# Problem in deriving Schrodinger and Pauli equation from Dirac's

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

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

Question:

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

Thank you for any help in advance

• Commented Jan 9, 2017 at 18:59
• 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. Commented Jan 9, 2017 at 22:50
• @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? Commented Jan 9, 2017 at 23:07
• 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). Commented Jan 9, 2017 at 23:07
• Ok, I'll try to figure it out in the while. Thanks for your time and help. Commented Jan 9, 2017 at 23:09

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.
• 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? Commented Jan 10, 2017 at 17:29
• 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. Commented Jan 11, 2017 at 2:08