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Two atoms in an optical cavity.

  1. Jun 6, 2014 #1

    I have a question about 2 (or more) atoms coupled to an optical cavity.
    The equations of motion in the interaction picture for 2 atoms in resonnance with an optical cavity starting from 1 excited atom (C_{eg,0}(t_0) = 1 )
    (g is the coupling constant)

    \dot{C}_{eg,0} &= - ig e^{i(\vec{k}\cdot \vec{r}_1)}C_{gg,1}\\
    \dot{C}_{ge,0} &= - ig e^{i(\vec{k}\cdot \vec{r}_2)}C_{gg,1}\\

    \dot{C}_{gg,1} &= -ig e^{-i(\vec{k}\cdot \vec{r}_1)}C_{eg,0} - ig e^{-i(\vec{k}\cdot \vec{r}_2)}C_{ge,0}

    Coming from an interaction picture Hamiltonian:
    V_2 = \left[
    0 & 0 & g \\
    0 &0 & g \\
    g & g & 0 \\

    But if i put this system in a damped cavity, something weird happens.
    Instead of the system just dying down, the probability of the atoms being excited converges to 1/4.
    The problem is that C_{eg0} and C_{ge0} have a phase difference of π and (C_{eg0}+C_{ge0}) \rightarrow 0 and therefore \dot{C}_{gg,1} \rightarrow 0 and the system stalls.

    This is the evolution, starting from a state |eg0> (meaning first atom excited, second ground and 0 photons in the cavity) Pruduced with a standard Master equation.

    As you can see, the purple line presents the state with the first atom being excited, and the yellow line the second. (the blue line is the ground state which increases as the photons leave the cavity)

    For the rest, the Master equation behaves exactly as it should, a photon without atoms decays exponentially with the correct decay constant.
    A closed cavity produces the correct Tavis-Cummings evolution,... everything else works just fine.

    The atoms are coupled to a damped cavity so, the atoms should de-excite completely just like they do if I put them in the cavity alone. I don't see a physical reason why and how two atoms could remain somewhat excited.

    Can it be that there is something wrong with the Hamiltonian? Although this Hamiltonian produces the correct evolution in all cases if there is no damping.

    I must be overlooking something, but what?

    Thank you for any insight.
    Last edited: Jun 6, 2014
  2. jcsd
  3. Jun 6, 2014 #2


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    ##(C_{eg0}+C_{ge0}) \rightarrow 0##, but ##(C_{eg0}-C_{ge0}) \rightarrow (C_{eg0}-C_{ge0})##, right? Or at least there is some stable linear combination of both.

    How did you include your damping?
  4. Jun 7, 2014 #3

    [itex]C_{eg0}[/itex] has always the opposite sign of [itex]C_{ge0}[/itex] (because of the two (-i) factors)
    And indeed, the system has an eigenstate [itex] [1/\sqrt{2}, -1/\sqrt{2},0] [/itex] so the system just damps to this eigen state.

    For example in the case of 1 atom, this "stationary" state does not exist so, in that case there is no problem.

    And the damping is included with 3 different methods.
    A standard Master equation. So, a general Lindbladian term added to the density operator equation of motion.
    This master equation should be correct since it describes for example the exponential damping of a photon in a cavity without atoms.

    And, with a quantum-jump Monte Carlo method.
    In every timestep, I predict the probability for a photon to be emitted, and apply the quantum Jump depending on a generated random number compared to that probability.
    Averaged over about a 1000 evolutions, the system behaves exactly the same as the Master equation.

    The effective non-hermitian hamiltonian is:
    [tex] H_{sys} - i\hbar \frac{\gamma}{2} a^{\dagger}a [/tex]

    As you can see [itex]a^{\dagger}a [/itex] is just a counting operator, which is not 0 only for the photon state. So, it effectively is just a damping of the cavity photon.

    And then there is a third method, (a dirty complicated method I won't explain it here, but it uses that effective hamiltonian and behaves exactly similar)

    All these methods work perfectly, as long as [itex]C_{ge0}[/itex] has the SAME sign as [itex]C_{eg0}[/itex] (for example when the initial condition is [itex]C_{gg0} = 1[/itex])
  5. Jun 7, 2014 #4


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    So where is the problem?
    You have a physical model where one part gets "damped away" (because it has a contribution with a photon in the cavity), the other part is stable as a superposition of the two excitations and does not decay.
  6. Jun 7, 2014 #5
    Thank you for helping out!

    The problem is that this result is unphysical.

    The cavity is damped, so it's spectrum is broadened. Therefore, the atoms should completely decay to the ground state. Just as described by Weisskopf-Wigner theory And this happens in the case of 1 atom, but not here, with two atoms.
    And the only reason is the different sign of those coefficients.

    I seem to be missing a piece of the dynamics, but I have no clue what.
    The same Hamiltonian of atoms coupling to a cavity is used everywhere in literature, but then they always add the damping of the atoms to the vacuum directly.

    But then again, I don't see why something is missing since in a closed system, all my simulations match perfectly with equations I've encountered in the literature.

    I'm a bit worried, although I'm quite sure that all the results I use are good. (i'm mainly interested a specific case that does behave correctly (with C_eg0 and C_ge0 the same sign))
    It will become very hard to prove that they actually are good if my simulations also produce this freak case.
    Last edited: Jun 7, 2014
  7. Jun 7, 2014 #6


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    I think your model in unphysical in the same way.

    You have a state that does not decay to the ground state at all. It does not matter if the cavity is damped if there is never a photon inside.

    I guess that is the important difference. This allows the (otherwise) stable excited state to decay as well.
  8. Jun 26, 2014 #7

    I just want to say that indeed it was correct after all.
    The system gets "entangled by dissipation".

    I thought it was something unphysical, but it makes more sense now.

    Thanks for the replies!
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