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While studying the Dirac equation, $$\left(i\gamma^{\mu} \partial_{\mu} - m\right)\psi = 0.$$ I have been finding difficulty understanding the following summarisation of the algebra that the $\gamma$-matrices follow, $$\{\gamma^{\mu},\gamma^{\nu}\} = 2 \eta^{\mu \nu}$$ Where $\eta^{\mu \nu}$ is the inverse Minkowski metric, and the curly brackets denote an anticommutator.

Now, $\{\gamma^{0},\gamma^{0} \} = 2I$, so does this mean $\eta^{00} = I$? What does it mean (if anything) to substitute numbers for $\mu$ and $\nu$ here? My main question is, how does this "summary" relate to $\left(\gamma^{0}\right)^2 = I$, $(\gamma^k)^2 = -I$ for $k = 1,2,3$, and $\{\gamma^\mu,\gamma^\nu \} = 0$ for $\mu \neq \nu$?

Something tells me it relates to the matrix elements of $\eta^{\mu \nu}$, as the off-diagonal elements are zero, and the top diagonal element is 1. So one can look at the entries of $\eta^{\mu \nu}$ to predict what the anticommutators of the $\gamma$-matrices will evaluate to.

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  • $\begingroup$ I honestly don't know much about the $\gamma$-matrices but looking up on wikipedia tells me the formula should be $\{\gamma^u,\gamma^v\} = 2\eta^{\mu \nu} I_4$ which I think should alleviate some of the confusion. $\endgroup$ – jgerber Aug 11 '18 at 6:09
  • $\begingroup$ @jgerber That makes sense if one interprets $\eta^{\mu \nu}$ as the matrix elements of what we write as, $\eta^{\mu \nu}$... Is distinction between matrix elements and the matrix itself based on context? $\endgroup$ – user154080 Aug 11 '18 at 6:11
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    $\begingroup$ When I see $\eta^{\mu \nu}$ I consider that to be an object which is a scalar which corresponds to an element of the metric tensor. Generally (in relativistic contexts) indexed objects are components of tensors/vectors. Of course, when talking about the Pauli or gamma matrices we have sets of objects where the indexed objects are matrices rather than components of matrices. For my work if I see an indexed object I consider it to be a matrix unless it is a Pauli matrix or a gamma matrix.. Someone who has spent more time with the Dirac equation could perhaps give a better general rule. $\endgroup$ – jgerber Aug 11 '18 at 6:18
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It can be easier to write everything in terms of explicit indices, $$\gamma^\mu_{\alpha \beta} \gamma^\nu_{\beta \delta} + \gamma^\nu_{\alpha \beta} \gamma^\mu_{\beta \delta} = 2 \eta^{\mu\nu} \delta_{\alpha \delta}.$$ There is a sum over $\beta$ on the left-hand side. When you plug in explicit values for $\alpha$, $\delta$, $\mu$, and $\nu$, both sides are just numbers. The indices $\mu$ and $\nu$ are Lorentz indices, describing the Lorentz transformation properties of both sides. The indices $\alpha$, $\beta$, and $\delta$ are spinor indices, which transform in a different way. It is a complete coincidence they both range from $1$ to $4$.

You can see how this looks clunky, so we could just factor the spinor indices out, $$(\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu)_{\alpha \delta} = 2 \eta^{\mu\nu} \delta_{\alpha \delta}.$$ Then the matrix multiplication on the left-hand side becomes implicit. We can go one step further and suppress the spinor indices entirely, giving $$\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu\nu}.$$ The problem is that the left-hand side is still clearly a $4 \times 4$ matrix in spinor space, but the right-hand side doesn't seem to have any spinor indices at all. They're there, but there isn't a simple way to show it. You could write $\eta^{\mu\nu} 1_4$ on the right-hand side, but that invites some confusion as it looks like the $1_4$ multiplies $\eta^{\mu\nu}$. You can't indicate it's an identity matrix in spinor space. The usual abbreviated notation is not perfect, but it's one of the best options we have.

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Let's try to summarize the characteristics of the $\gamma$-matrices and the metric tensor for flat Minkowski space. First our $\eta^{\mu\nu}$ can be represented by the matrix

$$\eta^{\mu\nu} = \begin{pmatrix} 1&0&0&0 \newline 0&-1&0&0\newline 0&0&-1&0\newline 0&0&0&-1 \\ \end{pmatrix}.$$

You now interpret the $\eta^{\mu\nu}$ as the $\mu\nu$-component of that matrix. So it doesn't make sense to say that $\eta^{00} = I_4$ but you can conclude that the $00$-component of your metric tensor is identical to the $00$-component of $I_4$ ($I^{00} = 1$).

If you look at $(\gamma^\mu)^2$ then you can show following identity

$$2\eta^{\mu\mu}I_4=\{\gamma^\mu;\gamma^\mu\} = \gamma^\mu \gamma^\mu + \gamma^\mu \gamma^\mu = 2(\gamma^\mu)^2.$$

For the off-diagonal elements you will obtain similar that

$$2\eta^{\mu\nu}I_4=\{\gamma^\mu;\gamma^\nu\} = \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = \gamma^\mu\gamma^\nu - \gamma^\mu\gamma^\nu = 0 \cdot I_4.$$

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FWIW, a similar issue arises in the CCR

$$ [\hat{q}^j, \hat{p}_k]~=~i\hbar ~\delta^j_k \hat{\bf 1} $$

of the Heisenberg algebra, where authors often do not write the identity operator $\hat{\bf 1}$ explicitly.

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