How to solve 2nd order TDSE for a Gaussian-kicked harmonic oscillator?

[skynelson]

TL;DR

The 2nd order time-dependent Schrodinger equation is a recursive integral over time, for which there are some readily available solved cases. Is the "kicked" harmonic oscillator one of these?

Consider the gaussian kick potential,
$\hat{V}(t) = \hat{x} \exp{(\frac{-t^2}{2 \tau^2})}$
where
$\hat{x} = a+a^\dagger$ in terms of creation and annihilation operators.

Then we define the potential in the interaction picture,
$\hat{V}_I(t) = e^{i\hat{H}t}\hat{V}(t)e^{-i\hat{H}t}$

I want to find the amplitude for transition from the ground state energy $|0>$ to the $|2>$ energy eigenstate, using a second order calculation of the TDSE.
The form for the time-dependent Schrödinger equation is
$c_f^{(2)}=\sum_{n} \int_{t_0}^{t'} dt_1 \int_{t_0}^{t_1} dt_2 \bra{\omega_f}\hat{V}_I(t_1)\ket{\omega_n}\bra{\omega_n} \hat{V}_I(t_2)\ket{\omega_i} c_0 e^{i\omega_0 t'}$

Expanding $V_I$ gives a recursive time integral which have Fourier transform kernels but whose domain is not infinite, so it is not a Fourier transform. The first order calculation is usually solved by assuming an infinite time domain approximation, which is OK for a gaussian perturbation because it decays quickly.

But for second order calculations, one cannot simply extend the nested integration domain to infinity, because the finite limits of that integral become the integration parameter for the outer integral.

Defining $\omega_{ab} = \omega_a - \omega_b$, I wrote the transition from 0->1->2 energy eigenstates as
$c_{0\rightarrow1\rightarrow2}^{(2)} = (-i/\hbar)^2 <2|\hat{x}|1><1|\hat{x}|0>\int_{t_0}^{t'} dt_1 e^{i\omega_{01}t_1}e^{-t_1^2/\tau^2} \int_{t_0}^{t_1} dt_2 e^{i\omega_{12}t_2}e^{-t_2^2/\tau^2}$

(my indices may be a little off here) My question is about evaluating the time integrals. I cannot find a reference for this. I am guessing the integral over $t_2$ can be expressed as an error function, erf, and then the integral over $t_1$ is usually assumed to extend to $t\rightarrow\infty$ so we can express the integral as a Fourier transform of a gaussian times erf.

But I don't know how to do that, and it seemed like this case may be a standard case and I wondered if anyone has advice on where to find a standard solution?

 

[skynelson]

Note that the first order solution is simply $\int_{-\infty}^{\infty} dt_1 e^{i \omega_{10}t_1} e^{-t_1^2/2\tau^2} = e^{-\tau^2 \omega_{10}^2/2}$

 

[jambaugh]

You are correct in that the finite integrals will be expressible only as error functions. However, consider combining the two integrations into an equivalent double integral in $(t_1, t_2)$ parameter space.
The integration may then be more tractable in polar coordinates. The problem is also that when you complete the square you will be skipping off the real axis for each variable of integration.

My thinking, (shooting from the hip here) is to first express each integral as a complex contour integral. You can use the analyticity of the exponentials and choose a path and change of variable for each integral so that you can then combine them in the way I mentioned i.e. as a double integral you can then express in polar coordinates. It sounds like a tricky problem if at all tractable. But that's the first stab I would take at it.

Does that make any sense to you?

 

[skynelson]

Hi Jambaugh, yes, that makes some sense...not that I am particularly good at contour integrals but I do understand a bit about why it would be useful in this case, since your exponents are not purely real or purely imaginary. You've also implicitly answered that to your knowledge this is not a standard example and is indeed quite tricky, which is a part of the info I was seeking. Thanks!

 

[WernerQH]

This problem is more easily solved in the holomorphic representation (using coherent states). You can then derive the desired coefficients by expanding in the unperturbed energy eigenstates.

Schwinger ("Brownian Motion of a Quantum Oscillator", J. Math. Phys. 2, 407-432) finds for a general (not necessarily Gaussian) driving force $K(t)$: $$
p(n_f,n_i,K) = \frac {n!} {m!} (|\gamma|^2)^{m-n}
\left[ L^{(m-n)}_n( |\gamma|^2) \right]^2 \exp (- |\gamma|^2)
$$ where $m = {\rm max}(n_i,n_f), n = {\rm min}(n_i,n_f)$ and
$$
\gamma = \int_{t_i}^{t_f} e^{i \omega t} K(t) \, dt
$$ (p. 412). In your particular case with $n_i = 0$ and $n_f = 2$, the generalized Laguerre polynomial $L_0^{(2)}$ is just 1, if I haven't misread Abramowitz/Stegun.

 

[skynelson]

Thank you @WernerQH that's an interesting set of ideas.

By the way, here is a sample problem in an anonymous textbook that gave me the indication that this problem could be addressed in some standard way by undergraduates. See the "Exercise" at the bottom.
The solutions we've come to so far seem beyond what is being suggested in this example problem. Any thoughts?

 

[WernerQH]

This is a nice problem, but perhaps a tad too difficult for undergraduates.

One has to change variables, and do the integration over $t_2 - t_1$ first, and then over $\frac 1 2 (t_1 + t_2)$. The first integral is well defined, because the two transitions occur in close succession (within the width of the gaussian). And the product of two gaussians is again a gaussian. I haven't worked it out, but I feel that one should be able to find a formula for the integral from 0 to $\infty$ in chapter 7.4 of Abramowitz-Stegun. The other integral should then be like the one in the first order calculation.

Coming to think of it, perhaps it is better to do the $\frac 1 2 (t_1 + t_2)$ integral first, if it permits the limits to be taken to infinity and helps to control the nasty oscillations of the integrand. But it's late now, and I'm too lazy to actually examine the equations. :-)