Stationary state

Remarks

The time evolution of a stationary state is $e^{-itH/\hbar}\psi = e^{-iEt/\hbar}\psi$. So $\psi(t)$ stays on the ray of $\psi$ in the projective Hilbert space, and every probability and expectation value is constant in time. The equation $H\psi = E\psi$ is the time-independent Schrödinger equation.

Example: let $\hbar = m = \omega = 1$ and $H = -\frac12 \frac{d^2}{dx^2} + \frac12 x^2$ on $L^2(\R)$, the harmonic oscillator. For $\psi_0(x) = \pi^{-1/4} e^{-x^2/2}$ one has $\psi_0'' = (x^2 - 1)\psi_0$, so $H\psi_0 = -\frac12 (x^2 - 1)\psi_0 + \frac12 x^2 \psi_0 = \frac12 \psi_0$. So $\psi_0$ is a stationary state with energy $\frac12$.

Non-example: with the same $H$, let $\psi_1(x) = \sqrt{2}\, \pi^{-1/4} x e^{-x^2/2}$, a stationary state with energy $\frac32$, and $\psi = (\psi_0 + \psi_1)/\sqrt{2}$. Then $\psi(t) = (e^{-it/2}\psi_0 + e^{-3it/2}\psi_1)/\sqrt{2}$, and the relative phase $e^{-it}$ changes with $t$. So $\psi(t)$ leaves the ray of $\psi$, and $\psi$ is not an eigenvector of $H$.