Why are momentum eigenstates in QFT plane waves? In Weinberg's The Quantum Theory of Fields, Chapter 2, he works out the Wigner classification of the unitary representations of the Poincare group. In particular, one finds that the possible one-particle Hilbert spaces are spanned by $|p,\sigma\rangle$ where $p$ is the eigenvalue of the momentum operators $P^\mu$, restricted by $p^2=-m^2$ in the irreducible representations, and $\sigma$ is the spin projection/helicity depending on whether the particle is massive or massless.
In that sense, states $|\psi\rangle$ of relativistic particles are naturally represented by momentum space wavefunctions $\psi_\sigma(p)$, where $$|\psi\rangle=\int \dfrac{d^3p}{(2\pi)^32\omega_p}\psi_\sigma(p)|p,\sigma\rangle,\quad \langle p,\sigma|\psi\rangle=\psi_\sigma(p).$$
That said, I often hear people talking about the momentum eigenstates as plane waves like $e^{ipx}$ in the scalar case or $\epsilon_\mu(p) e^{ipx}$ in the vector case with a polarization vector included, that being just some examples. I've also seem people taking linear combinations of these plane waves and saying that these are other possible states of the relativistic particles.
This is confusing for me for two main reasons:

*

*As far as I know we have no useful position operator in QFT. While we do have the momentum operators $P^\mu$ which naturally arise out of the requirement that there be a unitary representation of the Poincare group on the relevant Hilbert space, it looks like the matter about position operators is far more subtle. Without position operator I see no way in which we can have a position basis $|x,\sigma\rangle$ and without it I can't see how one is able to say that $|p,\sigma\rangle$ is a plane wave. Contrast this to non-relativistic QM: there we have position operators, these give us a basis $|x\rangle$ and if we evaluate $\langle x|p\rangle$ we do recover a plane wave. I can't see this happening in the relativistic setting.


*The next best thing we could try is to take our free fields $\psi_\ell(x)$ and try to construct states $\psi_\ell(x)|0\rangle$. If we do that already in the scalar field case and tentatively define $|x\rangle = \phi(x)|0\rangle$ we find after a bit of algebra that $$\langle x'|x\rangle=\dfrac{1}{(2\pi)^3}\int\dfrac{d^3p}{2p^0}e^{ip(x-x')},$$
and therefore these states are not orthonormal. In that setting they do not form an orthonormal basis and it would be very hard to view $\langle x|p\rangle$ as some "position representation" of the momentum eigenstates $|p\rangle$ which turns out to be a plane wave.
In that case I really can't see in which sense the abstract momentum eigenstates are plane waves. Given these two points I raised above, what is really behind saying that momentum eigenstates are plane waves? How do we really make sense of this here?
 A: I can't be sure that the people using the plane-wave language always mean the same thing, but here's one way to justify that language.
Consider a free scalar field. You're right that single-particle states cannot be strictly localized, as I'll review below. However, if $|p\rangle$ is a single-particle state with momentum $p$, then we can define a function of $x$ by
$$
 f(x)\equiv \langle 0|\phi(x)|p\rangle,
\tag{1}
$$
and this function is proportional to $e^{ip\cdot x}$. Proof:
\begin{align}
 \langle 0|\phi(x)|p\rangle
 &=
 \langle 0|e^{-iP\cdot x}\phi(0)e^{iP\cdot x}|p\rangle
\\
 &=
 \langle 0|\phi(0)e^{ip\cdot x}|p\rangle
\\
 &\propto e^{ip\cdot x}.
\end{align}
The state-vectors in the definition (1) are not normalizable, so they don't really belong to the Hilbert space, but that's an unimportant technicality here. We can fix it by smearing things a little without changing the message.
A function proportional to $e^{ip\cdot x}$ is often called a plane wave without regard for its heritage, as long as $x$ is a point in space. And in the present example, $x$ certainly is a point in space. It's not the location of a particle (the particle is not strictly localized at $x$), but it is the location of the field operator $\phi(x)$ that we're using to probe the state $|p\rangle$. Since the field operator is an observable in this model, calling (1) a plane wave is justifiable.
Calling (1) a plane wave in this sense doesn't imply that single-particle states can be strictly localized. The plane-wave language might have that connotation in other contexts, but it can't have that connotation here. To clarify this, consider the free scalar field again:

*

*The field operator $\phi(x)$ is localized at $x$ by definition. (In QFT, the field operators define what "location" means.)


*We can write it as $\phi(x)=\phi_+(x)+\phi_-(x)$, where $\phi_\pm(x)$ are its positive/negative-frequency parts.


*The operator $\phi_-(x)$ creates an individual particle, but it's not strictly localized. We can diagnose this from the fact that the equal-time commutator $[\phi(x),\phi_-(y)]$ is not zero when $x\neq y$. The field operator $\phi(x)$ is localized by definition, so the creation operator $\phi_-(x)$ cannot be localized. In particular, if we define $|x\rangle\equiv \phi_-(x)|0\rangle=\phi(x)|0\rangle$, then $\langle 0|\phi(x)|y\rangle$ is nonzero even when $x\neq y$.
Altogether, the same argument that we used to justify calling $\langle 0|\phi(x)|p\rangle$ a plane wave can also be used to show that single-particle states cannot be localized! They're resoundingly consistent with each other.
If we take off our let's-be-precise hats and put on our let's-be-practical hats, then we should emphasize that the creation operator $\phi_-(x)$ is approximately localized: the equal-time commutator $[\phi(x),\phi_-(y)]$ approaches zero exponentially as a function of $|x-y|$, with characteristic scale $1/m$, where $m$ is the single-particle mass. Even if $m=0$, then the commutator still falls off like a power of $1/|x-y|$, a high enough power to justify neglecting it compared to radiation. The particle in the state $\phi_-(x)|0\rangle$ is localized enough for many practical purposes, even though it's not strictly localized.
