Any physical meaning that the electric field operator consists of creation and annihilation operators?

Question

Quantum optics textbooks say that the electric field operator is written as a superposition as the positive frequency part and negative frequency part, e.g., for a single mode field, \begin{align} \hat{E}={\cal E}_{\bf k}({\bf r})\hat{a}(t)+{\cal E}_{\bf k}^{*}({\bf r})\hat{a}^{\dagger}(t) \end{align} where ${\cal E}_{\bf k}({\bf r})$ is the position-dependent classical mode function determined by a given physical structure and $\hat{a} (\hat{a}^{\dagger})$ is the annihilation (creation) operator.

Then, one question immediately came to my mind, what does the fact that the electric field consists of creation and annihilation operator mean? Is it like the electric field is the superposition of two physical processes such as absorption (by $\hat{a}$) and emission (by $\hat{a}^{\dagger}$)?

Someone said to me that they correspond, when compared to the classical complex representation, to a classical amplitude and its complex conjugate. However, I see this statement would be true only when a coherent state of light $\vert \alpha\rangle$ is considered. Is there any better way to understand the terms $\hat{a}$ and $\hat{a}^{\dagger}$ in the electric field operator? or do they represent any actual physical processes that eventually constitute the electric field? or is it related to the process to measure the electric field?

Answer 1

You need both parts for the operator to be hermitian( observable). Similarly if you had a two photon process, the sum of two photon absorption and annhilation would be hermitian and the sum of Rabi (or Rayleigh) cross products would be hermitian.

Answer 2

I will try to give a more physical explanation of the usefulness of having both terms containing an annihilation and a creation operator in the operator of the electrical field. From a pure formalistic view one would require both terms to make the electric field operator hermitian because its classical counterpart, the electric field, is an observable, so its operator has to be hermitian as it was already said in Dogson's post.

The impact on the physics is the following: With the electrical field operator we can build up the following interaction operator $V(t) = -e \cal{\hat{E}}\mathbf{r}$ which corresponds to an interaction between the dipole moment (or an atom or molecule) and the electrical field. Transitions between different quantum states can be described by matrix elements of this operator, in particular the absorption of a photon (with a wave vector $\mathbf{k}$):

$$e\mathbf{\epsilon}<N_\mathbf{k}-1| \hat{a}_\mathbf{k} | N_k><f|\mathbf{r}|i> $$

and the emission of a photon

$$e\mathbf{\epsilon}<N_\mathbf{k}+1| \hat{a}^\dagger_\mathbf{k} | N_\mathbf{k}><f|\mathbf{r}|i> $$

combined with the electron state transition in an atom or molecule whose amplitude is expressed by $<f|\mathbf{r}|i>$. $\mathbf{\epsilon}$ is a polarisation vector of the electrical field. So having an annihilation and a creation operator together in one interaction operator it is possible to describe with this unique operator an absorption of a photon as well as an emission of photon. That is exactly what happens if an electron in an atom or molecule interacts with the photon field, so dipolar radiation is absorbed or emitted.

Answer 3

"they represent an actual physical processes that ... constitute the electric field"

Yes, in the exact solutions. Perturbatively it only is a bla-bla.