Solving the Schroedinger equation with the initial condition as an energy eigenstate I was studying quantum mechanics by watching a video lecture series. In the lecture https://youtu.be/TWpyhsPAK14?list=PLUl4u3cNGP61-9PEhRognw5vryrSEVLPr&t=2784 , the professor tries to solve the Schroedinger equation with the initial condition of being in an energy eigen state.
$$\hat{E} \psi(x,0) = E \psi(x,0) \ \ \ \ \ \ \   eqn (1)$$
$$i\hbar\partial_t\psi(x,0) = \hat{E} \psi(x,0) = E \psi(x,0)\ \ \ \ \ \ \   eqn (2)$$
$$\partial_t\psi = \frac{-iE}{\hbar}\psi\ \ \ \ \ \ \   eqn (3)$$
$$\psi(x,t) = e^{-i\frac{Et}{\hbar}}\psi(x,0)\ \ \ \ \ \ \   eqn (4)$$
The above were the series of equations the professor wrote on the blackboard. I don't understand how you get from equation (2) to equation (3). Eqn (2) is valid only at $t=0$. How do you then get equation (3) which looks like it is valid for all $t$?
 A: By equation 2, $\psi$ at some infinitesimal time $\epsilon$ will be in the same state, multiplied by a phase $e^{-iE\epsilon/\hbar}$. Since it's the same state, you can use equation 1 again. And you can keep doing this forever.
But I feel like there should be a cleaner answer than this.
A: The intuitive reason this is true is simply that at time $t+\delta t$ the state is
$$|\psi(t)\rangle-\frac{i}{\hbar}\delta t \hat{H}|\psi\rangle.$$
If $|\psi\rangle$ is an eigenvector the this is just a multiple of $|\psi\rangle$ and so still an eigenvector. Another way to phrase this is that $\hat{H}$ can be written as a linear superposition of projection operators:
$$\hat{H}=\sum E_\alpha |\alpha\rangle\langle\alpha|,$$
and if you pick an initial state that lies already in one of the subspaces the projectors project onto, then the rest of the projectors just act like the $0$ operator. Then we can replace the Hamiltonian with just the one projector and the problem becomes $1D$.
Likewise, in classical mechanics, if there are only forces in the $x$ direction, and the initial velocity is along $x$, then you can say that the motion is restricted to $1D$. This is the same situation.

Formally, we could prove this by expanding in a basis of energy eigenstates
$$|\psi(t)\rangle = \sum c_\alpha(t) |\phi_\alpha \rangle, $$
then we have the (simultaneous, ordinary differential) equations
$$\frac{d c_\alpha}{dt} = -\frac{iE_\alpha t}{\hbar}c_\alpha(t),$$
for all $\alpha$, and then our initial condition is that $c_\alpha(0)=0$ for all but one energy eigenstate.
A: The reason it's true for all time is that if you start with an energy eigenstate you remain in that eigenstate (for all time) as the hamiltonian is the time evolution operator.
$$
i\hbar\frac{\partial|\phi\rangle}{\partial t}=\hat{H}|\phi\rangle
$$
and if $|\phi\rangle$ is an eigenstate,
$$
i\hbar\frac{\partial|\phi\rangle}{\partial t}=E|\phi\rangle
$$
which can be easily solved to get the $e^{-i\frac{Et}{\hbar}}$ phase factor you got.
