# Eigenstate of field operator in QFT

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 $\lvert p\rangle$.

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. I.e. eigenstate of $$\hat{\phi}$$ evolves to a non-eigenstate of $$\hat{\phi}$$. These fluctuations increase and spread with time. Practically, to obtain this state you need to measure $$\phi$$ over all space with considerable precision in relation to the Compton's wavelength of the field, and we know that in this scale fluctuations on the field start to show up as the evolution through time starts.

The fields that we probe in classical mechanics are actually 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. Actually this state has minimal fluctuations, and the uncertainty is constant in time. 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 $$\phi_{cl}(x)=0$$.