Field operator eigenstate vs. single-particle state
I would like to make sure I understand some basic QFT. My understanding so far is that field operators measure field intensity and their Fourier transform measure intensity of field oscillation. In the usual decoupled harmonic oscillators picture in momentum space a single particle state is $a_k^{\dagger}\mid 0\rangle$ and this is usually written $\mid k\rangle$. Now what seems a little paradoxical to me and therefore what I would like to confirm is that the expectation of the momentum $k$ field operator $\phi(k)$ is 0 on that single particle state instead of some positive number. 669
My understanding is that $\mid k\rangle$ is a sum of states in the $k$ summand of the momentum decomposition of the Hilbert space (a direct sum/integral of harmonic oscillator Hilbert spaces indexed by $k$), which is symmetric about 0. There is no definite intensity in field oscillation with momentum $k$ but a 1-particle excitation at that momentum.
On the contrary an eigenstate $\mid\phi(k)\rangle$ of $\phi(k)$ has momentum $k=\langle\phi(k)\mid\phi(k)\mid\phi(k)\rangle=\langle\phi(k)\mid k\mid\phi(k)\rangle$ (a 4-vector of expectations). That is, $\phi$ is a multiplication operator on fields and $\phi(k)$ is a multiplication operator (on $L^2(S'(\mathbb R^4))$ of square integrable functions on distributions) times a delta function in $\mathbb R^4$ at the $k$ factor, that multiplication by the value of distributions in $S'(\mathbb R^4)$ at $k$, so eigenstates $\phi(k)$ are delta functions at 1 of the $k$th harmonic oscillator, in its corresponding Hilbert space factor (once again the field Hilbert space seen as a sum/integral of a continuum of harmonic oscillator Hilbert spaces).
To summarize: am I right to believe that $\langle k\mid\phi(k)\mid k\rangle=0$ and $\langle\phi(k)\mid\phi(k)\mid\phi(k)\rangle=k$?
On a related matter I would like to confirm that coherent states are not field eigenstates but only field annihilation eigenstates. That is they are eigenstates of $\phi^+(x)=\sum_ke^{2i\pi xk}a_k$, not of $\phi(x)\simeq\sum_ke^{2i\pi xk}(a_k-a_k^\dagger)$. They have average field momentum $k$, that is expectation $k$ for $\phi(k)$ but they are not $\phi(k)$ eigenstates.
Thanks alot for any comment. 669