- #1
Malamala
- 309
- 27
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!
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!