Confused about the Purcell effect in a cavity

[Malamala]

I thought that the Purcell effect was simply an enhancement of the decay rate of an atom when placed inside the cavity (the mode enhanced is the one that the cavity amplifies, and I assume here that the light emitted by the atom and the cavity are resonant). However, I read this paper: https://arxiv.org/abs/1104.3594 and I am confused now. I want to emphasize that the paper is written by one of the leading experts in the field, Vladan Vuletic, and the same equations presented here are used in many of his other papers, both theoretical and experimental. Thus I am sure that everything is correct in terms of the equations, my question is about the interpretation of the equations. So here it is:

In Fig. 1 in that paper (I will only refer to that paper) they define the field (up to some normalization) emitted in a given Gaussian mode (in the absence of a cavity), $E_M$ (I will use normal uppercase letters) and they show (eq. 10) that the rate (or power) emitted in this mode in both direction, over the $4\pi$ scattering is given by the free space cooperativity, $\eta_{fs}$ (eq. 8). Now, if we put the atom in a cavity, and drive the atom from the side (fig. 6), the power leaving the cavity over the power emitted in $4\pi$ is (eq. 39), on resonance, given by the cooperativity, $\eta$, which is roughly the free space cooperativity times the finesse of the cavity. However, the field emitted by the atom itself doesn't seem to change. This is seen in the paragraph right before eq. 39, where they state $P_{4\pi} = |E_M|^2/\eta_{fs}$, given that the power emitted in the mode M is $|E_M|^2$ (I might have missed a factor of 2 in some equations, but it won't affect my question). So based on this (at least to my understanding), the atom scattering rate in the Gaussian mode defined by the cavity is not changed relative to the free space scattering, with or without the cavity being present. Thus from the point of view of the atom, the cavity doesn't matter. However, a photodetector place after the cavity mirror, would see the enhanced emission (by a factor of $\eta$) in the cavity mode. However, I thought that the Purcell effect means that the atom itself sees the enhancement, which would be equivalent to $P_{4\pi} = |E_M|^2/\eta$. What am I missing here? Why is the emission of the atom itself in the cavity mode not enhanced with respect to the $4\pi$ emission?

Also, this interpretation (which I assume it's wrong on my side) violates the conservation of energy, as it looks as if the atom emits in the cavity mode with a rate proportional to $\eta_{fs}$, but the photons leave the cavity at a rate proportional to $\eta$, despite the system being in steady state. I would appreciate if someone can help me understand this (mainly what is going on with this $E_M$). Thank you!

 

[Fred Wright]

In the paper you refer to the authors state in equ. 1 that the complex polarizability is given by,
$$
\alpha=6\pi \epsilon_0 c^3 \frac{\frac{\Gamma}{\omega_0^2}}{(\omega_0^2-\omega^2)-i\Gamma \frac{\omega^3}{\omega_0^2}}
$$
They state in equ. 2 that,
$$
\left| \alpha \right|^2=\frac{6\pi \epsilon_0}{k^3}Im(\alpha)
$$
I did the calculation and I find,
$$
\left| \alpha \right|^2=(\frac{6\pi \epsilon_0}{k^3}Im(\alpha))^2
$$
They make the same error in equ. 7. This raises serious doubt in my mind about the subsequent analysis.
Concerning your first question, in free space the atom radiates in space like a dipole oscillator, the density of final states after emission of the cavity mode M is small compared to the final density of states when the atom is centered in the cavity. The atom sees the enhancement, but only in the sense of density of states. The photo detector responds to field magnitude, not power. Neither the atom nor the detector integrate over time which is required for power.

 

[Malamala]

In the paper you refer to the authors state in equ. 1 that the complex polarizability is given by,
$$
\alpha=6\pi \epsilon_0 c^3 \frac{\frac{\Gamma}{\omega_0^2}}{(\omega_0^2-\omega^2)-i\Gamma \frac{\omega^3}{\omega_0^2}}
$$
They state in equ. 2 that,
$$
\left| \alpha \right|^2=\frac{6\pi \epsilon_0}{k^3}Im(\alpha)
$$
I did the calculation and I find,
$$
\left| \alpha \right|^2=(\frac{6\pi \epsilon_0}{k^3}Im(\alpha))^2
$$
They make the same error in equ. 7. This raises serious doubt in my mind about the subsequent analysis.
Concerning your first question, in free space the atom radiates in space like a dipole oscillator, the density of final states after emission of the cavity mode M is small compared to the final density of states when the atom is centered in the cavity. The atom sees the enhancement, but only in the sense of density of states. The photo detector responds to field magnitude, not power. Neither the atom nor the detector integrate over time which is required for power.

That equation is correct (it's just a form of the optical theorem). You probably did a mistake in your calculation, and this can be seen from a dimensional analysis. According to your formula, $(\epsilon_0/k^3)^2$ is unitless, which is obviously not the case ($|\alpha|^2$ and $Im(\alpha)^2$ have the same units).

As for the last part of the answer I am not sure I understand. People have done cavity QED experiments in the past showing that the lifetime of an atom is changed when inside a cavity. If the cavity is resonant with the atomic transition, the lifetime gets shorter and this is due to the fact that the atom is more likely to emit in the cavity mode compared to the case where no cavity is present. So an ensemble of excited atoms will physically emit more photons in the cavity mode compared to the same mode in the absence of the cavity. The power is simply the number of photons over time (given that each photon carries a fixed amount of energy). So if you count the photons with a photodetector, you would see more when the cavity is present (again this was shown experimentally already). I am not sure what you mean by not being able to measure power. I am personally measuring power on a regular basis in the lab using a powermeter. Also for power measurement, you don't need to integrate over time. It is the energy that is the integral of power over time.