Argument for spin-separated spinor bases

Since the Pauli Hamiltonian in a simple magnetic field has a simplified diagonal form, the \(\alpha\)- and \(\beta\)-components of Pauli spinors cannot couple: \[\begin{align} \require{physics} h_{PQ} &= \int \dd{\vb{r}} \phi_P^\dagger(\vb{r}) \thinspace \mathcal{H}^c \thinspace \phi_Q(\vb{r}) \\ &= \int \dd{\vb{r}} \phi_{P \alpha}^*(\vb{r}) \thinspace \mathcal{H}^{c, \alpha \alpha} \thinspace \phi_{Q \alpha}(\vb{r}) + \int \dd{\vb{r}} \phi_{P \beta}^*(\vb{r}) \thinspace \mathcal{H}^{c, \beta \beta} \thinspace \phi_{Q \beta}(\vb{r}) \thinspace . \end{align}\]

This seemingly small observation has many simplificational consequenses. Due to this diagonal form, we might as well restructure the spinor basis of the \(M\) spinors in a simplified way as \(M = K_\alpha + K_\beta\) spinors in which the first \(K_\alpha\) spinors have a zero \(\beta\) component and the last \(K_\beta\) spinors have a zero \(\alpha\) component: \[\begin{equation} \require{physics} \begin{cases} \phi_{p}(\vb{r}) = \begin{pmatrix} \phi_{p \alpha}(\vb{r}) \\ 0 \end{pmatrix} & \qquad \qquad 1 \leq p \leq K_\alpha \\ % \phi_{p}(\vb{r}) = \begin{pmatrix} 0 \\ \phi_{p \beta}(\vb{r}) \end{pmatrix} & \qquad \qquad K_\alpha < p \leq M \thinspace , \end{cases} \end{equation}\] which is a situation that we will refer to as a spin-separated spinor basis.

This all has profound rammifications for the integrals of the second-quantized Pauli Hamiltonian. The Pauli Hamiltonian matrix elements between two spinors with indices lower than or equal to \(K_\alpha\) or larger than \(K_\alpha\) now reduce to integrals over scalar functions. Similarly, the matrix elements between spinors in these two disjoint sets vanish. If \(p\) and \(q\) both belong to the \(\alpha\)-set, we have: \[\begin{equation} \matrixel{\phi_{p}}{\mathcal{H}^c}{\phi_{q}} = \int \dd{\vb{r}} \phi_{p \alpha}^*(\vb{r}) \qty( \frac{ % \norm{\boldsymbol{\pi}^c}^2 }{2} - \phi_{\text{ext}}(\vb{r}) + \frac{ % B_{\text{ext}, \thinspace z}(\vb{r}) }{2} ) \phi_{q \alpha}(\vb{r}) \thinspace , \end{equation}\] while if \(p\) and \(q\) both belong to the \(\beta\)-spinors, we have: \[\begin{equation} \matrixel{\phi_{p}}{\mathcal{H}^c}{\phi_{q}} = \int \dd{\vb{r}} \phi_{p \beta}^*(\vb{r}) \qty( \frac{ \norm{\boldsymbol{\pi}^c}^2 }{2} - \phi_{\text{ext}}(\vb{r}) - \frac{ B_{\text{ext}, \thinspace z}(\vb{r}) }{2} ) \phi_{q \beta}(\vb{r}) \thinspace . \end{equation}\] If the indices \(p\) and \(q\) correspond to spinors of different type, the one-electron integrals vanish: \[\begin{equation} \matrixel{\phi_{p}}{\mathcal{H}^c}{\phi_{q}} = 0 \thinspace . \end{equation}\]

The one-electron operator that occurs in the second-quantized Hamiltonian \[\begin{equation} \hat{h} = \sum_{pq}^M h_{pq} \hat{a}^\dagger_{p} \hat{a}_{q} \end{equation}\] can then be simplified to: \[\begin{align} \hat{h} &= \sum_{pq}^{K_\alpha} h_{p\alpha, q\alpha} \hat{a}^\dagger_{p \alpha} \hat{a}_{q \alpha} + \sum_{pq}^{K_\beta} h_{p\beta, q\beta} \hat{a}^\dagger_{p \beta} \hat{a}_{q \beta} \\ & = \sum_{\sigma} \sum_{pq}^{K_\sigma} h_{p\sigma, q\sigma} \hat{a}^\dagger_{p \sigma} \hat{a}_{q \sigma} \thinspace , \end{align}\] in which the creation operator \(\hat{a}^\dagger_{p \alpha}\) creates a spinor which has an \(\alpha\)-component consisting of the spatial orbital \(\phi_{p\alpha}(\vb{r})\) and a \(\beta\)-component that is zero. We have now also introduced the labels \(\sigma\) and \(\tau\), which label either \(\alpha\) or \(\beta\).