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I am currently reading through Itzyskon and Zuber for my quantum field theory class, and I came across this regarding the unitary transformations of the Dirac bispinors in chapter 2. They show that the transformation from one frame to another: $\psi'(x') = S(\Lambda)\psi(x)$ has the property:

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

Which I understand. Then, if one were to make an infinitesimal transformation:

$$ \Lambda^{\mu}_{\nu} = \delta^{\mu}_{\nu} + \omega^{\mu}_{\nu} \hspace{1mm} , \hspace{1mm} (\Lambda^{-1})^{\mu}_{\nu} = \delta^{\mu}_{\nu} - \omega^{\mu}_{\nu} $$

Then we can say that the unitary transformations have the form:

$$S(\Lambda) = 1 - \frac{i}{4}\sigma_{\mu \nu} \omega^{\mu \nu} \hspace{1mm} , \hspace{1mm} S(\Lambda^{-1}) = 1 + \frac{i}{4}\sigma_{\mu \nu} \omega^{\mu \nu} $$

I am somewhat perplexed by the $\frac{i}{4}$ factors, but I figure that they're a result of a convention in anticipation of some result. My real problem comes in when attempting to use these with the first equation to derive the commutation relation $[\gamma^{\mu},\sigma_{\alpha \beta}]$. In the book they give this as the following:

$$ [\gamma^{\mu},\sigma_{\alpha \beta}] = 2i(\delta^{\mu}_{\alpha}\gamma_{\beta} - \delta^{\mu}_{\beta}\gamma_{\alpha}) $$

In an attempt to recover these relations, I did the following:

$$S(\Lambda) \gamma^{\mu} S(\Lambda^{-1}) = (\Lambda^{-1})^{\mu}_{\nu} \gamma^{\nu} \rightarrow $$

$$(1 - \frac{i}{4}\sigma_{\alpha \beta} \omega^{\alpha \beta}) \hspace{1mm} \gamma^{\mu} \hspace{1mm} (1 + \frac{i}{4}\sigma_{\alpha \beta} \omega^{\alpha \beta} = (\delta^{\mu}_{\nu} - \omega^{\mu}_{\nu}) \gamma^{\nu} $$

Multiplying the terms, discarding higher than order 1 terms, and subtracting the $\gamma^{\mu}$ from each side, we get:

$$ \frac{i}{4} \big( \gamma^{\mu} \sigma_{\alpha \beta} \omega^{\alpha \beta} - \sigma_{\alpha \beta} \omega^{\alpha \beta} \gamma^{\mu} \big) = - \omega^{\mu}_{\nu} \gamma^{\nu} $$

We can change the above equation to get:

$$ \frac{i}{4} \big( \gamma^{\mu} \sigma_{\alpha \beta} - \sigma_{\alpha \beta} \gamma^{\mu} \big) \omega^{\alpha \beta} = \delta^{\mu}_{\beta} \gamma_{\alpha} \omega^{\alpha \beta} $$

If we switch the indices $\alpha$ and $\beta$ and then use the antisymmetric property of $\omega^{\alpha \beta}$, then we get:

$$ \frac{i}{4} \big( \gamma^{\mu} \sigma_{\alpha \beta} - \sigma_{\alpha \beta} \gamma^{\mu} \big) \omega^{\alpha \beta} = -\delta^{\mu}_{\alpha} \gamma_{\beta} \omega^{\alpha \beta} $$

So if I add these two and do the appropriate algebra, then I arrive at the above commutation relation:

$$ [\gamma^{\mu},\sigma_{\alpha \beta}] = 2i(\delta^{\mu}_{\alpha}\gamma_{\beta} - \delta^{\mu}_{\beta}\gamma_{\alpha}) $$

However, this can all only be done if I assume that the $\omega^{\alpha \beta}$ commute with the $\gamma^{\mu}$. But, on the other hand, it is obvious from the result of this derivation that the $\sigma_{\alpha \beta}$ don't commute with $\gamma^{\mu}$. Why is this the case? Why should I assume that? It seems like a magical property that the $\omega^{\alpha \beta}$ have to get some relations that we like. Does it have to do with them being infinitesimal?

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    $\begingroup$ $\omega$ is a set of 6 numbers whereas $\sigma$ is a set of 6 matrices. $\endgroup$
    – Prahar
    Commented Feb 14, 2022 at 5:00

1 Answer 1

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Look again at the defining equation for the $\sigma_{\mu\nu}$ and $\omega^{\mu\nu}$: $$ S = 1 -\frac{\mathrm{i}}{4}\sigma_{\mu\nu}\omega^{\mu\nu}$$ Here, $S$ is a matrix, and we're expanding it in term of some basis of matrices $\sigma_{\mu\nu}$ with coefficients $\omega^{\mu\nu}$. So the $\omega^{\mu\nu}$ are just real numbers, while the $\sigma_{\mu\nu}$ are matrices. Real numbers commute with everything, matrices don't.

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