Given a time dependent Hamiltonian which commutes at different times, we have the time evolution operator given by $$\mathcal{U}(t,0) = \text{exp}\bigg[-\bigg(\frac{i}{h}\bigg)\int_{0}^{t}dt' H(t')\bigg],$$ for some general state $| \Psi, t_0 = 0 \rangle$ at $t = 0$ we then get: $$\Psi(x,t) = \langle x| \mathcal{U}(t,0)| \Psi, t_0 =0 \rangle = \langle x | \mathcal{U(t,0) \sum_n |n\rangle \langle n | \Psi \rangle} \\ = \sum_{n} \langle x | n \rangle \langle n | \Psi \rangle \text{exp}\bigg[-\bigg(\frac{i}{\hbar}\bigg)\int_{0}^{t}dt' E_n (t')\bigg] = \sum_{n} c_{n} \psi_n\text{exp}\bigg[-\bigg(\frac{i}{\hbar}\bigg)\int_{0}^{t}dt' E_n (t')\bigg] = \sum_n c_n \psi_n e^{i \theta_n(t)}$$
Question:
Can anyone see why this does not agree with what Griffiths (in book "Introduction to Quantum Mechanics" page 372) got in the attachment below, where he starts proving the adiabatic theorem. He gets that $c_n$ and $\psi_n$ are both functions of time. Where have I gone wrong in what I have written? Thanks.
Best Answer
The OP's derivation assumes that the eigenstates of the Hamiltonian do not depend on time, which is a strong assumption, not done by Griffin.
Indeed, in general, the instantaneous eigenstates of $\hat H(t)$ will dependend on $t$.
A counter-example, where the derivation of the OP is fine, is the case of a free particle with a time-dependent mass : $$ \hat H(t) = \frac{\hat P^2}{2m(t)}.$$