Working out the non relativistic limit of the Dirac equation, we encounter this quantity: $(\vec{\sigma} \cdot \vec{p})$ and in my notes it says that $$ (\vec{\sigma} \cdot \vec{p})^2 = p^i p^j\sigma^i\sigma^j=\vec p^{\,2} \tag{1} $$

When we couple the Dirac equation and we write $$\vec{p} \rightarrow \vec{p} - \frac{e}{c} \vec{A} \equiv \vec{\pi} $$

we obtain a similar quantity: $(\vec{\sigma} \cdot \vec{\pi})$, but to calculate its square we now use the fact that $\sigma^i \sigma^j= \delta^{ij} + i \epsilon^{ijk}\sigma^k$ and we obtain

$$ (\vec{\sigma} \cdot \vec{\pi})^2= \pi^i \pi^j \sigma^i \sigma^j= \vec{\pi}^{\,2} + i \epsilon^{ijk} \pi^i \pi^j\sigma^k \tag{2}$$


Why does the $\epsilon^{ijk}$ term vanish in $(1)$ but it does not vanish in $(2)$?

Thank you for any help in advance

  • $\begingroup$ Related : Need help with solution of the Dirac equation. $\endgroup$
    – Frobenius
    Commented Jan 9, 2017 at 18:59
  • $\begingroup$ It turns out that most of what I said above was wrong :-/ sorry for wasting your time. The last term for $\vec p$ vanishes, but for $\vec \pi$ does not vanish. $\endgroup$ Commented Jan 9, 2017 at 22:50
  • $\begingroup$ @AccidentalFourierTransform Well you didn't actually waste my time, you corrected my proof and made me think about it, your arguments were convincing. I don't really see why they are wrong actually. Ps: should I delete the other comments like you did? $\endgroup$ Commented Jan 9, 2017 at 23:07
  • $\begingroup$ I really didn't think it through. The problem is that I thought that $\varepsilon^{ijk}\pi^i\pi^j=0$, but this is wrong! I cannot properly explain the reason in a comment, but if you don't get an answer by tomorrow I'll write it myself. Its a bit late for me and I should leave, but maybe tomorrow I'll write an answer, or maybe someone else will do it (PS yeah, delete the comments to keep the post clean if you don't mind). $\endgroup$ Commented Jan 9, 2017 at 23:07
  • $\begingroup$ Ok, I'll try to figure it out in the while. Thanks for your time and help. $\endgroup$ Commented Jan 9, 2017 at 23:09

1 Answer 1


Because $$\epsilon^{ijk} = -\epsilon^{jik}$$ is an antisymmetric tensor (it changes sign when you flip two of its consecutive indices). But, it is contracted with a symmetric tensor: $$p^ip^j = p^jp^i$$ (because momentum operator commutes with itself) and thus, the contraction is zero. Here is an explicit proof:

$$ p^ip^j\epsilon^{ijk} = \frac{1}{2}2p^ip^j\epsilon^{ijk} \\ = \frac{1}{2}(p^ip^j\epsilon^{ijk} + p^ip^j\epsilon^{ijk}) \\ = \frac{1}{2}(p^ip^j\epsilon^{ijk} + p^jp^i\epsilon^{jik}) \text{(renaming indices)}\\ = \frac{1}{2}(p^ip^j\epsilon^{ijk} - p^jp^i\epsilon^{ijk}) \text{(antisymmetric tensor)}\\ = \frac{1}{2}(p^ip^j\epsilon^{ijk} - p^ip^j\epsilon^{ijk}) \text{(symmetric tensor)}\\ =0 $$

But, for the $\vec{\pi}$ operator, its components does not necessarily commutes with one another since it depends of $\vec{A}$. Thus the $\epsilon^{ijk}$ term does not vanish.

  • $\begingroup$ Thank you for the detailed answer, I don't really get why the components of $\vec pi$ doesn't necessarily commute with one another. Beside the $e/c$ and its square is $\pi^i \pi^j = p^i p^j - A^i p^j - p^i A^j + A^i A^j$ And if I consider their components being real numbers they do commute. But can I do that? Or I always have tot think that $\vec p$ "is" $-i \hbar \nabla$ ? I really don't get why sometimes we just write $p$ and other times we write it in explicit operatorial form. Should I just leave the classical concept of $p$ and consider it just an operator? $\endgroup$ Commented Jan 10, 2017 at 17:29
  • $\begingroup$ Of course if they are just numbers, everything commutes and the tensor will be symmetric like the case with $p^ip^j$. Now, if you consider operators, generally, the components of $\vec{A}$ don't necessarily commute with themselves or with the ones of $\vec{p}$ and thus, the term remains because $\pi^i\pi^j\neq\pi^j\pi^i$. In quantum mechanics, you'll always consider operators so you can leave the concept of $p$ being a simple number. $\endgroup$
    – fgoudra
    Commented Jan 11, 2017 at 2:08

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