# Breaking of a commutator involving Dirac spinors and gamma matrices

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

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

• So what happens if you try to finish the calculation in your way? Eventually, all the terms should turn out to be proportional to the anti-commutator of the spinor fields... Jul 5, 2018 at 9:05
• @mavzolej Maybe, I'll try. Have you done this before? But independently of that, the second term of my step (2) is different from the second term of their step (2). Or can both terms be shown to be the same? Jul 5, 2018 at 9:09
• @mavzolej This starts putting out so many terms that I wonder if there isn't a cleverer way to do it... Jul 5, 2018 at 11:12

Using the formula $$[AB,CD] = A\{B,C\}D - AC \{B,D\} + \{A,C\}DB - C \{A,D\}B \quad,$$ and setting $$A = \bar{\psi}(x) \quad,\quad B = \gamma^\mu \psi(x) \quad,\quad C = \bar{\psi}(y) \quad,\quad D = \gamma^\nu \psi(y) \quad,$$ one gets (spinor indices suppressed): \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}
Actually, a more general statement is true $-$ see the bottom line on page 5 here.