# How does one describe individual particles in QFT in practice?

A similar question was posted on this site at least ten times, but not quite in this formulation, and with no satisfactory answers, so I give it another try.

Quantum field theory textbooks almost always describe particles as eigenstates of the free Hamiltonian, and their interaction is described in the following way: there are "in-states" of the infinite past, and "out states" of the infinite future, and there is the scattering matrix, which is the evolution operator from $$t=-\infty$$ to $$t=\infty$$, describing the amplitudes of transitions between in- and out-states. These in- and out-states are unphysical, they essentially represent non-localized "constant flows of particles", analogously to the description of scattering in classical quantum mechanics. Sometimes textbooks mention that one can use wave packets to describe particles as linear combinations of states with fixed momenta or positions, but this is more or less all that is said.

My question is: what do I do, in practice, to describe the scattering of an actual single particle? For instance, imagine the Coulumb scattering of two electrons emitted at times with given probability distributions, having a specific probability distribution of momenta. I should be able to calculate the distribution of times at which electrons hit detectors, right? How are these pretty pictures from LHC showing cascades of scattering byproducts compared with theoretical predictions?

• What does your QFT text say? Jan 14 at 17:06
• Please note that the experiments of particle physics measure the four vectors $(p_x,p_y,p_z, E)$ of the particles leaving the interaction vertex, and those distributions are compared to the QFT models predictions. Jan 14 at 19:27
• I have never seen a QFT book that claims that it can make any predictions about an individual quantum measurement. All of quantum mechanics, including QFT, only makes statements about ensemble distributions. What we call "a state" is a physical description of the possible outcomes of an infinite number of measurements. It's not a statement about a single quantum. That includes the time signatures of events. If you look at the time distribution of detections in LHC, it's dominated by the engineering of the accelerator and its beam bunch structure. Jan 15 at 21:58
• @FlatterMann, you confuse predictions about individual particles with predictions about individual measurements. Jan 16 at 11:01
• The OP was asking about a single particle (better quantum), which is a single, individual energy transfer. In case of an optical detector it's literally a piece of matter being heated by 2e-19J. A single quantum state, OTOH, does not have any physical meaning unless we also define the measurement in addition. Once we do, the Born rule drops a probability distribution on us. That distribution has a frequentist interpretation that we can test with counters in an experiment. QFT says nothing about the individual quantum and it makes a trivial statement (plane wave) about a one-quantum state. Jan 16 at 11:11

The general answer is that you can't: QFT predicts that such description is typically inconsistent. For most QFTs, there is no valid description of particles as individual entities. The "wave-packet" story is a pedagogical device to keep students from asking too many questions too soon; you need to develop a lot of machinery before you can actually begin thinking about these matters.

That being said, in some situations, it is doable. Specifically, whenever you have a weakly coupled theory and you are interested in low-energy processes, you can describe individual particles by simply writing down a Schrödinger equation for them.

The standard reference is ref.1. The rough (and oversimplified) idea is as follows. A basis of states for your Hilbert space is $$|p^\mu,\lambda\rangle$$ where $$p^\mu$$ are the eigenvalues of the momentum operators $$P^\mu$$ and $$\lambda$$ is a set of labels for other quantum numbers, such as spin, charge etc. We call these states particles.

We say an operator $$\phi$$ creates the state $$|p^\mu,\lambda\rangle$$ if $$\langle 0|\phi|p^\mu,\lambda\rangle\neq0$$. The operator $$\phi$$ usually satisfies some differential equation of the form $$\mathcal D\phi=0$$, where $$\mathcal D$$ is some differential operator, such as $$\mathcal D=\gamma^\mu\partial_\mu+A(x)$$ when $$\phi$$ is a spinor field, and $$\mathcal D=(\partial_\mu+iA_\mu)^2$$ when $$\phi$$ is a scalar field. Then, it follows that the function $$\Psi_{p,\lambda}(x)=\langle 0|\phi|p_\mu,\lambda\rangle$$ satisfies a Schrödinger-like equation $$\mathcal D\Psi=0$$.

If we are only interested in low-energy processes (such that, for example, multiparticle production is highly suppressed) then we can drop all the excited states $$|p^\mu,\lambda\rangle$$ except for the one with lowest mass and we end up with a truncated (but self-consistent) description of the dynamics of $$|p^\mu,\lambda\rangle$$ entirely via the wave-function equation $$\mathcal D\Psi=0$$.

So, in this setting, the effective description of the "one-particle" sector of your QFT is in terms of a Schrödinger-like equation for a single wave-function $$\Psi$$, entirely equivalent to the quantum-mechanical description we are used to from our QM 101 course. This description is only valid for the lightest state in our theory. We can of course add a few more particles by working with the next few wavefunctions $$\Psi_i$$, associated to the next few states with lowest mass $$p^2$$. But at some point this description stops being valid, since it is only consistent to the extent that we can ignore pair-production and the like.

References

1. Weinberg S. - Quantum theory of fields, Vol.1.§14.
• @FlatterMann : In free field theory, one-particle states like that are perfectly OK. But once you have two interacting fields, think of this. An electron cannot exist without its associated EM field. Hence, whatever operator excites the electron field, must also excite the EM field. That means a state with at least "fuzzily" many, but not zero, photons. But now ... the interaction goes both ways. The EM field also excites in a similarly fuzzy manner, the electron field! The back-and-forth thus produces a fuzzy state on BOTH fields! Hence, how can you have a 1-electron state any more? Jan 16 at 7:37
• Thus, I'd suggest, it would be better to say the base units of the interactive field are not electrons or photons, but these hybrid entities containing aspects of both. And that, one might argue, is the gist behind Haag's theorem, and why the Hilbert space of the interactive field theory is not the same as the free field theory. Jan 16 at 7:39
• So what is "the world" made of? I'd suggest that, from this perspective, "the world" is not made of electrons, photons, quarks etc. but this fuzzy, fluidy, quantumy gooey stuff that under various circumstances can look "mostly electrony", "mostly photony", etc. but when you have a fireball like in the LHC, you have a seething, bubbling, roiling mess that is none and all of them at once. Jan 16 at 7:45
• @The_Sympathizer If we have two interacting fields then we have more than one quantum, even if virtual. The virtual ones can not modify the energy and momentum of an incoming state, however. At most they add a mass term if I am not mistaken, so we can still express the single quantum theory as an effective free field theory, can we not? Again, I am not a theorist. I know how to measure real quanta but I am woefully uninformed about the intricacies of the theory at this level. I appreciate the reply, though. Jan 16 at 8:25
• @The_Sympathizer With regards to your last comment about the fuzzy mess. I think that's an easy one. The only self-consistent answer is "none". Quanta only "exist" at the edges, where emission at the source and absorption at the detector take place, or at infinity, to where the stable endproducts disappear. That's completely consistent with the Born Rule in Copenhagen and it even has classical analogs (What's the state of rolling dice? None. Rolling dice don't have a state that can be described by the elements of their outcome space. Only resting dice can be described that way.). Jan 16 at 8:33