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Measuring Rabi Oscillations in a Two State Atom

  1. Jun 17, 2015 #1
    Can someone give me a simple explanation of how to measure the Rabi probability oscillations in a two state atom?
     
  2. jcsd
  3. Jun 18, 2015 #2

    Cthugha

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    One possible method is deifferential transmission in pump-probe geometry.

    Measure the transmission of some resonant pulse through the two-level system (probe pulse). Now add a second stronger pulse slightly earlier, which also resonantly drives the transition (pump pulse). Now the measured differential transmission (with and without pump) of the probe pulse through the two-level system will show Rabi oscillations if you plot it against pulse area (usually proportional to the intensity) of the pump pulse.
     
  4. Jun 19, 2015 #3
    Thanks Cthugha, you're the first person to reply with an experimental procedure. Unfortunately, I don't completely understand it! By a "stronger pulse" do you mean one at a higher frequency, or do you mean a longer pulse at the resonant frequency? It seems that you are measuring the transmission of the pulses, i.e. how much of the pulse is not absorbed, no? Can you explain exactly how one would measure the transmission or can you refer me to a paper where this is done?
    I'm especially interested in measuring a qubit in quantum computing.
    Thanks, again
     
  5. Jun 19, 2015 #4

    Cthugha

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    By stronger I just mean higher intensity.

    The idea is to measure the transmission of the weaker probe pulse. It is usually delayed a bit or is directed to the system at an angle with respect to the pump pulse, so that there is no overlap with the stronger pump pulse. Fortunately you do not need absolute transmission, so the procedure to measure it is quite easy. Essentially you just put a photo diode there and measure the photocurrent created by the probe beam with and without the pump beam being present. From these two values you can calculate the differential transmission. Now you do that for several pump beam intensities and you are done. Of course you may need higher sensitivity, e.g. if you are measuring on a single qubit. In that case you may want to take homodyne detection or lock-in detection into account additionally, but the basic technique is the same.

    You could for example have a look at the following paper and references therein:
    T.H. Stievater et al., "Rabi Oscillations of Excitons in Single Quantum Dots", Phys. Rev. Lett. 87, 133603 (2001)
     
  6. Jun 20, 2015 #5
    Thanks again Cthugha! Unfortunately, I'm not at all well versed in optical quantum physics, so I don't really know some of your terms. I looked at the paper you mentioned and it was definitely doing what I wanted to understand, but was way over my head in the details!!
    I assume that "higher intensity" means more photons per second. Is there some way you could explain for me (more simply!) in terms of the probabilities,
    Snapshot.jpg of the energy levels of the 2 level atom, how these are affected by the weak pulse and the pump pulse, so that the Rabi probabilities are detected?

    Thanks Again!!
     
  7. Jun 21, 2015 #6

    Cthugha

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    Usually it is easier to consider the mean number of photons per pulse, but that of course also means more photons per second.

    Ok, first consider the case of shooting a single resonant photon at the two level system (TLS). We are interested in the transmission of the system. This will of course depend on the state of the two-level system. Consider just a single run of the experiment first. If the TLS is in the ground state assuming ideal interaction the photon will be absorbed. Zero photons will be transmitted. If the system is in the upper state, the photon cannot be absorbed as there is no transition to higher energies left. However, there is a transition to lower energies resonant with the incoming photon, so stimulated emission will occur. The system will go to its ground state and will simultaneously emit a photon. So in that case, the transmitted number of photons when firing a single photon at the TLS will be two photons. If the TLS is in a superposition of the ground and the excited state, you may get different outcomes when you do the experiment. Either 0 photons or two photons and the relative probability of each of these events taking place is given by the probabilities you mention. This weak is essentially the role of the probe pulse. It is there to check the state of the system.

    The role of the pump pulse is to prepare the state of the system. The situation is the same as described above, but you start from a well defined TLS in the ground state, now you shoot many photons at the TLS - and of course you do not have ideal interaction conditions in reality. However, what will happen is similar to the case above. For low photon numbers, not much will happen. The system is likely to stay in the ground state. For larger photon numbers, you will get a reasonable probability to find the system in the upper state. Adding even more photons, however, will not cause saturation at the upper level, but will cause stimulated emission to kick in. If you check the system now, it is more likely to find it in the ground state again. At some point you have gone full cycle and adding more photons will make it more likely to find the TLS in the upper state again and so on. The idea is that the pump pulse is a strong pulse that prepares the system and afterwards you check the state of the system with a probe pulse that is so weak that it (ideally) does not change the state of the system.
     
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