Stationary States: Difference between revisions
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:<math>|\Psi(\textbf{r},t)|^2=|\psi(\textbf{r})|^2</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 Schrödinger equation now becomes | The Schrödinger equation now becomes | ||
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Something similar happens when calculating the expectation value of any dynamical variable. | Something similar happens when calculating the expectation value of any dynamical variable. | ||
For any time-independent operator <math>Q(x,p),</math> | 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> | :<math> \langle Q(x,p)\rangle = \int \psi^{\ast}(x) Q\left(x,\frac{\hbar}{i} \frac{d}{dx}\right) \psi(x)\,dx </math> | ||
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</math> | </math> | ||
Typically, one lets <math> E = \frac{\hbar^2 k^2}{2m} | 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. | ||
</math> | </math> | ||
Show that (a) a plane wave <math> \psi\left(\mathbf{r} \right) = e^{ikz} \!</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
- 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
- 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. \! }