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I`m supposed to show as an exercises that for the Dirac field's associated current:

$$j^\mu=\bar{\Psi}\gamma^\mu\Psi$$

The microcausality relation holds:

$$ [j^\mu(x),j^\nu(y)]=0 \text{ for } (x-y)^2<0$$

I've worked out the commutator above until I reach the expression bellow:

$$\gamma^\mu\gamma^\nu\bar{\Psi}(x)\Psi(x)\bar{\Psi}(y)\Psi(y)-\gamma^\nu\gamma^\mu\bar{\Psi}(y)\Psi(y)\bar{\Psi}(x)\Psi(x)$$

Question: Am I in the right track? Any obvious reason for this to be zero exploiting that the space-time interval is spacelike.

**I`m using (+,-,-,-)

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  • $\begingroup$ Question: how did you move the $\gamma$'s past the $\Psi$'s? $\endgroup$
    – MannyC
    Apr 15, 2019 at 21:04
  • $\begingroup$ Actually I`m not very confident in the way I manipulated this guys. What I did was basically consider gammas and psi's as operators in a large tensor product space, each one acting in different subspaces. Then I just used commutator properties. In that manner I treated them as commuting but I guess this might be wrong. How should I manipulate it properly? $\endgroup$
    – Janov
    Apr 15, 2019 at 21:11
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    $\begingroup$ The $\gamma$ act on the finite dimensional spin $\frac12$ representation and $\Psi$ acts both in the big Fock space and in the spin $\frac12$ representation. But keep it simple: $\Psi$ has spinor indices $\bar{\Psi} \gamma^\mu \Psi = \bar{\Psi}^\alpha (\gamma^\mu)^\alpha_\beta \Psi^\beta$. $\endgroup$
    – MannyC
    Apr 15, 2019 at 21:27
  • $\begingroup$ I think this exercise wants you to either use microcausality of $\Psi$ or decompose in annihilation and creation operators. You'll never obtain zero identically just by manipulating the commutator. $\endgroup$
    – MannyC
    Apr 15, 2019 at 21:30
  • $\begingroup$ I was thinking exactly about using microcausality for $\Psi$ $\endgroup$
    – Janov
    Apr 15, 2019 at 23:16

1 Answer 1

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You can work out the commutator as follow:

\begin{equation} [j^\mu(x) , j^\nu(y) ] = [ \bar{\Psi}_\alpha(x) \gamma{^\mu}{_\alpha }{_\beta} \Psi_\beta(x) , \bar{\Psi}_\sigma(y) \gamma{^\nu}{_\sigma }{_\rho} \Psi_\rho(y)] \end{equation} \begin{equation} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \gamma{^\mu}{_\alpha }{_\beta} \gamma{^\nu}{_\sigma }{_\rho} [ \bar{\Psi}_\alpha(x) \Psi_\beta(x) , \bar{\Psi}_\sigma(y) \Psi_\rho(y)] \end{equation}

since $\gamma{^\mu}{_\alpha }{_\beta}$ and $\gamma{^\nu}{_\sigma }{_\rho}$ are complex numbers. Using the relations $[AB,C]=A[B,C] + [B,C]A$ and $[A,BC]=\{A,B\}C-B\{A,C\}$ and the fact that both $\bar{\Psi}$ and $\Psi$ satisfies microcauslity for a spacelike interval, you should get:

\begin{equation} [j^\mu(x) , j^\nu(y) ] = \gamma{^\mu}{_\alpha }{_\beta} \gamma{^\nu}{_\sigma }{_\rho} [ \ \bar{\Psi}_\alpha(x) \ \{\Psi_\beta(x) , \bar{\Psi}_\sigma(y)\ \} \ \Psi_\rho(y) \ - \ \bar{\Psi}_\sigma(y) \ \{\bar{\Psi}_\alpha(x) ,\Psi_\rho(y) \ \} \ \Psi_\beta(x) \ ] \end{equation}

Now impose microcausality to $\bar{\Psi}$ and $\Psi$ simultaneously you have $ \{ \bar{\Psi}_\epsilon(x) ,\Psi_\delta(y) \} = 0 $ since $(x-y)^2<0$. Then the commutator vanishes and the desired result follows.

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  • $\begingroup$ I did it precisely in this way at the end, it went ok. Writing down the spinor indicies basically solved any confusion. Thanks though :) $\endgroup$
    – Janov
    Apr 27, 2019 at 13:13

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