# Schrodinger basis kets with Time-dependent Hamiltonian

I was reading through the proof of the Adiabatic Theorem (in Sakurai) and I realised I'm not quite sure how Schrodinger Basis kets behave when we have a time-dependent Hamiltonian. I know that with a time-independent Hamiltonian the basis kets don't change in the Schrodinger Picture.

So if $|n;t\rangle$ are the energy eigenkets of $H(t)$ at time $t$ and $|\alpha;t\rangle$ is an arbitrary state at time $t$, is the following at all true? \begin{align*} |\alpha;t\rangle = \sum_n c_n(t)|n;t\rangle = \sum_n c_n(t) e^{i\theta_n(t)}|n,t_0\rangle \end{align*} where $\theta_n(t) = -\frac{1}{\hbar}\int_{t_0}^t H(t')\,dt'$ and $e^{i\theta_n(t)}$ is a time-evolution operator

Wikipedia and Sakurai both have (each in different notation): \begin{align*} |\alpha;t\rangle = \sum_n c_n(t) e^{i\theta_n(t)}|n;t\rangle \end{align*} I feel like I'm not understanding this properly at all

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Both equations you write down only express the fact that the basis of eigenstates of $H(t)$ is still a basis, so a general ket vector, including the actual state vector of the system, may be expanded as a linear superposition of these basic vectors with some general complex coefficients $c_n(t)$. The two expansions only differ by the phase one includes into the coefficients $c_n(t)$ or into the basis vectors $|n;t\rangle$. One convention includes the phase $\exp(i\theta_n(t))$, another one doesn't, and so on. Obviously, there is no "universally mandatory" rule that would dictate the right phase of these vectors so there's some freedom about the notation. Note that a phase factor times an eigenstate is still an eigenstate.