How do we extract the classical description of a particle from the QFT description? I'm coming off this post.
The free field QFT of an elementary particle moving at a relativistic speed can be approximated by the model of a classical particle.
The QFT description of a free particle is $a_{p}^{\dagger } |0 \rangle$. The time evolution is $a_{p}^{\dagger } e^{-iE_p t}|0\rangle $
The classical description of a free particle is two numbers : $(x,p')$ position and momentum. The time evolution is a straight line.
I'm using $p'$ instead of $p$ to not confuse it with the quantum field momentum eigenvalues $p$.
There's supposed to be some correspondence between these two descriptions. So I'm looking for some procedure to extract the approximate classical description from the QFT description.
How do we get the numbers $(x,p')$ of the classical model? Is $p'$ somehow equal to $p$? What about $x$ then?
 A: I'm kind of guessing. Feel free to downvote if I'm wrong.
Forget QFT for a moment. In ordinary QM, high speed elementary particles have wavefunctions that have a small variance from the mean, in the both the position and the momentum space representations. We identify the mean position and mean momentum with the variables of the approximate classical model of the particle.
The time evolution of these "extracted" classical variables is a straight line because of the Etherfest theorem.
When we bring QFT, virtually nothing changes. There's still a quantity called the "wavefunction" of a free particle.
The only difference is that the momentum eigenstates $|p\rangle$ of the particle acquire an additional meaning in terms of the field vacuum state, by the equation $|p \rangle = a_p^{\dagger} |0\rangle$. But this equation is completely irrelevant in the study of a single free particle. In the absence of any interaction, there won't be any particle creation or annihilation. So we don't need the full QFT description.
We can simply focus on the ordinary QM wavefunction description of the particle : $\psi = \int \psi (p) |p\rangle | dp$. We can forget about the irrelevant information $|p\rangle =a_p^{\dagger} |0\rangle$.
Same as in QM, we can conclude that, for high speed particles, both $\psi (p)$ and its Fourier transform will have a small variance.
We can extract both their mean values and call them the variables of our approximate classical model.
One important thing that remains is to show that the trajectory of "mean $x$" is still a straight line, same as in ordinary QM. The $E_p$ of ordinary QM was $\frac{p^2}{2m}$, while now it is $\sqrt {p^2+m^2}$.
The corresponding new Etherfest's theorem should resemble the Hamilton's equations of free particle special relativity. So I think the trajectory of "mean $x$" should still be a straight line, because SR predicts straight lines for free particles.
