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Is it possible to retrieve the matrix elements of the $\gamma$s by simply knowing their anti-commutation relation: $$ \{\gamma^\mu, \gamma^\nu\}=2\,g^{\mu\nu}\,\mathbb{I}_{4} $$ I'm just trying to reconstruct Dirac's reasoning when he first encountered these relations (previous to his equation). The analogy with the Pauli matrices is evident, which at that time were newly introduced, so it's reasonable to assume some kind of analogy. But it's still too early to guess correctly. He might have recognized the Clifford algebra... but again I think it's not enough.

ric.san
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    You actually can't deduce the matrices that way; you can only prove a specific choice of representation works. I think what you mean to ask is what Ansatz Dirac used to obtain one. I'd also recommend saying $\eta$, not $g$. – J.G. Nov 13 '23 at 10:26
  • I don't know, they taught me their expression as if they came out of the blue. You can't just guess them, there must have been some kind of reasoning behind – ric.san Nov 13 '23 at 10:31
  • If you're after the first representation here, for example, try a $2\times2$ block matrix representation, with $\gamma^0$ having vanishing off-diagonal blocks, while the $\gamma^k$ vanish in their diagonal blocks. – J.G. Nov 13 '23 at 10:34
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    The way I see it, the anti-commutation relation is fundamental. You can find it by requiring space and time to be on equal footing in the Schrödinger equation and also requiring that it reproduces the relativistic dispersion relation $E^2=p^2+m^2$ (but with units). The gamma matrices are just instances of matrices that satisfy these relations. To exemplify, on wikipedia you can find 4 different instances of gamma matrices that satisfy these relations, each with slightly different benefits. – AccidentalTaylorExpansion Nov 13 '23 at 11:04
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    The argument you can use is that the anti commutation relations and the fact that the Dirac matrices need to be Hermitian and traceless constrain the possible options you have. At that point, extending the Pauli matrices to 4-D is the most natural option available (since there are not enough Pauli matrices to make the Dirac equation work). – Matt Hanson Nov 13 '23 at 12:04
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    For the dimension of the Dirac matrices, see https://physics.stackexchange.com/q/53318/2451 – Qmechanic Nov 13 '23 at 13:41

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Yes, you can find a set of $\gamma$ matrices just by knowing their commutation relations $$\{\gamma^\mu, \gamma^\nu\}=2\,g^{\mu\nu}\,\mathbb{I}_{4}$$

One such set is the Dirac basis: $$\gamma^0=\begin{pmatrix}I_2&0\\0&-I_2\end{pmatrix},\quad \gamma^k=\begin{pmatrix}0&\sigma^k\\-\sigma^k&0\end{pmatrix} \text{ with }k=1,2,3$$ But this solution is not unique. Another widely used set is the Weyl basis: $$\gamma^0=\begin{pmatrix}0&I_2\\I_2&0\end{pmatrix},\quad \gamma^k=\begin{pmatrix}0&\sigma^k\\-\sigma^k&0\end{pmatrix} \text{ with }k=1,2,3$$

Furthermore, from a set of $\gamma$ matrices you can (with any unitary $4\times 4$ matrix $U$) construct many more sets of $\gamma$ matrices by $$\gamma'^\mu = U^\dagger\gamma^\mu U$$ which will automatically satisfy the required commutation relations as well.

And for the $\gamma^\mu$ you are not restricted to $4\times 4$ matrices. You can also find $8\times 8$ matrices or even bigger matrices. But, since the physical results got from these will be the same, there is no need to bother with them.

Thomas Fritsch
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