Probability and charge densities

The \(N\)-electron probability density \(\require{physics} \rho_{\text{prob}}(\vb{r})\) integrates to the number of electrons: \[\begin{equation} \int \dd{\vb{r}} \rho_{\text{prob}}(\vb{r}) = N \thinspace . \end{equation}\] By multiplying both sides of this equation by the electronic charge \((-e)\), which becomes \((-1)\) in atomic units, we find that the \(N\)-electron charge density integrates to the total electronic charge: \[\begin{equation} \int \dd{\vb{r}} \rho_{\text{charge}}(\vb{r}) = -Ne \thinspace , \end{equation}\] where we have adequately defined the charge density in terms of the probability density: \[\begin{equation} \rho_{\text{charge}}(\vb{r}) = - \rho_{\text{prob}}(\vb{r}) \thinspace . \end{equation}\]

In order for the (charge) continuity equation \[\begin{equation} \pdv{\rho_{\text{charge}}(\vb{r})}{t} + \boldsymbol{\nabla} \vdot{\vb{j}_{\text{charge}}(\vb{r})} = 0 \end{equation}\] to hold, we multiply the (probability) continuity equation \[\begin{equation} \pdv{\rho_{\text{prob}}(\vb{r})}{t} + \boldsymbol{\nabla} \vdot{\vb{j}_{\text{prob}}(\vb{r})} = 0 \end{equation}\] with the electronic charge \((-1)\), we must then make the following connection between the charge current density and the probability current density: \[\begin{equation} \vb{j}_{\text{charge}}(\vb{r}) = - \vb{j}_{\text{prob}}(\vb{r}) \thinspace . \end{equation}\]

It is often the case that, for notational convenience, the symbol \(\vb{j}\) is used for either the charge current density or the probability current density, leaving the reader to the context to determine which current density is used.