# I Question about the Pauli equation

#### Airton Rampim

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#### vanhees71

Gold Member
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

#### 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

Gold Member
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

#### 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

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

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

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

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

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

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

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