# Missing identity element in the Clifford relation

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.

• 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. Aug 11 '18 at 6:09
• @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?
– user154080
Aug 11 '18 at 6:11
• 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. Aug 11 '18 at 6:18

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.

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

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

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.