The one-electron Pauli Hamiltonian
One thing that we are still missing in the Schrödinger treatment of the molecular Hamiltonian is the interaction of the electron spin with the electromagnetic field. Following Dyall (G. Dyall and Faegri 2007), we see that Lévy-Leblond (Lévy-Leblond 1967) has noted that formally substituting \[\begin{align} \require{physics} &\boldsymbol{\pi}^c \rightarrow \boldsymbol{\sigma} \vdot \boldsymbol{\pi}^c \\ % & \phi(\vb{r}) \rightarrow \phi(\vb{r}) \vb{I}_2 \end{align}\] leads to the Pauli Hamiltonian, which recovers the interaction of electron spin with the external electromagnetic field through the introduction of the Pauli matrices. By using this Lévy-Leblond substitution, we recover the one-electron Pauli Hamiltonian \[\begin{equation} \mathcal{H}^c(\phi, \vb{A}) = \frac{1}{2} ( \boldsymbol{\sigma} \vdot \boldsymbol{\pi}^c(\vb{A}))^2 ) - \phi(\vb{r}) \vb{I}_2 \thinspace , \end{equation}\] which describes the time-independent interaction of an electron with an electromagnetic field characterized by the potentials \((\phi, \vb{A})\).
Working out the square through the application of Dirac’s relation and by using \[\begin{equation} \boldsymbol{\pi}^c(\vb{A}) \cross \boldsymbol{\pi}^c(\vb{A}) = -i \thinspace \vb{B}(\vb{r}) \thinspace , \end{equation}\] we find that the Pauli Hamiltonian consists of two parts: \[\begin{equation} \mathcal{H}^c(\phi, \vb{A}) = \qty[ k^c(\vb{A}) - \phi(\vb{r}) ] \vb{I}_2 + M(\vb{B}) \thinspace , \end{equation}\] in which the spin-magnetic field interaction \(M(\vb{B})\) is a \((2 \times 2)\)-matrix operator given by: \[\begin{equation} M(\vb{B}) = \frac{1}{2} \boldsymbol{\sigma} \vdot \vb{B}(\vb{r}) \thinspace , \end{equation}\] which is a matrix paramagnetic term. The Pauli Hamiltonian (for only one electron) is thus a \((2 \times 2)\)-matrix operator: \[\begin{equation} \mathcal{H}^c(\phi, \vb{A}) = \begin{pmatrix} k^c(\vb{A}) - \phi(\vb{r}) + B_z(\vb{r}) / 2 % & B_x(\vb{r}) / 2 - i B_y(\vb{r}) / 2 \\ % B_x(\vb{r}) / 2 + i B_y(\vb{r}) / 2 % & k^c(\vb{A}) - \phi(\vb{r}) - B_z(\vb{r}) / 2 \end{pmatrix} \thinspace , \end{equation}\] whose diagonal contributions consists of the Schrödinger Hamiltonian, modified with the interaction of the spin of the electron with the magnetic field. Spin can thus be regarded as internal property of the electron that is only apparent when it interacts with an (external) magnetic field.
If an external uniform magnetic field \(\vb{B}_{\text{ext}}\) is applied, the spin-magnetic field interaction term becomes a constant matrix operator: \[\begin{equation} M_{\text{ext}}(\vb{B}_{\text{ext}}) = \frac{1}{2} \boldsymbol{\sigma} \vdot \vb{B}_{\text{ext}} \thinspace . \end{equation}\] If, in a molecular system, we also want to include the effects of the nuclear permanent magnetic dipole moments \(\qty{ \vb{M}_K }\), we must recognize that the total vector potential \(\vb{A}\) consists of two parts: \[\begin{equation} \vb{A} ( \vb{r}; \vb{B}_{\text{ext}}, \qty{ \vb{M}_K }, \vb{G} ) = \vb{A}_{\text{ext}} ( \vb{r}; \vb{B}_{\text{ext}}, \vb{G} ) + \vb{A}_{\text{nuc}} ( \vb{r}; \qty{ \vb{M}_K } ) \thinspace , \end{equation}\] such that the total magnetic field \(\vb{B}\) is given by: \[\begin{equation} \vb{B} ( \vb{r}; \vb{B}_{\text{ext}}, \qty{ \vb{M}_K } ) = \vb{B}_{\text{ext}} + \vb{B}_{\text{nuc}}( \vb{r}, \qty{ \vb{M}_K } ) % \thinspace , \end{equation}\] in which the part that is generated by the nuclear magnetic (dipole) moments \(\qty{ \vb{M}_K }\) can be divided in two contributions (T. Helgaker et al. 2012): \[\begin{equation} \vb{B}_{\text{nuc}}( \vb{r}, \qty{ \vb{M}_K } ) = \vb{B}_{\text{FC}} ( \vb{r}, \qty{ \vb{M}_K } ) + \vb{B}_{\text{SD}} ( \vb{r}, \qty{ \vb{M}_K } ) \thinspace , \end{equation}\] in which the first part \[\begin{equation} \vb{B}_{\text{FC}} ( \vb{r}, \qty{ \vb{M}_K } ) = \frac{8 \pi}{3 c^2} \sum_K \delta(\vb{r} - \vb{R}_K) \vb{M}_K \end{equation}\] gives rise to the Fermi contact operator: \[\begin{equation} M_{\text{FC}} ( \qty{ \vb{M}_K } ) = \frac{1}{2} \boldsymbol{\sigma} \vdot \vb{B}_{\text{FC}} ( \vb{r}, \qty{ \vb{M}_K } ) \end{equation}\] and the second part: \[\begin{equation} \vb{B}_{\text{SD}} ( \vb{r}, \qty{ \vb{M}_K } ) = \frac{1}{c^2} \sum_K \qty( \frac{ 3 (\vb{r} - \vb{R}_K) \qty[ \vb{M}_K \vdot (\vb{r} - \vb{R}_K) ] - \vb{M}_K \norm{\vb{r}- \vb{R}_K}^2 }{ \norm{\vb{r} - \vb{R}_K}^5 } ) \end{equation}\] gives rise to the spin-electric dipole operator: \[\begin{equation} M_{\text{SD}} ( \qty{ \vb{M}_K } ) = \frac{1}{2} \boldsymbol{\sigma} \vdot \vb{B}_{\text{SD}} ( \vb{r}, \qty{ \vb{M}_K } ) \thinspace . \end{equation}\] The total paramagnetic spin-magnetic field interaction term is then given by: \[\begin{equation} M ( \vb{B}_{\text{ext}}, \qty{ \vb{M}_K } ) = M_{\text{ext}} ( \vb{B}_{\text{ext}} ) + M_{\text{FC}} ( \qty{ \vb{M}_K } ) + M_{\text{SD}} ( \qty{ \vb{M}_K } ) \thinspace . \end{equation}\]
Since the Pauli Hamiltonian is a \((2 \times 2)\)-matrix operator, it should accordingly act on a 2-component wave function: \[\begin{equation} \begin{split} \Psi(\vb{r}) : \qquad & \mathbb{R}^3 \rightarrow \mathbb{C}^2 \\ % & \Psi(\vb{r}) = \begin{pmatrix} \psi_{\alpha}(\vb{r}) \\ \psi_{\beta}(\vb{r}) \end{pmatrix} \thinspace , \end{split} \end{equation}\] which is called a (Pauli) spinor.
If the external magnetic field \(\require{physics} \vb{B}_{\text{ext}}\) only has a component in the \(z\)-direction (which is often the case, as most of the magnetic field calculations consist of applying a uniform magnetic field along the \(z\)-axis), the one-electron Pauli Hamiltonian \(\eqref{eq:Pauli_Hamiltonian_2_component}\) reduces to a diagonal form: \[\begin{equation} \mathcal{H}^c(\phi, \vb{A}) = \begin{pmatrix} \norm{\boldsymbol{\pi}^c}^2 / 2 - \phi_{\text{ext}}(\vb{r}) + B_{\text{ext}, \thinspace z}(\vb{r}) / 2 & 0 \\ 0 & \norm{\boldsymbol{\pi}^c}^2 / 2 - \phi_{\text{ext}}(\vb{r}) - B_{\text{ext}, \thinspace z}(\vb{r}) / 2 \end{pmatrix} \thinspace . \end{equation}\]
Due to this diagonal form, it will commute with \(\sigma_z\), such that we may choose to quantize the one- and two-electron operators in a spin-separated spinor bases.
Furthermore, if no magnetic fields are present, the Pauli Hamiltonian reduces to a diagonal form: \[\begin{equation} \mathcal{H}^c(\phi, \vb{A}_{0}) = \begin{pmatrix} T^c - \phi(\vb{r}) & 0 \\ % 0 & T^c - \phi(\vb{r}) % \end{pmatrix} \thinspace , \end{equation}\] such that the Pauli Hamiltonian eigenvalue problem (for one electron) for a \(2\)-component spinor reduces to the Schrödinger Hamiltonian eigenvalue problem for a scalar state function: \[\begin{equation} \qty[T^c - \phi(\vb{r})] \psi(\vb{r}) = E \psi(\vb{r}) % \thinspace , \end{equation}\] where \(\psi\) represents either component of the \(2\)-spinor: \[\begin{equation} \psi(\vb{r}) = \psi_\alpha(\vb{r}) = \psi_\beta(\vb{r}) \thinspace , \end{equation}\] whose eigenvalues \(E\) are two-fold degenerate.