2
$\begingroup$

It is well-known that a hydrogen atom in free space prepared in an excited state, for example $\left| 2p_y \right\rangle$, will over time decay into the ground state $\left| 1s \right\rangle$. If not perturbed by any measurement of the atom or the emitted photon it will undergo an evolution through a superposition of the excited and ground state $$A(t) \left| 1s \right\rangle + e^{-i \omega t} B(t) \left| 2p_y \right\rangle$$ which leads to an oscillatory movement of the charge density along the $y$-axis. See for example this answer or this applet. This oscillating dipole causes the emission of a photon with a polarization given by the orientation of the dipole ($\hat{E} = \hat{d} = \hat{y}$ in this case).

This principle also applies to molecules, just with more complicated orbitals. Unlike the hydrogen atom they aren't spherically symmetric, hence the dipole moment is given by the geometry of the molecule. If you know how a molecule is embedded in a crystal you also know the polarization of the photons emitted by it.

Things start to get more complicated when the excited molecule is forced to interact with its own radiation inside a Fabry-Pérot cavity. If the cavity is resonant with the molecule's transition, the emitted field builds up in the cavity. The relative phase between the field and the charge oscillation is such that the molecule transfers energy to the intracavity field, making the molecule decay to the ground state faster than in free space. This is similar to stimulated emission, just that here one part of a photon triggers the emission of another part. [There is also a less intuitive explanation of the Purcell effect via the density of final states, for reference see Mark Fox – Quantum Optics: An Introduction chapter 10.]

My question is about the case when the cavity mode dictates the polarization of the field. This happens for example if there is a birefringent crystal in the cavity, see for example Wang et al. PRX 7(2) 021014 (2017). Light polarized along the fast axis of the crystal sees accumulates a different round trip phase than light polarized along the slow axis. Therefore the resonance frequency for the these two polarization eigenmodes is different.

Now imagine the following situation: The cavity mode with polarization along the fast axis (= $y$-axis here) is resonant with the $\left| e \right\rangle \to \left| g \right\rangle$ transition of the molecule, but the transition dipole moment of the molecule is not aligned with the cavity mode polarization:

There is a correction factor $\xi^2 = \left| \hat{E} \circ \hat{d} \right|^2$ for the Purcell factor $F_{\text{P}}$ (equation 10.35 in Mark Fox' book) which takes this into account, but it doesn't talk about how the charges oscillate. I could imagine two different possibilities, both not convincing me:

  1. The charge density starts oscillating along the dipole axis. The field of the cavity mode (polarized along $y$-axis) builds up and decelerates the charge movement. But it can only decelerate the movement along the $y$-axis. The $x$ component is dampened as weakly as in free space. So after some time the movement becomes elliptical and finally parallel to the $x$-axis.

    What I don't like about this scenario is that it would involve additional orbitals to explain the motion off the $\vec{d} = \left\langle e | \vec{x} | g \right\rangle$ axis, hence involving other orbitals, which have completely different energies. Furthermore the molecule would reach the ground state after measureably longer time, because the oscillation along the $x$-axis decays as slow as without cavity.

  2. The charge density moves only along the $y$-axis from the beginning, which means the field dampens the full movement of the charge density.

    This situation also requires additional orbitals. Furthermore it assumes that the molecule knows about its environment already at the start of the emission and adjusts its behavior accordingly.

How does the center of charge actually evolve in the molecule in the birefringent cavity? For simplicity let's assume that the cavity effectively only supports this single mode and the molecule decays on the Purcell-enhanced transition $| e \rangle \to | g \rangle$ while al other transitions are much slower, like in the follow-up of the cited paper: Wang et al. Nat. Phys. 15(5) 483-489 (2019).
$\endgroup$
2
  • $\begingroup$ 1) seems more general doesn't it? Why would molecule instantly orient its orbitals to align the expected average of dipole moment vector to external field? This process should take some non-zero time, similarly to relaxation of spin components when interacting with magnetic field. $\endgroup$ – Ján Lalinský Dec 16 '20 at 21:36
  • $\begingroup$ @JánLalinský True, 1) requires breaking fewer "dogmas". But additionally to the problem with the additional orbitals it would even predict a measurable effect, which I haven't seen reported anywhere: The occupation of the ground state would be dominated by the slow decay rate of the $x$-oscillation. $\endgroup$ – A. P. Dec 16 '20 at 22:00
1
$\begingroup$

EDIT: I will provide some details on the dynamics in Scenario 1 mentioned below, since this seems to be the OP's central question, as clarified in the comments.

In the single transition case, one can write down a Master equation for the system, which reads

$$ \dot{\rho}_t = -\frac{i}{\hbar} [H_t, \rho_t] + \mathcal{L}_t[\rho_t] \,. $$

Here, $\rho_t$ is the density matrix of the ground and excited state of the single transition ($t$ for "transition"). This Master equation therefore only contains the degrees of freedom of the transition, because we have traced out the field degrees of freedom within the Markov approximation, as usual in weakly coupled quantum optics.

The effective Hamiltonian and Lindblad term are given by [1]

$$ H_s = \hbar (\omega_t + J_t) \frac{\hat{\sigma}^z}{2} \,,$$

and

$$ \mathcal{L}_t[\rho_t] = \frac{\Gamma_t}{2}\left( 2\hat{\sigma}^-\rho_t\hat{\sigma}^+ - \hat{\sigma}^+\hat{\sigma}^-\rho_t - \rho_t\hat{\sigma}^+\hat{\sigma}^-\right) \,.$$

This is about the simplest Master equation you can get, with a decay rate $\Gamma_t$ and a corresponding frequency shift $J_t$ and it generally applies for a single transition in a dielectric environment, as long as we are at weak coupling (see [1] for details. $\omega_t$ is the bare transition frequency in free space.

The main question is: what are the values of $\Gamma_t$ and $J_t$ in the electromagnetic environment? This is what is influenced by structure of the environment, such as the cavity or material properties. As also stated below [1]

$$\Gamma_t = \frac{2\omega_t^2}{\hbar\epsilon_0c^2} \mathbf{d}_t\cdot \mathrm{Im}[\mathbf{G}(\mathbf{r}_t, \mathbf{r}_t, \omega_t] \cdot \mathbf{d}_t^* \,,$$

and similarly

$$J_t = \frac{\omega_t^2}{\hbar\epsilon_0c^2} \mathbf{d}_t\cdot \mathrm{Re}[\mathbf{G}(\mathbf{r}_t, \mathbf{r}_t, \omega_t] \cdot \mathbf{d}_t^* \,.$$

While in the OP's geometry, the Green's function might be slightly complicated, the above shows that the dynamics qualitatively stay the same, no matter the geometry. There is decay and a frequency shift, that's it. The axis changing charge dynamics do not happen. The reason is as already syspected by the OP: due to the single transition, the dynamics have been confined along a single axis given by the dipole moment vector.

Note, however, that the above situation can be physical. Depending on the natural line width and the frequencies involved, the dynamics may naturally be restricted to a single transition by resonance conditions.


I'll start with the last question.

What actually happens in the described situation?

It depends what precisely is meant by "the described situation". When I read through the question, I see a few different scenarios, mixing quantum mechanical state decay via dipole transitions and classical charge oscillation concepts. In the following, I will try to address each of them separetely.

Scenario 1: Single transition model

We consider the following model: A single dipole transition with its dipole moment in the described direction relative to the slow and fast axis of the cavity. In this case, the most general expression for the Purcell enhancement is given in terms of the Green's tensor of the surrounding dielectric structure. The full cavity enhanced line width is (see [1])

$$\Gamma = \frac{2\omega_A^2}{\hbar\epsilon_0c^2} \mathbf{d}\cdot \mathrm{Im}[\mathbf{G}(\mathbf{r}_A, \mathbf{r}_A, \omega_A] \cdot \mathbf{d}^* \,,$$

where $\omega_A$ is the transition frequency of the atom and $\mathbf{r}_A$ its position. Compared to the formula in Mark Fox' book, this formula is more general, not making assumptions such as single mode cavities etc. The polarization response is fully included in the Green's tensor.

Most notably, the Purcell enhancement is a statement about the decay rate of the excited state via the transition. This means that the slow and fast direction are in a sense summed over in the above formula (due to the dot product on both sides). The birefringence property will therefore only cause an interesting dependence on the dipole moment direction, since the Green's tensor is not isotropic in this case. However, it will not change the decay dynamics of the excited state, which still just decays with a modified rate.

Scenario 2: Single excited state model

First to answer the obvious question: How is this different from a single transition model? The point is that in general, in particular for degenerate systems, the excited state can decay via multiple channels to the different degenerate ground states. In that case, we have to sum over multiple transitions $i$, such that

$$\Gamma = \sum_i \frac{2\omega_A^2}{\hbar\epsilon_0c^2} \mathbf{d}_i \cdot \mathrm{Im}[\mathbf{G}(\mathbf{r}_A, \mathbf{r}_A, \omega_A] \cdot \mathbf{d}^*_i \,,$$

but the statement above remains true.

Scenario 3: A realistic atom in an excited state

For the realistiic case, everything depends on the parameters and physical setting. Which states participate in the decay dynamics? Which transition channels are important? Maybe we even have strong coupling which fundamentally changes the dynamics from pure decay to oscillatory behavior? This scenario is to broad to address in general.

In this general case, however, dynamics such as described by the OP are conceivable.

Scenario 4: Classical charge oscillation in the field

For classical charge oscillation, we do not have a notion of quantized orbitals or the state structure defined by the atom. In a complicated environment as the birefringent cavity, a rotation of the charge oscillation can happen. Even a separation into a multi-modal oscillation is possible.

Summary

  • In Scenario 1 and 2, not much will change, other than a quantitative modification of the decay rate.
  • In Scenario 3 and 4, complex decay dynamics can occur, which require further specificiation of the specific physical situation.
$\endgroup$
6
  • $\begingroup$ Thank you for the answer. I've clarified my question a bit to account for the details you've asked. The reason I started out with the hydrogen atom is because I'm trying to understand the fundamental dynamics involved without being sidetracked by the complicated energy level structure of a molecule in a crystal. $\endgroup$ – A. P. Dec 22 '20 at 17:25
  • $\begingroup$ @A.P. thanks, that’s great! If I understand correctly you are interested in scenario 1 then? Does that part answer your question already or should I add some detail on any particular aspect? $\endgroup$ – Wolpertinger Dec 22 '20 at 19:10
  • $\begingroup$ Indeed, it's your scenario 1 in which I'm interested. I unfortunately don't have much experience with the Green's tensor, therefore I wouldn't be able to simply write it down for a birefringent cavity. But I see in the linked reference that it connects the electric field and the current density, so it's definitely the mathematical tool to answer my question. Coming back to what I actually asked: Did you ever do this calculation? If so, what can you tell me about the movement of the charge? Is it one of the 2 possibilities in my question or something different? $\endgroup$ – A. P. Dec 23 '20 at 2:47
  • $\begingroup$ @A.P. Thanks for clarifying! I've added some details regarding the dynamics in Scenario 1. Hope that helps! $\endgroup$ – Wolpertinger Dec 25 '20 at 19:05
  • $\begingroup$ Thank you for the update. I got it. What helped me the most was your argument that the charge movement is dictated by the orbitals in the first step of the logic, basically confining the dipole to a 'rail'. This is probably equivalent to saying that the polarizability of the molecule is extremely anisotropic close to its transition frequency. $\endgroup$ – A. P. Dec 26 '20 at 19:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.