As $\phi(f)$ and $\pi(f)$, which are self adjoint, satisfy the same commutation relations of $X$ and $P$, the closure of the space generated by polynomials of the former pair of operators applied to $\lvert 0\rangle$ is isomorphic to $L^2(R)$, Therefore the spectrum of $\phi(f)$ and $\pi(f)$, is purely continuous and coincides to $R$ and there are no proper eigenvectors, but they are just formal ones and isomorphic to $\lvert x\rangle$ and $\lvert p\rangle$.

This post imported from StackExchange Physics at 2014-04-21 16:24 (UCT), posted by SE-user V. Moretti