Eigenstate of field operator in QFT Why don't people discuss the eigenstate of the field operator? For example, the real scalar field the field operator is Hermitian, so its eigenstate is an observable quantity.
 A: One thing missing in the other answers... people actually do discuss eigenstates of the field operator, or at least they are important in QFT. A complete set of field eigenstates are used to prove that n-point functions can be written in terms of a path integral, which is a critical result. But they are not used as "states after a field measurement", as $|x\rangle$ and $|p\rangle$ are used in quantum mechanics. At least not in mainstream applications / that I know of.
In case you're curious to know how exactly they are used, I'll sketch it out below, with the real scalar field as an example. In deriving the path integral, it is necessary to write the identity in a basis of field eigenstates
$$1 = \int \mathcal{D}\phi \, |\phi_t(\vec{x})\rangle \langle \phi_t(\vec{x}) |$$
with (operators have hats, numbers do not)
$$\hat{\phi}(t, \vec{x})|\phi_t(\vec{x})\rangle = \phi_t(\vec{x})|\phi_t(\vec{x})\rangle 
\,\,\,\,\,\,\,\,\, \forall \vec{x}$$
So at a given time $t$, each eigenvector $|\phi_t(\vec{x})\rangle$ is a simultaneous eigenstate of all field operators of different $\vec{x}$ but equal $t$. The eigenvalue $\phi_t(\vec{x})$ depends on which $\vec{x}$ is chosen in the field argument, so we write it as a function of $\vec{x}$. This defines a classical field (a map from $\mathbb{R}^3\to \mathbb{R})$. Operators can be simultaneously diagonalized if they commute*, and luckily the usual QFT commutation relations give exactly this for equal times:
$$[ \phi(t,\vec{x}), \phi(t, \vec{y})] = 0$$
The diagonalization process can be repeated for any time because the commutation relation above holds for any $t$.
*I think that strictly there are more things to worry about for infinite-dimensional operators, so you might want to take the commutation relations as an indication that they can be simultaneously diagonalized, rather than a proof. I don't know enough to expand on this.
A: As $\phi(f)$ and $\pi(f)$, which are self adjoint, satisfy the same commutation relations as $X$ and $P$, the closure of the space generated by polynomials of the former pair of operators applied to $\lvert 0\rangle$ is isomorphic to $L^2(\mathbb R)$. Therefore the spectrum of  $\phi(f)$ and $\pi(f)$, is purely continuous and coincides to $\mathbb R$ and there are no proper eigenvectors, but they are just formal ones and isomorphic to $\lvert x\rangle$ and $\lvert p\rangle$.
A: The answer is simple. They are not realized in nature too often. And more, they are not stationary states, so such a state evolves in time to a state that contains fluctuations of the field variable $\phi$ over all the space. The eigenstate of $\hat{\phi}$ evolves to a non-eigenstate of $\hat{\phi}$. These fluctuations increase and spread with time. To obtain this state, you need to measure $\phi$ over the space with considerable precision about Compton's field wavelength. We know that in this scale, fluctuations on the field start as the evolution through time begins.
The fields that we probe in classical mechanics are the coherent state:
$$
|\phi_{cl}\rangle=\exp\left(\int \phi_{cl}(x)\hat{\phi}(x)\right)|0\rangle.
$$
This state is not an eigenstate of $\hat{\phi}$, so the problems of the first paragraph can be avoided. This state has minimal fluctuations, and the uncertainty is constant in time. Therefore, the expectation value of the field operator is:
$$
\langle\hat{\phi}(x)\rangle=\langle \phi_{cl}|\hat{\phi}(x)|\phi_{cl}\rangle=\phi_{cl}(x).
$$
You can check this by the definition of $|\phi_{cl}\rangle$.
Note that the vacuum state $|0\rangle$ is also a coherent state associated with the trivial classical solution $\phi_{cl}(x)=0$.
