The course comprises two pen-and-paper exercises which are based on materials of the morning-session lectures of Day 1.

# Dirac’s relation¶

A relation that is often exploited throughout the course, e.g. to achieve a separation of spin-dependent and independent terms in the Dirac equation (see lecture on Day 2), is Dirac’s relation, which can be written for two arbitrary vector operators $$\vec{u}$$ and $$\vec{v}$$ as:

$(\vec{\sigma} \cdot \vec{u})(\vec{\sigma} \cdot \vec{v}) = \vec{u} \cdot \vec{v} I_{2} + i \vec{\sigma} \cdot (\vec{u} \times \vec{v})$

where $$\vec{\sigma}$$ are the Pauli spin matrices and $$I_{2}$$ is a $$2 \times 2$$ unit matrix. Note that $$\vec{u}$$ and $$\vec{v}$$ do not necessarily commute.

• Problem 1: Verify the relation expanding the left-hand side of Dirac’s relation. Hint: use the relations for the Pauli spin matrices:
$\vec{\sigma} = (\sigma_x, \sigma_y, \sigma_z)$$\sigma_x^2 = \sigma_y^2 = \sigma_z^2 = I_{2}$$\sigma_i \sigma_j = \delta_{ij} I_{2} + i \sum_{k=1}^{3} \epsilon_{ijk} \sigma_k$
• Problem 2: derive a final expression inserting for $$\vec{u} = \vec{v}$$ the kinematical momentum operator $$\vec{\pi} = \vec{p} - \frac{q_e}{c}\vec{A}$$ (where $$\vec{A}$$ is an external electromagnetic vector potential). Hint:
$[\pi_i,\pi_j] = \frac{q}{c} i \hbar \sum\limits_{k=1}^{3} \epsilon_{ijk} B_k$

# Two-component Pauli equation (0th order Pauli equation)¶

From the one-electron Dirac equation in external magnetic fields we may derive the Pauli equation (also known as 0th order Pauli equation) by considering the non-relativistic limit $$c \rightarrow \infty$$.

$\left[ \frac{\vec{p}^2}{2m_e} + \frac{q_e^2 \vec{A}^2}{2m_e c^2} - \frac{q_e}{2m_ec}(\bf{l} + 2 \bf{s}) \cdot \vec{B} + V \right] \Psi^L = i \hbar \frac{\partial}{\partial t} \Psi^L$
• Problem 3: Derive the above equation starting from the one-electron Dirac equation in external magnetic fields written in two-spinor form:
$\begin{split}i \hbar \frac{\partial}{\partial t} \left( \begin{array}{c} \Psi^L \\ \Psi^S \end{array} \right) = c \left( \begin{array}{c} (\vec{\sigma} \cdot \vec{\pi}) \Psi^S \\ (\vec{\sigma} \cdot \vec{\pi}) \Psi^L \end{array} \right) + m_ec^2 \left( \begin{array}{c} \Psi^L \\ -\Psi^S \end{array} \right) + V \left( \begin{array}{c} \Psi^L \\ \Psi^S \end{array} \right)\end{split}$

The following hints may be useful:

1. shift the energy limit to the non-relativistic limit (E = 0 rather than E = $$m_ec^2$$):
$V \rightarrow V - m_ec^2$
1. elimininate the small-component $$\Psi^S$$ from the upper component using the kinetic balance condition:
$\Psi^S \approx \frac{\vec{\sigma} \cdot \vec{\pi}}{2m_ec^2} \Psi^L$
1. use $$\vec{A} \cdot \vec{p} = \vec{p} \cdot \vec{A} - (\vec{p} \cdot \vec{A})$$
2. remember for the Coulomb gauge:
$a div(\vec{A}) = 0$
1. for a constant and homogeneous magnetic field $$\bf{B} = curl(\bf{A})$$, we may write $$\bf{A} = \frac{1}{2}(\bf{B} \times \bf{r})$$
• Problem 4: define the Bohr magneton and the gyromagnetic ratio g of the electron (Lande factor) according to the Pauli Hamiltonian.