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Hello! I thought that in spontaneous emission (say for an atom with 2 energy levels) we have the electron in the excited state and then it decays to the ground state emitting a photon at the resonance frequency. However I saw the attached figure, which introduces Mollow triplet. I understand the math of it, and how you go from normal energy levels to dressed states (within Jaynes-Cummings model), but I am not sure I understand physically what is going on. If we start in an atomic excited state, we are in a linear combination of dressed states. The atom can spontaneously decay in one of the 4 ways shown in the figure. But this means that after the atom decays (and we can measure the emitted photon), we still have 50% chances to find it in an excited atomic state (it will decay to a dressed state, which on resonance is an equal linear combination of ground and excited atomic states). I am not sure I understand this. How can we still have a chance of finding the atom in the excited state, even after a decay?

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Cthugha
Consider a single atom in a cavity of very high, but not unlimited quality factor. If the atom decays and emits a photon into the cavity mode, it becomes trapped inside the cavity. Then it may excite the atom again, which may in turn decay again by emitting a photon into the cavity mode, which may in turn.... and so on. This is vacuum Rabi oscillations in a nutshell. You may have loss, either due to a photon leaving the cavity or the atom emitting into a non-cavity mode. In that case, the probability to find the atom in the excited state afterwards will be 0.

Now consider again the same atom emitting into a cavity mode which contains already 10000 photons. The atom may decay, so 10001 photons will be present in the cavity. The 10001 photons may excite the atom again, which may reemit and so on. Due to the presence of way more photons, the excitation and emission cycle will of course speed up. If you now have 10001 photons inside the cavity and 1 gets lost, you still have 10000 photons inside, which may drive the atom to the excited state and you can certainly find it in the excited state.

In a typical Mollow triplet experiment, you do not necessarily have a cavity (exactly to avoid these losses), but drive the atom with a laser of constant intensity, so that a constant flux of photons arrives at the atom in a certain mode. Here, the equivalent of the photon loss channel is inelastic scattering off the atom.

Consider a single atom in a cavity of very high, but not unlimited quality factor. If the atom decays and emits a photon into the cavity mode, it becomes trapped inside the cavity. Then it may excite the atom again, which may in turn decay again by emitting a photon into the cavity mode, which may in turn.... and so on. This is vacuum Rabi oscillations in a nutshell. You may have loss, either due to a photon leaving the cavity or the atom emitting into a non-cavity mode. In that case, the probability to find the atom in the excited state afterwards will be 0.

Now consider again the same atom emitting into a cavity mode which contains already 10000 photons. The atom may decay, so 10001 photons will be present in the cavity. The 10001 photons may excite the atom again, which may reemit and so on. Due to the presence of way more photons, the excitation and emission cycle will of course speed up. If you now have 10001 photons inside the cavity and 1 gets lost, you still have 10000 photons inside, which may drive the atom to the excited state and you can certainly find it in the excited state.

In a typical Mollow triplet experiment, you do not necessarily have a cavity (exactly to avoid these losses), but drive the atom with a laser of constant intensity, so that a constant flux of photons arrives at the atom in a certain mode. Here, the equivalent of the photon loss channel is inelastic scattering off the atom.
Thank you for your reply! I understand your explanation of Rabi oscillations. The higher the number of photons, the higher the power and hence the Rabi frequency and the higher the frequency of going back and forth between the ground and excited state. In that case the energy contained in the atom + field is conserved. Based on the figure I attached, if we start in ##|g,n+1>## we will keep going between ##|g,n+1>## and ##|e,n>## indefinitely (assuming we had no loses, such as spontaneous decay). However I am still confused about the situation with losses. Assume we start in ##|e,n> = \frac{1}{\sqrt{2}}(|+,n>-|-,n>)##. The only place we can end up in, after losing one photon from the cavity is ##|+,n-1>## or ##|-,n-1>##. Assume we end up in ##|+,n-1>##. But ##|+,n-1>=\frac{1}{\sqrt{2}}(##|g,n>## -##|e,n-1>## )##, so we still have ##50%## chances of being in the excited state. What am I missing?

Cthugha
I am not sure I get your problem. Let me rephrase it. As you correctly point out, in a strongly coupled system the total number of excitations in the atom and in the field is the relevant quantity of interest. Now assume that you go from a total of 10000 excitations to 9999 excitations being present. In most cases this corresponds to a photon leaking out of the cavity. Why should a photon that leaks out of the cavity put the atom into its ground state?

The spontaneous emission indeed comes from the coupled system, not from the bare atom. In the strong driving regime, where the Mollow triplet occurs, the decay of the coupled system towards a lower number of excitations does not correspond to the atom going from the excited to the ground state at all. It really is the coupled system that "emits" a photon, not the bare atom. Actually, as most of the photons in the system are brought in by the driving laser, it is often better to visualize the resonance fluorescence as laser light scattered elastically (for the central peak) or inelastically (for the sidebands) off the atom. The emission really corresponds to the loss of one excitation from the system and it does not matter whether this excitation was initially brought into the system via the excited state of the atom or as one of the 10000 photons in the laser beam interacting with the atom.

I am not sure I get your problem. Let me rephrase it. As you correctly point out, in a strongly coupled system the total number of excitations in the atom and in the field is the relevant quantity of interest. Now assume that you go from a total of 10000 excitations to 9999 excitations being present. In most cases this corresponds to a photon leaking out of the cavity. Why should a photon that leaks out of the cavity put the atom into its ground state?

The spontaneous emission indeed comes from the coupled system, not from the bare atom. In the strong driving regime, where the Mollow triplet occurs, the decay of the coupled system towards a lower number of excitations does not correspond to the atom going from the excited to the ground state at all. It really is the coupled system that "emits" a photon, not the bare atom. Actually, as most of the photons in the system are brought in by the driving laser, it is often better to visualize the resonance fluorescence as laser light scattered elastically (for the central peak) or inelastically (for the sidebands) off the atom. The emission really corresponds to the loss of one excitation from the system and it does not matter whether this excitation was initially brought into the system via the excited state of the atom or as one of the 10000 photons in the laser beam interacting with the atom.
Oh, so you mean that if a photon comes out of the system, as in the case of Mollow triplet, I should think of it as coming out of the coupled atom-field system. So going down in energy, I have to move along the dressed energy levels, not along the atomic energy levels. And the idea of spontaneous emission with the atom going from the excited state to the ground state would work only if I have an isolated atom initially in the excited state, without any EM field interacting with it?

One more question. Going back to my example in the previous post, assuming that after the system emits a photon we end up in, say, ##|+,n-1>## and we turn off the EM right after that emission. If we were to measure the state of the atom (basically measuring several separate atoms prepared in this way), will we find the atom 50% of the time in the ground state and 50% in the excited state? So basically, turning off the field after I measured a scattered photon, is equivalent to applying a ##\pi/2## pulse to the atom (i.e. it places it in a linear combination of ground and excited state)?

hutchphd
Homework Helper
I believe an alternate way to consider this is that the coupling is to a very high temperature bath. High temperature means that the excited state and the ground state are almost equally populated. Turning off the interaction just maintains this population.

Cthugha
Oh, so you mean that if a photon comes out of the system, as in the case of Mollow triplet, I should think of it as coming out of the coupled atom-field system. So going down in energy, I have to move along the dressed energy levels, not along the atomic energy levels.

Yes, indeed.

And the idea of spontaneous emission with the atom going from the excited state to the ground state would work only if I have an isolated atom initially in the excited state, without any EM field interacting with it?

Well, of course there are not only the cases of no EM field and a dominant EM field, but in between you get some transition. If you use a rather weak laser beam, you end up in the so called Heitler regime, where the resonance fluorescence consists of a single peak which is as narrow as the laser line. This light shows antibunching, so the atomic character is quite dominant in this case. Let me put it this way: if the mean number of excitations is 10000 and the atom contributes 1 excitation to this, then the picture you mention is not a very reasonable one.

One more question. Going back to my example in the previous post, assuming that after the system emits a photon we end up in, say, ##|+,n-1>## and we turn off the EM right after that emission. If we were to measure the state of the atom (basically measuring several separate atoms prepared in this way), will we find the atom 50% of the time in the ground state and 50% in the excited state? So basically, turning off the field after I measured a scattered photon, is equivalent to applying a ##\pi/2## pulse to the atom (i.e. it places it in a linear combination of ground and excited state)?

This is nontrivial as it depends on a lot of things. First, the state you end up depends on the photon that is emitted. If it is emitted from the low energy sideband, you know that you are in the + state. If it is emitted from the high energy sideband, you know that you are in the - state. If it is emitted from the central peak, both states are possible.

Anyway, assuming that the initial preparation of the atom has taken place so long ago that the memory about the initial state is not present anymore (which means that on average a bit more than one loss process has indeed taken place for the excited state of the atom), then even in the steady state, the you will find the system in the excited or ground state with 50% probability. If a photon is emitted, this just does not change (assuming that we are in the limit of a large number of excitations). If you think about the comment by @hutchphd , you will find that the analogy to the high temperature bath is a good one. For that reason, lossy quantum optics systems are usually treated as open systems coupled to baths.