Consider a qubit-cavity system governed by the usual Jaynes-Cummings Hamiltonian in the dispersive regime $$ H_0 = \omega_q \frac{\sigma_z}{2} + \omega_a a^\dagger a + \chi a^\dagger a \frac{\sigma_z}{2} $$ where $a$ is the annihilation operator of the cavity, and $\sigma_z$ is the z Pauli operator on the qubit. We drive the system with a time-dependent Hamiltonian $$ H_d(t) = \Omega \mathrm{e}^{i t \omega_q} \sigma_+ + \Omega^* \mathrm{e}^{-i t \omega_q} \sigma_- $$ where $\sigma_\pm$ are the Pauli creation and annihilation operators. For $|\Omega| \ll \chi$, since the drive is resonant with the qubit at frequency $\omega_d = \omega_q - 0 \chi$, the system will see Rabi oscillations for the photonless cavity state $|0\rangle$. In other words, if the initial state of the system is $|\psi(t=0)\rangle = |0,g\rangle$, the system will oscillate between $|0, g\rangle$ and $|0, e\rangle$ such that $$ |\langle 0,g| \psi(t) \rangle|^2 = \cos^2\left(\frac{\Omega t}{2}\right) $$ $$ |\langle 0,e| \psi(t) \rangle|^2 = \sin^2\left(\frac{\Omega t}{2}\right) $$

Question: How can I compute the exact form of $|\psi(t)\rangle$ starting from any initial state $|\psi(t=0)\rangle$ under this time-dependent drive?


1 Answer 1


First point: The undriven version of this Hamiltonian is not at all like the Jaynes-Cummings Hamiltonian

Solutions of the Jaynes-Cummings dynamics are usually based on the observation that the interaction term only couples two states. This is not the case for the Hamiltonian given by the OP, since

$$a^\dagger a \sigma_z |n,g\rangle = a^\dagger a(|e\rangle\langle e|-|g\rangle\langle g|)|n,g\rangle = -a^\dagger a |n,g\rangle = -n |n,g\rangle \,$$


$$a^\dagger a \sigma_z |n,e\rangle = a^\dagger a(|e\rangle\langle e|-|g\rangle\langle g|)|n,e\rangle = +a^\dagger a |n,e\rangle = +n |n,e\rangle \,,$$

where $|n,e/g\rangle = |n\rangle_\mathrm{photon} \otimes |e/g\rangle$, with $|n\rangle$ a photon number Fock state and $|e/g\rangle$ the excited/ground state, respectively, of the qubit.

This Hamiltonian therefore only yields a phase evolution of these states. You can check this by writing down the equation of motion $|\dot{\psi}\rangle \propto iH|\psi\rangle$ for these states.

We can also understand this, since the $a^\dagger a$-term simply counts the photon number and the $\sigma_z$-term measures the excitedness/inversion of the qubit. This term is therefore diagonal in the Fock-state basis.

In the standard Jaynes-Cummings model, we instead have the interaction term $(a^\dagger \sigma_- + a \sigma_+)$. This term couples $|n,e\rangle$ to $|n+1,g\rangle$.

Second point: The time-dependent solution can be found in the same way as for the JC model

In the question, Rabi oscillations are obtained through the driving term, which contains $\sigma_\pm$ and therefore couples the states. You can then simply set up the equations of motion $|\dot{\psi}\rangle \propto iH|\psi\rangle$ in the $\{|n,g\rangle, |n,e\rangle\}$-basis and solve the resulting equations of motion for your initial states. This method is the same as for the Jaynes-Cummings model and can be found in any standard textbook on quantum optics. If the time-dependent drive is a problem, one can go to an interaction picture rotating at the drive frequency.

  • $\begingroup$ Thank you for your answer. Concerning your first point, the Hamiltonian I mention is a Jaynes-Cummings, but in the dispersive regime (see for instance equations (3.1) to (3.3) in arxiv.org/abs/0810.1336 for the derivation). $\endgroup$
    – Ronan
    Feb 2, 2021 at 14:18
  • $\begingroup$ Concerning your second point, I have a hard time actually solving in the $|n,g\rangle, |n,e\rangle$ basis, even by going in the frame rotating at $\omega_q$. The driving term $H_d$ couples both the ground and excited states, which actually do not rotate at the same frequency due to the dispersive term in $\chi a^\dagger a \sigma_z / 2$. Do you have any idea on how to deal with this? Many thanks again $\endgroup$
    – Ronan
    Feb 2, 2021 at 14:20
  • $\begingroup$ @Ronan ah I see! In that case, could you include the resulting equations of motion in your question to pinpoint where exactly the problem lies? This would be helpful, since for the standard (non-dispersive) JC model, the solution is rather simple. So it would be good to show why you're stuck for this Hamiltonian and not for the JC model. $\endgroup$ Feb 2, 2021 at 15:15

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