Probability density and probability current density operators
If we wish to find the operator whose expectation value is exactly the Pauli probability density, Helgaker (T. Helgaker, Jørgensen, and Olsen 2000) gives a nice reasoning that we can use Dirac-delta functions \(\require{physics} \delta(\vb{r} - \vb{r}_1)\) to probe the presence of electron 1 at the position \(\vb{r}\). Indeed, we can write the coordinate representation of the density operator as \[\begin{equation} \rho^c(\vb{r}; \vb{r}_1) = \delta(\vb{r} - \vb{r}_1) \thinspace \vb{I}_2 \thinspace , \end{equation}\] where \(\vb{r}_1\) is the dummy variable that will be integrated over when taking the expectation value, and is thus often just left away in the operator symbol to simplify notation. The identity matrix \(\vb{I}_2\) is included because in Pauli theory, any operator must be a \((2 \times 2)\)-matrix, but often it is left away for notational convenience. \(\rho^c(\vb{r})\) is an operator (in coordinate representation) in the sense that the expectation value of this operator (with respect to the Pauli spinor \(\Psi(\vb{r}_1, t)\)) gives the Pauli probability density at the position \(\vb{r}\): \[\begin{align} \rho(\vb{r}, t) &= \int \dd{\vb{r}_1} \Psi^\dagger(\vb{r}_1, t) \thinspace \rho^c(\vb{r}) \thinspace \Psi(\vb{r}_1, t) \\ &= \Psi^\dagger(\vb{r}, t) \Psi(\vb{r}, t) \thinspace . \end{align}\] We can furthermore show that the (Pauli) probability current density operator \(\vb{j}^c(\vb{r})\) is the vector operator \[\begin{equation} \vb{j}^c(\vb{r}) = - \frac{i}{2} \Bigg( \delta(\vb{r} - \vb{r}_1) \thinspace \grad + \grad \thinspace \delta(\vb{r} - \vb{r}_1) + 2 i \thinspace \vb{A}(\vb{r}) \thinspace \delta(\vb{r} - \vb{r}_1) \Bigg) \vb{I}_2 \thinspace . \end{equation}\] When plugging this into the expectation value, we encounter following integral: \[\begin{equation} I = \int \dd{\vb{r}_1} \Psi^\dagger(\vb{r}_1, t) \grad \thinspace \delta(\vb{r} - \vb{r}_1) \Psi(\vb{r}_1, t) \thinspace , \end{equation}\] which, at first sight, seems to be very difficult to calculate because the gradient operator acts on both the Dirac delta function and the Pauli spinor. Part of this difficulty could also arise from being unfamiliar with how to deal with the gradient operator, the Hermitian adjoint of the gradient operator, and all this in conjunction with the Hermitian adjoint of the wave function. Since we are most comfortable with working with scalar functions instead of two-component spinors, let us expand the spinors in its components: \[\begin{equation} I = \sum_\sigma \int \dd{\vb{r}_1} \Psi^*_\sigma(\vb{r}_1, t) \grad \thinspace \delta(\vb{r} - \vb{r}_1) \Psi_\sigma(\vb{r}_1, t) \thinspace . \end{equation}\] We should now note that the gradient operator is not Hermitian, but we have the following relation: \[\begin{equation} \grad^\dagger = - \grad \thinspace . \end{equation}\] Therefore, we can simplify the integral to \[\begin{align} I &= - \sum_\sigma \int \dd{\vb{r}_1} \grad{ \Psi_\sigma(\vb{r}_1, t) } \thinspace \delta(\vb{r} - \vb{r}') \Psi_\sigma(\vb{r}_1, t) \\ &= - [ \grad{ \Psi^\dagger(\vb{r}, t) } ] \thinspace \Psi(\vb{r}, t) \thinspace , \end{align}\] in which we should emphasize that the gradient of the Hermitian adjoint of a spinor is the Hermitian adjoint of the two-vector composed of the gradients of the scalar components: \[\begin{equation} \grad{ \Psi^\dagger(\vb{r}, t) } \equiv \begin{pmatrix} \grad{ \Psi^*_\alpha(\vb{r}, t) } & \grad{ \Psi^*_\beta(\vb{r}, t) } \end{pmatrix} \thinspace . \end{equation}\] Indeed, we then find for the expectation value of probability current density operator: \[\begin{align} \vb{j}(\vb{r}, t) &= \int \dd{\vb{r}_1} \Psi^\dagger(\vb{r}_1, t) \thinspace \vb{j}^c(\vb{r}) \thinspace \Psi(\vb{r}_1, t) \\ &= -\frac{i}{2} \Bigg( \Psi^\dagger(\vb{r}, t) \thinspace \qty[ \grad{\Psi(\vb{r}, t)} ] - [ \grad{\Psi^\dagger(\vb{r}, t)} ] \thinspace \Psi(\vb{r}, t) % no qty + 2i \thinspace \Psi^\dagger(\vb{r}, t) \Psi(\vb{r}, t) \thinspace \vb{A}(\vb{r}, t) \Bigg) \thinspace , \end{align}\]