Phy5645/schrodingerequationhomework2

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

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

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