# Stationary States

Stationary states are the energy eigenstates of the Hamiltonian operator. These states are called "stationary" because their probability distributions are independent of time.

For a conservative system with a time independent potential, $V(\textbf{r})$, the Schrödinger equation takes the form:

$i\hbar\frac{\partial \Psi(\textbf{r},t)}{\partial t} = \left[ -\frac{\hbar^2}{2m}\nabla^2 + V(\textbf{r})\right]\Psi(\textbf{r},t)$

Since the potential and the Hamiltonian do not depend on time, solutions to this equation can be written as

$\Psi(\textbf{r},t)=e^{-iEt/\hbar}\psi(\textbf{r})$.

Obviously, for such state the probability density is

$|\Psi(\textbf{r},t)|^2=|\psi(\textbf{r})|^2$

which is independent of time, hence the term, "stationary state".

The Schrödinger equation now becomes

$\left[ -\frac{\hbar^2}{2m}\nabla^2 + V(\textbf{r})\right]\psi(\textbf{r})=E\psi(\textbf{r})$

which is an eigenvalue equation with eigenfunction $\psi(\textbf{r})$ and eigenvalue $E\!$. This equation is known as the time-independent Schrödinger equation.

Something similar happens when calculating the expectation value of any dynamical variable.

For any time-independent operator $Q(x,p),\!$

$\langle Q(x,p)\rangle = \int \psi^{\ast}(x) Q\left(x,\frac{\hbar}{i} \frac{d}{dx}\right) \psi(x)\,dx$

## Problem

The time-independent Schrodinger equation for a free particle is given by

$-\frac{\hbar^2}{2m} \nabla^2 \psi \left(\mathbf{r} \right) = E \psi\left(\mathbf{r} \right)$

Typically, one lets $E = \frac{\hbar^2 k^2}{2m},$ obtaining

$\left( \nabla^2 + k^2 \right) \psi \left( \mathbf{r} \right) = 0.$

Show that

(a) a plane wave $\psi\left(\mathbf{r} \right) = e^{ikz}, \!$ and

(b) a spherical wave $\psi\left(\mathbf{r} \right) = \frac{e^{ikr}}{r}, \!$ where $r = \sqrt{x^2 + y^2 + z^2}, \!$

satisfy the equation. In either case, the wave length of the solution is given by $\lambda = \frac{2\pi}{k} \!$ and the momentum by de Broglie's relation $p = \hbar k. \!$