$\newcommand{\ket}[1]{\left \lvert #1 \right \rangle}$


Consider a system described by

$$H(t) = H_0 + V_0 v(t) \mathcal{O}$$

where $V_0$ defines the strength of a time dependent perturbation, $v(t)$ is a dimensionless function of time, and $\mathcal{O}$ is a dimensionless operator. Supposing the system starts in state $\ket{i}$, the second order contribution to the probability to transition to state $\ket{f}$ is

\begin{align} \langle f | \Psi(t) \rangle =& - \left( \frac{V_0^2}{\hbar}\right)^2 \sum_n \langle f | \mathcal{O} | n \rangle \langle n | \mathcal{O} | i \rangle \\ \times & \int_0^t dt' v(t') \exp(i(\omega_f - \omega_n)t') \int_0^{t'} dt'' v(t'') \exp(i(\omega_n - \omega_i)t'') \, . \end{align} I would like to interpret this in terms of the so-called "virtual transitions" to and from the intermediate states $\ket{n}$.

What I understand

It's obvious that the system goes from $\ket{i} \rightarrow \ket{n} \rightarrow \ket{f}$ and that the matrix elements of $\mathcal{O}$ play into the amplitudes for those transitions. It's also clear that the drive function $v$ has to have spectral content near the $\omega_f - \omega_n$ and $\omega_n - \omega_i$ transitions.


How do we interpret the limits on the integrals? I want to think of the process as two sequential first order processes, one for $\ket{i} \rightarrow \ket{n}$ going from time 0 to time $t'$ and then another for $\ket{n} \rightarrow \ket{f}$ going from time $t'$ to time $t$. With that thinking, we'd have something like \begin{equation} \int_0^t dt' \left( \int_{t'}^t dt'' v(t'') e^{i (\omega_f - \omega_n)t''} \int_0^{t'} dt'' v(t'') e^{i (\omega_n - \omega_i)t''} \right) \end{equation} which is incorrect.


1 Answer 1


Your problem comes down to the limits of integration. $\int_0^t\mathrm dt^\prime\int_0^{t^\prime}\mathrm dt^{\prime\prime}$ is correct, and is similar to the interpretation that you are meaning. In it, $0<t^{\prime\prime}<t^\prime<t$ and that gives the ordering you were expecting. But if you look at the Dyson series, it really is $0\to t^{\prime\prime}$ as $\left|i\right>$, changed to $\left|n\right>$, then evolved back to 0, before the 2nd part evolving from $0\to t^\prime$, where it then changes to $\left|f\right>$, evolved back to 0 before evolving to $t$ in the end as the final state.

  • $\begingroup$ "Your problem comes down to the limits of integration." Yes, that's why I wrote "How do we interpret the limits on the integrals?" :-) $\endgroup$
    – DanielSank
    Jul 17 at 16:42
  • $\begingroup$ How do you see an "evolved back to 0" in the expression? $\endgroup$
    – DanielSank
    Jul 17 at 16:47
  • $\begingroup$ If you look at the Wikipedia page on this time dependent perturbation series, where it derives the Dyson equations, you can see the form of the expression, after inserting the resolution of identity (i.e. add the $\left|n\right>\left<n\right|$, but before you get the denominator terms, the evolution is easy to read as this evolution. Turns out, the denominator as the difference of energy eigenvalues, is really coming from evolving forward in time, changing, and then evolving backwards in time. $\endgroup$ Jul 17 at 17:37

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