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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.

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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 : $$ \hat H(t) = \frac{\hat P^2}{2m(t)}.$$

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  • $\begingroup$ Okay thanks but I am trying to correctly derive the last equation in the attachment using time evolution operator. Please advise if the following is correct when now considering energy eigenstates which are a function of time? $\endgroup$ – user101311 May 3 '17 at 17:49
  • $\begingroup$ $$\Psi(x,t) = \langle x| \mathcal{U}(t,0)|\Psi, t_0 = 0 \rangle = \langle x| \mathcal{U}(t,0)\sum |n(t \rangle \langle n(t)| \Psi \rangle \\= \sum \langle x| \text{exp} \bigg[- \frac{i}{\hbar}\int_{0}^{t}dt'H(t')\bigg] | n(t) \rangle \langle n(t)| \Psi \rangle \\= \sum \langle x | n(t) \rangle \langle n(t)| \Psi \rangle \text{exp}[-\frac{i}{\hbar}\int^{t}_{0} dt' E_{n}(t')] = \sum_{n}c_{n}(t)\psi_{n}(t)e^{i \theta_n(t)}.$$ $\endgroup$ – user101311 May 3 '17 at 17:49
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The fact that Hamiltonian is changing with time implies that the energy eigenstates of the system are varying. This is because if they didn't vary, they wouldn't probably be energy eigenstates of evolving Hamiltonian. As you've written already, the correction in your equations should be $$ |n\rangle\ \to |n(t)\rangle\ $$ $$ c_n \to c_n(t) $$

In addition, the proof doesn't assume that Hamiltonian commutes at different times. Even though your assumption is not sort of complete, your steps are correct. In this case, time evolution operator should be of the form, $$ U(t,0) = T\space exp\Bigg[-\frac{i}{\hbar}\int^{t}_{0}dt'H(t')\Bigg] $$ where T indicates that it is a time-ordered exponential. When acting on an energy eigenstate, this operator becomes,

$$ U(t,0) = exp\Bigg[-\frac{i}{\hbar}\int^{t}_{0}dt'E_n(t')\Bigg] $$

which gives the correct form at the end.

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