# What do commutators have to do with causality?

In (all?) QFT textbooks, it is stated that if two events \(x\) and \(y\) are space-like separated, then causality requires that $$ [A(x),B(y)] = 0 $$ for any local operators \(A\) and \(B\). It was not obvious to me what this requirement had to do with causality, and I ignored the issue for years. This is to briefly record the cleanest answer I know of.

Commutators come up *naturally* when considering linear response. Suppose at \(t=0\) a system is described by the state \(|\psi\rangle\). We could measure some operator \(A\), or we could wait some time \(T>0\) and measure then:
$$
\langle A(T) \rangle_0 =
\langle \psi | e^{i H T} A e^{-i H T}| \psi \rangle
\text.
$$
Suppose at time \(t=0\), we hit the system with a small hammer, evolving for a brief time under the Hamiltonian \(B\). (Note that \(B\), being a Hermitian operator, is both an observable and a legal Hamiltonian – the concepts are one and the same.) For arbitrarily large perturbations, of course, the expectation value of \(A(T)\) is
$$
\langle A(T) \rangle_\epsilon =
\langle \psi | e^{i B \epsilon} e^{i H T} A e^{-i H T} e^{-i B \epsilon}| \psi \rangle
\text.
$$

The expression simplifies somewhat if we consider only *linear* response; that is, we drop all terms nonlinear in \(\epsilon\). In particular, the first derivative of the above expression is
$$
\frac{\mathrm d}{\mathrm d\epsilon} \langle A(T) \rangle_\epsilon =
-i\langle \psi |
[A(T), B(0)]
| \psi \rangle
\text.
$$
If \(A(x)\) and \(B(y)\) commute, then, than a small whack of operator \(B(y)\) has no effect on the expectation value of \(A(x)\) (no matter the initial state). If this holds for *all* operators at \(x\) and \(y\), then this (repeated to all orders in \(\epsilon\)) suffices to show that no message can be sent from \(y\) to \(x\). That’s causality!