The system of interest is a two level system placed inside a cavity. This can be an atom and a three dimensional cavity, a superconducting qubit and a coplanar waveguide resonator, but the details do not matter. We will simply assume that we have a two level atom coupled to a cavity and write the Hamiltonian as \begin{equation} \mathcal{H/\hbar} = \frac{\omega_a}{2} \hat{\sigma_z}+ \omega_r \hat{ a}^\dagger \hat{a} + g\left(\hat\sigma^+\hat{a} + \hat a \hat\sigma^- \right) \end{equation} which is just the Jaynes-Cummings Hamiltonian for a two level system coupled to a cavity.

In 1946 E. Purcell derived that when such a two level system is placed inside the cavity (as we have done above), it has an altered spontaneous emission rate $\Gamma_{1a}$, where the $a$ is used to denote that we are talking about the two level system[1]. It is known as the Purcell effect, and it has been observed in a few different systems such as electrical circuits ([2], arXiv at [3]), and most interestingly for me also in transmon qubits coupled to coplanar waveguide resonators ([4], arXiv at [5]).

What I found in literature [2] is that there are two main regimes for this: the first is that the cavity is resonant with the atom, in which case they hybridize into two states with $\Gamma_{1\{r,a\}}^{'} = \frac{\Gamma_{1r}+\Gamma_{1a}}{2}$ where I used the prime to differentiate the bare loss rates and the effective loss rates due to the interaction of the two systems. The second scenario is the so called dispersive regime, in which $\Delta^2 = \left|\omega_r-\omega_a \right|^2 \gg g^2$, for which one has that \begin{equation} \Gamma_{1a}^{'} \approx [1-\left(\frac{g}{\Delta}\right)^2]\Gamma_{1a} + \left(\frac{g}{\Delta}\right)^2\Gamma_{1r} \end{equation}

My question is now how one can find an expression for $\Gamma_{1a}^{'}$ in the intermediate regime; between the dispersive part and the fully hybridized part. The paper by Koch et al. [4] writes something about this in section IV B, noting that one can use Fermi's golden rule, but I don't feel as if the way it is phrased there can be used for a full expression. Similarly [2] shows a theoretical curve in figure 2b, but contains no reference to how it is computed.

So that is what my question comes down to; given the situation sketched above, knowing all of the parameters defined, how do I compute $\Gamma_{1a}^{'}(\Delta)$ for arbitrary detuning $\Delta$? I understand that one might do this numerically rather than with a nice analytic equation; that is fine by me.

Finally, if it helps I can give some orders of magnitude for the quantities in question. Let us take $\Delta = [0 - 400]$ MHz/2$\pi$, $g = 80$ MHz/2$\pi$, $\Gamma_{1a} = 60$ MHz/2$\pi$ and $\Gamma_{1r} = 1$ MHz/2$\pi$. It's just an estimate of the scenario I am interested in, it shouldn't matter for the problem.


1 Answer 1


I believe the equation you are looking for is

$$\Gamma = \frac{\kappa}{2} - \frac{\sqrt{2}}{2} \sqrt{-A + \sqrt{A^2 + (\kappa \Delta)^2}}$$


$$A \equiv \Delta^2 + 4 g^2 - \frac{\kappa^2}{4}$$

and $\kappa$ is the resonator decay rate, $\Delta$ is the qubit-resonator detuning, and $g$ is the qubit-resonator coupling. This equation was taken from Eqs. (10) and (12) from Ref. [a]. That equation was derived for the case of a single excitation in the system. At larger excitation number, the result changes, interestingly decreasing the decay rate. However, at sufficiently large number of resonator photons, roughly $n \gtrsim (\Delta/g)^2/4$, the Jaynes-Cumming model fails to accurately describe the system: the two level approximation of the qubit and the rotating wave approximation for the couplng fail in a way that allow the qubit to transition upward to levels $\left \lvert 3 \right\rangle$ and above$^{[b]}$.

[a]: Eyob Sete et al., 2014. Purcell effect with microwave drive: Suppression of qubit relaxation rate

[b]: Daniel Sank et al., 2016. Measurement-induced state transitions in a superconducting qubit: Beyond the rotating wave approximation

  • $\begingroup$ Very nice indeed. It is interesting to see that the loss rate goes down at larger excitation numbers, I wonder what the intuitive picture is. Perhaps the resonator is so occupied that it is less favorable to add yet another excitation, originating from the qubit. Although, it is a linear system, so I do not see why. In any case, I am interested in the single excitation case, which your answer (and the paper) do answer. Great! $\endgroup$
    – user129412
    Aug 29, 2016 at 11:46
  • $\begingroup$ Glad this helped. I mentioned the effect when $n \gtrsim (\Delta/g)^2$ because the failure of dispersive readout at large $n$ was a mystery in our field for quite some time and is surrounded by misinformation. For a long time, everyone thought that the loss of QNDness at large $n$ came from the failure of the dispersive approximation. It turns out the dispersive approximation has nothing to do with it; it's the failure of the RWA that causes the qubit state to go nuts. In fact, if you stay in the RWA, you can go to really big $n$ and nothing interesting happens to the qubit. $\endgroup$
    – DanielSank
    Aug 29, 2016 at 16:25

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