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

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?

• There is no reasonable pictorial view of current QM or QFT theories. – image Oct 20 '16 at 21:51
• At the very least, I would think you would need to destroy the particle too, so as to keep it localized in time. – Hurkyl Oct 20 '16 at 22:11
• Why the wave packet in the form of creation and annihilation operators is not sufficient? page 33 tcm.phy.cam.ac.uk/~bds10/aqp/lec3_compressed.pdf – anna v Oct 21 '16 at 4:42
• @annav To define the meaning of a wavepacket in position space in the usual way, one needs to have basis of localized position states. – Sjorszini Oct 22 '16 at 8:28
• @SjorsHeefer: You are not surely saying that the wave packet description does not exist in QM and QFT, are you? – Vladimir Kalitvianski Oct 22 '16 at 8:37

## 2 Answers

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.

• Small technical detail - there are parts of the detector that measure position and not momentum. The transition radiation tracker in ATLAS, for instance, measures $x$ and $y$ by which straws in the detector lit up. I'm less certain how they measure position along the tube, but I think it can be done if you use signal time delay since bunch crossing. The momentum is then inferred from path curvature, if charged, other detectors if not. – Sean E. Lake Nov 1 '16 at 7:40
• @SeanLake All detectors measure positions, but these are tracking devises, within the HUP uncertainties which are much smaller . They are secondary interactions. They lead to calculating energy and momentum leaving the interaction vertex.That is where the QFT predictions are made . – anna v Nov 1 '16 at 7:46

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.