Relativistic probability amplitude of a particle to be in certain position 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?
 A: 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.
A: 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.
