Phy5645/Energy conservation: Difference between revisions

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==  Example 1   ==
(1) The energy operator in three dimensions is: <math>H=-\frac{\hbar^2}{2m}\nabla^2+V</math>
Consider a particle moving in a potential field <math>V(\textbf{r})</math>,
so the average energy in state <math> \Psi </math> is:
<math>\left\langle E\right\rangle=\iiint \Psi^{\ast}H\Psi\,d^3\textbf{r}=\iiint \Psi^{\ast}\left (-\frac{\hbar^2}{2m}\nabla^2\Psi + V\Psi\right )\,d^3\textbf{r}</math>


(1) Prove the average energy equation: <math><E>=\int W d^3x=\int\left[\frac{\hbar^2}{2m}\nabla\psi^*\cdot\nabla\psi\right]d^3x</math>, where W is energy density,
Using the identity, <math>\Psi^*\nabla^2\Psi=\nabla\cdot\left(\Psi^*\nabla\Psi\right)-\nabla\Psi^{\ast}\cdot\nabla\Psi,</math> we obtain


(2) Prove the energy conservation equation: <math>\frac{\partial W}{\partial t}+\nabla \cdot \textbf{S}=0</math>, where <math>\textbf{S}</math> is energy flux density: <math>\textbf{S}=-\frac{\hbar^2}{2m}\left(\frac{\partial\psi^*}{\partial t}\nabla\psi + \frac{\partial\psi}{\partial t}\nabla\psi^*\right)</math>
<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>


Proof:
If we apply Gauss' Theorem to the first term,


(1):the energy operator in three dimensions is: <math>H=-\frac{\hbar^2}{2m}\nabla^2+V</math>
<math>-\frac{\hbar^2}{2m}\iiint\nabla\left (\Psi^{\ast}\nabla\Psi\right )\,d^3\textbf{r}=\iint\Psi^{\ast}\nabla\Psi\cdot d\textbf{S},</math>
so the average energy in state <math> \psi </math> is:
<math><E>=\iiint \psi^*H\psi d^3x=\iiint \psi^*\left(-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi\right) d^3x </math>,
Using: <math>\psi^*\nabla^2\psi=\nabla\left(\psi^*\nabla\psi\right)-\nabla\psi^*\nabla\psi </math>,
hence:
<math><E>=\iiint\left(-\frac{\hbar^2}{2m}\right)\left(\nabla\left(\psi^*\nabla\psi\right)-\nabla\psi^*\nabla\psi\right)d^3x +\iiint\psi^*\nabla\psi d^3x </math>
<math>=-\frac{\hbar^2}{2m}\iiint\nabla\left(\psi^*\nabla\psi\right)d^3x + \frac{\hbar^2}{2m}\iiint\nabla\psi^*\nabla\psi d^3x + \iiint\psi^*V\psi d^3x</math>,


Using Gauss Theorem for the last term:
as well as the condition, <math>\lim_{r \to \infty}\Psi^{\ast}\nabla\Psi=0,</math> we obtain
<math>-\frac{\hbar^2}{2m}\iiint\nabla\left(\psi^*\nabla\psi\right) d^3x=\iint\psi^*\nabla\psi\cdot d\textbf{S}</math>,
with the condition: <math>\lim_{r \to \infty}\psi^*\nabla\psi=0</math>, for infinite surface.


Hence:<math><E>=\int W d^3x=\int\left[\frac{\hbar^2}{2m}\nabla\psi^*\cdot\nabla\psi\right]d^3x</math>
<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):first we find the time derivative of energy density:
(2) We first find the time derivative of energy density:


<math>\frac{\partial W}{\partial t}=\frac{\partial}{\partial t}\left(\nabla\psi^*\nabla\psi+\psi^*\nabla\psi\right)
<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\psi^*\nabla\frac{\partial\psi}{\partial t} + \nabla\frac{\partial\psi^*}{\partial t}\nabla\psi\right) + \frac{\partial\psi^*}{\partial t}\nabla\psi+\psi^*\nabla\frac{\partial\psi}{\partial t}</math>,
=\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(\nabla\cdot\left(\nabla\psi^*\cdot\frac{\partial\psi}{\partial t} + \frac{\partial\psi^*}{\partial t}\cdot\nabla\psi\right) - \left(\frac{\partial\psi}{\partial t}\nabla^2\psi^*+\frac{\partial\psi^*}{\partial t}\nabla^2\psi\right)\right)+\frac{\partial\psi^*}{\partial t}\nabla\psi+\psi^*\nabla\frac{\partial\psi}{\partial t}</math>
<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\psi^*\cdot\frac{\partial\psi}{\partial t} + \frac{\partial\psi^*}{\partial t}\cdot\nabla\psi\right)+\frac{\partial\psi^*}{\partial t}\left(-\frac{\hbar^2}{2m}\nabla^2\psi+\nabla\psi\right)+\frac{\partial\psi}{\partial t}\left(-\frac{\hbar^2}{2m}\nabla^2\psi^*+\nabla\psi^*\right)</math>,
<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 Schrodinger Equations:
Using the Schrödinger equation,
<math>i\hbar\frac{\partial\psi}{\partial t}=-\frac{\hbar^2}{2m}\nabla^2\psi+\nabla\psi</math>,
<math>i\hbar\frac{\partial\Psi}{\partial t}=-\frac{\hbar^2}{2m}\nabla^2\Psi+V\Psi,</math>
and, <math>-i\hbar\frac{\partial\psi^*}{\partial t}=-\frac{\hbar^2}{2m}\nabla^2\psi^*+\nabla\psi^*</math>,


Also the energy flux density is:
and its complex conjugate,
<math>\textbf{S}=-\frac{\hbar^2}{2m}\left(\frac{\partial\psi^*}{\partial t}\nabla\psi + \frac{\partial\psi}{\partial t}\nabla\psi^*\right)</math>,
<math>-i\hbar\frac{\partial\Psi^{\ast}}{\partial t}=-\frac{\hbar^2}{2m}\nabla^2\Psi^{\ast}+V\Psi^{\ast},</math>


So:<math>\frac{\partial W}{\partial t}=-\nabla\cdot\textbf{S}+\frac{\partial\psi^*}{\partial t}\frac{\partial\psi}{\partial t}-\frac{\partial\psi}{\partial t}\frac{\partial\psi^*}{\partial t}=-\nabla\cdot\textbf{S}</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>
Hence:
 
<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#Problems|Relation Between the Wave Function and Probability Density]]

Latest revision as of 13:21, 18 January 2014

(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