Phy5645/schrodingerequationhomework2: Difference between revisions

Revision as of 16:49, 16 April 2013

By definition:

${\frac {\partial \rho }{\partial t}}={\frac {\partial }{\partial t}}\sum _{i}\rho _{i}(\mathbf {r} ,t)$

$=\left.\sum _{i}\int \cdots \int d^{3}\mathbf {r} _{1}\,\cdots \,d^{3}\mathbf {r} _{i-1}\,d^{3}\mathbf {r} _{i+1}\,\cdots \,d^{3}\mathbf {r} _{N}\left(\Psi ^{\star }{\frac {\partial \Psi }{\partial t}}+{\frac {\partial \Psi ^{\star }}{\partial t}}\Psi \right)\right|_{\mathbf {r} _{i}=\mathbf {r} }$

$=\sum _{i}\left.\rho _{i}(\mathbf {r} _{i},t)\right|_{\mathbf {r} _{i}=\mathbf {r} }\quad (1)$

The wave function of the many-particle system $\Psi ({\textbf {r}}_{1},{\textbf {r}}_{2},\ldots ,{\textbf {r}}_{N};t)$ satisfies the following Schrödinger equation:

${\begin{cases}i\hbar {\frac {\partial \Psi }{\partial t}}=\sum _{k}(-{\frac {\hbar ^{2}}{2m}}\nabla ^{2})\Psi +\sum _{jk}v_{jk}\Psi \\-i\hbar {\frac {\partial \Psi ^{\star }}{\partial t}}=\sum _{k}(-{\frac {\hbar ^{2}}{2m}}\nabla _{k}^{2})\Psi ^{\star }+\sum _{jk}v_{jk}\Psi ^{\star }\end{cases}}$

If we substitute ${\frac {\partial \Psi }{\partial t}}$ and ${\frac {\partial \Psi ^{\star }}{\partial t}}$ into formula $(1)$ , we obtain

${\frac {\partial \rho _{i}}{\partial t}}={\frac {i\hbar }{2m}}\int \cdots \int d^{3}\mathbf {r} _{1}\,\cdots \,d^{3}\mathbf {r} _{i-1}\,d^{3}\mathbf {r} _{i+1}\,\cdots \,d^{3}\mathbf {r} _{N}\sum _{k}(\Psi ^{\star }\nabla _{k}^{2}\Psi -\Psi \nabla _{k}^{2}\Psi ^{\star })$

$={\frac {i\hbar }{2m}}\int \cdots \int d^{3}\mathbf {r} _{1}\,\cdots \,d^{3}\mathbf {r} _{i-1}\,d^{3}\mathbf {r} _{i+1}\,\cdots \,d^{3}\mathbf {r} _{N}\sum _{k}\nabla _{k}\cdot (\Psi ^{\star }\nabla _{k}\Psi -\Psi \nabla _{k}\Psi ^{\star }),$

or, taking the sum over $i$ ,

${\frac {\partial \rho }{\partial t}}={\frac {i\hbar }{2m}}\sum _{i}\left.\int \cdots \int d^{3}\mathbf {r} _{1}\,\cdots \,d^{3}\mathbf {r} _{i-1}\,d^{3}\mathbf {r} _{i+1}\,\cdots \,d^{3}\mathbf {r} _{N}\sum _{k}\nabla _{k}\cdot (\Psi ^{\star }\nabla _{k}\Psi -\Psi \nabla _{k}\Psi ^{\star })\right|_{\mathbf {r} _{i}=\mathbf {r} }.$

Let us now consider terms for which $i\neq k.$ In these cases, we may use Gauss' Theorem, along with the requirement that $\lim _{r_{k}\rightarrow \infty }\Psi ^{\ast }\nabla _{k}\Psi =0$ for all $k,$ to show that these terms must vanish. Therefore,

${\frac {\partial \rho }{\partial t}}={\frac {i\hbar }{2m}}\sum _{i}\left.\int \cdots \int d^{3}\mathbf {r} _{1}\,\cdots \,d^{3}\mathbf {r} _{i-1}\,d^{3}\mathbf {r} _{i+1}\,\cdots \,d^{3}\mathbf {r} _{N}\nabla _{i}\cdot (\Psi ^{\star }\nabla _{i}\Psi -\Psi \nabla _{i}\Psi ^{\star })\right|_{\mathbf {r} _{i}=\mathbf {r} }$

$=-\sum _{i}\nabla \cdot \mathbf {j} _{i}(\mathbf {r} ,t)=-\nabla \cdot \mathbf {j} (\mathbf {r} ,t),$

or

${\frac {\partial \rho }{\partial t}}+\nabla \cdot \mathbf {j} =0.$

Back to Relation Between the Wave Function and Probability Density

@@ Line 1: / Line 1: @@
 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}
@@ Line 13: / Line 13: @@
 -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_{i}\cdot(\Psi^{\star}\nabla_{i}\Psi-\Psi\nabla_{i}\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 Probability Density]]

Phy5645/schrodingerequationhomework2: Difference between revisions

Revision as of 16:49, 16 April 2013

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