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

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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.

This post imported from StackExchange Physics at 2015-11-01 20:59 (UTC), posted by SE-user JLA
asked Jul 11, 2015
retagged Nov 1, 2015
Observe that for the real KG field the symbol $\phi(x)$ is selfadjoint hence $(\phi(y)\Omega,\phi(x)\Omega) = (\Omega,\phi(y)\phi(x)\Omega)$

This post imported from StackExchange Physics at 2015-11-01 20:59 (UTC), posted by SE-user Phoenix87
@Phoenix87 I know, I only wrote the adjoint because I am interested in other fields too.

This post imported from StackExchange Physics at 2015-11-01 20:59 (UTC), posted by SE-user JLA
@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.

This post imported from StackExchange Physics at 2015-11-01 20:59 (UTC), posted by SE-user JLA
@JLA I messed up badly, now the comment is deleted.

This post imported from StackExchange Physics at 2015-11-01 20:59 (UTC), posted by SE-user Weather Report

## 1 Answer

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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.

This post imported from StackExchange Physics at 2015-11-01 20:59 (UTC), posted by SE-user Arnold Neumaier
answered Jul 11, 2015 by (14,547 points)
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...

This post imported from StackExchange Physics at 2015-11-01 20:59 (UTC), posted by SE-user JLA
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.

This post imported from StackExchange Physics at 2015-11-01 20:59 (UTC), posted by SE-user Arnold Neumaier
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.

This post imported from StackExchange Physics at 2015-11-01 20:59 (UTC), posted by SE-user Arnold Neumaier
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 $c\,.$

This post imported from StackExchange Physics at 2015-11-01 20:59 (UTC), posted by SE-user JLA
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.

This post imported from StackExchange Physics at 2015-11-01 20:59 (UTC), posted by SE-user Arnold Neumaier
The main point to take home from this discussion is that the particle concept is a semiclassical notion, valid only in an approximation that fails at short distances. For example, the photon concept works intuitively correct for the quantized electromagnetic field in exactly those situations where geometric optics is applicable.

This post imported from StackExchange Physics at 2015-11-01 20:59 (UTC), posted by SE-user Arnold Neumaier

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