# Spinor transformation matrix derivation

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.

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

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}
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.
• @StijnBoshoven Yes, since $\omega$ is antisymmetric. May 18, 2020 at 23:10
• 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 $\eta^{\mu\nu}=-\eta^{\nu\mu}$ May 18, 2020 at 23:45
• @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 $\omega^{\mu\nu}$. May 19, 2020 at 0:08
• Ahh genius, when assuming step $(6)$ we basically assumed that the $\beta$'s are antisymmetric! Thank you so much, I've got it now May 19, 2020 at 0:12