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By definition:
By definition:


<math>\frac{\partial\rho}{\partial t}=\frac{\partial}{\partial t}\sum_{i}\rho_{i}(\overrightarrow{r},t)</math>
<math>\frac{\partial\rho}{\partial t}=\frac{\partial}{\partial t}\sum_{i}\rho_{i}(\mathbf{r},t)</math>


<math>=\sum_{i}\int\cdots\int d^{3}r_{1}\cdots d^{3}r_{i-1}d^{3}r_{i+1}\cdots d^{3}r_{N}(\Psi^{\star}\frac{\partial\Psi}{\partial t}+\frac{\partial\Psi^{\star}}{\partial t}\Psi)</math>
<math>=\left.\sum_{i}\int\cdots\int d^{3}\mathbf{r}_{1}\,\cdots\,d^{3}\mathbf{r}_{i-1}\,d^{3}\mathbf{r}_{i+1}\,\cdots\,d^{3}\mathbf{r}_{N}\left (\Psi^{\star}\frac{\partial\Psi}{\partial t}+\frac{\partial\Psi^{\star}}{\partial t}\Psi\right )\right |_{\mathbf{r}_i=\mathbf{r}}</math>


<math>=\sum_{i}\rho_{i}(\overrightarrow{r_{i}},t)      \quad (1)</math>
<math>=\sum_{i}\left. \rho_{i}(\mathbf{r}_{i},t)\right |_{\mathbf{r}_i=\mathbf{r}}     \quad (1)</math>


The wave function of many particles system <math>\Psi(\overrightarrow{r_{1}}\overrightarrow{r_{2}}\cdots\overrightarrow{r_{N}},t)</math> satisfies the Schrodinger equation for many particles system:
The wave function of the many-particle system <math>\Psi(\textbf{r}_{1},\textbf{r}_{2},\ldots,\textbf{r}_{N};t)</math> satisfies the following Schrödinger equation:


<math>\begin{cases}
<math>\begin{cases}
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-i\hbar\frac{\partial\Psi^{\star}}{\partial t}=\sum_{k}(-\frac{\hbar^{2}}{2m}\nabla_{k}^{2})\Psi^{\star}+\sum_{jk}v_{jk}\Psi^{\star}\end{cases}</math>
-i\hbar\frac{\partial\Psi^{\star}}{\partial t}=\sum_{k}(-\frac{\hbar^{2}}{2m}\nabla_{k}^{2})\Psi^{\star}+\sum_{jk}v_{jk}\Psi^{\star}\end{cases}</math>


Substitute <math>\frac{\partial\Psi}{\partial t}</math> and <math>\frac{\partial\Psi^{\star}}{\partial t}</math> in to formula <math>(1)</math>, we get:
If we substitute <math>\frac{\partial\Psi}{\partial t}</math> and <math>\frac{\partial\Psi^{\star}}{\partial t}</math> into formula <math>(1)</math>, we obtain


<math>\frac{\partial\rho_{i}}{\partial t}=-\int\cdots\int d^{3}r_{1}\cdots d^{3}r_{i-1}d^{3}r_{i+1}\cdots d^{3}r_{N}\cdot\sum_{k}\frac{\hbar}{2im}(\Psi^{\star}\nabla_{k}^{2}\Psi-\Psi\nabla_{k}^{2}\Psi^{\star})</math>
<math>\frac{\partial\rho_{i}}{\partial t}=\frac{i\hbar}{2m}\int\cdots\int d^{3}\mathbf{r}_{1}\,\cdots\,d^{3}\mathbf{r}_{i-1}\,d^{3}\mathbf{r}_{i+1}\,\cdots\,d^{3}\mathbf{r}_{N}\sum_{k}(\Psi^{\star}\nabla_{k}^{2}\Psi-\Psi\nabla_{k}^{2}\Psi^{\star})</math>


<math>=-\int\cdots\int d^{3}r_{1}\cdots d^{3}r_{i-1}d^{3}r_{i+1}\cdots d^{3}r_{N}\cdot\sum_{k}\frac{\hbar}{2im}\nabla_{k}\cdot(\Psi^{\star}\nabla_{k}\Psi-\Psi\nabla_{k}\Psi^{\star})</math>
<math>=\frac{i\hbar}{2m}\int\cdots\int d^{3}\mathbf{r}_{1}\,\cdots\,d^{3}\mathbf{r}_{i-1}\,d^{3}\mathbf{r}_{i+1}\,\cdots\,d^{3}\mathbf{r}_{N}\sum_{k}\nabla_{k}\cdot(\Psi^{\star}\nabla_{k}\Psi-\Psi\nabla_{k}\Psi^{\star}),</math>


We can also have:
or, taking the sum over <math>i</math>,


<math>\nabla\cdot\overrightarrow{j}\equiv\sum_{i}\nabla_{i}\cdot\sum_{i}j_{i}(\overrightarrow{r_{i}},t)</math>
<math>\frac{\partial\rho}{\partial t}=\frac{i\hbar}{2m}\sum_{i}\left.\int\cdots\int d^{3}\mathbf{r}_{1}\,\cdots\,d^{3}\mathbf{r}_{i-1}\,d^{3}\mathbf{r}_{i+1}\,\cdots\,d^{3}\mathbf{r}_{N}\sum_{k}\nabla_{k}\cdot(\Psi^{\star}\nabla_{k}\Psi-\Psi\nabla_{k}\Psi^{\star})\right |_{\mathbf{r}_i=\mathbf{r}}.</math>


<math>=\nabla_{1}\cdot\overrightarrow{j_{1}}(\overrightarrow{r_{1}},t)+\nabla_{2}\cdot\overrightarrow{j_{2}}(\overrightarrow{r_{2}},t)+\cdots\nabla_{i}\cdot\overrightarrow{j_{i}}(\overrightarrow{r_{i}},t)\cdots</math>
Let us now consider terms for which <math>i\neq k.</math>  In these cases, we may use Gauss' Theorem, along with the requirement that <math>\lim_{r_k\rightarrow\infty}\Psi^{\ast}\nabla_{k}\Psi=0</math> for all <math>k,</math> to show that these terms must vanish.  Therefore,


<math>=\sum_{i}\nabla_{i}\cdot\overrightarrow{j_{i}}(\overrightarrow{r_{i}},t)</math>
<math>\frac{\partial\rho}{\partial t}=\frac{i\hbar}{2m}\sum_{i}\left.\int\cdots\int d^{3}\mathbf{r}_{1}\,\cdots\,d^{3}\mathbf{r}_{i-1}\,d^{3}\mathbf{r}_{i+1}\,\cdots\,d^{3}\mathbf{r}_{N}\nabla_{i}\cdot(\Psi^{\star}\nabla_{i}\Psi-\Psi\nabla_{i}\Psi^{\star})\right |_{\mathbf{r}_i=\mathbf{r}}</math>


<math>=\frac{\hbar}{2im}\sum_{i}\int\cdots\int d^{3}r_{1}\cdots d^{3}r_{i-1}d^{3}r_{i+1}\cdots d^{3}r_{N}\times\nabla_{j}\cdot(\Psi^{\star}\nabla_{k}\Psi-\Psi\nabla_{k}\Psi^{\star}) \quad (2)</math>
<math>=-\sum_{i}\nabla\cdot\mathbf{j}_{i}(\mathbf{r},t)=-\nabla\cdot\mathbf{j}(\mathbf{r},t),</math>


<math>\frac{\partial\rho}{\partial t}=\sum_{i}\frac{\partial\rho}{\partial t}=\sum_{i}\int\cdots\int d^{3}r_{1}\cdots d^{3}r_{i-1}d^{3}r_{i+1}\cdots d^{3}r_{N}\times\sum_{k}\frac{\hbar}{2im}\nabla_{k}\cdot(\Psi^{\star}\nabla_{k}\Psi-\Psi\nabla_{k}\Psi^{\star}) (3) </math>
or


Combine the sum over in equation <math>(3)</math>, we find that the terms for <math>i\neq k</math> do not exist any more, so equation <math>(2)</math> is the same as equation <math>(3)</math>, so we get <math>\frac{\partial\rho}{\partial t}+\nabla\cdot\overrightarrow{j}=0</math>
<math>\frac{\partial\rho}{\partial t}+\nabla\cdot\mathbf{j}=0.</math>


Back to [[Relation Between the Wave Function and the Probability Density]]
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

By definition:

The wave function of the many-particle system satisfies the following Schrödinger equation:

If we substitute and into formula , we obtain

or, taking the sum over ,

Let us now consider terms for which In these cases, we may use Gauss' Theorem, along with the requirement that for all to show that these terms must vanish. Therefore,

or

Back to Relation Between the Wave Function and Probability Density