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Mar14-12, 03:21 PM
P: 22
Quote Quote by lugita15 View Post
In QM books, they often do an informal derivation where they say that the Hamiltonian is constant on each infinitesimal interval (t,t+dt), and so you can define U(t,dt) quite easily. Then just take an infinite product to find the time evolution operator for a finite time interval. Is there anything wrong with that kind of approach?
Hamiltonian is still defined by the Schrodinger equation, so it satisfies the equation [itex] i \hbar \frac{dU}{dt} (t) = H(t) U(t)[/itex].
The solution involves something like the Dyson operator, but I only know that is (rigorously) obtained by iterative method, as you say.

The evolution from time t0 to time t1 [itex] U(t_0, t_1)[/itex] depends in general on t0 and t1 separately, not only on their difference [itex] U(t_1 - t_0)[/itex].
All you can say is that [itex] U(0, t_1 + t_2) = U(t_1, t_1 + t_2) U(0, t_1)[/itex] and not that [itex] U(0, t_1 + t_2) = U(0, t_2) U(0, t_1)[/itex].
If [itex] U(t_2, t_1) =U(t_1 - t_2) [/itex] holds for any t1,t2 you can apply Stone's theorem, so you have a time-independent Hamiltonian.