I'm trying to figure out what the driving term in the Hamiltonian is when you drive an emitter (say a transmon qubit) through a coupled cavity (such as a CPW resonator).

For this I am considering a typical Jaynes-Cummings Hamiltonian of the type \begin{equation} H = \omega_r a^\dagger a + \omega_q \sigma^\dagger\sigma +i g (a^\dagger\sigma - \sigma^\dagger a) \end{equation} where $\omega_q, \omega_r$ are the natural frequencies of the emitter mode $\sigma$ and the cavity mode $a$ respectively and $g$ represents their coupling strength.

Now, if one inputs a coherent state into the cavity at some frequency $\omega_d$ you add a term to the Hamiltonian of the type

\begin{equation} H(t) = \epsilon_\text{drive} \, \left( e^{i\omega_d t} a + e^{-i\omega_d t}a^{\dagger} \right) \end{equation}

I imagine that this should result in an effective drive term acting on the emitter $\Omega \sigma_x$ but I have not been able to find a paper that discusses this. What I'm mainly interested in is how $\Omega$, the effective drive, depends on the parameters $g$, $\Delta_{rq} = \vert \omega_q - \omega_r \vert$, $\Delta_{dr} = \vert \omega_d - \omega_r \vert$, and $\Delta_{dq} = \vert \omega_d - \omega_q \vert$. Perhaps what should also be included is the cavity decay rate $\kappa$ as this sets the linewidth of the cavity.

Because I can imagine two scenarios: either you build up photons inside the cavity (for which you have to be close to resonance i.e. $\Delta_{dr}$ is small), or you directly drive the emitter 'through' the filter function of the resonator (if $\Delta_{rq}$ is large and $\Delta_{dq}$ is small).

In my quest to figuring this out, I first looked at Input-output theory for waveguide QED with an ensemble of inhomogeneous atoms by Lalumiere et al. Here they consider a slightly different system; a transmon coupled to a waveguide. Here equations 9 and 10 give an effective $\Omega$ (called $d$ in the paper) in terms of the coupling term $g$. But since the transmission line has no frequency dependence the expression is not exactly what I am looking for.

  • $\begingroup$ Interesting question! I can write an answer for it, just one clarifiication: are you intersted in weak or strong emitter-cavity coupling? Or both? $\endgroup$ – Wolpertinger Jun 10 '20 at 8:03
  • $\begingroup$ I'm interested in the situation where the coupling constant of cavity and emitter modes exceeds their decay rates, so that would be strong coupling, right? $\endgroup$ – user129412 Jun 10 '20 at 9:13
  • $\begingroup$ Thanks for the clarification and yes to the question! $\endgroup$ – Wolpertinger Jun 10 '20 at 9:55

Disclaimer: I wrote this answer a bit hastily, but will try to supplement it with some references in the future. I hope it is useful nevertheless.

Some notes on the question:

  1. There is a standard way to derive such an effective driving term for the emitter that can be applied to the Jaynes-Cummings Hamiltonian with a driven cavity mode that the OP is considering. It is know as adiabatic elimination (there is a lot of material on this in standard quantum optics books and in the form of papers). The idea is to simply set $\dot{a}(t) = 0$ and simplify the operator equations accordingly. The resulting equations can be expressed in terms of an effective emitter-only Hamiltonian which includes such a direct driving term.

    However, it is important to note that the underlying assumption there is that the cavity mode is a fast degree of freedom, that is it dissipates quickly on the time scale of the interaction. So we require $\kappa\gg g$, which is the weak coupling regime. In addition, the driving pulse has to be long compared to $1/\kappa$.

  2. At strong coupling, the situation is not so simple and I will argue that in general one needs to perform approximations to do what the OP asks. To see this, let us look at the operator equations of motion for the Hamiltonian (only considering the interaction and driving term for simplicity and probably getting minus signs wrong):

    $$\frac{d}{dt} a = g \sigma + i \epsilon_\mathrm{drive} e^{-i\omega_d t}\,,$$ $$\frac{d}{dt} \sigma = g \sigma^z a \,,$$ $$\frac{d}{dt} \sigma^z = 2g \sigma^\dagger a + 2g \sigma a^\dagger \,.$$

    If a loss-term $\kappa$ is wanted it can easily be added. My point is that we already know the driving term, for the atom, it is $ga(t)$. In order to know its functional form, we have to solve for the cavity dynamics $a(t)$. But the equation for $a(t)$ depends on $\sigma$, such that we have to solve the entire coupled equation system. In many cases approximations can be performed to make this possible. However, in general, an analytic solution of these equations is not known. The semi-classical version can be tackled in the particular case given above, but with decay that gets difficult already as well.

  3. In the paper linked by the OP, the situation is rather different. The emitter is directly coupled to a transmission line/waveguide, which is not driven via an external interaction, but by populating the wave guide modes with a wave packet far from the atom. This allows the input-output formalism that is developed in the paper to be applied.

  • $\begingroup$ Thanks for this! Would the situation change if we consider an emitter that is in the dispersive regime ($\Delta_{rq}\gg g$) and that we also drive at the emitter frequency far away from the cavity resonance? In that case I would imagine that there is perhaps more of a link to the waveguide type situation, with the resonator acting as some sort of filter, and the drive ending up almost directly on the emitter. Is that a situation that can be studied more easily? $\endgroup$ – user129412 Jun 11 '20 at 7:29
  • 1
    $\begingroup$ @user129412 Quite possible! I'm not that familiar with common approximations in the dispersive regime, but I can imagine that one might be able to simplifiy the situation in that case. As far as I know one can make some approximate unitary transformations in that regime, so maybe one does not even need an adiabatic elimination then to get an effective driving term. I'm afraid I don't have a complete answer to that question though. $\endgroup$ – Wolpertinger Jun 11 '20 at 8:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.