Consider the Helicity projectors $$\Pi_{\pm}:=\frac{\mathbb{1}\pm\gamma^{5}\gamma_{\nu}n^{\nu}}{2}$$ where I define the versor $n^{\mu}\equiv(\frac{|\stackrel{\rightarrow}{p}|}{m},\frac{ω_{p}}{m}\frac{\stackrel{\rightarrow}{p}}{|\stackrel{\rightarrow}{p}|})$. It's easy to show that in the rest frame of our particle the following relations hold $$\Pi_{+}u_{1}(0)=u_{1}(0)\hspace{1cm}\Pi_{-}u_{2}(0)=u_{2}(0)\hspace{1cm}\Pi_{+}v_{1}(0)=v_{1}(0)\hspace{1cm}\Pi_{-}v_{2}(0)=v_{2}(0)$$ where I denoted with $u,v$ the usual Dirac spinors at rest $$u_{1}(0)=\begin{pmatrix}1\\ 0\\ 0\\ 0\\ \end{pmatrix}\hspace{0.5cm}u_{2}(0)=\begin{pmatrix}0\\ 1\\ 0\\ 0\\ \end{pmatrix}\hspace{0.5cm}v_{1}(0)=\begin{pmatrix}0\\ 0\\ 1\\ 0\\ \end{pmatrix}\hspace{0.5cm}v_{2}(0)=\begin{pmatrix}0\\ 0\\ 0\\ 1\\ \end{pmatrix} $$ This can be easily shown by using for simplicity $n^{\mu}=(0,0,0,1)$ and then by direct computation.

Now I would like to prove that those relations are also true in whatever frame I consider, and usually this is skipped because the expression $\gamma^{5}\gamma^{\mu}n_{\mu}$ is covariant. My question is about that:- If I consider a Lorentz transformation $\Lambda$ then each gamma matrix should keep unchanged, while the versor should transform as $n^{\mu}\rightarrow n'^{\mu}=\Lambda^{\mu}_{\nu}n^{\nu}$. On the other hand Dirac spinors should transform as $S(\Lambda)u$ as like $v$, how can I show explicitly the covariance of those relations? My guess is that one should show that $$S^{-1}(\Lambda)\gamma^{5}\gamma_{\mu}\Lambda^{\mu}_{\nu}S(\Lambda)=\gamma^{5}\gamma_{\nu}$$

How can I do that? I thought about using $S^{-1}(\Lambda)\gamma^{\mu}S(\Lambda)=\Lambda^{\mu}_{\nu}\gamma^{\nu}$ but I didn't succeed.

  • 1
    $\begingroup$ Your last equation is correct for proper (det $\Lambda$ =1) Lorentz transoformations, as $\gamma^\mu$ always transforms as a 4-vector and $\gamma^5$ is invariant under such transformations. $\endgroup$
    – mike stone
    Commented Sep 22, 2023 at 16:33
  • $\begingroup$ Mm do you know some reference that shows why $\gamma^{5}$ is invariant? I tried implementing the identity at each place and using $\gamma^{5}=i\gamma^{0}\gamma^{1}\gamma^{2}\gamma^{3}$ but I get stuck $\endgroup$
    – Filippo
    Commented Sep 22, 2023 at 16:36

1 Answer 1


The proof is the following:

$$\Pi_\pm = \frac{1}{2} \pm \frac{1}{2} \gamma_5 \gamma^\mu n_\mu$$

Under Lorentz transformations:

  • $$\frac{1}{2} \underset{Lorentz}{\longrightarrow}\frac{1}{2}$$

  • $$\gamma^\mu \underset{Lorentz}{\longrightarrow} S^{-1}(\Lambda) \gamma^\mu S(\Lambda) = \Lambda^\mu _\nu \gamma^\nu$$

  • $$n_\mu \underset{Lorentz}{\longrightarrow} \Lambda_\mu ^\nu n_\nu$$

For $\gamma_5$ is more subtle: remember that:

$$\gamma_5 = \frac{i}{4!}\epsilon_{\alpha \beta \rho \sigma} \gamma^\alpha \gamma^\beta \gamma^\rho \gamma^\sigma$$

So under Lorentz transformations:

$$\gamma_5 \underset{Lorentz}{\longrightarrow} \frac{i}{4!}\epsilon_{\alpha \beta \rho \sigma} S^{-1}(\Lambda) \gamma^\alpha S(\Lambda) S^{-1}(\Lambda) \gamma^\beta S(\Lambda) S^{-1}(\Lambda)\gamma^\rho S(\Lambda) S^{-1}(\Lambda)\gamma^\sigma S(\Lambda) = $$

$$\gamma_5 \underset{Lorentz}{\longrightarrow} \frac{i}{4!}\epsilon_{\alpha \beta \rho \sigma} \Lambda_{\alpha'}^\alpha \Lambda_{\beta'}^\beta \Lambda_{\rho'}^\rho \Lambda_{\sigma'}^\sigma \gamma^{\alpha'} \gamma^{\beta'} \gamma^{\rho'}\gamma^{\sigma'}$$

Since: $$\epsilon_{\alpha \beta \rho \sigma} \Lambda_{\alpha'}^\alpha \Lambda_{\beta'}^\beta \Lambda_{\rho'}^\rho \Lambda_{\sigma'}^\sigma = \epsilon_{\alpha' \beta' \rho' \sigma'} Det(\Lambda) $$ And that $Det(\Lambda) = +1$


$$\gamma_5 \underset{Lorentz}{\longrightarrow} \frac{i}{4!}\epsilon_{\alpha' \beta' \rho' \sigma'} \gamma^{\alpha'} \gamma^{\beta'} \gamma^{\rho'}\gamma^{\sigma'} = \gamma_5$$

Then the full projector:

$$\Pi_\pm \underset{Lorentz}{\longrightarrow} \frac{1}{2} \pm \frac{1}{2} \gamma_5 \Lambda^\mu_\alpha \gamma^\alpha \Lambda^\beta_\mu n_\beta$$

The $\Lambda$ gives you $\delta^\alpha_\beta$ so:

$$\Pi_\pm \underset{Lorentz}{\longrightarrow} \frac{1}{2} \pm \frac{1}{2} \gamma_5 \delta_\alpha^\beta \gamma^\alpha n_\beta = \frac{1}{2} \pm \frac{1}{2} \gamma_5 \gamma^\alpha n_\alpha = \Pi_\pm$$

Note that depending on your signature choice and if you are working in Minkowski or Euclidean+Wick rotation minus signs could pop out in the derivation, but cancel out in the end. The steps are the same nonetheless

  • $\begingroup$ Excellent, thank you! $\endgroup$
    – Filippo
    Commented Sep 22, 2023 at 17:36
  • $\begingroup$ You're welcome :) The invariance of $\gamma_5$ is one of the first non-trivial demonstrations of QFT one encounters $\endgroup$
    – LolloBoldo
    Commented Sep 22, 2023 at 17:58

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