# (Anti)commutators at different times

Why does the commutator of two operators evaluated at different times vanish? Take for instance a fermonic field $$\psi_\sigma (\vec{x},t)$$, which satisfies the well known anti-commutation relations at equal times

$$\begin{equation} [\psi_\sigma (\vec{x},t),\psi^\dagger_{\sigma'} (\vec{x}',t)]_+ = \delta^{(3)}(\vec{x}-\vec{x}') \, \delta_{\sigma \sigma'} \end{equation}$$ where $$\sigma$$ is a flavour index.

Is that correct to state that the commutator at different times vanishes? In other words $$\begin{equation} [\psi_\sigma (\vec{x},t),\psi_{\sigma} (\vec{x},t')]_- = 0\quad ? \end{equation}$$

## 1 Answer

No. (Anti)commutators do not necessarily vanish at different times. In order to compute the (anti)commutator at different times you have to solve the dynamics of the system with $$\psi(x,t)\to e^{iHt}\psi(x,0) e^{-iHt}$$. What is true in a relativistic field theory, with Bose (Fermi) fields satisfying the spin-statistics relation however, is that the (anti)commutator will vanish if $$x$$ and $$x'$$ are spacelike separated, so that $$(x-x')^2<0$$ in the $$(+,-,-,\ldots)$$ metric. This ensures that no signals can travel faster than light.