Open Quantum System

Atom Hamiltonian Setup§

For a single atom, it's Hamiltonian should easily be

$$\hat{H}=\hbar\omega_{eg}\hat{\sigma}_e$$

Where we set the ground state energy to 0, and since light is the only thing that can interact with it, we match up it's energy notation with the choise for light thus the $\hbar \omega_{eg}$, either to emmit light or to couple with light, this is the resonant frequency of it.

Since we're treating this as a two-level system, all we need are the transitions between $|g\rangle$ and $|e\rangle$ — near resonance, every other atomic transition is so far detuned that it just doesn't contribute. So the two relevant operators are the raising and lowering operators

$$\hat{\sigma}_+ = |e\rangle\langle g|, \qquad \hat{\sigma}_- = |g\rangle\langle e|$$

these will come up immediately once we write out the interaction.

Bath§

The bath is the random uncertainty fluctuation noise of the background. Which thus

$$\hat{H}=\sum_{\mathbf{k},\lambda}\hbar\omega_{\mathbf{k}}\hat{b}_{\mathbf{k}, \lambda}^\dagger\hat{b}_{\mathbf{k}, \lambda}$$

Where $\hat{b}^\dagger\hat{b}$ the number of photon in that $\mathbf{k}$ and polarization $\lambda$ and the energy for that state is $\hbar\omega_\mathbf{k}$. And we use $b$ is just to avoid notation clash, where $b$ is for resovior / bath and $a$ is for photon in the cavity.

The corresponding electric field built from these bath modes is

$$\hat{E}(\mathbf{r}) = \sum_{\mathbf{k},\lambda} \mathcal{E}_\mathbf{k}\left(\hat{b}_{\mathbf{k},\lambda}\,e^{i\mathbf{k}\cdot\mathbf{r}} + \hat{b}^\dagger_{\mathbf{k},\lambda}\,e^{-i\mathbf{k}\cdot\mathbf{r}}\right)\hat{\varepsilon}_{\mathbf{k},\lambda}$$

where $\mathcal{E}_\mathbf{k} = \sqrt{\hbar\omega_\mathbf{k}/2\epsilon_0 V}$ is the per-mode vacuum field amplitude from quantizing in a box of volume $V$. This is what makes $\hat{H}_{int}$ concrete rather than just a formal symbol.

Interaction§

The only way in our inteded model that an atom can be interacted is by the electric dipole moment as monopole is assumed to be macroscopingly 0 and quadrupole term isn't dominant. ( Which is indeed a very good approximation ). Then

$$\hat{H}=-\hat{d}\cdot\hat{E}(\mathbf{r})$$

One more thing before substituting in: since the atom is tiny compared to an optical wavelength — $a_0 \sim 0.5\,\text{Å}$ versus $\lambda \sim 10^3\,\text{Å}$ — the field barely varies across the atom. So we can just evaluate it at the atomic position and freeze it there

$$e^{i\mathbf{k}\cdot\mathbf{r}_{atom}} \approx 1 \implies \hat{E}(\mathbf{r}_{atom}) \approx \hat{E}(\mathbf{0})$$

This is the dipole approximation, and it's actually what makes writing $-\hat{d}\cdot\hat{E}$ sensible to begin with.

For the dipole operator itself, since we only have two levels, the only matrix element that survives is between $|g\rangle$ and $|e\rangle$, so

$$\hat{d} = d_{eg}\hat{\sigma}_x\hat{\varepsilon}$$

where $\hat{\varepsilon}$ is the polarization direction of the atom's dipole.

Plugging both in, the full interaction Hamiltonian becomes

$$\hat{H}_{int} = -\sum_{\mathbf{k},\lambda} \hbar g_{\mathbf{k},\lambda}\,(\hat{\sigma}_+ + \hat{\sigma}_-)(\hat{b}_{\mathbf{k},\lambda} + \hat{b}^\dagger_{\mathbf{k},\lambda})$$

where we define the coupling constant as

$$g_{\mathbf{k},\lambda} = \frac{1}{\hbar}d_{eg}\,\mathcal{E}_\mathbf{k}\,\left(\hat{\varepsilon}_{\mathbf{k},\lambda}\cdot\hat{d}\right)$$

which absorbs all the geometry. Expanding the product out gives four terms: $\hat{\sigma}_+\hat{b},\,\hat{\sigma}_+\hat{b}^\dagger,\,\hat{\sigma}_-\hat{b},\,\hat{\sigma}_-\hat{b}^\dagger$. Two conserve energy, two don't — and dropping the non-conserving ones is exactly what RWA does, which is the next note.

Total Hamiltonian§

Putting all three pieces together, the full Hamiltonian is just

$$\hat{H} = \hat{H}_A + \hat{H}_B + \hat{H}_{int}$$

This split — system, reservoir, coupling — is the standard scaffolding of any open quantum system treatment. Everything downstream (master equation, Lindblad operators, decay rates) starts from exactly this decomposition.