Dispersive regime and resonator frequency pull

Question

I have come across terms such as "dispersive regime" and "dispersive measurement" often.

To understand this, I have been reading this, this, and this.

In the first paper, the gist I got is the following:

• In the dispersive regime, the qubit is detuned from the resonator by an amount $\Delta$ and thus induces a shift $g^2/\Delta$ in the resonance frequency (I believe the shift is happening to the resonator's resonance frequency).

In the second paper, what I got out of is that dispersive regime is a perturbative approximation of the Jaynes-Cummings Hamiltonian. Also, I got the following:

• In the limit that detuning between the cavity and the qubit is large, no energy is exchanged. In this situation, the interaction is said to be dispersive.

To improve my understanding, Fig 5 in the third paper was useful as it shows the resonator's frequency shift in either direction, dependent on the state of the qubit.

My question centers around the equation in the second paper (shown below). It seems you can understand the effect of $$2g \lambda \left( a^+ a +\frac{1}{2} \right) \frac{\sigma_z}{2}$$ in two ways, as stated below the equation. For this though, are we ignoring $\frac{1}{2}$ term after $a^+ a$ term?

Additionally, I would appreciate it if someone could provide a more holistic picture of the so-called dispersive regime. I feel that I haven't internalized it yet.

to second order in $\lambda$, it is simple to obtain the effective Hamiltonian describing the dispersive regime \begin{align} H_\mathrm{eff} & = \mathbf D_\mathrm{Linear}^\dagger H_s\mathbf D_\mathrm{Linear} \\ & = \hbar \omega_r a^\dagger a + \hbar \left( \omega_a + 2g\lambda \left[a^\dagger a+\frac12\right]\right) + \mathcal O(\lambda^2). \tag{3.3} \end{align}

The qubit transition frequency is shifted by a quantity proportional to the photon population $2g\lambda\langle a^\dagger a\rangle$. Alternatively, this shift can be seen as a qubit dependent pull of the resonator frequency $\omega_r \to \omega_r \pm g\lambda$. As a result, shinning microwaves at the input port of the resonator at a frequency close to $\omega_r$ and measuring the transmitted signal using standard homodyne

Answer 1

You are correct that the dispersive approximation is just a perturbative expansion of the Jaynes-Cummings Hamiltonian in the parameter $g/\Delta$ following unitary transformation. Instead of focusing on the mathematics, I'd like to give you an intuitive feel for what these terms mean. Let us rewrite our effective Hamiltonian as $$ \begin{align}H_{\text{eff}}/\hbar &= \omega_ra^\dagger a+\omega_{a}\frac{\sigma_z}{2}+\frac{2g^2}{\Delta}(a^\dagger a)(\frac{\sigma_z}{2}) \\&= (\omega_r+\frac{2g^2}{\Delta}\frac{\sigma_z}{2})a^\dagger +\omega_{a}\frac{\sigma_z}{2} \\&=\omega_ra^\dagger a +(\omega_a+\frac{2g^2}{\Delta}a^\dagger a)\frac{\sigma_z}{2}\end{align}$$ where we have neglected the $\frac{1}{2}$ term that you mentioned above (note that the $\frac{1}{2}$ term is actually the ac Stark shift of the atom due to vacuum fluctuations in the resonator). Let us define $$\omega_{r}'=(\omega_r+\frac{2g^2}{\Delta}\frac{\sigma_z}{2})$$ and $$\omega_a'=(\omega_a+\frac{2g^2}{\Delta}a^\dagger a).$$ Notice that our primed 'resonance frequencies' contain operators. Now our Hamiltonian reads $$H_{\text{eff}}/\hbar = \omega_r' a^\dagger a + +\omega_{a}\frac{\sigma_z}{2}$$ when written in terms of $\omega_r'$, or alternatively $$H_{\text{eff}}/\hbar = \omega_ra^\dagger a + \omega_a'\frac{\sigma_z}{2}$$ when written in terms of $\omega_a'$.

Let's look at our effective Hamiltonian from two different perspectives. First, assume we know the state of the qubit (atom) with exquisite accuracy. We can then evaluate $\omega_r'$ and find that the resonator frequency depends on the qubit state (since $\sigma_z$ gives $\pm1$ depending on whether we are in the state $|0\rangle$ or $|1\rangle$). Now if we drive our resonator with photons in the vicinity of $\omega_r'$, we would find that the accumulated phase shift of these drive photons depends on the precise value of $\omega_r'$ (and thus the state of the qubit). One might say that the drive photons are 'dispersed' in a way that depends on the state of the qubit. Now let's switch things up and suppose instead that we know the photon number in the resonator very well. In that case, I would claim that we could evaluate $\omega_a'$ (which depends on $a^\dagger a$ and thus the photon number) and would find that the qubit frequency had shifted due to the presence of photons in the resonator.

At the end of the day, these are just two different ways of looking at the same term. For more details on dispersive measurement in superconducting qubits, I encourage you to read the first three chapters of Daniel Sank's PhD thesis.