Stationary States: Difference between revisions
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Stationary states are the energy eigenstates of the Hamiltonian operator. These states are called "stationary" because their probability distributions are independent of time. | {{Quantum Mechanics A}} | ||
Stationary states are the energy eigenstates of the [[Commutation relations and simultaneous eigenvalues#Hamiltonian|Hamiltonian operator]]<nowiki />. These states are called "stationary" because their probability distributions are independent of time. | |||
For a conservative system with a time independent potential, <math>V(\textbf{r})</math>, the Schrödinger equation takes the form: | For a conservative system with a time independent potential, <math>V(\textbf{r})</math>, the [[Schrödinger equation]] takes the form: | ||
:<math> i\hbar\frac{\partial \ | :<math> 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)</math> | ||
Since the potential and the Hamiltonian do not depend on time, solutions to this equation can be written as | Since the potential and the Hamiltonian do not depend on time, solutions to this equation can be written as | ||
:<math>\ | :<math>\Psi(\textbf{r},t)=e^{-iEt/\hbar}\psi(\textbf{r})</math>. | ||
Obviously, for such state the probability density is | Obviously, for such state the probability density is | ||
:<math>|\ | :<math>|\Psi(\textbf{r},t)|^2=|\psi(\textbf{r})|^2</math> | ||
which is independent of time | which is independent of time, hence the term, "stationary state". | ||
The | The Schrödinger equation now becomes | ||
:<math>\left[ -\frac{\hbar^2}{2m}\nabla^2 + V(\textbf{r})\right]\psi(\textbf{r})=E\psi(\textbf{r})</math> | |||
:<math> | |||
which is an eigenvalue equation with eigenfunction <math>\psi(\textbf{r})</math> and eigenvalue <math>E\!</math>. 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 <math>Q(x,p),\!</math> | |||
:<math> \langle Q(x,p)\rangle = \int \psi^{\ast}(x) Q\left(x,\frac{\hbar}{i} \frac{d}{dx}\right) \psi(x)\,dx </math> | |||
==Problem== | |||
The time-independent Schrodinger equation for a free particle is given by | |||
:<math> | |||
-\frac{\hbar^2}{2m} \nabla^2 \psi \left(\mathbf{r} \right) = E \psi\left(\mathbf{r} \right) | |||
</math> | |||
Typically, one lets <math> E = \frac{\hbar^2 k^2}{2m},</math> obtaining | |||
:<math> | |||
\left( \nabla^2 + k^2 \right) \psi \left( \mathbf{r} \right) = 0. | |||
</math> | |||
Show that | |||
'''(a)''' a plane wave <math> \psi\left(\mathbf{r} \right) = e^{ikz}, \!</math> and | |||
'''(b)''' a spherical wave <math> \psi\left(\mathbf{r} \right) = \frac{e^{ikr}}{r}, \! </math> where <math> r = \sqrt{x^2 + y^2 + z^2}, \! </math> | |||
satisfy the equation. In either case, the wave length of the solution is given by <math> \lambda = \frac{2\pi}{k} \!</math> and the momentum by de Broglie's relation <math> p = \hbar k. \! </math> | |||
[[Phy5645/Free particle SE problem|Solution]] |
Latest revision as of 16:49, 12 August 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:
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 term, "stationary state".
The Schrödinger equation now becomes
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