Transition amplitudes

We have a forced simple harmonic oscillator Lagrangian $$L = \frac{\dot{\phi}^2}{2} - \frac{m^2{\phi}^2}{2} + f(t)\phi \, .$$ The external force goes to $$0$$ as $$t \to \pm \infty$$.

I'm trying to compute $$\langle 0|T \exp( -i \int_{-\infty}^{+\infty}\mathrm dt \,V_I(t) )|n\rangle$$ Where $$V_I(t)$$ is the potential in the interaction picture.

I know how to carry the calculations exactly when $$n=0$$ using Wick's theorem but How do I proceed in the general case?

• Express $\vert n \rangle$ in terms of the field $\phi$ acting on the vacuum. Then you have an expression that can be dealt with using Wick's theorem. – loewe Oct 17 '18 at 9:30
• I'm not sure how things would simplify, could you show me the steps ? – QHarmony Oct 17 '18 at 13:09
• You are right, I think it makes more sense to express $V$ in terms of creation and annihilation operators. – loewe Oct 17 '18 at 16:59

Operators in the interaction picture follow the free equation of motion. Thus, we can express the field in $$V_I(t)$$ (I will drop the index now) in terms of creation and annihilation operators of the free QHO, that is $$\begin{equation} V(t) = \frac{f(t)}{\sqrt{2m}} \left( a(t) + a^{\dagger}(t) \right). \end{equation}$$ The initial state of n excitations is $$\begin{equation} \vert n \rangle = \frac{\left(a^\dagger\right)^n}{\sqrt{n!}}\vert 0 \rangle. \end{equation}$$ The way I understand your question, the initial state is prepared before the driving starts, i.e. it is $$a$$ here is also an interaction picture operator but at time $$-\infty$$. Then your expression becomes $$\begin{equation} \frac{1}{\sqrt{n!}} \sum_{\alpha = 0}^{\infty} \frac{(-i)^\alpha}{\alpha!(2m)^{\alpha/2}} \int \textrm{d}t_1 ... \int \textrm{d}t_\alpha f(t_1)...f(t_\alpha) \langle 0 \vert T \left\{ \left( a(t_1) + a^{\dagger}(t_1) \right) ... \left( a(t_\alpha) + a^{\dagger}(t_\alpha) \right) \left(a^\dagger (-\infty)\right)^n \right\} \vert 0 \rangle, \end{equation}$$ where I pulled the creation operators in the infinite past into the time ordering. The trick is to realize that only terms that have equal number of creation an annihilation operators survive (time evolution in the interaction picture conserves the number of excitations). To lowest order in $$f$$, this gives corresponds to $$\alpha = n$$ and thus $$\begin{equation} \frac{1}{\sqrt{n!}} \frac{(-i)^n}{n!(2m)^{n/2}} \int \textrm{d}t_1 ... \int \textrm{d}t_n f(t_1)...f(t_n) \langle 0 \vert T \left\{ a(t_1) ... a(t_n) \left(a^\dagger (-\infty)\right)^n \right\} \rangle . \end{equation}$$ By Wick's theorem, the expression in the integral is just $$n!$$ copies of the same diagram, i.e. $$\begin{equation} \frac{(-i)^n}{\sqrt{n!}(2m)^{n/2}} \lim_{t'\to-\infty} \left[ \int_{-\infty}^{\infty} \textrm{d}t\ f(t)\ G_0 (t - t') \right]^n, \end{equation}$$ where $$G_0 (t - t')$$ is the free time ordered propagator of the theory.