I considered light-atom interaction in nonlinear cavity case. This situation is described by Buck-Sukumar model: $$ H = {\omega}_0 \sigma_z + {\omega}(a^\dagger a) + g \left(\sqrt{(a^\dagger a)}a^\dagger \sigma^- + \sigma^+ a \sqrt{(a^\dagger a)}\right) $$ This is special case of Jaynes-Cummings Hamiltonian. First of all I decided to use interaction picture: $$ H_0 = {\omega}_0 \sigma_z + {\omega}(a^\dagger a) $$ $$ H_{int}= g \left(\sqrt{(a^\dagger a)}a^\dagger \sigma^- e^{i\delta t} + \sigma^+ a \sqrt{(a^\dagger a)}e^{-i\delta t}\right) $$ where $\delta = 2\omega_0 - \omega$. I used potating wave approximation to obtain this form. Now I need to deal with wavefunction. I thinnk it is outer product of wavefunction of atom and field: $$ |\Psi(t)\rangle = c_g(t)|g,\alpha\rangle + c_e(t)|e,\alpha\rangle $$ I considered that initial state of a system: TLS in ground state and photon in a coherent state. And at this point I got stuck. I need to turn my wavefunction into Interaction picture and it seems like too complicated to act on my w-f with $exp(iH_0t)$ because using BCH formula taking into account all of the terms I get irreducible sum. After that plugging this expression into time-dependent Schrödinger equation should give me a system of two differential equations for coefficients. Could someone help me please and advice what to do with this appearing infinite summation? All in all after obtaining equations for coefficients (I think I will get some sines and cosines with Rabi frequency) my aim will be to find inverse population and discuss its evolution at the timescale of Rabi oscillations (seems like it should be something like $exp(-\alpha^2)cos(2\Omega_Rt)$), but I am not sure because in this case I was expecting to see collapse and revival somehow. Thanks in advance!
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$\begingroup$ Have you tried numerical integration? Have you tried looking for eigenstates of the interaction Hamiltonian? Do you know what happens when $\alpha$ is large and you can approximately make the semiclassical replacement $a\to\alpha$ in the Hamiltonian? $\endgroup$– Quantum MechanicCommented May 15, 2023 at 21:29
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$\begingroup$ 1) I want to obtain result analytically because it was made in a bit different way but still without numerical integrations. 2) Yes, but it doesn't lead to obtaining coefficients, may be I made a mistake. $\endgroup$– AndrewCommented May 15, 2023 at 21:36
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$\begingroup$ 3) Semiclassical approach in my opinion will not work here because in this case we will observe only action of field on our two-level system, while the Hamiltonian (which is written in quantum approach) describes two sided interaction: Field on a TLS and vise versa. $\endgroup$– AndrewCommented May 15, 2023 at 21:42
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$\begingroup$ Ok, I was just throwing out ideas. Another option is to try an expansion like $a\approx \alpha+\delta a$ for c-number $\alpha$ and bosonic operator $\delta a$, then keep only the lowest order terms in $\delta a$ and $\delta a^\dagger$ in the Hamiltonian, that would linearize the problem $\endgroup$– Quantum MechanicCommented May 15, 2023 at 21:54
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The Hamiltonian commutes with total number of excitations defined as $\hat{n} = \sigma_z + \hat{a}^\dagger \hat{a}$. Therefore, the eigenfunctions can be found using an ansatz \begin{equation} |\psi\rangle = c_1 |g, n+1\rangle + c_2 |e, n\rangle, \end{equation} where $|g, n\rangle = |g\rangle\otimes |n\rangle$, and $|n\rangle$ are the Fock states. After plugging the ansatz into the Schroedinger equation $\hat{H}|\psi\rangle = E|\psi\rangle$, you will get a $2\times 2$ eigenvalue problem. Then, you can use the set of eigenfunctions to solve time-dependent Schroednger equation with any initial condition.
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$\begingroup$ Still there is a question how should I solve eigenvalue problem? In my understanding it is not analytically solvable because action of square root of operators on your ansatz is non trivial to me. I can use Tailor series expansion, but for this I should assume that operator under root is small, which is not reasoned enough. $\endgroup$– AndrewCommented May 16, 2023 at 11:56
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$\begingroup$ @Andrew you can use the definition $f(a^\dagger a)|n\rangle= f(n)|n\rangle$; that's what a function of an operator is defined as (easy to see with analytic operators) $\endgroup$ Commented May 16, 2023 at 12:21
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$\begingroup$ Yes, I am silly, solved problem. Thanks for your support. $\endgroup$– AndrewCommented May 16, 2023 at 12:31