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


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:


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 (+,-,-,-)

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.