In An Introduction to Quantum Field Theory, by Peskin and Schroeder, when discussing the quantized real Klein-Gordon field ($\phi=\phi^\dagger$), they show the commutator $[\phi(x),\phi(y)]$ vanishes when $y-x$ is space-like. They then say on p. 28-29

Thus we conclude that no measurement in the Klein-Gordon theory can affect another measurement outside the light-cone.

However, when I tried verifying this claim, I ran into problems. I tried using the operators $\phi(x)|0\rangle\langle 0|\phi(x)$ and $\phi(y)|0\rangle\langle 0|\phi(y)$, which I believe correspond to measuring whether there is a particle at space-time position $x$ and $y$ respectively. Then the commutator of these two operators is $$\phi(x)|0\rangle\langle 0|\phi(x)\phi(y)|0\rangle \langle 0|\phi(y)-\phi(y)|0\rangle \langle 0|\phi(y)\phi(x)|0\rangle \langle 0|\phi(x).$$ Now I know $\langle 0|\phi(x)\phi(y)|0\rangle$ doesn't vanish outside the light-cone (P&S equation 2.52). Furthermore, as far as I can tell, $\phi(x)|0\rangle\langle 0|\phi(y)$ is not proportional to $\phi(y)|0\rangle\langle 0|\phi(x)$, so it seems to me that this commutator is non-zero (a measurement at $x$ can affect a measurement made outside the light-cone of $x$). I'm not sure what I did wrong. I suspect it may have something to do with choosing incorrect operators for position measurement. I'd appreciate any help! There are many related questions (specifically, this one was the closest I could find). However, none of them address this point.


1 Answer 1


The operator $\phi(x)|0\rangle\langle 0|\phi(x)$ doesn't correspond to measuring whether there is a particle at $x$, and in fact this operator is not local at all, because $|0\rangle\langle 0|$ is not local: it projects onto the state of lowest total energy, and "total energy" is non-local.

A strict particle-position observable does not exist in relativistic QFT. This is reviewed in my answer here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.