Undergrad Question about the Pauli equation

[Airton Rampim]

I have a question about this note: https://ocw.mit.edu/courses/physics...i-spring-2018/lecture-notes/MIT8_06S18ch2.pdf
I don't understand the expression (2.2.15). The complete relation would be

$$ \pi_i \pi_j = \frac{1}{2}\left(\left[\pi_i, \pi_j\right] + \left\{\pi_i, \pi_j\right\}\right), $$
where
$$ \vec{\pi} = \vec{p} - \frac{q}{c}\vec{A} $$
in gaussian unit. How can I prove that {πi, πj} = 0?

 

[vanhees71]

In (2.2.15) you don't need the anticommutator, because the $\epsilon_{ijk}$ cancels this contribution identically. The commutator is most easily calculated in the position representation, where $\hat{\vec{p}}=-\mathrm{i} \hbar \vec{\nabla}$ or $\hat{p}_i=-\mathrm{i} \partial_i$.

 

[Airton Rampim]

In (2.2.15) you don't need the anticommutator, because the $\epsilon_{ijk}$ cancels this contribution identically. The commutator is most easily calculated in the position representation, where $\hat{\vec{p}}=-\mathrm{i} \hbar \vec{\nabla}$ or $\hat{p}_i=-\mathrm{i} \partial_i$.

Sorry, but I can't see how the Levi-Civita tensor cancels the anticommutator. I calculated the commutator using the position representation, as you mentioned. What I can't figure out is how I relate the commutator with the vector product $\vec{\pi}\times\vec{\pi}$.

 

[vanhees71]

You have
$$(\vec{\pi} \times \vec{\pi})_k=\epsilon_{ijk} \pi_i \pi_j=\frac{1}{2} \epsilon_{ijk} (\pi_i \pi_j-\pi_j \pi_i)=\frac{1}{2} \epsilon_{ijk}[\pi_i,\pi_j],$$
because $\epsilon_{ijk}=-\epsilon_{jik}$.

 

[Airton Rampim]

You have
$$(\vec{\pi} \times \vec{\pi})_k=\epsilon_{ijk} \pi_i \pi_j=\frac{1}{2} \epsilon_{ijk} (\pi_i \pi_j-\pi_j \pi_i)=\frac{1}{2} \epsilon_{ijk}[\pi_i,\pi_j],$$
because $\epsilon_{ijk}=-\epsilon_{jik}$.

Hmmm, so this is valid for any operator, right? Now I got it. I did in this way before, but I thought that was wrong, because it wasn't working with the $\vec{L}$ operator. But I forgot an extra $\epsilon_{ijk}$ that appears in $\left[L_{i},L_{j}\right]$. So this gives

$$ {\displaystyle \left(\vec{L}\times\vec{L}\right)_{k}=\sum_{i,j}\frac{\epsilon_{ijk}}{2}\underbrace{\left[L_{i},L_{j}\right]}_{{\displaystyle i\hbar\sum_{k}\epsilon_{ijk}L_{k}}}=\frac{i\hbar}{2}\left(\sum_{i,j}\epsilon_{ijk}\epsilon_{ij1}L_{1}+\sum_{i,j}\epsilon_{ijk}\epsilon_{ij2}L_{2}+\sum_{i,j}\epsilon_{ijk}\epsilon_{ij3}L_{3}\right)} $$

$$ {\displaystyle =\frac{i\hbar}{2}\left[\left(\epsilon_{23k}\epsilon_{231}+\epsilon_{32k}\epsilon_{321}\right)L_{1}+\left(\epsilon_{13k}\epsilon_{132}+\epsilon_{31k}\epsilon_{312}\right)L_{2}+\left(\epsilon_{12k}\epsilon_{123}+\epsilon_{21k}\epsilon_{213}\right)L_{3}\right]} $$

$$ {\displaystyle =\frac{i\hbar}{2}\left[\left(\epsilon_{23k}-\epsilon_{32k}\right)L_{1}+\left(\epsilon_{31k}-\epsilon_{13k}\right)L_{2}+\left(\epsilon_{12k}-\epsilon_{21k}\right)L_{3}\right]} $$

$$ {\displaystyle =\frac{i\hbar}{2}\left[2\epsilon_{23k}L_{1}+2\epsilon_{31k}L_{2}+2\epsilon_{12k}L_{3}\right]} $$

$$ {\displaystyle =i\hbar\epsilon_{k23}L_{1}+i\hbar\epsilon_{1k3}L_{2}+i\hbar\epsilon_{12k}L_{3}} $$

$$ {\displaystyle =i\hbar L_{k}}, $$

which is the correct answer. It makes sense to me now. Thank you very much for your help :)