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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?

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  • $\begingroup$ 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... $\endgroup$
    – mavzolej
    Jul 5 '18 at 9:05
  • $\begingroup$ @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? $\endgroup$
    – johani
    Jul 5 '18 at 9:09
  • $\begingroup$ @mavzolej This starts putting out so many terms that I wonder if there isn't a cleverer way to do it... $\endgroup$
    – johani
    Jul 5 '18 at 11:12
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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.

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  • $\begingroup$ Since you obviously know latex, I suggest you doing such calculations not on paper but right in latex, with a lot of copy-pasting. This prevents you from making silly mistakes and, in the long run, saves your time. $\endgroup$
    – mavzolej
    Jul 7 '18 at 0:00
  • $\begingroup$ Which identities did you use in going from step 2 to 3? Now I got the time to look at this properly and I actually didn't understand that step. $\endgroup$
    – johani
    Jul 10 '18 at 1:50
  • $\begingroup$ Just the fact that gamma matrices commute with spinors - their indices are totally independent of each other. You can freely move gamma matrices to the left or right until you reach another gamma matrix. $\endgroup$
    – mavzolej
    Jul 11 '18 at 14:13
  • $\begingroup$ Look, I asked that question here, physics.stackexchange.com/questions/414440/… , and the answers were different. $\endgroup$
    – johani
    Jul 11 '18 at 15:09
  • $\begingroup$ I was probably wrong when used the word 'commute'. As it is explained in that post, this word is not very suitable in this case. Please see the link to the detailed solution which I have added to the answer. $\endgroup$
    – mavzolej
    Jul 12 '18 at 6:43

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