The spinor in the Dirac equation should transform via a 4x4 matrix $S$ that depends on the specific Lorentz boost/rotation:

$\psi '(x')=S(\Lambda )\psi(x)\tag1$

Where S satisfies:

$S^{-1}\gamma ^{\mu }S=\gamma ^{\nu }\Lambda _{\nu }^{\mu }\tag2$

(Where the $\gamma^{\mu}$ are the gamma matrices) As to ensure Lorentz invariance of the Dirac equation.

If we start by considering an infinitesimal Lorentz transformation:

$\Lambda _{\nu }^{\mu }=\delta_{\nu }^{\mu }+\omega _{\nu }^{\mu }\tag3$

That, upon lowering indices, gives:

$\Lambda _{ \mu\nu }^{}=\eta_{\mu\nu }+\omega _{\mu\nu }\tag4$

Where $\omega _{\mu\nu }$ is antisymmetric.

$\omega _{\mu\nu }=-\omega _{\nu\mu }\tag5$

We shall make the assumption that:

$S=I-\frac{i}{4}\beta ^{\mu\nu}\omega_{\mu\nu}\tag6$

Where the $\beta ^{\mu\nu}$ are all 4x4 matrices.

In for example this paper: https://physicspages.com/pdf/Lahiri%20QFT/Lahiri%20&%20Pal%20Problems%2004.04.pdf

It is stated that these $\beta ^{\mu\nu}$ satisfy:

$\left [\gamma ^{\mu},\beta^{\lambda\rho} \right ]=2i(\eta ^{\mu\lambda}\gamma^{\rho}-\eta ^{\mu\rho}\gamma^{\lambda})\tag6$

For which I cannot find a proof anywhere...

My attempt is to try and stick this $S$ into $(2)$:

$S^{-1}=I+\frac{i}{4}\beta ^{\mu\nu}\omega_{\mu\nu}\tag7$

So $2$ becomes:

$(I+\frac{i}{4}\beta ^{\alpha\nu}\omega_{\alpha\nu})\gamma^{\mu}(I-\frac{i}{4}\beta ^{\phi\theta}\omega_{\phi\theta})=\gamma^{\nu}(\delta_{\nu }^{\mu }+\omega _{\nu }^{\mu })\tag8$

Expanding and ignoring the term that would be second-order in $\omega$:

$\gamma^{\mu}+\frac{i}{4}\beta ^{\alpha\nu}\gamma^{\mu}\omega_{\alpha\nu}-\frac{i}{4}\gamma^{\mu}\beta ^{\phi\theta}\omega_{\phi\theta}=\gamma^{\mu}+\gamma^{\nu}\omega _{\nu }^{\mu }\tag9$

Cancelling the $\gamma$'s and lower the $\mu$ on the $\omega$ on the right-hand side:

$\frac{i}{4}\beta ^{\alpha\nu}\gamma^{\mu}\omega_{\alpha\nu}-\frac{i}{4}\gamma^{\mu}\beta ^{\phi\theta}\omega_{\phi\theta}=\gamma^{\nu}\eta^{\alpha\mu}\omega _{\alpha\nu }\tag{10}$

Relabelling and seperating out the $\omega$'s and multiplying by $4i$:

$-\beta ^{\alpha\nu}\gamma^{\mu}\omega_{\alpha\nu}+\gamma^{\mu}\beta ^{\alpha\nu}\omega_{\alpha\nu}=4i\gamma^{\nu}\eta^{\alpha\mu}\omega _{\alpha\nu }\tag{11}$

Which is where I am stuck... I'm quite sure I have to use the antisymmetry but I can't see how exactly.

  • $\begingroup$ Are you trying to prove : $\left [\gamma ^{\mu},\beta^{\lambda\rho} \right ]=2i(\eta ^{\mu\lambda}\gamma^{\rho}-\eta ^{\mu\rho}\gamma^{\lambda})$ ? $\endgroup$ Commented May 18, 2020 at 18:46
  • $\begingroup$ Yes I am indeed $\endgroup$
    – Quanta
    Commented May 18, 2020 at 19:28

1 Answer 1


You are just missing a little thing : using the anti-symmetric property of $\omega_{\mu\nu}$ as shown below $\rightarrow$ $$ \begin{aligned} -\beta ^{\alpha\nu}\gamma^{\nu}\omega_{\alpha\nu}+\gamma^{\nu}\beta ^{\alpha\nu}\omega_{\alpha\nu}&=4i\gamma^{\nu}\eta^{\alpha\mu}\omega _{\alpha\nu} \\ &= 2i \gamma^\nu \eta^{\alpha\mu}(\omega_{\alpha\nu} - \omega_{\nu\alpha}) \end{aligned} $$

I am sure you can derive the result from here. If not, please get back to me and I will update my answer with full solution.

As an aside, do note that since $\eta$ is symmetric and $\mu$ is anti-symmetric the term $\sim \eta^{\alpha\mu}\omega _{\alpha\nu}$ can only be non zero if both of them appear in this term with same behaviour of sign under exchange of indices.

  • $\begingroup$ Wait, are the beta's antisymmetric? $\endgroup$
    – Quanta
    Commented May 18, 2020 at 22:54
  • $\begingroup$ @StijnBoshoven Yes, since $\omega$ is antisymmetric. $\endgroup$ Commented May 18, 2020 at 23:10
  • $\begingroup$ but $\beta^{\mu\nu}\omega_{\mu\nu}=-\beta^{\nu\mu}\omega_{\mu\nu}$ does not mean that $\beta^{\mu\nu}=-\beta^{\nu\mu}$ right? In the same way that $\eta^{\mu\nu}\omega_{\mu\nu}=-\eta^{\nu\mu}\omega_{\mu\nu}$ does not mean that $\eta^{\mu\nu}=-\eta^{\nu\mu}$ $\endgroup$
    – Quanta
    Commented May 18, 2020 at 23:45
  • $\begingroup$ @StijnBoshoven You can write $\beta^{\mu\nu}$ as a sum of symmetric part and an antisymmetric part. Since $\omega^{\mu\nu}$ is antisymmetric the symmetric part of $\beta^{\mu\nu}$ will give zero when contracted with $\omega^{\mu\nu}$. $\endgroup$ Commented May 19, 2020 at 0:08
  • $\begingroup$ Ahh genius, when assuming step $(6)$ we basically assumed that the $\beta$'s are antisymmetric! Thank you so much, I've got it now $\endgroup$
    – Quanta
    Commented May 19, 2020 at 0:12

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