How does QFT account for localization (into a finite volume of space), in practice?

Question

In Quantum Field Theory there is until yet no agreement (as far as I know) on the issue of localization of particles. When one talks about a 'particle' in QFT, one usually means a single-particle state of definite momentum, or a wavepacket made out of such states. It is not clear, however, what (if any) are the states that correspond to something that is localized in space, or even something that is localized into a finite region of space.

Some textbooks on QFT (e.g. Peskin and Schroeder, page 24) suggest that (at least in the case of the free Klein-Gordon theory) the field operator $\phi(\vec{x})$ creates a particle at position $\vec{x}$, i.e., the state \begin{equation} |\vec{x}\rangle := \phi(\vec{x})|0\rangle \end{equation} would correspond to a particle localized at $\vec{x}$. However, it can be easily shown that such states are not mutually orthogonal, i.e., $\langle \vec{y}|\vec{x}\rangle\neq 0$ if $\vec{y}\neq \vec{x}$. So these states cannot possibly correspond to localized particles.

This bothers me and I would gladly hear other people's views on this. Still, I can imagine, for instance, that these states do actually correspond to effectively localized states, by which I mean that in practice it makes sense to regard them as localized states, even if they technically aren't. But this is only a shot in the dark; I have no idea whether that makes any sense. And if this is the case, then what is the justification for this view?

Other references advocate that one should use the eigenstates of the so-called Newton-Wigner position operator, which is explained in detail in this excellent answer. Although these states also have their peculiarities, they seem to be preferable over the states $\phi(\vec{x})|0\rangle$.

So theoretically it is not clear how we should describe localized particles. Nevertheless, in collider experiments, for instance, the particles (or perhaps I should say the quantum fields) clearly are effectively localized into a finite region of space. And there the theory really works! So apparently we are able to describe localized particles. So how does one describe this spatial dependence, in practice? I imagine one uses some kind of wavepackets? And does this give any insight into the theoretical problem?

Answer 1

Nevertheless, in collider experiments, for instance, the particles (or perhaps I should say the quantum fields) clearly are effectively localized into a finite region of space. And there the theory really works!

It works because collider experiments do not measure (x,y,z,t). They measure (p_x,p_y,p_z,E). The calculations are done for point particles entering Feynman diagrams but the numbers that predict measurements are not dependent on space time, but on energy momentum.

No experiment can measure the localization of an individual interaction with the accuracy necessary to see effects of spatial uncertainty: the incoming protons have the Heisenberg uncertainty even if they were measured individually and not as a beam, and the same would be true for the outgoing particles that would have to be extrapolated back to the vertex. Any predictions on the localization of the interaction in the beam crossing region would fall within these combined HUP uncertainties, imo of course.

Answer 2

Viewing $\hat{\phi}(x)|0\rangle$ as a "quasi-localized state" centered at $x$ makes perfect sense, and in fact we can define a "wave function" for it that is analogous to the one in non-relativistic QM. One has the following relation (if I did the integral right!):

$$\langle 0| \hat{\phi}^{\dagger}(y) \hat{\phi}(x) |0\rangle = \frac{\pi \lambda_C}{4} e^{-|x - y|/\lambda_C}$$

where $\lambda_C$ is a quantity known as the (reduced) Compton wavelength,

$$\lambda_C = \frac{\hbar}{mc}.$$

As the "epicenters" of the effects of $\hat{\phi}(x)$ and $\hat{\phi}(y)$, get farther apart, the states' "angle" increases asymptotically, even if not exactly, to orthogonality.

This is not any different from how if we take, say, a Gaussian, or other semilocalized, non-relativistic one-particle state

$$\psi_x(x) = A_0 e^{-kx^2}$$

in ordinary NRQM, then consider the inner product of it with itself translated by some $x'$ and $y'$. You will find a similar decay law and, in fact, using Fourier transforms, one can recover the form of the wave function from this, though I haven't worked it out totally in the RQFT case.

Hence, RQFT can, in some sense, be regarded as mandating a "strengthening" of the Heisenberg uncertainty principle so that $\Delta x$ has an absolute floor, and not merely its product with $\Delta p$.

But this says nothing about when we consider measurements with resolution well above this floor! For an electron, $\lambda_C \approx 3.87 \times 10^{-10}\ \mathrm{mm}$. If a droplet in a cloud chamber is, say, $0.1\ \mathrm{mm}$, you can see that the effective position measurement must thus be far, far too imprecise to have to worry about the intrinsic uncertainty due to the relativistic effects.

(In general quantum mechanics, this is called a "weak measurement", I believe; i.e. a measurement that reveals only partial information about an observable.)

Answer 3

Here is a partial answer to your question, it concerns the transition from QFT to a non-relativistic limit : https://arxiv.org/abs/1407.8050. In the relativistic regime below the Compton wavelength, one can always define regions of space at an instant of time as subsystems and study spin or other degrees of freedom therein defined, but I guess one simply needs a trade-off in defining such subsystems between respecting causality and having a finite entanglement.