We can now define the so-called charge operator as $$ \frac{\hat{Q}}{e} = \frac{1}{2}\left(1-\hat{\tau}_z\right)=\begin{Bmatrix} 0 & 0 \\ 0 & 1 \end{Bmatrix}, $$ which results in $$ \frac{\hat{Q}}{e}\psi^p(\mathbf{r})=\psi^p(\mathbf{r}), $$ and $$ \frac{\hat{Q}}{e}\psi^n(\mathbf{r})=0, $$ as it should be.