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This might be a rookie mistake.
For a magnetic field oriented in the z-direction of form, $B = B_0 \cos(\omega t) \hat{k}$.
The Hamiltonian in this case will be $H = \omega_0 \cos(\omega t) \hat{S_z}$.
For an initial state of, $|{\psi(0)} \rangle = |+z \rangle $.
By solving the Schrodinger's equation, I obtain $$|\psi(t) \rangle = e^{-\frac{i \omega_0}{2} \sin(\omega t)} |+z\rangle $$.
But when I tried to find the time evolution of the state by using the relation between
the time evolution operator and the Hamiltonian, of form $$\hat{U} = e^{- \frac{i \hat{H}t}{\hbar}}$$ I got the state to become into, $$|\psi(t)\rangle = e^{- \frac{i \omega_0 \cos(\omega t) t}{2}} |+z\rangle $$ The second solution does not seem to agree with the Schrodinger's equation, is this because the Hamiltonian is explicitly dependent on time?

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  • $\begingroup$ I guess this happens because the relation of the provided relation for time evolution operator with the Hamiltonian is only applicable for time independent Hamiltonian! $\endgroup$ – adwait naravane Dec 19 '20 at 14:59
  • $\begingroup$ For this case you only need to integrate the time-dependent Hamiltonian and don't need that long series because the Hamiltonian at different time commute. $\endgroup$ – Frank Dec 19 '20 at 16:29
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Recall that the Schrodinger equation

$$ i \frac{\partial}{\partial t} | \psi \rangle = H |\psi \rangle $$

yields the time evolution

$$ |\psi(t)\rangle = e^{-i H t} | \psi(0) \rangle $$

only when the Hamiltonian is time independent.

There are two other situations:

(1) If the Hamiltonian commutes with itself at all times, then the solution for the time evolution operator is given by

$$ |\psi(t)\rangle = e^{-i \int_0^t H(t') dt'} | \psi(0) \rangle $$

(2) If the Hamiltonian does not commute with itself at different times then the formal time evolution is a Dyson series

\begin{align*} |\psi(t)\rangle = \left(\mathbb{1} + \sum_{n=1}^\infty\int_0^t dt_1 \int_0^{t_1} dt_2\cdots \int_0^{t_{n-1}} dt_n H(t_1) H(t_2)\cdots H(t_n)\right)|\psi(0)\rangle \end{align*}

For more on to handle these situation see pg. 72 of Sakurai.

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  • $\begingroup$ although in the case of a spin-1/2 system the solution is easily found without the perturbative series. $\endgroup$ – ZeroTheHero Dec 25 '20 at 18:28

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