The Virial Theorem: Difference between revisions
(New page: Consider <math> \begin{align} &\frac{d}{dt}<xp>\\ &=\frac{1}{i\hbar}<[xp,H]> \\ &=\frac{ 2<p^{2}> }{2m}+\frac{1}{i\hbar}<xpV-xVp>\\ &=\frac{2<p^{2}>}{2m}+ \frac{1}{i\hbar}\int_{-\infty}...) |
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{{Quantum Mechanics A}} | |||
We will now derive the quantum mechanical virial theorem. For a Hamiltonian of the form, | |||
<math>\hat{H}=\frac{\hat{\mathbf{p}}^2}{2m}+\hat{V}(\hat{\mathbf{r}})=\hat{K}+\hat{V}(\hat{\mathbf{r}}),</math> | |||
this theorem gives the expectation value of the kinetic energy in a stationary state in terms of the potential energy. To derive this relation, we consider the expectation value of <math>\hat{\mathbf{r}}\cdot\hat{\mathbf{p}}.</math> The time derivative of this expectation value is | |||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
\frac{d}{dt}\langle\hat{\mathbf{r}}\cdot\hat{\mathbf{p}}\rangle&=-\frac{i}{\hbar}<[\hat{\mathbf{r}}\cdot\hat{\mathbf{p}},\hat{H}]>=-\frac{i}{2m\hbar}\langle[\hat{\mathbf{r}}\cdot\hat{\mathbf{p}},\hat{\mathbf{p}}^2]\rangle-\frac{i}{\hbar}\langle[\hat{\mathbf{r}}\cdot\hat{\mathbf{p}},\hat{V}(\hat{\mathbf{r}})]\rangle \\ | |||
&=\frac{\langle\hat{\mathbf{p}}^2\rangle}{m}-\langle\hat{\mathbf{r}}\cdot\nabla\hat{V}(\hat{\mathbf{r}})\rangle=2\langle\hat{K}\rangle-\langle\hat{\mathbf{r}}\cdot\nabla\hat{V}(\hat{\mathbf{r}})\rangle. | |||
\end{align} | |||
</math> | |||
For a stationary state, the expectation value of <math>\hat{\mathbf{r}}\cdot\hat{\mathbf{p}}</math> is constant in time. This gives us the relation, | |||
\ | <math>2\langle\hat{K}\rangle=\langle\hat{\mathbf{r}}\cdot\nabla\hat{V}(\hat{\mathbf{r}})\rangle.</math> | ||
</math> | |||
This is the virial theorem. | |||
As an example of its application, let us consider the isotropic three-dimensional harmonic oscillator, | |||
<math>\hat{V}(\hat{\mathbf{r}})=\tfrac{1}{2}m\omega^2\hat{r}^2.</math> | |||
</math> | |||
The right-hand side of the virial theorem is given by | |||
<math>\hat{\mathbf{r}}\cdot\nabla\hat{V}(\hat{\mathbf{r}})=m\omega^2\hat{r}^2=2\hat{V}(\hat{\mathbf{r}}).</math> | |||
<math>2 | |||
Therefore, | |||
<math> | <math>\langle\hat{K}\rangle=\langle\hat{V}\rangle.</math> |
Latest revision as of 10:43, 16 August 2013
We will now derive the quantum mechanical virial theorem. For a Hamiltonian of 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 \hat{H}=\frac{\hat{\mathbf{p}}^2}{2m}+\hat{V}(\hat{\mathbf{r}})=\hat{K}+\hat{V}(\hat{\mathbf{r}}),}
this theorem gives the expectation value of the kinetic energy in a stationary state in terms of the potential energy. To derive this relation, we consider the expectation value of 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 \hat{\mathbf{r}}\cdot\hat{\mathbf{p}}.} The time derivative of this expectation value 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 \begin{align} \frac{d}{dt}\langle\hat{\mathbf{r}}\cdot\hat{\mathbf{p}}\rangle&=-\frac{i}{\hbar}<[\hat{\mathbf{r}}\cdot\hat{\mathbf{p}},\hat{H}]>=-\frac{i}{2m\hbar}\langle[\hat{\mathbf{r}}\cdot\hat{\mathbf{p}},\hat{\mathbf{p}}^2]\rangle-\frac{i}{\hbar}\langle[\hat{\mathbf{r}}\cdot\hat{\mathbf{p}},\hat{V}(\hat{\mathbf{r}})]\rangle \\ &=\frac{\langle\hat{\mathbf{p}}^2\rangle}{m}-\langle\hat{\mathbf{r}}\cdot\nabla\hat{V}(\hat{\mathbf{r}})\rangle=2\langle\hat{K}\rangle-\langle\hat{\mathbf{r}}\cdot\nabla\hat{V}(\hat{\mathbf{r}})\rangle. \end{align} }
For a stationary state, the expectation value of 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 \hat{\mathbf{r}}\cdot\hat{\mathbf{p}}} is constant in time. This gives us the 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 2\langle\hat{K}\rangle=\langle\hat{\mathbf{r}}\cdot\nabla\hat{V}(\hat{\mathbf{r}})\rangle.}
This is the virial theorem.
As an example of its application, let us consider the isotropic three-dimensional harmonic oscillator,
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 \hat{V}(\hat{\mathbf{r}})=\tfrac{1}{2}m\omega^2\hat{r}^2.}
The right-hand side of the virial theorem 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 \hat{\mathbf{r}}\cdot\nabla\hat{V}(\hat{\mathbf{r}})=m\omega^2\hat{r}^2=2\hat{V}(\hat{\mathbf{r}}).}
Therefore,
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\hat{K}\rangle=\langle\hat{V}\rangle.}