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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}

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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.

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