# Understanding the Dirac equation

I am reading a physics book where the Dirac equation is being introduced in the form: $$\left[c \boldsymbol{\alpha} \cdot\left(\boldsymbol{p}+\frac{e \boldsymbol{A}}{c}\right)-e \phi+\beta m c^{2}\right] \Psi=\mathrm{i} \hbar \frac{\partial \Psi}{\partial \mathrm{t}}$$ where $$\boldsymbol{\alpha}_{k}=\left[\begin{array}{cc} 0_{2} & \sigma_{k} \\ \sigma_{k} & 0_{2} \end{array}\right], \quad \beta=\left[\begin{array}{cc} \mathbb{I}_{2} & 0_{2} \\ 0_{2} & -\mathbb{I}_{2} \end{array}\right].$$ The author then wants to show the solutions for a constant magnetic field, with $$\phi\equiv0$$. He then writes: $$\left[\begin{array}{ll} m c^{2} & c \boldsymbol{\sigma} \cdot \boldsymbol{\pi} \\ c \boldsymbol{\sigma} \cdot \boldsymbol{\pi} & -m c^{2} \end{array}\right]\left[\begin{array}{l} \psi_{u} \\ \psi_{l} \end{array}\right]=E\left[\begin{array}{l} \psi_{u} \\ \psi_{l} \end{array}\right]$$ With no explanation anywhere of what $$\boldsymbol{\pi}$$ is, however I can see that it must be connected to $$\boldsymbol{p}$$. Next, he states the following equation, with no explanation of where it comes from: $$$$(\boldsymbol{\sigma} \cdot \boldsymbol{\pi})^{2}=\pi^{2}+i \boldsymbol{\sigma} \cdot \boldsymbol{\pi} \times \boldsymbol{\pi}\tag{*}$$$$ And I do not know why these terms are equal. Can someone explain

1. what is $$\boldsymbol{\pi}$$ exactly?

2. Where does the result in eq. $$(*)$$ come from?

• Please state the book title and author. Apr 15, 2020 at 11:49
• It is from Theoretical Foundations of Electron Spin Resonance by John Harriman.
– user260769
Apr 15, 2020 at 14:01

## 1 Answer

I think the author wants to find solutions of the form: $$\Psi(t) = e^{-i\frac{1}{\hbar }E \;t} \begin{pmatrix} \psi_u \\ \psi_l \end{pmatrix}$$ If you insert that and then substitute the values of $$\beta$$ and $$\alpha$$ you'll get, $$\begin{pmatrix} I_2 mc^2 & c\sum_k\sigma_k (p_k+ e\frac{A_k}{c}) \\ c\sum_k \sigma_k(p_k+ e\frac{A_k}{c}) & -I_2 mc^2 \end{pmatrix} \begin{pmatrix} \psi_u \\ \psi_l \end{pmatrix} = E \begin{pmatrix} \psi_u \\ \psi_l \end{pmatrix}$$ So you've obtained the value of $$\mathbf{\pi} =(\textbf{p}+ e\frac{\textbf{A}}{c})$$ . As for where (*) comes from, it's very straightforward if you know the Pauli matrix identities. In particular: $$\sigma_i \sigma_j = 2I\delta_{ij} + i \varepsilon_{ijk}\sigma_k$$ Apply that to $$(\sigma\cdot\pi)^2$$ and remember that $$(a \times b)_k= \varepsilon_{ijk}a_i b_j$$. You will see that you obtain (*)