Phy5645/Energy conservation

Example 1

Consider a particle moving in a potential field $V({\textbf {r}})$,

(1) Prove the average energy equation: $<E>=\int Wd^{3}x=\int \left[{\frac {\hbar ^{2}}{2m}}\nabla \psi ^{*}\cdot \nabla \psi \right]d^{3}x$, where W is energy density,

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

Proof: (1):the energy operator in three dimensions is: $H=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V$ so the average energy in state $\psi$ is: $<E>=\iiint \psi ^{*}H\psi d^{3}x=\iiint \psi ^{*}\left(-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi +V\psi \right)d^{3}x$, Using: $\psi ^{*}\nabla ^{2}\psi =\nabla \left(\psi ^{*}\nabla \psi \right)-\nabla \psi ^{*}\nabla \psi$, hence: $<E>=\iiint \left(-{\frac {\hbar ^{2}}{2m}}\right)\left(\nabla \left(\psi ^{*}\nabla \psi \right)-\nabla \psi ^{*}\nabla \psi \right)d^{3}x+\iiint \psi ^{*}\nabla \psi d^{3}x$ $=-{\frac {\hbar ^{2}}{2m}}\iiint \nabla \left(\psi ^{*}\nabla \psi \right)d^{3}x+{\frac {\hbar ^{2}}{2m}}\iiint \nabla \psi ^{*}\nabla \psi d^{3}x+\iiint \psi ^{*}V\psi d^{3}x$,

Using Gauss Theorem for the last term: $-{\frac {\hbar ^{2}}{2m}}\iiint \nabla \left(\psi ^{*}\nabla \psi \right)d^{3}x=\iint \psi ^{*}\nabla \psi \cdot d{\textbf {S}}$, with the condition: $\lim _{r\to \infty }\psi ^{*}\nabla \psi =0$, for infinite surface.

Hence:$<E>=\int Wd^{3}x=\int \left[{\frac {\hbar ^{2}}{2m}}\nabla \psi ^{*}\cdot \nabla \psi \right]d^{3}x$

(2):first we find the time derivative of energy density:

${\frac {\partial W}{\partial t}}={\frac {\partial }{\partial t}}\left(\nabla \psi ^{*}\nabla \psi +\psi ^{*}\nabla \psi \right)={\frac {\hbar ^{2}}{2m}}\left(\nabla \psi ^{*}\nabla {\frac {\partial \psi }{\partial t}}+\nabla {\frac {\partial \psi ^{*}}{\partial t}}\nabla \psi \right)+{\frac {\partial \psi ^{*}}{\partial t}}\nabla \psi +\psi ^{*}\nabla {\frac {\partial \psi }{\partial t}}$, $={\frac {\hbar ^{2}}{2m}}\left(\nabla \cdot \left(\nabla \psi ^{*}\cdot {\frac {\partial \psi }{\partial t}}+{\frac {\partial \psi ^{*}}{\partial t}}\cdot \nabla \psi \right)-\left({\frac {\partial \psi }{\partial t}}\nabla ^{2}\psi ^{*}+{\frac {\partial \psi ^{*}}{\partial t}}\nabla ^{2}\psi \right)\right)+{\frac {\partial \psi ^{*}}{\partial t}}\nabla \psi +\psi ^{*}\nabla {\frac {\partial \psi }{\partial t}}$ $={\frac {\hbar ^{2}}{2m}}\nabla \cdot \left(\nabla \psi ^{*}\cdot {\frac {\partial \psi }{\partial t}}+{\frac {\partial \psi ^{*}}{\partial t}}\cdot \nabla \psi \right)+{\frac {\partial \psi ^{*}}{\partial t}}\left(-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi +\nabla \psi \right)+{\frac {\partial \psi }{\partial t}}\left(-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi ^{*}+\nabla \psi ^{*}\right)$,

Using Schrodinger Equations: $i\hbar {\frac {\partial \psi }{\partial t}}=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi +\nabla \psi$, and, $-i\hbar {\frac {\partial \psi ^{*}}{\partial t}}=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi ^{*}+\nabla \psi ^{*}$,

Also the energy flux density is: ${\textbf {S}}=-{\frac {\hbar ^{2}}{2m}}\left({\frac {\partial \psi ^{*}}{\partial t}}\nabla \psi +{\frac {\partial \psi }{\partial t}}\nabla \psi ^{*}\right)$,

So:${\frac {\partial W}{\partial t}}=-\nabla \cdot {\textbf {S}}+{\frac {\partial \psi ^{*}}{\partial t}}{\frac {\partial \psi }{\partial t}}-{\frac {\partial \psi }{\partial t}}{\frac {\partial \psi ^{*}}{\partial t}}=-\nabla \cdot {\textbf {S}}$, Hence: ${\frac {\partial W}{\partial t}}+\nabla \cdot {\textbf {S}}=0$