Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My text, when solving hydrogen in the Dirac equation, makes the claim

$\varphi_{j m_j}^{(+)} = \frac{\mathbf{\sigma} \cdot \mathbf{x}}{r} \varphi_{j m_j}^{(-)}$

where $\varphi_{j m_j}^{(\pm)}$ are the Pauli spinors for $j=\ell \pm \frac{1}{2}$:

$\varphi_{j m_j}^{(+)} = \pmatrix{\sqrt{\frac{\ell + m_j + 1/2}{2 \ell +1}} Y_{\ell}^{m_j-1/2} \\ \sqrt{\frac{\ell - m_j + 1/2}{2 \ell +1}} Y_{\ell}^{m_j+1/2}}$

$\varphi_{j m_j}^{(-)} = \pmatrix{\sqrt{\frac{\ell - m_j + 1/2}{2 \ell +1}} Y_{\ell}^{m_j-1/2} \\ -\sqrt{\frac{\ell + m_j + 1/2}{2 \ell +1}} Y_{\ell}^{m_j+1/2}}$

Leaving the proof as an exercise, the author suggests that one first note that $\varphi_{j m_j}^{(-)}$ is an eigenfunction of $\mathbf{\sigma} \cdot \mathbf{L}$ (indeed, the eigenvalue is in fact $-\ell-1$) and then show that the commutator $[\mathbf{\sigma} \cdot \mathbf{L}, \frac{\mathbf{\sigma} \cdot \mathbf{x}}{r}]$ is

$\frac{2}{r} (r^2 \mathbf{\sigma}\cdot \nabla -(\sigma \cdot \mathbf{x}) ( \mathbf{x} \cdot \nabla) - \mathbf{\sigma} \cdot \mathbf{x}))$

This commutator is not hard to compute, but I do not know how to proceed. Could somebody point me in the right direction?

share|cite|improve this question
The text suggests that, alternatively, I could use the anticommutator instead of the commutator, and I found this to be easier since (assuming my algebra was right) there aren't any gradients to worry about. I found the anticommutator to be $-\frac{2\mathbf{\sigma}\cdot\mathbf{x}}{r}$. This, combined with the the fact that $\varphi_{j m_j}^{(-)}$ is an eigenfunction of $\mathbf{\sigma} \cdot \mathbf{L}$ with eigenvalue $-(\ell+1)$ sets up an eigenvalue equation for $\frac{\mathbf{\sigma}\cdot\mathbf{x}}{r} \varphi_{j m_j}^{(-)}$...but I get the wrong eigenvalue ($\ell-1$ instead of $\ell$) – mikefallopian Dec 9 '13 at 22:31
and even if I made some stupid algebra mistake (for which I'm currently looking) and the eigenvalue of $\frac{\mathbf{\sigma}\cdot\mathbf{x}}{r}$ does come out to $\ell$, that of $\varphi_{j m_j}^{(+)}$, the correspondence of eigenvalues doesn't guarantee equality, since the eigenspace might be degenerate – mikefallopian Dec 9 '13 at 22:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.