Time-dependence of ladder operators in quantized EM fields

Question

My Question

Are the operators for the $A$, $E$ and $B$ field to be treated as operators in a Heisenberg description or is their time dependence explicit when performing a textbook EM quantization as in Sakurai?

The following two sections document the chain of thoughts that brought me to this point and define the quantities I am talking about.

Quantization of the EM Field

In the standard textbook procedure of quantizing the EM field (like the following from Sakurai) one usually starts with a Coulomb gauge in vacuum, which leads to an expression for the $A$-field which has the form

$\mathbf{A}(\mathbf{x};t) = \sum_{\mathbf{k},\lambda} A_{\mathbf{k},\lambda}(\mathbf{x};t) \ \mathbf{e}_{\mathbf{k},\lambda}$,

where $\mathbf{e}_{\mathbf{k},\lambda}$ are the unit vectors of circular polarizations and

$A_{\mathbf{k},\lambda}(\mathbf{x};t) = A_{\mathbf{k},\lambda} e^{- i(\omega_k t - \mathbf{k} \cdot \mathbf{x})} + c.c.$

The procedure is now to find an expression for the energy density of the field and identify the appropriate choice for annihilation and creation operators. In Gaussian units this would be

$A_{\mathbf{k},\lambda} = \sqrt{\frac{4 \pi\hbar c^2}{2 \omega_k V}} \ a_{\lambda}(\mathbf{k})$,

where $a_{\lambda}(\mathbf{k})$ are boson operators. So far, so good. Now my conceptional problems start. If I now consider the quantized version of a single mode $E$ field, it has the form

$\mathbf{E}(\mathbf{x};t) = i E^{0}_k \left(a_{\lambda}(\mathbf{k}) e^{- i(\omega_k t - \mathbf{k} \cdot \mathbf{x})} - h.c.\right) \mathbf{e}_{\mathbf{k},\lambda}$,

where $E^{0}_k = \sqrt{\frac{4 \pi \hbar \omega_k}{2 V}}$.

My Problem

In the description of a two-level system with ladder operators $\sigma$ and $\sigma^\dagger$ which is interacting with a single mode cavity one usually considers a coupling of the form

$H = \hbar g\ (\sigma+\sigma^\dagger)(a+a^\dagger)$

with coupling parameter $g$. This is surprising for me, since my naive assumption would be to start from a dipole interaction with the quantized field that I derived above and write

$H = - \boldsymbol{\mu} \cdot \mathbf{E}(\mathbf{x};t)$

I understand how to get rid of the space dependency by considering a long-wavelength limit but I have no idea how to treat the time dependence of the field. My feeling is that the operators which emerged in this quantization procedure should be treated as being Heisenberg operators evolving in time but I am not sure about this. The way of performing the quantization in this way offers less insight to me whether the result should be understood in the Heisenberg picture or not. Especially because usual Schrödinger operators are allowed to have explicit time-dependence.

Answer 1

Time-dependence of operators when quantizing a free field

When quantizing EM field, the QM picture used is about choice. All different pictures of QM are unitary equivalent. But, surely there are some pictures that seems to be more convenient than others.

Here, since the ladder operator is the basic bloc of a quantized theory of EM-field, it seems to be more appropriate to stick the time-dependence to it, i.e. to use Heisenberg picture.

Let us decompose the vector potential in modes : $$ \textbf{A}(\textbf{x},t)=\sum_\textbf{k}\textbf{a}_\textbf{k}(t)\,e^{\mathrm{i}\textbf{k}\cdot\textbf{x}}+\textbf{a}^*_\textbf{k}(t)\,e^{-\mathrm{i}\textbf{k}\cdot\textbf{x}} $$ which is the solution of the equation of motion : $$ \left[\Delta-\frac{1}{c^2}\partial^2_t\right]\textbf{A}(\textbf{x},t)=0 $$ This equation then provides two things :

• the dispersion relation of the modes : $\quad\omega^2_\textbf{k}=\textbf{k}^2\,c^2\quad\text{i.e.}\quad\omega_\textbf{k}=\pm kc$

• this leads to two possible solutions for the mode amplitudes : $\quad\textbf{a}_\textbf{k}(t)=\textbf{a}_\textbf{k}e^{\pm\mathrm{i}\omega_\textbf{k}t}$

Generally, the $\textbf{a}_\textbf{k}e^{-\mathrm{i}\omega_\textbf{k}t}$ solution is prefered since it allows to see $\textbf{A}(\textbf{x},t)$ as a sum of "wave-like propagating modes" described with an $e^{\mathrm{i}(\textbf{k}\cdot\textbf{x}-\omega_\textbf{k}t)}$ term.

Then, when quantizing the field, the $\textbf{a}_\textbf{k}$ are associted to ladder operators : $$ \textbf{a}_\textbf{k}\rightarrow\hat{a}_\textbf{k} $$ Now, and only now, it is possible to choose the QM picture. And here it's easy to see that the Heisenberg picture goes quite well with the quantization process : $$ \textbf{a}_\textbf{k}(t)=\textbf{a}_\textbf{k}e^{-\mathrm{i}\omega_\textbf{k}t}\rightarrow\hat{a}_\textbf{k}(t)=\hat{a}_\textbf{k}e^{-\mathrm{i}\omega_\textbf{k}t} $$

Time-dependence of operators in the interacting case

Since now you are dealing with time dependant couplings (coupling between an atom and the cavity), it's often neater to express your operators in the interaction picture. If you do so, you will find that : $$ a(t)=a\,e^{-\mathrm{i}\omega t}\quad\text{and}\quad \sigma_\pm(t)=\sigma_\pm\,e^{\pm\mathrm{i}\omega_0 t} $$ where the Pauli matrices $\sigma_\pm$ are : $$ \sigma_+=|e\rangle\langle g|\quad\text{and}\quad\sigma_-=|g\rangle\langle e| $$ Here, $|g\rangle$ and $|e\rangle$ stands respectively for the ground state and the excited state of your atom. $\omega_0$ is the resonance frequency of your atomic transition, and $\omega$ is the cavity mode frequency.

Then by putting all of this in the interaction hamiltonian : $$ \mathcal{H}_{int}(t)=\hbar g(\sigma_+(t) + \sigma_-(t))(a(t)+a^\dagger(t)) $$ which can be expanded as : $$ \mathcal{H}_{int}(t)=\hbar g(\sigma_+ a\,e^{\mathrm{i}(\omega_0-\omega) t} +\sigma_+ a^\dagger\,e^{\mathrm{i}(\omega_0+\omega) t}+\sigma_- a\,e^{-\mathrm{i}(\omega_0+\omega) t}+\sigma_- a^\dagger\,e^{-\mathrm{i}(\omega_0-\omega) t}) $$ Then generally the physics is studied close to resonance, i.e. $|\omega-\omega_0|<<\omega_0$ so that very rapid oscillating terms in $\omega+\omega_0\sim 2\omega_0$ can be neglected compared to the slow terms in $\omega-\omega_0$. Such approximation is oftenly refered as the rotating wave approximation.

A fortiori, if you are exactly at resonance $\omega=\omega_0$, the interaction hamiltonian reads : $$ \mathcal{H}_{int}=\hbar g(\sigma_+ a\ +\sigma_- a^\dagger) $$ which is "removing" the time dependence.

And as you suggested, $\mathcal{H}_{int}$ is a dipolar coupling between the atom and the field and is derived from a $-\hat{\textbf{d}}\cdot\hat{\textbf{E}}$ term.