QM: Magnus expansion, Gaussian integral

[Bapelsin]

Homework Statement

The time-evolution operator \hat{U}(t,t_0) for a time-dependent Hamiltonian can be expressed using a Magnus expansion, which can be written

\hat{U}(t,t_0)=e^{\hat{A}(t,t_0)}=exp\left(\sum_{n=1}^{\infty}\frac{1}{n!}\left(\frac{-\imath}{\hbar}\right)^n\hat{F}_n(t,t_0)\right)

The first two terms are given by

\hat{F}_1=\int_{t_0}^t\hat{H}(t_1)dt_1
\hat{F}_2=\int_{t_0}^t\int_{t_0}^{t_1}[\hat{H}(t_2),\hat{H}(t_1)]dt_2dt_1

Consider a Harmonic oscillator with \hat{H}=\hat{H}_0 + \hat{V}(t) where
\hat{H}_0=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega^2\hat{x}^2

It interacts with an external field of the form \hat{V}=\hat{V}_0f(t)

Evaluate the Magnus expansion to second order for the case \hat{V}_0=V_0\hat{x} and f(t)=e^{-t^2/\sigma^2}

Homework Equations

See above.

The Attempt at a Solution

First term: \hat{F}_1=\int_{t_0}^t\left(\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega^2\hat{x}^2+V_0\hat{x}e^{-t^2/\sigma^2}\right)dt_1= \left(\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega^2\hat{x}^2\right)\cdot(t-t_0)+ ?

Second term (after solving the commutator specified above): \hat{F}_2=\frac{1}{m}V_0\imath\hbar\hat{p}\int_{t_0}^t\int_{t_0}^{t_1}\left(e^{-t_2^2/\sigma^2}-e^{-t_1^2/\sigma^2}\right)dt_1dt_2 = ? + ?

How do I integrate the Gaussian functions above? It seems like the integrals only are easily solvable when integrated from -\infty to \infty, is there any trick to get past this problem?

Any kind of help appreciated!

 

[gabbagabbahey]

Just make use of the error function:

\text{erf}(x)\equiv\frac{2}{\sqrt{\pi}}\int_0^x e^{-u^2}du

And remember ,

\int_{t_0}^{t} f(t_1)dt_1=\int_{0}^{t} f(t_1)dt_1-\int_{0}^{t_0} f(t_1)dt_1