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.


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