Instantaneous eigenstate and time dependent Schrodinger equation Instantaneous eigenstate $\psi(t)$ is defined as
$\hat H(t)\psi(t)=E(t)\psi(t)\tag{1}$
But in the lecture notes of Quantum Physics III MIT (in the section of adiabatic approximation), it is written that $\psi(t)$ may not be the solution of time dependent Schrodinger equation.
I am not able to understand why is it that.
At $t=0$, $(1)$ becomes $\hat H(0)\psi(0)=E(0)\psi(0)\tag{2}$
This means $\psi(0)$ is the eigenstate of $\hat H(0)$
We know that TDSE is first order in time. If we know the state at time $t=0$, then we get state vector at other times also.
$i\hbar\frac{\partial}{\partial t}\psi(t)=\hat H(t)\psi(t),\;\text{At t=0 we have }\psi(0)\tag{3}$
Now from $(3)$, $\psi(t)$ evolves and $\hat H$ also changes with time as it depends explicitly on time.
So can't $\psi(t)$ will evolve such that $(1)$ and $(3)$ holds simultaneously?
I am not able to clearly understand why both $(1)$ amd $(3)$ won't hold simultaneously.
 A: I think the easiest way to see this is with an example. Consider a Hamiltonian that changes suddenly at $t=0$
\begin{align}
H(t) = \begin{cases} H_1 & t \le 0 \\ H_2 & t > 0  \end{cases}
\end{align}
Now the instantaneous eigenstates will, correspondingly, change suddenly at $t=0$ from eigenstates of $H_1$ to eigenstates of $H_2$. If we consider the solutions to the Scrhodinger equation, however, there is no way they can jump in this way. They do not know what Hamiltonian $H_2$ the system is going to jump to, or that it going to jump at all.
The instantaneous eigenstates are just that; they are entirely determined by the instantaneous Hamiltonian. The result of integrating the Schrodinger equation, however it determined by the entire history of $H(t)$ for all times back to whatever time we set our boundary conditions.
A: The point made by Zweibach in those notes is true even for time-independent Hamiltonians.
Let $\psi:\mathbb R\rightarrow \mathscr H$ be a trajectory through the Hilbert space $\mathscr H$, where $\psi(t)$ is understood to be the state vector of the system at time $t$.  The Schrodinger equation is
$$i\hbar \psi'(t) = H(t) \psi(t)\tag{1}$$
Let us assume that for each $t$, $\psi(t)$ is an eigenvector of $H(t)$ with eigenvalue $E(t)$.  That is,
$$H (t)\psi(t) = E(t)\psi(t)\tag{2}$$
Zweibach's point is that it does not follow that $\psi(t)$ is a solution to $(1)$. Plugging $(2)$ into $(1)$ yields
$$i\hbar \psi'(t) = E(t) \psi(t) \implies \psi'(t) = -\frac{i E(t)}{\hbar} \psi(t)$$
Obviously this equation is not satisfied by arbitrary trajectories $\psi$; even if the Hamiltonian is time-independent and $E(t) \equiv E_0$ is constant, in order for $\psi$ to be a solution to $(1)$ we must have that
$$\psi(t) = e^{-iE_0 t/\hbar} \psi(0)$$
As an explicit example, consider the elementary particle in a box of length $L$.  The ground state eigenvector for this system is $\psi_0 = \sqrt{2/L} \sin\big(\pi x/L\big)$ with eigenvalue $E_0 = \pi^2\hbar^2/2mL^2$.   If we let $\psi(t)=\psi_0$, we see that for each $t$, $\psi(t)$ is an instantaneous eigenvector of the Hamiltonian (meaning that $H\psi(t) = E_0 \psi(t)$); however, it is also obvious that $\psi(t)$ does not solve the (time-dependent) Schrodinger equation $(1)$.
