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  Proof of equivalence of different representations of the $\gamma$-matrices in the Dirac equation

+ 3 like - 0 dislike

This question concerns the Dirac equation and the $4\times4$ $\gamma$-matrices. The task is to prove that a similarity transformation of the standard $\gamma$-matrix conserves the commutation relation

$$ \{\gamma^\mu,\gamma^\nu\} ~=~ 2g^{\mu\nu}, $$

where $2g^{\mu\nu}$ is the metric tensor $\text{diag}(1,-1,-1,-1)$, and the similarity transformation is defined as

$$ \tilde{\gamma}^\mu = S \gamma^\mu S^\dagger, $$

and $S$ is a unitary matrix. I will write down the start of my proof to show where I stop. First of all, we can use that $S$ is unitary and show that $\gamma^\mu = S^\dagger\tilde{\gamma}^\mu S$, and insert this into the commutator. This leaves us, again using that $SS^\dagger = I$, with

$$ S^\dagger\{\tilde{\gamma}^\mu,\tilde{\gamma}^\nu\}S = 2g^{\mu\nu} $$

which again gives us

$$ \{\tilde{\gamma}^\mu,\tilde{\gamma}^\nu\} = 2Sg^{\mu\nu}S^\dagger. $$

In order for the proof to hold, it requires that $g^{\mu\nu}$ and $S$ commute so that

$$ 2Sg^{\mu\nu}S^\dagger = 2g^{\mu\nu}SS^\dagger = 2g^{\mu\nu}. $$

So my question is: Do all unitary matrices commute with the metric tensor $g^{\mu\nu}$? If yes, how can I show this easily?

This post imported from StackExchange Physics at 2014-05-04 11:17 (UCT), posted by SE-user camzor00
asked May 2, 2014 in Theoretical Physics by camzor00 (15 points) [ no revision ]
retagged May 4, 2014
Hint to the question (v2): The commutation relation has an (implicit) identity matrix on the rhs: $\{\gamma^\mu,\gamma^\nu\} ~=~ 2g^{\mu\nu}~{\bf1}_{4\times 4}$.

This post imported from StackExchange Physics at 2014-05-04 11:17 (UCT), posted by SE-user Qmechanic
I don't see it, even with the hint..

This post imported from StackExchange Physics at 2014-05-04 11:17 (UCT), posted by SE-user camzor00
If you fix $\mu$ and $\nu$ $g^{\mu\nu}$ is a number, so it commutes with every matrix.

This post imported from StackExchange Physics at 2014-05-04 11:17 (UCT), posted by SE-user V. Moretti

Your argument holds for any fixed pair of indices, hence $g^{\mu\nu}$ is just a number, which can be taken out of your expression to complete the proof. 

1 Answer

+ 0 like - 1 dislike

Dirac $\gamma$ matrices are 4x4 matrices, they differ from each other by index $\mu$. And there is a 4x4 unity matrix amongst all 4x4 matrices, usually denoted by $I$. It stays (often implicitly for simplicity) at  $g^{\mu\nu}$. (A product of two 4x4 matrices is also a 4x4 matrix; a sum of 4x4 matrices is also a 4x4 matrix. So no wonder that the anticommutator is some 4x4 matrix too. If you anticommute, say, $\gamma^3$ with itself, you will obtain $2I$ since $g^{33}=1$.)

If some set of 4x4 marices $\gamma'$ may be obtained from an "original representation" of $\gamma$ with a unitary transformation, then it is as good as the original set of $\gamma$, because it obeys the same Dirac algebra and leads to the Klein-Gordon equation, as usual.

In practice there are more or less convenient particular $\gamma$ matrix choices, but it is subjective and related to the volume of calculations for a human being.

answered Oct 26, 2016 by Vladimir Kalitvianski (102 points) [ revision history ]
edited Oct 26, 2016 by Vladimir Kalitvianski

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