the spinors, I cannot find any other way to prove equation (1).

Did I make any mistakes in the above derivations?

If not, how to prove (1)?

Answer 1

tldr; It seems to me your are counting the legs of a millipede and dividing by a thousand. (1) holds because the l.h.s. is minus half itself.

That is, you use the second line of the Fierz matrix, $VV\to -SS+VV/2 +AA/2+PP$. Now ,for Majorana spinors, only V is transposition-antisymmetric, $\bar \psi \gamma_\mu \chi = - \bar \chi \gamma_\mu \psi $; while A,S,P are symmetric, so they collapse by the antisymmetrizing action of $f_{abc}$. Thus,

$$f_{abc}\left(\bar{\epsilon}\gamma_{\mu}\psi^{a}\right)\left(\bar{\psi}^{b}\gamma^{\mu}\psi^{c}\right)=\frac{1}{2} f_{abc}\left(\bar{\epsilon}\gamma_{\mu}\psi^{c}\right)\left(\bar{\psi}^{b}\gamma^{\mu}\psi^{a}\right)=-\frac{1}{2} f_{abc}\left(\bar{\epsilon}\gamma_{\mu}\psi^{a}\right)\left(\bar{\psi}^{b}\gamma^{\mu}\psi^{c}\right), $$ so the expression must be 0 , your identity (1).