Time evolution in an oscillating magnetic field for spin-1/2 particles

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?

• 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! Dec 19, 2020 at 14:59
• 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. Dec 19, 2020 at 16:29

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.

• although in the case of a spin-1/2 system the solution is easily found without the perturbative series. Dec 25, 2020 at 18:28