# Time-dependent perturbation theory in a harmonic oscillator with a time-dependent force

Lets assume the classic quantum harmonic oscillator (HO) with a Hamiltonian $$\hat{H}(t) = - \frac{\hbar^2}{2m} \frac{d^2}{d x^2} + \frac{m \omega^2 \hat{x}^2}{2} + F(t) \hat{x}$$ where $F(t)$ is a time-dependent force defined via $$F(t)=\frac{F_0 \tau / \omega}{\tau^2 + t^2}$$ At time $t \to -\infty$ the particle with mass $m$ is in the ground state $| 0 \rangle$ of the HO potential. Since this is a time-dependent problem, I try to use time-dependent perturbation theory to first order to obtain the probability that at $t \to \infty$ the particle is in the first excited state $|1\rangle$. From the lecture we have obtained the (slightly more general) formula $$|c_f (t)|^2 = \frac{1}{\hbar^2} \left| \int_{t_0}^t V_{fi} (t') e^{i \omega_{fi} t'} dt' \right|^2$$ which we already have applied to the kicked oscillator as shown here (I used the same notation to make life easier for the readers here). But, how can I (can I?) apply this approach to this problem? As I see it, I need to somehow use $F(t)$ to get an expression for $V_{fi}(t')$ that gives me a calculable expression.

• Just do the integral. Is the problem that you don't know how to do the integral? – DanielSank Mar 30 '17 at 16:15
• Is your problem that you don't know how to relate $V$ and $F$? If that's your question, back up and review where that integral formula came from. – DanielSank Mar 30 '17 at 16:17
• Also, there's an error in the question. The drive term in $H$ should be $-F \hat{x}$. – DanielSank Mar 30 '17 at 16:20
• I'm not really sure how to relate $V$ and $F$. You're right, the $\hat{x}$ was missing on the problem sheet, which threw me off. I'll fix the Hamiltonian accordingly. – John W. Mar 30 '17 at 16:22

Your hamiltonian looks pretty wonky, and the notation should give itself away. $F(t)$ is a force, so what is it doing inside the hamiltonian without an $x$?
Instead, you might want to consider the hamiltonian $$H(t) = - \frac{\hbar^2}{2m} \frac{d^2}{d x^2} + \frac{1}{2}m \omega^2 x^2 + F(t)x,$$ where $V(t) = F(t)x$ is a bona fide potential. This will save you from some embarrassing features of your original hamiltonian, like the fact that if you just set $V(t) = F(t)$ then its matrix elements will just be $$V_{fi}(t) = \langle f|V(t)|i\rangle = \langle f|F(t)|i\rangle = F(t)\langle f|i\rangle = \delta_{fi}F(t),$$ and you will not get any transitions at all. If, instead, you use the actual potential $V(t)=F(t)x$, the matrix elements are $$V_{fi}(t) = \langle f|V(t)|i\rangle = \langle f|F(t)x|i\rangle = F(t)\langle f|x|i\rangle,$$ and they can be calculated relatively easily by expressing $x$ as a sum of $a$ and $a^\dagger$, giving you an expression proportional to $\delta_{f,i-1}+\delta_{f,i+1}$. In the end, this will reduce rather easily to some prefactor multiplying the integral $$\int_{-\infty}^\infty F(t) e^{i\omega t}\mathrm dt,$$ which requires some handling of the exponential integral Ei function (see also here) (probably after some partial-fractions handling of the denominator? e.g. you can split $$\frac{2}{\tau^2+t^2}=\frac{1}{t+i\tau}+\frac{1}{t-i\tau}$$ or something) but which should not be particularly challenging.
• I set $\hat{x}=\sqrt{\hbar/2m \omega} (a + a^\dagger)$, which does yield the two expressions $\delta_{1,1}=1$ and $0$ (from applying $\hat{a}$ to $|0\rangle$. So the root $\sqrt{\hbar/2m \omega}$ is the prefactor mentioned by you? – John W. Mar 30 '17 at 16:41