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- Thread starter BillKet
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I am not sure what you mean. In the spectra, you would get 3 peaks, a main one and 2 secondary (with amplitudes half of the main one), which are stationary (in terms of position and amplitude). This is assuming the laser is on resonance (this is a Mollow triplet). So I am not sure I understand how can you get 3 peaks with only 2 energy levels. As I said, an actual clean derivation/explanation would really help me understand.

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Cthugha

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I assume that you are already familiar with the standard Jaynes-Cummings model, right? If you couple a light field on resonance to a two-level system, you naively get two states: state 1 consists of n photons in the light field and the two-level system being in the excited state. state 2 consists of n+1 photons in the light field and the two-level system being in the ground state. These states are coupled by the light-matter interaction rate. If this rate exceeds the loss rates, you end up in the strong coupling regime and you can diagonalize the Hamiltonian and will end up with dressed states, which both carry partly two-level system character and partly light field character. The lower energy dressed state corresponds to in-phase energy exchange between these two systems and the higher energy dressed state corresponds to out-of-phase energy exchange between these two modes. This might sound esoteric, but the physics is actually not really different from having two coupled spring pendulums and calculating the normal modes of the coupled system from the modes of the simple oscillators.

Now the interesting point is that the interaction term depends on the magnitude of the light field, which is the number of photons present. This interaction term determines the energy splitting between the dressed states. Simply put: The splitting will be larger if the light field switches back and forth between 9 and 10 photons (so the total number of excitation in the light field and in the two-level system is 10) as compared to when it goes back and forth between 0 and 1 photon (so the total number of excitations is 1). It scales as the square root of the number of photons present.

Now the relevant question is what emission from such a system will look like. Any emission will be a loss channel. So imagine that you start with a state with 2 excitations present, a photon gets lost and you end up in a state with one excitation present. The energy splitting between the dressed states is sqrt(2) times some constant for the initial state (as two excitations are present in total) and sqrt(1) times the same constant for the final state (as there is only one excitation present in the final state). Now you have two possible initial states and two possible final states, which gives four possible transitions. However, two of them are at the same energy and will form the main peak, so only three transition energies remain - the Mollow triplet.

The point is that it is not sufficient to consider the two levels of the two-level atom, but you also need to consider the states of the light field.

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Cthugha

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Interestingly, from my limited experience I can say that students often find the explanation of the Mollow triplet in terms of a quantized light field more intuitive as compared to the semiclassical approach, which is most likely due to time-dependent Hamiltonians not being discussed in too much detail in lectures. Maybe this will change again as Floquet physics becomes more and more relevant in current research.

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