# 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)

• Hint: decompose your classical field into a sum of monochromatic plane waves. – Mark Mitchison Jan 22 '15 at 10:42
• @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. – Sofia Jan 22 '15 at 11:16
• – 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 $$\hat A_a(x) | \psi \rangle = \int d^{3} k' e^{-ik'x} \hat a(k')\otimes_k |\alpha(k) \rangle_k =\int d^{3} k' e^{-ik'x} \alpha(k') \otimes_k|\alpha(k) \rangle_k = \psi(x) | \psi \rangle$$

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).