Fock representation of an electromagnetic wave

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).