# On time-evolution of a quantum state

Suppose I have a quantum system governed by a time-independent Hamiltonian $$H$$. Its eigenvectors $$\{|\varphi_n\rangle\}_{n\in\mathbb{N}}$$ form a complete orthonormal set (or basis) for the Hilbert space $$\mathscr{H}$$ hosting the state $$|\psi\rangle$$ of the system.

If $${\displaystyle \left|\psi (t)\right\rangle }$$ is a state at time $$t$$, then

$${\displaystyle H\left|\psi (t)\right\rangle =i\hbar {\partial \over \partial t}\left|\psi (t)\right\rangle .}$$

The eigenvectors of $$H$$ at a given time $$t_0$$ are $$\{|\varphi_n\rangle\}_{n\in\mathbb{N}}$$ (assuming discrete, non-degenerate spectrum). Then $$|\varphi_n(t)\rangle = \exp\left(-\frac{i}{\hbar}E_nt\right)|\varphi_n\rangle \qquad \forall\;t>t_0$$

Since $$\{|\varphi_n\rangle\}_{n\in\mathbb{N}}$$ form a complete orthonormal set (or basis) for the Hilbert space $$\mathscr{H}$$ any state $$|\psi(t_0)\rangle \in \mathscr{H}$$ at the time $$t_0$$ can be expressed as

$$|\psi(t_0)\rangle= \sum_{n\in\mathbb{N}}c_n|\varphi_n\rangle \qquad \operatorname{at} \ t=t_0$$

What will the general expression for $$|\psi(t)\rangle$$ at time $$t>t_0$$ be? Well, evidently $$\{|\varphi_n(t)\rangle\}_{n\in\mathbb{N}}$$ is still a basis for $$\mathscr{H}$$, at any time $$t$$. Therefore

$$|\psi(t)\rangle= \sum_{n\in\mathbb{N}}c_n(t)|\varphi_n(t)\rangle \qquad \operatorname{at} \ t>t_0$$

But actually what turns out is slightly different $$|\psi(t)\rangle= \sum_{n\in\mathbb{N}}c_n\exp\left(-\frac{i}{\hbar}E_n t\right)|\varphi_n\rangle \qquad \operatorname{at} \ t>t_0$$ In other words $$c_n(t)=c_n$$. Why do the coefficients remain the same?

Write $$\vert\Psi(t)\rangle= \sum_n c_n(t)e^{-iE_nt/\hbar}\vert\varphi_n\rangle$$ with $$H\vert\varphi_n\rangle=E_n\vert\varphi_n\rangle$$. Insert this into the TDSE: $$i\hbar\sum_n \dot{c}_n(t)e^{-iE_nt/\hbar}\vert\varphi_n\rangle +i\hbar \sum_n \frac{-iE_n}{\hbar}c_n(t)e^{-iE_nt/\hbar}\vert\varphi_n\rangle = \sum_n c_n(t) E_n e^{-iE_nt/\hbar}\vert\varphi_n\rangle\, ,$$ where $$\dot{c}_m(t)=\frac{d}{dt}c_m(t)$$ and where $$H\vert\varphi_n\rangle=E_n\vert\varphi_n\rangle$$ has been used.
Close with $$\langle\varphi_m\vert$$ and use orthogonality: $$i\hbar\dot{c}_m(t)e^{-iE_mt/\hbar} +i\hbar \frac{-iE_m}{\hbar}c_m(t)e^{-iE_mt/\hbar} = c_m(t) E_m e^{-iE_mt/\hbar}\, .$$ The terms in $$c_m(t)$$ cancel out and you're left with $$i\hbar\dot{c}_m(t)e^{-iE_mt/\hbar}=0$$ which implies $$c_m$$ is constant.