A Anti-symmetric tensor question

[DuckAmuck]

TL;DR

Can the gamma matrices really represent any anti-symmetric tensor?

The sigma tensor composed of the commutator of gamma matrices is said to be able to represent any anti-symmetric tensor.
\sigma_{\mu\nu} = i/2 [\gamma_\mu,\gamma_\nu]
However, it is not clear how one can arrive at something like the electromagnetic tensor.
F_{\mu\nu} = a \bar{\psi} \sigma_{\mu\nu} \psi ?
Any clarity will be appreciated.

 

[PeterDonis]

The sigma tensor composed of the commutator of gamma matrices is said to be able to represent any anti-symmetric tensor.

"Said" where? Do you have a reference?

 

[DuckAmuck]

"Said" where? Do you have a reference?

Chapter 7 of Griffith's particle book

 

[dextercioby]

OK, can you be more specific (page, edition)? I think your question can be answered as follows. The antisymmetric product (basically matrix commutator) of gammas bears the same index structure as a genuine antisymmetric tensor in spacetime. However, since these are constant matrices (thus can't be varied through a Lorentz transformation), you need them to be "sandwitched" between a product of Dirac spinors. To show that Psibar.sigma_munu.Psi truly transforms as an antimmetric tensor (2-form) under Lorentz transformations, is not an easy task, it's rather tedious.

And F (the Faraday tensor of electromagnetism) is not related to a mere product of Dirac spinors "intertwined" through sigma_munu. You can say that these two tensors are not identical, but covariant (i.e. transform the same way under a Lorentz transformation).

 

[haushofer]

TL;DR Summary: Can the gamma matrices really represent any anti-symmetric tensor?

The sigma tensor composed of the commutator of gamma matrices is said to be able to represent any anti-symmetric tensor.
\sigma_{\mu\nu} = i/2 [\gamma_\mu,\gamma_\nu]
However, it is not clear how one can arrive at something like the electromagnetic tensor.
F_{\mu\nu} = a \bar{\psi} \sigma_{\mu\nu} \psi ?
Any clarity will be appreciated.

To add to dextercioby: you can't in general. F has 6 independent components and psi, depending on the representation, max 4. There's no way you can express F always in this way. The equality is not in the components, but in their transformation under the Lorentz group.

 

[George Jones]

$$F_{\mu\nu} = a \bar{\psi} \sigma_{\mu\nu} \psi ?$$

This isn't quite how it is done. Treat
$$\bar{\psi} \sigma_{\mu\nu} \psi$$
as a set of 6 basis elements. To get a general element, write a linear combination of the basis elements, e.g.,
$$F = F^{\mu \nu} \bar{\psi} \sigma_{\mu\nu} \psi.$$
$F$ is an anti-symmetric tensor with components $F^{\mu \nu}$.

 

[malawi_glenn]

2nd edition, page 237

Note, that the text only states that $\bar \psi \sigma^{\mu \nu}\psi$ is an antisymmetric tensor (7.68). It does not say that is equal to the "electromagnetic" tensor.

The sigma tensor composed of the commutator of gamma matrices is said to be able to represent any anti-symmetric tensor

It does not say this either. We are working with bilinear products of Dirac-spinors in this subchapter.