# Commutation of field operators with differents times

We know that field operators in the Heisemberg (or interacting) picture satisfy the commutation relations

$$\{ \hat{\psi}(\textbf{r},t), \hat{\psi}^{\dagger}(\textbf{r}',t) \} = \delta( \textbf{r} - \textbf{r}')$$

$$\{ \hat{\psi}(\textbf{r},t), \hat{\psi}(\textbf{r}',t) \} = 0$$

But what happen if the times of the operators are not equal? What is $$\{ \hat{\psi}(\textbf{r},t), \hat{\psi}^{\dagger}(\textbf{r}',t') \}$$ to $$t \neq t'$$? Is it has a defined value in this case?

Not in general, as it depends on the dynamics of the system, as described by the Hamiltonian. The field operator at time $$t'$$ is given by $$\psi({\bf r},t) = U(t,t')\psi({\bf r},t')U^{\dagger}(t,t')$$ where $$U$$ the time evolution operator. As $$U$$ might be very complicated, $$\psi$$ might evolve into some many-body state that will have very complicated commutation relations with $$\psi^{\dagger}({\bf r},t')$$. This is the case in general.

Let us define the function $$i\Delta(x-y)=[\phi(x),\phi^\dagger(y)].$$

(I'm assuming a scalar field with commutators, rather than fermion fields with anticommutation relations, though similar considerations may be done). Note that we implicitly say that spacetime is homogeneous, since the commutator only depends on the difference $$x-y$$. If we expand the fields in plane-waves, we find

$$\Delta(x)=-\int\frac{d^3p}{(2\pi)^3} \frac{\sin p_\mu x^\mu}{\omega_p}.$$

From this, you find some properties, the interesting one for us being

$$\frac{\partial}{\partial x_0}\Delta(x_o,\mathbf{x})\big|_{x_0=0} = -\delta^3(\mathbf{x}),$$

from which we recover the equal-time commutation relations (check it!).

Intuitively, if the separation $$x-y$$ is time-like, you can always make a Lorentz boost to find a reference frame where $$x_0=y_0$$, also recovering the ETCR, since you know how the fields transform (even for fields with spin different than $$0$$). What if the separation is space-like? This relates to the idea of microcausality. In essence, if the separation is space-like the commutator vanishes, that is, if

$$[\phi(x),\phi^\dagger(y)] =0\ \text{ if }\ (x-y)^2<0,$$

which can be deduced from the property of $$\Delta(x)$$ I stated before. This has the consequence that measurements at two points not causally connected can't influence each other, which is very reasonable.

I recommend checking section 4.4 of Greiner's Field Quantization for more details, it is a great book.