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the_schroedinger_equation [2021/03/17 21:23] – [3.iii.3 The Continuity Equation] adminthe_schroedinger_equation [2021/03/17 21:25] (current) – [3.iii.3 The Continuity Equation] admin
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 The next step is to subtract equation \eqref{eq2} from equation \eqref{eq1}, which gives The next step is to subtract equation \eqref{eq2} from equation \eqref{eq1}, which gives
-\[i\hbar \left [ \psi^*(\vec{r},t)\frac{\partial \psi(\vec{r},t)}{\partial t} + frac{\partial \psi^*(\vec{r},t)}{\partial t}\psi(\vec{r},t) \right ] = -\frac{\hbar^2}{2m} \left [ \psi^*(\vec{r},t)\nabla^2 \psi(\vec{r},t) - \psi(\vec{r},t)\nabla^2 \psi^*(\vec{r},t) \right ].\]+\[i\hbar \left [ \psi^*(\vec{r},t)\frac{\partial \psi(\vec{r},t)}{\partial t} + \frac{\partial \psi^*(\vec{r},t)}{\partial t}\psi(\vec{r},t) \right ] = -\frac{\hbar^2}{2m} \left [ \psi^*(\vec{r},t)\nabla^2 \psi(\vec{r},t) - \psi(\vec{r},t)\nabla^2 \psi^*(\vec{r},t) \right ].\]
 Before doing anything else, let's just divide both sides of this equation by $i\hbar$, which gives Before doing anything else, let's just divide both sides of this equation by $i\hbar$, which gives
 \begin{equation} \begin{equation}
 \label{eq3} \label{eq3}
-\psi^*(\vec{r},t)\frac{\partial \psi(\vec{r},t)}{\partial t} + frac{\partial \psi^*(\vec{r},t)}{\partial t}\psi(\vec{r},t) = \frac{i\hbar}{2m} \left [ \psi^*(\vec{r},t)\nabla^2 \psi(\vec{r},t) - \psi(\vec{r},t)\nabla^2 \psi^*(\vec{r},t) \right ].+\psi^*(\vec{r},t)\frac{\partial \psi(\vec{r},t)}{\partial t} + \frac{\partial \psi^*(\vec{r},t)}{\partial t}\psi(\vec{r},t) = \frac{i\hbar}{2m} \left [ \psi^*(\vec{r},t)\nabla^2 \psi(\vec{r},t) - \psi(\vec{r},t)\nabla^2 \psi^*(\vec{r},t) \right ].
 \end{equation} \end{equation}
  
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 Again, by the product rule for derivatives, we have Again, by the product rule for derivatives, we have
 \begin{align*} \begin{align*}
-\vec{\nabla} \cdot \vec{J}(\vec{r},t) = \frac{i\hbar}{2m} \left [ \psi(\vec{r},t) \nabla^2 \psi^*(\vec{r},t) + \vec{\nabla}\psi(\vec{r},t) \cdot \vec{\nabla} \psi^*(\vec{r},t) - \vec{\nabla}\psi^*(\vec{r},t) \cdot \vec{\nabla} \psi(\vec{r},t) - \psi^*(\vec{r},t)\nabla^2 \psi(\vec{r},t)\right ] \\+\vec{\nabla} \cdot \vec{J}(\vec{r},t) & = \frac{i\hbar}{2m} \left [ \psi(\vec{r},t) \nabla^2 \psi^*(\vec{r},t) + \vec{\nabla}\psi(\vec{r},t) \cdot \vec{\nabla} \psi^*(\vec{r},t) - \vec{\nabla}\psi^*(\vec{r},t) \cdot \vec{\nabla} \psi(\vec{r},t) - \psi^*(\vec{r},t)\nabla^2 \psi(\vec{r},t)\right ] \\
 & = \frac{i\hbar}{2m} \left [ \psi(\vec{r},t) \nabla^2 \psi^*(\vec{r},t) +  - \psi^*(\vec{r},t)\nabla^2 \psi(\vec{r},t)\right ]. & = \frac{i\hbar}{2m} \left [ \psi(\vec{r},t) \nabla^2 \psi^*(\vec{r},t) +  - \psi^*(\vec{r},t)\nabla^2 \psi(\vec{r},t)\right ].
 \end{align*} \end{align*}