In the text full units are used for clarity, but in practice one can employ SI-based atomic units, setting \(\hbar=m_{e}=e=4\pi\varepsilon_{0}=1\).

Pauli spin matrices¶

The Pauli spin matrices

\[\begin{split}\sigma_{x}=\left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right];\quad\sigma_{y}=\left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right];\quad\sigma_{z}=\left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right]\end{split}\]

are representation matrices of of the operator \(2\mathbf{s}\) in the basis \(\left\{ \left|\alpha\right\rangle ,\left|\beta\right\rangle \right\}\) of spin-1/2 functions (spin-up and spin-down, respectively), that is \(2\mathbf{s}\rightarrow\hbar\boldsymbol{\sigma}\). These spin functions obey the following relations

\[\begin{split}\begin{array}{lclclcl} \hat{s}_{z}\left|\alpha\right\rangle & = & +\frac{1}{2}\hbar\left|\alpha\right\rangle & & \hat{s}_{z}\left|\beta\right\rangle & = & -\frac{1}{2}\hbar\left|\beta\right\rangle \\ \hat{s}_{+}\left|\alpha\right\rangle & = & 0 & & \hat{s}_{+}\left|\beta\right\rangle & = & \hbar\left|\alpha\right\rangle \\ \hat{s}_{-}\left|\alpha\right\rangle & = & \hbar\left|\beta\right\rangle & & \hat{s}_{-}\left|\beta\right\rangle & = & 0 \end{array}\end{split}\]

where appears the ladder operators \(s_{\pm}=s_{x}\pm is_{y}\). The spin-1/2 functions are furthermore orthonormal. Let us see how the Pauli \(\sigma_{z}\) matrix is generated:

Element (1,1): \(\left\langle \alpha\left|2s_{z}\right|\alpha\right\rangle /\hbar=\left\langle \alpha\left|\right.\alpha\right\rangle =1\)

Element (1,2): \(\left\langle \alpha\left|2s_{z}\right|\beta\right\rangle /\hbar=\left\langle \alpha\left|\right.\beta\right\rangle =0\)

Element (2,1): \(\left\langle \beta\left|2s_{z}\right|\alpha\right\rangle \hbar=\left\langle \beta\left|\right.\alpha\right\rangle =0\)

Element (2,2): \(\left\langle \beta\left|2s_{z}\right|\beta\right\rangle /\hbar=-\left\langle \beta\left|\right.\beta\right\rangle =-1\)

In the same manner construct representation matrices for the ladder operators \(s_{+}\) and \(s_{-}\) and from this \(\sigma_{x}\) and \(\sigma_{y}\).

Demonstrate, for general vector operators \(\boldsymbol{A}\) and \(\boldsymbol{B}\), the Dirac identity

\[\left(\boldsymbol{\sigma}\cdot\boldsymbol{A}\right)\left(\boldsymbol{\sigma} \cdot\boldsymbol{B}\right)=\boldsymbol{A}\cdot\boldsymbol{B}+i\boldsymbol{\sigma} \cdot\left(\boldsymbol{A}\times\boldsymbol{B}\right)\]

Use the Dirac identity to calculate

\(\left(\boldsymbol{\sigma}\cdot\hat{\boldsymbol{p}}\right)\left(\boldsymbol{\sigma}\cdot\hat{\boldsymbol{p}}\right)\)

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

Show that \(\left(\boldsymbol{\sigma}\cdot\hat{\boldsymbol{p}}\right)V_{eN}\left(\boldsymbol{\sigma}\cdot\hat{\boldsymbol{p}}\right)=\boldsymbol{p}V_{eN}\cdot\boldsymbol{p}+\hbar\left(\frac{Ze^{2}}{2\pi\varepsilon_{0}r^{3}}\right)\hat{\boldsymbol{s}}\cdot\hat{\boldsymbol{\ell}}\), where \(V_{eN}\) is the electron-nucleus interaction \(V_{en}=-\frac{Ze^{2}}{4\pi\varepsilon_{0}r}\) in an atom