Breaking of a commutator involving Dirac spinors and gamma matrices

Question

I'm trying to understand a particular step in the solution to problem 27 in THIS solution sheet. By the middle of the page, they start with the simplification of this expression

$$\left[s^{\mu}\left(x\right),s^{\nu}\left(y\right)\right]\overset{(1)}{=}e^{2}\left[\overline{\psi}\left(x\right)\gamma^{\mu}\psi\left(x\right),\overline{\psi}\left(y\right)\gamma^{\nu}\psi\left(y\right)\right]\overset{(2)}{=}\cdots$$

and I don't understand how they break this commutator at step (2). Is their algebra correct? Using $\left[AB,C\right]=A\left[B,C\right]+\left[A,C\right]B$, and identifying $A=\overline{\psi}\left(x\right)\gamma^{\mu}$ and $B=\psi\left(x\right)$, shouldn't it be

$$ \overset{2}{=}e^{2}\overline{\psi}\left(x\right)\gamma^{\mu}\left[\psi\left(x\right),\overline{\psi}\left(y\right)\gamma^{\nu}\psi\left(y\right)\right]+e^{2}\left[\overline{\psi}\left(x\right)\gamma^{\mu},\overline{\psi}\left(y\right)\gamma^{\nu}\psi\left(y\right)\right]\psi\left(x\right) $$

The same kind of misplacement of $\gamma^{\nu}$ seems to happen again at step (3) when they break the first commutator. In here, if they were using $\left[A,BC\right]=\left\{ A,B\right\} C-B\left\{ A,C\right\} $ with $B=\overline{\psi}\left(y\right)$, $C=\gamma^{\nu}\psi\left(y\right)$, I guess one would have

$$ \left[\psi\left(x\right),\overline{\psi}\left(y\right)\gamma^{\nu}\psi\left(y\right)\right]=\left\{ \psi\left(x\right),\overline{\psi}\left(y\right)\right\} \gamma^{\nu}\psi\left(y\right)-\overline{\psi}\left(y\right)\left\{ \psi\left(x\right),\gamma^{\nu}\psi\left(y\right)\right\} $$

instead of what's there. I must be missing something, but I can't see what! Could you please help?

Answer 1

Using the formula \begin{equation} [AB,CD] = A\{B,C\}D - AC \{B,D\} + \{A,C\}DB - C \{A,D\}B \quad, \end{equation} and setting \begin{equation} A = \bar{\psi}(x) \quad,\quad B = \gamma^\mu \psi(x) \quad,\quad C = \bar{\psi}(y) \quad,\quad D = \gamma^\nu \psi(y) \quad, \end{equation} one gets (spinor indices suppressed): \begin{equation}\begin{alignedat}{9} &[\bar{\psi}(x) \gamma^\mu \psi(x), \bar{\psi}(y) \gamma^\nu \psi(y)] \\&= \bar{\psi}(x)\{ \gamma^\mu \psi(x),\bar{\psi}(y)\}\gamma^\nu \psi(y) - \bar{\psi}(x)\bar{\psi}(y) \{ \gamma^\mu \psi(x),\gamma^\nu \psi(y)\} \\&+ \{\bar{\psi}(x),\bar{\psi}(y)\}\gamma^\nu \psi(y) \gamma^\mu \psi(x) - \bar{\psi}(y) \{\bar{\psi}(x),\gamma^\nu \psi(y)\} \gamma^\mu \psi(x) \\&= \gamma^\mu\bar{\psi}(x)\{ \psi(x),\bar{\psi}(y)\}\gamma^\nu \psi(y) - \gamma^\mu\bar{\psi}(x)\bar{\psi}(y) \{ \psi(x), \psi(y)\}\gamma^\nu \\&+ \{\bar{\psi}(x),\bar{\psi}(y)\}\gamma^\nu \psi(y) \gamma^\mu \psi(x) - \bar{\psi}(y) \{\bar{\psi}(x), \psi(y)\} \gamma^\nu\gamma^\mu \psi(x)=0\quad. \end{alignedat}\end{equation}

Actually, a more general statement is true $-$ see the bottom line on page 5 here.

For the detailed solution see Problem 11 here.