Physics Asked by S-low on December 31, 2020
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’)]_- = 0quad ?
end{equation}
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.
Answered by mike stone on December 31, 2020
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