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In the book “The story of spin” by Tomonaga on page 110, it says

They insisted that a concept like "the probability of a particle to be at $x$ in space" is meaningless for relativistic particles—be they electrons, photons, or Klein-Gordon particles—and therefore it is meaningless to interpret $\psi (x)$ as the probability amplitude.

He is talking about Pauli and Weisskopf ideas on relativistic wave functions.

I would like to understand that argument. Why the concept of "probability of a particle to be at $x$ in space" is meaningless for a relativistic particle? Are those arguments still valid?

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My understanding is "the probability of a particle to be at $x$ in space" is meaningless for measurements of relativistic particles which are more precise than (roughly) the compton wavelength, for the following reason. If the uncertainty of position of a particle (say, electron) is low, the momentum uncertainty and, therefore, energy uncertainty is high, therefore, besides the initial electron there can be additional electron-positron pairs at (almost) the same position, so you don't really know which electron has that position (I read something like that in a book by Landau). I tend to think this argument is still valid.

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  • $\begingroup$ Hmm. But even in the relativistic case, one can do an experiment with a screen and find a particle to be in a particular place. So how to predict such an experiment in QFT? $\endgroup$
    – don't train ai on me
    Commented Jun 18, 2021 at 8:15
  • $\begingroup$ Can you point to which book you mean? $\endgroup$
    – don't train ai on me
    Commented Jun 18, 2021 at 8:16
  • $\begingroup$ @doublefelix : books.google.com/… , Introduction, para 1. So it is indeed Course of theoretical physics by Landau and Lifshitz, but Landau is not an author of this volume (Quantum Electrodynamics) $\endgroup$
    – akhmeteli
    Commented Jun 18, 2021 at 11:16
  • $\begingroup$ @doublefelix : One can find a particle in a particular place, but which particle is that? The original one or a particle from an electron-positron pair created in a precise coordinate measurement, which requires high energy? So in QFT one uses, e.g., the Fock space, and the system can contain states with more than one particle. $\endgroup$
    – akhmeteli
    Commented Jun 18, 2021 at 11:22
  • $\begingroup$ To me, the big picture is: we can do position measurements on particles moving at relativistic speeds. They give a definite position on a screen, experimentally. Whether this is the original particle or not can be argued about, but the main point is: how do we predict the distribution of positions seen on the screen? I see nothing ill-defined about this. $\endgroup$
    – don't train ai on me
    Commented Jun 19, 2021 at 18:50
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I think the text in question is quite vague, but there are some issues that come up with relativistic theories.

For instance if something, say $p(x),$ is a probability density, then $\int p(x,t)dx=1.$ But if you have a relativistic theory then you should also be able to compute $\int p(x',t')dx'=1,$ where $x'$ various over a surface of $t'=$ constant for a different inertial frame. And it is generally impossible to have both.

An example where you can't have both is $p(x,t)=\frac{1}{\sigma\sqrt{2\pi}}e^{(x-vt)^2/2\sigma^2}.$ And it's not just because of the problem that appears when $v=0$ where density is frame dependent because of length contraction. It is the relativity of simultaneity for the case ($v\neq 0$) where when you integrate over a surface of constant $t'$ you can get regions of earlier $t$ to the left and regions of later $t$ to the right, thus integrating to larger than one even after you adjust the for length contraction.

But even in nonrelativistic theory it is generally wrong to assume that there simply is a probability that a particle has preexisting properties with a particular probability. Its a story some people tell and in some situations it won't bite you. But it isn't right and thinking it will get you in trouble.

All you really want is relative frequencies of different interaction outcomes. Since that is what you measure in the lab.

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    $\begingroup$ of course, you can have both. In the other frame $p(x)$ will change with a $\gamma$ factor. $\endgroup$
    – Anthonny
    Commented Jul 26, 2015 at 17:31
  • $\begingroup$ @Anthonny I edited the answer to include more details so you won't be able to think a gamma factor fixes it the way it would for a delta function probability distribution. There are cases where it can work, but it doesn't work in general. $\endgroup$
    – Timaeus
    Commented Jul 26, 2015 at 19:08
  • $\begingroup$ "But even in nonrelativistic theory it is generally wrong to assume that there simply is a probability that a particle has preexisting properties with a particular probability." Could you explain why that would be wrong? $\endgroup$
    – Ján Lalinský
    Commented Jul 26, 2015 at 23:19
  • $\begingroup$ @JánLalinský It's wrong because assuming it leads to experimentally falsified predictions. It is meaningless to say something has properties with probabilities if you can't reveal those properties with the right frequencies. But that doesn't happen. What you have instead is something that interacts in certain ways, which is different. If you assume it has a spin, momentum, kinetic energy, position, etc. And that measurements just reveal those properties preexisting by revealing the preexisting values then it leads to errors. For instance, the correlations in a bell type measurement. $\endgroup$
    – Timaeus
    Commented Jul 27, 2015 at 0:47
  • $\begingroup$ @JánLalinský You can look at an example of a Stern-Gerlach interaction. If you do it the correct way, you analyze the actual setup in an accurate and continuous manner. And you mathematically see the incoming beam split into two beams and the sizes of the beams are determined by the relative size of the spin's projection onto the eigenspaces and after the beam separates the two beams each have their spin completely polarized. If you spilt the beam first with a spin independent beam splitter then send each part through then different parts end up becoming spin up and spin down. $\endgroup$
    – Timaeus
    Commented Jul 27, 2015 at 0:55

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