A Anti-symmetric tensor question

DuckAmuck
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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.
 
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DuckAmuck said:
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?
 
PeterDonis said:
"Said" where? Do you have a reference?
Chapter 7 of Griffith's particle book
 
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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).
 
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DuckAmuck said:
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.
 
DuckAmuck said:
$$F_{\mu\nu} = a \bar{\psi} \sigma_{\mu\nu} \psi ?$$

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}##.
 
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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.

DuckAmuck said:
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.
 
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