Phy5645/Energy conservation: Difference between revisions
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Using the identity, <math>\Psi^*\nabla^2\Psi=\nabla\cdot\left(\Psi^*\nabla\Psi\right)-\nabla\Psi^{\ast}\cdot\nabla\Psi,</math> we obtain | Using the identity, <math>\Psi^*\nabla^2\Psi=\nabla\cdot\left(\Psi^*\nabla\Psi\right)-\nabla\Psi^{\ast}\cdot\nabla\Psi,</math> we obtain | ||
<math>\left\langle E\right\rangle=-\frac{\hbar^2}{2m}\iiint\left | <math>\left\langle E\right\rangle=-\frac{\hbar^2}{2m}\iiint\left [\nabla\cdot\left (\Psi^{\ast}\nabla\Psi\right)-\nabla\Psi^{\ast}\cdot\nabla\Psi\right ]\,d^3\textbf{r}+\iiint\Psi^{\ast}V\Psi\,d^3\textbf{r} </math> | ||
<math>=-\frac{\hbar^2}{2m}\iiint\nabla\cdot\left (\Psi^{\ast}\nabla\Psi\right)\,d^3\textbf{r}+\frac{\hbar^2}{2m}\iiint\nabla\Psi^{\ast}\cdot\nabla\Psi\,d^3\textbf{r}+\iiint\Psi^{\ast}V\Psi\,d^3\textbf{r}</math> | <math>=-\frac{\hbar^2}{2m}\iiint\nabla\cdot\left (\Psi^{\ast}\nabla\Psi\right)\,d^3\textbf{r}+\frac{\hbar^2}{2m}\iiint\nabla\Psi^{\ast}\cdot\nabla\Psi\,d^3\textbf{r}+\iiint\Psi^{\ast}V\Psi\,d^3\textbf{r}</math> | ||
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as well as the condition, <math>\lim_{r \to \infty}\Psi^{\ast}\nabla\Psi=0,</math> we obtain | as well as the condition, <math>\lim_{r \to \infty}\Psi^{\ast}\nabla\Psi=0,</math> we obtain | ||
<math>\left\langle E\right\rangle=\int W\,d^3\textbf{r}=\int\left | <math>\left\langle E\right\rangle=\int W\,d^3\textbf{r}=\int\left (\frac{\hbar^2}{2m}\nabla\Psi^{\ast}\cdot\nabla\Psi+\Psi^{\ast}V\Psi\right )d^3\textbf{r}</math> | ||
(2) | (2) We first find the time derivative of energy density: | ||
<math>\frac{\partial W}{\partial t}=\frac{\partial}{\partial t}\left(\nabla\ | <math>\frac{\partial W}{\partial t}=\frac{\partial}{\partial t}\left (\nabla\Psi^{\ast}\cdot\nabla\Psi+\Psi^{\ast}V\Psi\right ) | ||
=\frac{\hbar^2}{2m}\left(\nabla\ | =\frac{\hbar^2}{2m}\left (\nabla\Psi^{\ast}\cdot\nabla\frac{\partial\Psi}{\partial t} + \nabla\frac{\partial\Psi^{\ast}}{\partial t}\cdot\nabla\Psi\right ) + \frac{\partial\Psi^{\ast}}{\partial t}V\Psi+\Psi^{\ast}V\frac{\partial\Psi}{\partial t}</math> | ||
<math>=\frac{\hbar^2}{2m}\left | <math>=\frac{\hbar^2}{2m}\left [\nabla\cdot\left (\nabla\Psi^{\ast}\frac{\partial\psi}{\partial t} + \frac{\partial\Psi^{\ast}}{\partial t}\nabla\Psi\right) - \left (\frac{\partial\Psi}{\partial t}\nabla^2\Psi^{\ast}+\frac{\partial\Psi^{\ast}}{\partial t}\nabla^2\Psi\right )\right ]+\frac{\partial\Psi^{\ast}}{\partial t}V\Psi+\Psi^{\ast}V\frac{\partial\Psi}{\partial t}</math> | ||
<math>=\frac{\hbar^2}{2m}\nabla\cdot\left(\nabla\ | <math>=\frac{\hbar^2}{2m}\nabla\cdot\left (\nabla\Psi^{\ast}\frac{\partial\Psi}{\partial t}+\frac{\partial\Psi^{\ast}}{\partial t}\nabla\Psi\right)+\frac{\partial\Psi^{\ast}}{\partial t}\left (-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi\right )+\frac{\partial\Psi}{\partial t}\left (-\frac{\hbar^2}{2m}\nabla^2\Psi^{\ast}+\nabla\Psi^{\ast}\right )</math>, | ||
Using | Using the Schrödinger equation, | ||
<math>i\hbar\frac{\partial\ | <math>i\hbar\frac{\partial\Psi}{\partial t}=-\frac{\hbar^2}{2m}\nabla^2\Psi+V\Psi,</math> | ||
and its complex conjugate, | |||
<math> | <math>-i\hbar\frac{\partial\Psi^{\ast}}{\partial t}=-\frac{\hbar^2}{2m}\nabla^2\Psi^{\ast}+V\Psi^{\ast},</math> | ||
and defining the energy flux density as <math>\textbf{S}=-\frac{\hbar^2}{2m}\left(\frac{\partial\Psi^{\ast}}{\partial t}\nabla\Psi + \frac{\partial\Psi}{\partial t}\nabla\Psi^{\ast}\right ),</math> | |||
<math>\frac{\partial W}{\partial t}+\nabla \cdot \textbf{S}=0</math> | We obtain | ||
<math>\frac{\partial W}{\partial t}=-\nabla\cdot\textbf{S}+\frac{\partial\Psi^{\ast}}{\partial t}\frac{\partial\Psi}{\partial t}-\frac{\partial\Psi}{\partial t}\frac{\partial\Psi^{\ast}}{\partial t}=-\nabla\cdot\textbf{S},</math> | |||
or, rearranging, | |||
<math>\frac{\partial W}{\partial t}+\nabla \cdot \textbf{S}=0.</math> | |||
Back to [[Relation Between the Wave Function and Probability Density]] | Back to [[Relation Between the Wave Function and Probability Density]] |
Revision as of 15:36, 16 April 2013
(1) The energy operator in three dimensions is: so the average energy in state is:
Using the identity, we obtain
If we apply Gauss' Theorem to the first term,
as well as the condition, we obtain
(2) We first find the time derivative of energy density:
,
Using the Schrödinger equation,
and its complex conjugate,
and defining the energy flux density as
We obtain
or, rearranging,
Back to Relation Between the Wave Function and Probability Density