Stationary States: Difference between revisions
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'''(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> | '''(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. | 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]] | [[Phy5645/Free particle SE problem|Solution]] |
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
- 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.
Something similar happens when calculating the expectation value of any dynamical variable.
For any time-independent 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 Q(x,p),\!}
- 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 }
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{\hbar^2}{2m} \nabla^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},} 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 .