# Microcausality for Dirac's current

Im 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 (+,-,-,-)

• Question: how did you move the $\gamma$'s past the $\Psi$'s? Apr 15, 2019 at 21:04
• 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? Apr 15, 2019 at 21:11
• 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 $\bar{\Psi} \gamma^\mu \Psi = \bar{\Psi}^\alpha (\gamma^\mu)^\alpha_\beta \Psi^\beta$. Apr 15, 2019 at 21:27
• 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. Apr 15, 2019 at 21:30
• I was thinking exactly about using microcausality for $\Psi$ Apr 15, 2019 at 23:16

$$$$[j^\mu(x) , j^\nu(y) ] = [ \bar{\Psi}_\alpha(x) \gamma{^\mu}{_\alpha }{_\beta} \Psi_\beta(x) , \bar{\Psi}_\sigma(y) \gamma{^\nu}{_\sigma }{_\rho} \Psi_\rho(y)]$$$$ $$$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \gamma{^\mu}{_\alpha }{_\beta} \gamma{^\nu}{_\sigma }{_\rho} [ \bar{\Psi}_\alpha(x) \Psi_\beta(x) , \bar{\Psi}_\sigma(y) \Psi_\rho(y)]$$$$
since $$\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$$ satisfies microcauslity for a spacelike interval, you should get:
$$$$[j^\mu(x) , j^\nu(y) ] = \gamma{^\mu}{_\alpha }{_\beta} \gamma{^\nu}{_\sigma }{_\rho} [ \ \bar{\Psi}_\alpha(x) \ \{\Psi_\beta(x) , \bar{\Psi}_\sigma(y)\ \} \ \Psi_\rho(y) \ - \ \bar{\Psi}_\sigma(y) \ \{\bar{\Psi}_\alpha(x) ,\Psi_\rho(y) \ \} \ \Psi_\beta(x) \ ]$$$$
Now impose microcausality to $$\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.