Probability Current and Continuity Equation

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To set the stage we start with the Schrödinger Equation in Position Space. \begin{align} i \, \hbar \frac{\partial}{\partial t} \psi(\mathbf{r},t)& = \frac{-\hbar^2 }{2m} \nabla ^2 \psi(\mathbf{r},t)+ V(\mathbf{r}) \psi(\mathbf{r},t)\\ \tag{1} \end{align} and the complex conjugate of this equation \begin{align} i \, \hbar \frac{\partial}{\partial t} \psi^*(\mathbf{r},t)& = \frac{-\hbar^2 }{2m} \nabla ^2 \psi^*(\mathbf{r},t) + V(\mathbf{r}) \psi^*(\mathbf{r},t)\\ \tag{2} \end{align} The solution the Schrödinger Equation describes a particle with the normalised wave function $\psi(\mathbf{r},t)$. We define the probability density as \begin{align} \rho(\mathbf{r},t) = | \psi(\mathbf{r},t)|^2 \end{align} At time $t$ the probability $dP(\mathbf{r},t)$ to find the particle at position $\mathbf{r}$ in the infinitesimal volume element $dr^3$ is \begin{align} dP(\mathbf{r},t)=| \psi(\mathbf{r},t)|^2 dr^2=\rho(\mathbf{r},t)dr^3 \end{align}

Now we multiply (1) by $\psi^*(\mathbf{r},t)$ and (2) by $\psi(\mathbf{r},t)$ and subtract (2) from (1). This gives \begin{align} \psi^*(\mathbf{r},t)i \, \hbar \frac{\partial}{\partial t} \psi(\mathbf{r},t)-\psi(\mathbf{r},t)i \, \hbar \frac{d}{dt} \psi^*(\mathbf{r},t)& = \psi^*(\mathbf{r},t)\frac{-\hbar^2 }{2m} \nabla ^2 \psi(\mathbf{r},t)+ V(\mathbf{r}) \psi^*(\mathbf{r},t) \psi(\mathbf{r},t)-\psi(\mathbf{r},t)\frac{-\hbar^2 }{2m} \nabla ^2 \psi^*(\mathbf{r},t) - V(\mathbf{r}) \psi(\mathbf{r},t)\psi^*(\mathbf{r},t)\\ \tag{3} \end{align} On the LHS we can pull the derivative and $i\hbar$ in front to get $i \, \hbar \frac{d}{dt}\psi^(\mathbf{r},t)\psi^*(\mathbf{r},t)$. On the RHS the potential terms simply cancel. So we have \begin{align} i \, \hbar \frac{\partial}{\partial t} |\psi(\mathbf{r},t)|^2 & = \psi^*(\mathbf{r},t)\frac{-\hbar^2 }{2m} \nabla ^2 \psi(\mathbf{r},t)+ -\psi(\mathbf{r},t)\frac{-\hbar^2 }{2m} \nabla ^2 \psi^*(\mathbf{r},t) \\ \tag{4} \end{align} Now we pull out the factor $\frac{-\hbar^2 }{2m}$ and apply Greens first identity \begin{align} \nabla \cdot (\phi \,\nabla \psi)= \phi \, \nabla^2 \psi+ \nabla \phi \, \nabla \psi\\ \end{align}

on the RHS \begin{align} i \, \hbar \frac{\partial}{\partial t} |\psi(\mathbf{r},t)|^2 & = \frac{-\hbar^2 }{2m} \nabla \left[ \psi^*(\mathbf{r},t) \nabla \psi(\mathbf{r},t) - \psi(\mathbf{r},t) \nabla \psi^*(\mathbf{r},t) \right] \\ \tag{5} \end{align} bringing $i \, \hbar$ on the other side we have \begin{align} \frac{\partial}{\partial t} |\psi(\mathbf{r},t)|^2 & = \nabla \frac{\hbar }{2mi} \left[ \psi(\mathbf{r},t) \nabla \psi^*(\mathbf{r},t) - \psi^*(\mathbf{r},t) \nabla \psi(\mathbf{r},t)\right] \\ \tag{6} \end{align} We call the the expression \begin{align} \vec{j}(\mathbf{r},t)=\frac{\hbar }{2mi} \left[ \psi(\mathbf{r},t) \nabla \psi^*(\mathbf{r},t) - \psi^*(\mathbf{r},t) \nabla \psi(\mathbf{r},t)\right] \\ \tag{7} \end{align} is the probability current of the wave function $\psi(\mathbf{r},t)$. Notice, $\vec{j}(\mathbf{r},t)$ is a real quantity since $\psi(\mathbf{r},t) \nabla \psi^*(\mathbf{r},t)$ minus its complex conjugate gives an imaginary number, but we devide by $i$ so $\vec{j}(\mathbf{r},t)$ must be real.

rewriting equ (7) using the probability density we find

the Continuity Equation for the probability density \begin{align} \frac{\partial}{\partial t} \rho(\mathbf{r},t) & = \nabla \vec{j}(\mathbf{r},t) \tag{8}\\ \end{align} with the probability current \begin{align} \vec{j}(\mathbf{r},t)=\frac{\hbar }{2mi} \left[ \psi(\mathbf{r},t) \nabla \psi^*(\mathbf{r},t) - \psi^*(\mathbf{r},t) \nabla \psi(\mathbf{r},t)\right] \\ \end{align}

This has the profound implication that the probability density is conserved \begin{align} \frac{\partial}{\partial t} \rho(\mathbf{r},t) - \nabla \vec{j}(\mathbf{r},t)=0 \tag{9}\\ \end{align} This is just another fact underlining that the Schrödinger Equation transforms unitary in time and preserves the norm of the wave function $\psi(\mathbf{r},t)$ as we discussed in the Heisenberg Picture.