Rabi oscillation in the Jaynes-Cummings model

Question

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 $|0rangle$. In other words, if the initial state of the system is $|psi(t=0)rangle = |0,grangle$, the system will oscillate between $|0, grangle$ and $|0, erangle$ such that
$$
|langle 0,g| psi(t) rangle|^2 = cos^2left(frac{Omega t}{2}right)
$$
$$
|langle 0,e| psi(t) rangle|^2 = sin^2left(frac{Omega t}{2}right)
$$
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?

Wolpertinger · Answer

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
$$a^dagger a sigma_z |n,grangle = a^dagger a(|eranglelangle e|-|granglelangle g|)|n,grangle = -a^dagger a |n,grangle = -n |n,grangle ,$$
and
$$a^dagger a sigma_z |n,erangle = a^dagger a(|eranglelangle e|-|granglelangle g|)|n,erangle = +a^dagger a |n,erangle = +n |n,erangle ,,$$
where $|n,e/grangle = |nrangle_mathrm{photon} otimes |e/grangle$, with $|nrangle$ a photon number Fock state and $|e/grangle$ 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 $|dot{psi}rangle propto iH|psirangle$ 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 $|n,erangle$ to $|n+1,grangle$.
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 $|dot{psi}rangle propto iH|psirangle$ in the ${|n,grangle, |n,erangle}$-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.

Rabi oscillation in the Jaynes-Cummings model

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