Phy5645/schrodingerequationhomework2

Assume that the Hamiltonian for a system of N particles is ${\hat {H}}=-\sum _{i=1}^{N}{\frac {\hbar }{2m}}\nabla _{i}^{2}+\sum _{i=1}^{N}\rho _{ij}[|{\overrightarrow {r_{i}}}-{\overrightarrow {r_{j}}}|]$, and $\Psi ({\overrightarrow {r_{1}}}{\overrightarrow {r_{2}}}\cdots {\overrightarrow {r_{N}}},t)$ is the wave fuction.

We define:

$\rho ({\overrightarrow {r}},t)=\sum \rho _{i}({\overrightarrow {r}},t)$

${\overrightarrow {j}}({\overrightarrow {r}},t)=\sum {\overrightarrow {j_{i}}}({\overrightarrow {r}},t)$

$\rho _{1}({\overrightarrow {r_{1}}},t)=\int \cdots \int d^{3}r_{3}d^{3}r_{3}\cdots d^{3}r_{N}\Psi ^{\star }\Psi$

${\overrightarrow {j}}({\overrightarrow {r}},t)={\frac {\hbar }{2im}}\int \cdots \int d^{3}r_{3}d^{3}r_{3}\cdots d^{3}r_{N}(\Psi ^{\star }\nabla _{1}\Psi -\Psi \nabla _{1}\Psi ^{\star })$

Prove the following relation: ${\frac {\partial \rho }{\partial t}}+\nabla \cdot {\overrightarrow {j}}=0$

Solution:

By definition:

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

$=\sum _{i}\int \cdots \int d^{3}r_{1}\cdots d^{3}r_{i-1}d^{3}r_{i+1}\cdots d^{3}r_{N}(\Psi ^{\star }{\frac {\partial \Psi }{\partial t}}+{\frac {\partial \Psi ^{\star }}{\partial t}}\Psi )$

$=\sum _{i}\rho _{i}({\overrightarrow {r_{i}}},t)\quad (1)$

The wave function of many particles system $\Psi ({\overrightarrow {r_{1}}}{\overrightarrow {r_{2}}}\cdots {\overrightarrow {r_{N}}},t)$ satisfies the Schrodinger equation for many particles system:

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

Substitute ${\frac {\partial \Psi }{\partial t}}$ and ${\frac {\partial \Psi ^{\star }}{\partial t}}$ in to formula $(1)$, we get:

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

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

We can also have:

$\nabla \cdot {\overrightarrow {j}}\equiv \sum _{i}\nabla _{i}\cdot \sum _{i}j_{i}({\overrightarrow {r_{i}}},t)$

$=\nabla _{1}\cdot {\overrightarrow {j_{1}}}({\overrightarrow {r_{1}}},t)+\nabla _{2}\cdot {\overrightarrow {j_{2}}}({\overrightarrow {r_{2}}},t)+\cdots \nabla _{i}\cdot {\overrightarrow {j_{i}}}({\overrightarrow {r_{i}}},t)\cdots$

$=\sum _{i}\nabla _{i}\cdot {\overrightarrow {j_{i}}}({\overrightarrow {r_{i}}},t)$

$={\frac {\hbar }{2im}}\sum _{i}\int \cdots \int d^{3}r_{1}\cdots d^{3}r_{i-1}d^{3}r_{i+1}\cdots d^{3}r_{N}\times \nabla _{j}\cdot (\Psi ^{\star }\nabla _{k}\Psi -\Psi \nabla _{k}\Psi ^{\star })\quad (2)$

${\frac {\partial \rho }{\partial t}}=\sum _{i}{\frac {\partial \rho }{\partial t}}=\sum _{i}\int \cdots \int d^{3}r_{1}\cdots d^{3}r_{i-1}d^{3}r_{i+1}\cdots d^{3}r_{N}\times \sum _{k}{\frac {\hbar }{2im}}\nabla _{k}\cdot (\Psi ^{\star }\nabla _{k}\Psi -\Psi \nabla _{k}\Psi ^{\star })(3)$

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