# 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 $(\phi(y)\Omega,\phi(x)\Omega) = (\Omega,\phi(y)\phi(x)\Omega)$ Jul 11, 2015 at 15:13
• @Phoenix87 I know, I only wrote the adjoint because I am interested in other fields too.
– JLA
Jul 11, 2015 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, 2015 at 16:16
• @JLA I messed up badly, now the comment is deleted. Jul 12, 2015 at 16:39

The quantum mechanical interpretation in terms of probabilities of being at a point in space is intrinsically nonrelativistic. To get this interpretation for a relativistic particle, one needs to perform an additional Foldy-Wouthuysen transformation, which transforms the covariant measure in spacetime to the noncovariant Lebesgue measure in space. This is more or less done as in discussions of the Dirac equation. In the resulting Foldy-Wouthuysen coordinates (corresponding to the Newton-Wigner position operator), the probabilistic position interpretation is valid, and only in this representation. See the entry ''Particle positions and the position operator'' in Chapter B1: The Poincare group of my theoretical physics FAQ.

[added August 2021:] POVMs for position measurements are discussed in my paper Born's rule and measurement. The FW transformation turns the covariant metric into the standard 3D metric for calculating norms and inner products. The position operator in the Foldy representation is just $$q$$, with the standard inner product as in the nonrelativistic case. The FW transformation transforms the Dirac state $$\psi$$ into the Foldy state.

For the electromagnetic field, point localization is impossible; your question regarding it doesn't make sense because of gauge invariance.

• 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, 2015 at 16:46
• Position is never fixed, always uncertain with an uncertainty of the Compton length. This reconciles the different points of view. Note that probabilities are associated with observations, which always happen in the eigenframe of the observer. What you describe is a simplified version of the Unruh effect, which even says that the notion of particle is frame dependent. Jul 11, 2015 at 20:03
• I meant to say (but wasn't allowed to edit it): Covariant position is never fixed, always uncertain with an uncertainty of the Compton length, due to Zitterbewegung. Jul 11, 2015 at 20:09
• 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, 2015 at 22:00
• @jmg: I address this now in an addition to my answer. Aug 29, 2021 at 16:19