# Phase in time evolution operator for time-dependent Hamiltonian [duplicate]

In Quantum Mechanics, a state vector $$|\psi\rangle$$ will evolve in time according to $$|\psi(t)\rangle=e^{-\frac{i}{\hbar}\hat H t}|\psi(0)\rangle$$ Imagine we have a system such that, for a short period of time $$T$$, the Hamiltonian increases by a constant and then returns to normal, such that $$\hat H=\hat H_0+ \begin{cases} 0 & \text{(t\lt0,\, t\gt T)}\\ A & \text{(0\leq t\leq T)}\\ \end{cases}$$ At $$t=T$$ we will have $$|\psi(T)\rangle=e^{-\frac{i}{\hbar}AT}e^{-\frac{i}{\hbar}\hat H_0T}|\psi(0)\rangle$$ Now, following the first equation, since after $$t=T$$ there is no $$A$$, it should just become $$|\psi(t)\rangle=e^{-\frac{i}{\hbar}\hat H_0 t}|\psi(0)\rangle$$ But this seems strange, it's as if that period of interaction with whatever caused the extra energy had no effect on the particle whatsoever. I think it makes more sense to apply the time evolution operator separately and obtain $$|\psi(t)\rangle=\hat U(t-T)|\psi(T)\rangle=e^{-\frac{i}{\hbar}\hat H_0(t-T)}e^{-\frac{i}{\hbar}AT}e^{-\frac{i}{\hbar}\hat H_0T}|\psi(0)\rangle=e^{-\frac{i}{\hbar}\hat H_0t}e^{-\frac{i}{\hbar}AT}|\psi(0)\rangle$$ Is my idea wrong or is the time evolutikon operator different in this case? If so, the what would be the case for a time-dependent Hamiltonian?

## marked as duplicate by Aaron Stevens, ZeroTheHero, Buzz, Qmechanic♦ 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(); } ); }); }); Dec 18 '18 at 7:14

Your first equation does not hold for a time dependent Hamiltonian.

• "Is my idea wrong ... in this case? If so, then what would be the case for a time-dependent Hamiltonian?" – Aaron Stevens Dec 18 '18 at 3:38