Phy5645/Energy conservation

Example 1

(1) The energy operator in three dimensions is: $H=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V$ so the average energy in state $\psi$ is: $\left\langle E\right\rangle =\iiint \psi ^{\ast }H\psi \,d^{3}{\textbf {r}}=\iiint \psi ^{\ast }\left(-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi +V\psi \right)\,d^{3}{\textbf {r}}$

Using the identity, $\psi ^{*}\nabla ^{2}\psi =\nabla \cdot \left(\psi ^{*}\nabla \psi \right)-\nabla \psi ^{\ast }\cdot \nabla \psi ,$ we obtain

$\left\langle E\right\rangle =-{\frac {\hbar ^{2}}{2m}}\iiint \left(\nabla \cdot \left(\psi ^{\ast }\nabla \psi \right)-\nabla \psi ^{\ast }\cdot \nabla \psi \right)\,d^{3}{\textbf {r}}+\iiint \psi ^{\ast }V\psi \,d^{3}{\textbf {r}}$ $=-{\frac {\hbar ^{2}}{2m}}\iiint \nabla \cdot \left(\psi ^{\ast }\nabla \psi \right)\,d^{3}{\textbf {r}}+{\frac {\hbar ^{2}}{2m}}\iiint \nabla \psi ^{\ast }\cdot \nabla \psi \,d^{3}{\textbf {r}}+\iiint \psi ^{\ast }V\psi \,d^{3}{\textbf {r}}$

If we apply Gauss' Theorem to the first term,

$-{\frac {\hbar ^{2}}{2m}}\iiint \nabla \left(\psi ^{*}\nabla \psi \right)d^{3}x=\iint \psi ^{*}\nabla \psi \cdot d{\textbf {S}},$

as well as the condition, $\lim _{r\to \infty }\psi ^{*}\nabla \psi =0,$ we obtain

$\left\langle E\right\rangle =\int W\,d^{3}{\textbf {r}}=\int \left[{\frac {\hbar ^{2}}{2m}}\nabla \psi ^{\ast }\cdot \nabla \psi \right]d^{3}{\textbf {r}}$

(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: ${\frac {\partial W}{\partial t}}+\nabla \cdot {\textbf {S}}=0$

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