Fock representation of an electromagnetic wave Suppose an arbitrary classical (electromagnetic) wave package $E(x)$. What is its Fock space representation? I.e. I am looking for a state $| \psi \rangle$ such that $\langle \psi | \hat E(x) | \psi \rangle = E(x)$ , where $\hat E(x)$ is the quantum electric field. (For a monochromatic plane wave, the solution is a coherent single mode state)
 A: Thanks for your hints so far. The question was to some extend already discussed here: (Eigenstate of field operator in QFT). Bellow, I give a more verbose and down-to-earth answer to my question. There is also a remark why I was confused first...
Eigenstates of the annihilation field as classical states 
We assume a real valued scalar quantum field $A(x)$ of the form (neglecting normalization)
\begin{equation}
\hat A(x) = \int  \hat a^{*}(k)e^{ikx} \, d^{3} k + \int  \hat a(k)e^{-ikx}  \, d^{3} k = \hat A_c(x) + \hat A_a(x)
\end{equation}
which is split in a creation and annihilation part. We are looking for a eigenstate $| \psi \rangle$ of the annihilation operator $\hat A_a(x)$.
Such a state is the tensor product of single mode coherent states: $| \psi \rangle = \otimes_k |\alpha(k) \rangle_k$, since every single
mode state in the tensor product is an eigenstate of the corresponding annihilation operator $a(k)$ with eigenvalue $\alpha(k)$. Hence
$\hat A_a(x) | \psi \rangle = \int  d^{3} k' e^{-ik'x} \hat a(k')\otimes_k |\alpha(k) \rangle_k =\int  d^{3} k' e^{-ik'x} \alpha(k') \otimes_k|\alpha(k) \rangle_k = \psi(x) | \psi \rangle$
with the eigenvalue $\psi(x) = \int  d^{3} k e^{-ikx} \alpha(k)$. The eigenstate $| \psi \rangle$ is also called coherent and can be written as
\begin{equation}
| \psi \rangle = \otimes_k e^{\alpha(k) \hat a^{*}(k)} |0 \rangle = e^{\int d^{3}k \alpha(k) \hat a^{*}(k) } |0 \rangle =
e^{\int d^{3}x \hat A_c(x) \psi(x)} |0 \rangle = e^{\int d^{3}x \hat A(x) \psi(x)} |0 \rangle
\end{equation}
The first part of the above equation is more or less by definition true (refer to single mode coherent states)
the second part is valid since all $\hat a^{*}(k)$ commute, the third part is valid since (Fourier expansion of field and eigenvalue)
\begin{equation}
\int d^{3}x \hat A_c(x) \psi(x) = \int \int d^{3} k d^{3} q \, \int e^{i(k - q)x} d^{3}x \, \alpha(q) \hat a^{*} (k)  =
\int d^{3} k \alpha(k) \hat a^{*}(k) 
\end{equation}
(The space integral results in a delta function $\delta(k-q)$). The last part of the defining equation above is true since $\hat A_a |0 \rangle = 0$ and
hence $e^{\int d^{3}x \hat A_a(x) \psi(x)} |0 \rangle = 0$. 
Corollary: $\langle  \psi |\hat A_a(x) |\psi \rangle = \langle  \psi |\psi(x)|\psi \rangle  = \psi(x) \langle  \psi |\psi \rangle  = \psi(x)$.
$\langle \psi |\hat A_c(x) |\psi \rangle = \langle \hat A^{*}_c(x) \psi |\psi \rangle = \langle \hat A_a(x) \psi |\psi \rangle = \langle \psi(x) \psi |\psi \rangle = \psi^{*}(x) \langle \psi |\psi \rangle = \psi^{*}(x)$.
Finally: $\langle  \psi | \hat A(x) |\psi \rangle = \psi^{*}(x) + \psi(x) \propto  Re{\psi}(x)$. So the fact that a coherent state
is an eigenvector of the annihilation field together with the fact that the creation field is the hermitian conjugate of the annihilation field
shows that the quantum state $| \psi \rangle$ corresponds to the classical field $\psi(x)$. 
Remark: According the correspondence principle, there is a temptation to define $| \psi \rangle$ as $| \psi \rangle \propto \int d^{3}k |\alpha(k)\rangle$, however, this
is not the state looked for since $\langle \alpha(q) | \alpha(k) \rangle \ne \delta(k -q)$. (All single mode coherent states contain the vacuum).
