# 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. 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.

• "My understanding so far is that field operators measure field intensity and their Fourier transform measure intensity of field oscillation". What does this statement translate to for the harmonic oscillator, a zero-dimensional field theory? Re the last paragraph, operators which are not Hermitian-symmetric do not have eigenstates. – Keith McClary Sep 27 '15 at 3:15
• Hmm have not thought about it since I asked. There is only a single momentum in 0 dimensions: 0. The (intensity of the) field in 0 dimension would be the number of particles at that momentum's oscillator and the Fourier transform of the field would be 0 since it is a constant operator on spacetime -of course there's only 1 point. So field intensity would be quantized/discrete (0,1,...), while in more dimensions there is a continuum of momenta creation operators integrated over to define $\phi(x)$ at point $x$. – plm Nov 15 '15 at 18:42
• Possibly related to 312006. – Cosmas Zachos Feb 20 '17 at 23:27