# What's the evolution of a state under an adiabatic evolution? [duplicate]

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?

## marked as duplicate by ACuriousMind♦ quantum-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 23 '17 at 12:28

• @ 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)? – XXDD Aug 23 '17 at 9:59