0
$\begingroup$

This question already has an answer here:

For an initial state $|\Psi\rangle_0$ as the ground state of a Hamiltonian $H(0)$, if it undergoes an adiabatic evolution $H(t)$ to reach the ground state $|\Psi\rangle_1$ of $H(1)$.

Then what's the evolution from the state $|\Psi\rangle_0$ and $|\Psi\rangle_1$?

Can the evolution be written in a form given by $H(t)$ something as $exp(-i\int H(t))dt$? If this is the case, then what's the difference between an adiabatic evolution and a non-adiabatic evolution?

$\endgroup$

marked as duplicate by ACuriousMind quantum-mechanics Aug 23 '17 at 12:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Concerning the second question, see physics.stackexchange.com/questions/45455/… $\endgroup$ – Adam Aug 23 '17 at 8:16
  • $\begingroup$ @ Adam Thanks. Yes, the time-ordered integration seems to be the solution. But then there is no difference between the adiabatic and the non-adiabatic evolution since this applies to a normal non-adiabatic evolution too. Does the adiabatic evolution is just to control the Hamiltonian $H(t)$ so that the excited state will not be generated during the evolution (or just the evolution is SLOW enough)? $\endgroup$ – XXDD Aug 23 '17 at 9:59
  • $\begingroup$ An adiabatic evolution is just an evolution that is slow enough (and without level crossing) such that in good approximation, the system stays in the instantaneous ground-state of the Hamiltonian. $\endgroup$ – Adam Aug 23 '17 at 14:10