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, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(\textbf{r})} , the Schrödinger equation takes the form:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi(\textbf{r},t)=e^{-iEt/\hbar}\psi(\textbf{r})} .
Obviously, for such state the probability density is
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi(\textbf{r},t)|^2=|\psi(\textbf{r})|^2}
which is independent of time. Hence, the name is "stationary state".
The same thing happens in calculating the expectation value of any dynamical variable.
For some operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle Q(x,p,t) \rangle \!}
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle Q(x,p)\rangle = \int \psi^{\ast}(x) Q\left(x,\frac{\hbar}{i} \frac{d}{dx}\right) \psi(x) dx }
The Schrödinger equation now becomes
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(\textbf{r})} and eigenvalue Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E\!} . This equation is known as the time-independent Schrödinger equation.
Problem
The time-independent Schrodinger equation for a free particle is given by
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2m} \left( \frac{\hbar}{i} \frac{\partial}{\partial \mathbf{r}} \right)^2 \psi \left(\mathbf{r} \right) = E \psi\left(\mathbf{r} \right) }
Typically, one lets Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E = \frac{\hbar^2 k^2}{2m} \!} to simplify the equation
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( \nabla^2 + k^2 \right) \psi \left( \mathbf{r} \right) = 0. }
Show that (a) a plane wave Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi\left(\mathbf{r} \right) = e^{ikz} \!} , and (b) a spherical wave Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi\left(\mathbf{r} \right) = \frac{e^{ikr}}{r} \! } where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = \sqrt{x^2 + y^2 + z^2} \! } , satisfy the equation. (In either case, the wave length of the solution is given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda = \frac{2\pi}{k} \!} and the momentum by de Broglie's relation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p = \hbar k \! } . )
A sample problem: Free Particle SE Problem.