​An arbitrary quantum mechanical system is initially in the ground state |0). At t = 0, a perturbation of the form ​

Question

An arbitrary quantum mechanical system is initially in the ground state \(|0\rangle\). At t = 0, a perturbation of the form H'(t) = H₀e^-t/T is applied. Show that at large times the probability that the system is in state \(|1\rangle\) is given by

\(\frac{|\langle 0|H_0|1\rangle |^2}{h^2/T^2+(\Delta \varepsilon)^2}\)

where \(\Delta \varepsilon\) is the difference in energy of states \(|0\rangle\) and \(|1\rangle\). Be specific about what assumption, if any, were made arriving at your conclusion.

Answer 1

In the first order perturbation method the transition probability amplitude is given by

where \(\Delta \varepsilon = \omega _{10}h\) is the energy difference between the \(|0\rangle\) and \(|1\rangle\) states. Hence the probability that the system is in state \(|1\rangle\) at large times is

\(P_{10} = |c_{10}(t)|^2\frac{|\langle 0|H_0|1\rangle |^2}{h^2/T^2+(\Delta \varepsilon)^2}\)

It has been assumed in the above that H₀ is very small so that first order perturbation method is applicable.