# Probability and charge densities

The $N$-electron _probability_ density $\require{physics} \rho_{\text{prob}}(\vb{r})$ [integrates to the number of electrons](#n-electron-probability-densities-and-probability-current-densities):
\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](#maxwells-equations) 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](#maxwells-equations)
\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](#one-electron-probability-densities-and-probability-current-densities)
\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.
