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.

  • $\begingroup$ Observe that for the real KG field the symbol $\phi(x)$ is selfadjoint hence $(\phi(y)\Omega,\phi(x)\Omega) = (\Omega,\phi(y)\phi(x)\Omega)$ $\endgroup$ – Phoenix87 Jul 11 '15 at 15:13
  • $\begingroup$ @Phoenix87 I know, I only wrote the adjoint because I am interested in other fields too. $\endgroup$ – JLA Jul 11 '15 at 16:42
  • $\begingroup$ @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. $\endgroup$ – JLA Jul 12 '15 at 16:16
  • $\begingroup$ @JLA I messed up badly, now the comment is deleted. $\endgroup$ – Weather Report Jul 12 '15 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.

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

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  • $\begingroup$ 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... $\endgroup$ – JLA Jul 11 '15 at 16:46
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    $\begingroup$ 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. $\endgroup$ – Arnold Neumaier Jul 11 '15 at 20:03
  • $\begingroup$ 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. $\endgroup$ – Arnold Neumaier Jul 11 '15 at 20:09
  • $\begingroup$ 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 $c\,.$ $\endgroup$ – JLA Jul 11 '15 at 22:00
  • $\begingroup$ Feynman diagrams and their conventional interpretation as processes are only for building intuition for the perturbative series. They can in no way be identified with actual processes happening in space-time. That's why one talks about ''virtual'' particles and processes. Virtual means nonreal, unphysical, imagined. $\endgroup$ – Arnold Neumaier Jul 12 '15 at 13:10

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