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)

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    $\begingroup$ Hint: decompose your classical field into a sum of monochromatic plane waves. $\endgroup$ – Mark Mitchison Jan 22 '15 at 10:42
  • $\begingroup$ @alain See Fock states. For photons, these are symmetrized number states. But, is your wave-packet monochromatic? A wave-packet finite in time is not purely monochromatic. $\endgroup$ – Sofia Jan 22 '15 at 11:16
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    $\begingroup$ See en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field $\endgroup$ – Urgje Jan 22 '15 at 12:07

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 $\hat A_a(x) | \psi \rangle = \int d^{3} k e^{-ikx} \hat a(k) \otimes_k |\alpha(k) \rangle_k = \int d^{3} k e^{-ikx} \alpha(k) \otimes_k |\alpha(k) \rangle_k = \psi(x) | \psi \rangle$ 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).


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