In the text full units are used for clarity, but in practice one can employ SI-based atomic units, setting $\hslash =m_e=e=4\pi {\epsilon }_0=1$.

Pauli spin matrices

The Pauli spin matrices

$$\begin{array}{r}{\sigma }_x=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right];{\sigma }_y=\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right];{\sigma }_z=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]\end{array}$$

are representation matrices of of the operator $2\mathbf{s}$ in the basis $\{|\alpha ⟩,|\beta ⟩\}$ of spin-1/2 functions (spin-up and spin-down, respectively), that is $2\mathbf{s}\to \hslash \mathit{\sigma }$. These spin functions obey the following relations

$$\begin{array}{r}\begin{array}{lclclcl}{\hat{s}}_z|\alpha ⟩& =& +\frac{1}{2}\hslash |\alpha ⟩& & {\hat{s}}_z|\beta ⟩& =& -\frac{1}{2}\hslash |\beta ⟩\\ {\hat{s}}_{+}|\alpha ⟩& =& 0& & {\hat{s}}_{+}|\beta ⟩& =& \hslash |\alpha ⟩\\ {\hat{s}}_{-}|\alpha ⟩& =& \hslash |\beta ⟩& & {\hat{s}}_{-}|\beta ⟩& =& 0\end{array}\end{array}$$

where appears the ladder operators $s_{\pm }=s_x\pm i{s}_y$. The spin-1/2 functions are furthermore orthonormal. Let us see how the Pauli ${\sigma }_z$ matrix is generated:

• Element (1,1): $⟨\alpha \left|2s_z\right|\alpha ⟩/\hslash =⟨\alpha |\alpha ⟩=1$

• Element (1,2): $⟨\alpha \left|2s_z\right|\beta ⟩/\hslash =⟨\alpha |\beta ⟩=0$

• Element (2,1): $⟨\beta \left|2s_z\right|\alpha ⟩\hslash =⟨\beta |\alpha ⟩=0$

• Element (2,2): $⟨\beta \left|2s_z\right|\beta ⟩/\hslash =-⟨\beta |\beta ⟩=-1$

1. In the same manner construct representation matrices for the ladder operators $s_{+}$ and $s_{-}$ and from this ${\sigma }_x$ and ${\sigma }_y$.

2. Demonstrate, for general vector operators $\mathit{A}$ and $\mathit{B}$, the Dirac identity

$$(\mathit{\sigma }\cdot \mathit{A})(\mathit{\sigma }\cdot \mathit{B})=\mathit{A}\cdot \mathit{B}+i\mathit{\sigma }\cdot (\mathit{A}\times \mathit{B})$$

3. Use the Dirac identity to calculate

• $(\mathit{\sigma }\cdot \hat{\mathit{p}})(\mathit{\sigma }\cdot \hat{\mathit{p}})$

• $(\mathit{\sigma }\cdot \hat{\mathit{\pi }})(\mathit{\sigma }\cdot \hat{\mathit{\pi }})$, where $\hat{\mathit{\pi }}$ is mechanical momentum $\hat{\mathit{\pi }}=\hat{\mathit{p}}+e\mathit{A}$ and $\mathit{A}$ is a vector potential. Show how this expression is simplified in Coulomb gauge: $\mathbf{\nabla }\cdot \mathit{A}=0$.

• Show that $(\mathit{\sigma }\cdot \hat{\mathit{p}})V_{eN}(\mathit{\sigma }\cdot \hat{\mathit{p}})=\mathit{p}{V}_{eN}\cdot \mathit{p}+\hslash \left(\frac{Z{e}^2}{2\pi {\epsilon }_0{r}^3}\right)\hat{\mathit{s}}\cdot \hat{\mathit{\ell }}$, where $V_{eN}$ is the electron-nucleus interaction $V_{en}=-\frac{Z{e}^2}{4\pi {\epsilon }_{0}r}$ in an atom