# What is $\phi(x)|0\rangle$?

Suppose for instance that $\phi$ is the real Klein-Gordon field. As I understand it, $a^\dagger(k)|0\rangle=|k\rangle$ represents the state of a particle with momentum $k\,.$ I also learned that $\phi^\dagger(x)$ acts on the vacuum $\phi(x)^\dagger|0\rangle\,,$ creating a particle at $x\,.$ But it seems that $\phi^\dagger(x)|0\rangle\,,\phi^\dagger(y)|0\rangle$ are not even orthogonal at equal times, so I don't see how this is possible. So what is it exactly? And what about for fields that aren't Klein-Gordon, ie. electromagnetic potential.

Edit: As I now understand it, $\phi(x)|0\rangle$ doesn't represent a particle at $x$, but can be interpreted as a particle most likely to be found at $x$ upon measurement and which is unlikely to be found outside of a radius of one Compton wavelength (by analyzing $\langle 0|\phi(y)\phi(x)|0\rangle)$. So taking $c\to\infty\,,$ $\phi(x)|0\rangle$ represents a particle located at $x\,,$ and I suppose generally experiments are carried over distances much longer than the Compton wavelength so for experimental purposes we can regard $\phi(x)|0\rangle$ as a particle located at $x\,.$ Is this the case? If so it's interesting that this doesn't seem to be explained in any QFT books I've seen.

• Observe that for the real KG field the symbol $\phi(x)$ is selfadjoint hence $(\phi(y)\Omega,\phi(x)\Omega) = (\Omega,\phi(y)\phi(x)\Omega)$ – Phoenix87 Jul 11 '15 at 15:13
• @Phoenix87 I know, I only wrote the adjoint because I am interested in other fields too. – JLA Jul 11 '15 at 16:42
• @WeatherReport I'm not sure what the commutation relations have to do with this. But according to everything I've seen, the states aren't orthogonal at equal times. – JLA Jul 12 '15 at 16:16
• @JLA I messed up badly, now the comment is deleted. – Weather Report Jul 12 '15 at 16:39

• OK so to be clear, you are saying that despite the common occurrence of QFT books/notes claiming that $\phi(x)$ operating on the vacuum creates a particle at $x\,,$ it does not? I've read about the Newton-Wigner position operator, though it does seem a bit weird to me. So in one frame the particle can be located at a point, but at another its wave function is spread out... – JLA Jul 11 '15 at 16:46
• Do you know why then the Feynman propagator (say for the Klein-Gordon field) is described as the amplitude for a particle to travel from $x$ to $y$? It seems this isn't the case then; at best it's the amplitude for a particle which is most likely to be found near $x$ to end up in a state for which it is most likely to be found at $y$. This kind of kills the whole "mystery" of why the Feynman propagator is nonzero outside the light cone, since initially the particle had an amplitude to be outside of the light cone and so never needed to travel faster than $c\,.$ – JLA Jul 11 '15 at 22:00