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Edited after comments.

I can't seem to find (or derive) the exact solution of the two quantum state system absorption problem in the following formulation. Let's consider the general harmonic wave absorption:

$$i \hbar \partial_t \Psi(\mathbf{r},t)=\left(\hat{H_0}(\mathbf{r})+\hat{H}(\mathbf{r}) e^{-\gamma |t|} \cos \omega t \right)\Psi(\mathbf{r},t) \tag{1}$$

Where the optical pulse has the exponential decay both for positive and negative time (it's the model I decided to use for simplicity).

Using the eigenstates of $\hat{H_0}$, we expand the wavefunction:

$$\Psi(\mathbf{r},t)=\sum_n C_n(t) \psi_n(\mathbf{r}) e^{-i \omega_n t} \tag{2}$$

Using orthonormality ($H_{mn}$ are matrix elements of the harmonic wave amplitude $\hat{H}(\mathbf{r})$):

$$i \hbar \dot{C_m}(t)=\sum_n C_n(t) H_{mn} \cos \omega t~ e^{i (\omega_m-\omega_n) t-\gamma |t|} \tag{3}$$

Now assume that we have a simple two state system.

$$\begin{bmatrix} \dot{C_1} \\ \dot{C_2} \end{bmatrix}=\frac{e^{-\gamma | t|} \cos \omega t}{i\hbar}\begin{bmatrix} H_{11} & H_{12} e^{i \omega_{12} t} \\ H_{12} e^{-i \omega_{12} t} & H_{22} \end{bmatrix} \begin{bmatrix} C_1 \\ C_2 \end{bmatrix} \tag{4}$$

How to obtain the exact general solution for (4)? A reference would be enough for me, because I was only able to find either some particular case or a perturbation treatment of the problem.

Here's a related question but with a different matrix.

What's interesting, if we use Fourier transform:

$$C_n=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty S_n(u) e^{-i u t} du \tag{5}$$

Then we obtain a pair of coupled integral equations:

$$\frac{2 \pi}{\gamma}\hbar u S_1(u)=H_{11}\int_{-\infty}^\infty S_1(v) \left( \frac{dv}{(v-u-\omega)^2+\gamma^2}+\frac{dv}{(v-u+\omega)^2+\gamma^2} \right)+ \\ +H_{12}\int_{-\infty}^\infty S_2(v) \left( \frac{dv}{(v-u-\omega+\omega_{12})^2+\gamma^2}+\frac{dv}{(v-u+\omega+\omega_{12})^2+\gamma^2} \right) \tag{6}$$

$$ \frac{2 \pi}{\gamma}\hbar u S_2(u)=H_{22}\int_{-\infty}^\infty S_2(v) \left( \frac{dv}{(v-u-\omega)^2+\gamma^2}+\frac{dv}{(v-u+\omega)^2+\gamma^2} \right) + \\ + H_{12}\int_{-\infty}^\infty S_1(v) \left( \frac{dv}{(v-u-\omega-\omega_{12})^2+\gamma^2}+\frac{dv}{(v-u+\omega-\omega_{12})^2+\gamma^2} \right) \tag{7}$$

Where the initial conditions for $C_n$ uniquely determine the normalization, as could be seen from (5).

These should be easy enough to solve numerically and then use FFT to get back to $C_n(t)$.

If this problem was never treated exactly or numerically, then I suppose it could be fun to try it on my own and see how the results are different from the usual approximations.

Yuriy S
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  • A few issues. First, $\hat{H}(\boldsymbol{r})e^{-i\omega t - \gamma t}$ is non-Hermitian. you need to add the complex conjugate term as well. Second. you are trying to model in damping $\gamma$ but in general damping occurs because of a coupling to a large number of bath modes. Off-hand I know of three approaches to handle this. a) Explicitly include the bath modes in which case you can't use a simple two-level model. b) use a density matrix approach (can't just work with kets) c) use a stochastic schrodinger equation (different than the ordinary schrodinger equation). – Jagerber48 Dec 19 '19 at 15:29
  • @jgerber, I understand the first issue, though I want to model absorption and not emission, so it's not clear to me what would adding the complex conjugate mean. As for the damping, what if my optical pulse simply has a finite lifetime $\tau=1/ \gamma$? It would have nothing to do with bath modes. Thank you for your input all the same. – Yuriy S Dec 19 '19 at 15:32
  • In closed system quantum mechanics all dynamics are unitary. So if the Hamiltonian permits absorption it also must permit emission. The Hamiltonian must be Hermitian in a closed system. Yes, you can model what will happen with a time dependent pulse. You should probably include in the model that the pulse doesn't turn on until time equals zero. As written down your pulse is infinite in magnitude at large negative times. I'm not sure what you're trying to model about absorption but I don't think this approach will be analytically tractable. – Jagerber48 Dec 19 '19 at 16:10
  • @jgerber, thank you for the explanation. I will edit the question at the earliest opportunity – Yuriy S Dec 19 '19 at 16:41
  • Then, you can apply the rotating-wave approximation and finally, the problem becomes treatable. E.g. see https://physics.stackexchange.com/questions/27425/rigorous-justification-for-rotating-wave-approximation/27426#27426. – Jon Dec 20 '19 at 12:58
  • @Jon, thank you, but I'm trying to see if there's a way to treat the problem exactly – Yuriy S Dec 20 '19 at 13:00
  • The hamiltonian you've set up is called the Rabi hamiltonian (or a minor variation coming from the unphysical exponential decay, in any case) which goes down to the Jaynes-Cummings model if you do the rotating-wave approximation. The latter is long known to be exactly soluble, and it's standard material in quantum optics books. The former was shown to be soluble (in the sense that its eigenvectors and eigenvalues can be obtained in terms of a special function) by Daniel Braak in 2011, https://arxiv.org/abs/1103.2461. Whether that solution is useful, of course, is subjective. – Emilio Pisanty Dec 20 '19 at 13:51
  • Note also that the absolute value of time in the exponential seems to disappear quite early. I don't know why you claim that it's there "for simplicity" -- it's likely that its only effect will be to make the problem unsolvable, or at least just forcing you to solve the negative and positive times separately and then add an ugly bit of glue between them. – Emilio Pisanty Dec 20 '19 at 13:54
  • @EmilioPisanty, thank you for the link. If you don't mind elaborating, what's unphysical about the exponential factor? I thought it was a commonly used model of a decaying pulse – Yuriy S Dec 20 '19 at 13:54
  • @EmilioPisanty, I forgot to fix it, thank you. The absolute value should be everywhere – Yuriy S Dec 20 '19 at 13:56
  • It's not an analytical function, and it has a discontinuous derivative. I've never seen it used for that -- where did you see it used in this way? – Emilio Pisanty Dec 20 '19 at 13:56
  • @EmilioPisanty, ah, I see, you mean the absolute value part? I mean the simple exponential decay. I added the absolute value part to get a pretty expression after Fourier transform, I suppose I could use a Heavisde function instead to model the onset of the pulse. In any case, for $\gamma \to 0$ I should get the classic Rabi model, is that correct? – Yuriy S Dec 20 '19 at 13:58
  • You can model things however you want. Whether they'll give you a solvable model will then depend on the details. We are not going to chase down every variation you care to come up with. But the overall strategy is start simple: solve for an infinite monochromatic driver first, and only then see whether it admits a pulse. – Emilio Pisanty Dec 20 '19 at 14:01
  • Hi Yuriy, note that all SE posts are version controlled, so there is no reason to indicate that it was edited in the posts but writing "edited after comments"; the timestamp at the bottom of the post is a link to the edit history of the post so that interested users can see what's changed. – Kyle Kanos Dec 20 '19 at 14:17

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