In quantum mechanics, the probability density of a particle's position is $$\rho(x)=|\langle x|\psi\rangle|^2$$
What is the corresponding expression in QFT to predict this distribution? Since $\rho(x)$ can be measured (at least to some accuracy) in experiment, this seems like a fair question to ask.
I have asked this question to a few people at my university. To anticipate a few answers which in my opinion did not resolve this question, let me point out that
- In QFT, because particle number is not conserved, it has been suggested that the question I am asking might not make sense. A slightly reformulated version of the question does, though: "what is the probability density of finding (for example) exactly one electron at x?".
- I have read some answers to similar questions like this one, which mentions a position operator defined by Newton & Wigner, so that its eigenstates give the probability of a position outcome as in QM. According to the same post, this attempt at an X operator failed because the probability was not Lorentz invariant.
- U(1) symmetry in $\mathscr{L}$ suggests something like a probability density, but this quantity can be negative, which is why it is interpreted as charge density instead (as far as I have read).
- I am aware that QFT is not usually used to predict position probability densities. However such things are commonly measured, for example in the double slit experiment. Therefore if the theory is at least as general as QM, it is desirable that it has something to say here.
Of course if one of these bullet points somehow does resolve the question I would be very interested to better understand how.
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