0
$\begingroup$

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: \begin{equation}(\boldsymbol{\sigma} \cdot \boldsymbol{\pi})^{2}=\pi^{2}+i \boldsymbol{\sigma} \cdot \boldsymbol{\pi} \times \boldsymbol{\pi}\tag{*} \end{equation} 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?

$\endgroup$
2
  • $\begingroup$ Please state the book title and author. $\endgroup$ Apr 15, 2020 at 11:49
  • $\begingroup$ It is from Theoretical Foundations of Electron Spin Resonance by John Harriman. $\endgroup$
    – user260769
    Apr 15, 2020 at 14:01

1 Answer 1

1
$\begingroup$

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 (*)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.