(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

V= V_{0}(r) + V_{1}(r,t)

V_{0}(r) =-e^2/r

V_{1}(r,t) is a small perturbation which is being activated only in the interval 0<t<t_{f}

The system starts in the ground state, where l =0

1. If the change in the potential is very slow, what is the probability of finding the system at

t_{f}in the state where l = 1?

2. The same question only now consider the case where the change is fast.

3. What is the condition that t_{f}needs to satisfy in order to be small enough as to be considered as a fast perturbation?

2. The attempt at a solution

1. The change is adiabatic so the state of the system does not change. Thus it remains in the ground state and the probability to find the system in another state is zero.

2. Here the change very quick, so the wave function doesn't change at all, so again the transition probability is zero.

But here I need your help - I get the same transition probability in both cases. What is the difference between the two systems (the one that changed adiabatically and the one that changed rapidly) at t_{f}? they both remain in an eigenstate of the Hamiltonian operator, but the eigenenergies are different? Does the wavefunction in the first case change?

3. It is clear that t_{f}should be smaller than some characteristic time of the system, but how do I find it?

Thanks in advance!

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# Homework Help: QM - time dependent potential

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