# Massive particles and speed of their propagation

Can one show that in quantum field theory at least some example massive particles propagate with speed less than speed of light, while massless travel at speed of light? Well, motion is a different thing in QM than in classical mechanics, and question might be formulated differently.

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I am not sure what kind of argument do you want. There are many, each with different amount of hand-waving. The most simple is: in the long distance limit QFT has to reduce to classical physics, so the particles have to move according to special relativity.

With little less hand-waving, the propagator of a free scalar particle with mass $m$ is ${1 \over p^2 - m^2}$. This has a pole at $p^2 = m^2$ and therefore this is the major contribution to the propagation. So the particle needs to be close to being on-shell (note that nobody is saying that it needs to be precisely on-shell and indeed, it need not; this is especially clear in calculation of loop diagrams).

The same thing holds for any other (i.e. non-scalar) particle, the propagator will again contain a pole at mass $m$; the only difference is that it will carry some additional Lorentz/Dirac structure.

Now, the most non-trivial part is: what if the theory is not free? Well, then obviously all the usual problems of the QFT set in, one has to renormalize, etc. For one thing, mass of the particle gets screwed up by loop contributions and needs to be fixed by counterterms (indeed, if one computes mass of the electron in a naive QED, it comes out infinite, so it wouldn't be able to even move, contrary to the observations).

I am not sure any of this satisfies you. You should be more precise about what kind of argument you want to hear.

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can you elaborate why the not free theory behaves so different? if the mass needs to be computed from other loops, doesn't that correction affect the $\frac{1}{p^2 - m^2}$-like propagators on the free theory too? – lurscher Jun 20 '11 at 16:01
@lurscher: of course it affects them, that's what I was trying to say. The effective mass will depend on the energy cutoff (or some other parameter). It is precisely where running coupling comes from. – Marek Jun 20 '11 at 16:13

And here is an experimentalist's point of view:

When particles are virtual, anything goes as long as the equations allow it. QFT is a mathematical system that successfully predicts the crossections and angular distributions of particle interactions. It is full of virtual particles, i.e. off mass shell and whatever.

The only zero mass on shell particles we can experiment with are photons , and they do travel with the speed of light .

The nonzero stable (or semi) ones like electrons and protons and muons etc we have experimental evidence that they go with less than the speed of light. All the unstable ones with lifetimes of 10^-14 sec or less, are not accessible to measurement of the speed so the question is moot.

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And the as yet un-measured intermediate case is the mass (non-zero but small) and velocity (very, very nearly $c$ at the energies we can reliably detect the little scamps) of the neutrinos. – dmckee Jun 22 '11 at 1:55

Although people often talk about particles in Particle Physics, nonetheless it's as tendentious to say that there are particles in QFT as it is to say so in non-relativistic QM. There are only events. In general, we can't tie multiple events tightly together as caused by one particle unless the events are strung out close together on an obvious single track. In general, however, we can compute correlations between events.

The relevant mathematics here is the propagator in its real space representation instead of in its Fourier space representation. Formulae and graphs of the way in which correlations between events that are caused by a free quantum field change at various space-like and time-like separations can be found on Wikipedia as formulae and graphs for the Feynman propagator. There are correlations at space-like and time-like separation even for zero-mass propagators. If we say that such correlations are caused by particles even when they are less than 100%, then particles do travel at less than the speed of light, but then you may feel that you have to say that particles also travel faster than light because there are correlations at space-like as well as at time-like separation (but I'm suggesting that you resist talking about particles, talk about fields instead; talk about High Energy Physics instead of about Particle Physics). For interacting quantum fields, there are higher-order correlations that do not appear for free fields, which are typically small perturbations of the free field case for quantum electrodynamics, but which may be qualitatively different for hadrons. This, however, does not change there being correlations at space-like and at time-like separation.

QFT computes the evolution of probability distributions from an initial time to a final time, the S-matrix, as part of which there are various conserved integer charges. This discreteness justifies some kinds of particle thinking, but is not enough to justify fully classical ideas. What we observe at the final time are tracks in enormous detectors, but the presence of these tracks does not justify believing that when there is no enormous detector there still are tracks when we're not making them happen by putting somewhat exotic metastable states of matter in strategic places. Even for High Energy Physics, there are interference effects when there is no enormous detector present (although interference effects at high energy happen at small scales).

What is required to be causal in QFT are the effects of measurements. If we measure some quantity in a given region of space-time, it should not change the results of measurement in space-like separated regions of space-time. This, however, does not impose a hard and fast causality for states. In particular, vacuum states of quantum field theories require correlations between measurements at all space-like separations to be the same at all points of space-time.

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Good answer but nevertheless, one needs to talk about particles if one is to connect with classical notions everyone is familiar with. – Marek Jun 20 '11 at 18:19
@Marek, thanks for that. I hope you can see me trying to make contact with "particle" ways of speaking and understanding the world, though of course it's in a somewhat but not completely non-standard way, so it's prolix. I think too much acceptance of the standard ways of speaking about particles in QFT is not good for progress either in understanding or for the mathematics of QFT. One needs to talk about macroscopic events, one may choose to talk either of (quantum) fields or of (quantum) particles as the causes of those events, leading to significantly differing accounts. – Peter Morgan Jun 20 '11 at 18:48
@Peter: I couldn't agree more. Especially as there are actually no particles in theories I am interested in (strings, CFT). I was just trying to point out that the general public (most of the working physicists included) might not appreciate this view :) – Marek Jun 20 '11 at 18:56
@Marek, That's why I was thankful for your comment. I thought this was, and still is, a prime candidate to be downvoted, particularly because there's more sweat in it than in something more run of the mill. I have enough reps not to care much in the greater scheme of things, but it's always interesting to see what is downvoted, especially if someone chooses to say why. We can argue about strings or CFTs as causes of events sometime, but surely there are still various discrete structures, e.g. countable Hilbert space dimension. – Peter Morgan Jun 20 '11 at 19:12
Thank you for your answer, but, I m not sure what do you mean by hard and fast causality. Causality in QFT can be icorporated by imposing that commutators between space-like separated points vanish. I dont agree with you that there are only events and not particles in qft. After all how do you give mass to event ;) Being massive means that excitation of field has minimal energy in its rest frame which is different from zero. You cannot have arbitrary wavelengths of this field. Now, can we show that such field cannot propagate faster then speed of light, at least in classical field theory? – Newman Jun 21 '11 at 8:56

One can do the same thing you can do in ordinary quantum mechanics: show that $\frac{d}{dx}\langle\hat{X}\rangle \approx \frac{1}{m}\langle\hat{P}\rangle$ in the non-relativistic limit where $\hat{P}$ and $\hat{X}$ are the average momentum and average position operators respectively. You can do this for the free Klein-Gordon field with basic tools. I can elaborate if you want.

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