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| </math> | | </math> |
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| Typically, one lets <math> E = \frac{\hbar^2 k^2}{2m} \!</math> to simplify the equation | | Typically, one lets <math> E = \frac{\hbar^2 k^2}{2m},</math> obtaining |
| :<math> | | :<math> |
| \left( \nabla^2 + k^2 \right) \psi \left( \mathbf{r} \right) = 0. | | \left( \nabla^2 + k^2 \right) \psi \left( \mathbf{r} \right) = 0. |
Revision as of 10:43, 17 April 2013
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,
, the Schrödinger equation takes the form:
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ff4d2352b860ab251182602acebf87543b3b5a5)
Since the potential and the Hamiltonian do not depend on time, solutions to this equation can be written as
.
Obviously, for such state the probability density is

which is independent of time. Hence, the name is "stationary state".
The Schrödinger equation now becomes
![{\displaystyle \left[-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V({\textbf {r}})\right]\psi ({\textbf {r}})=E\psi ({\textbf {r}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e80d2d34fde6c1fa4546dae66011fff67105b81)
which is an eigenvalue equation with eigenfunction
and eigenvalue
. 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

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

Typically, one lets
obtaining

Show that (a) a plane wave
, and (b) a spherical wave
where
, satisfy the equation. (In either case, the wave length of the solution is given by
and the momentum by de Broglie's relation
. )
A sample problem: Free Particle SE Problem.