# Question about the Pauli equation

• I
• Airton Rampim
In summary, the expression (2.2.15) can be written as$$\pi_i \pi_j = \frac{1}{2}\left(\left[\pi_i, \pi_j\right] + \left\{\pi_i, \pi_j\right\}\right),$$where$$\vec{\pi} = \vec{p} - \frac{q}{c}\vec{A}$$in gaussian unit. The commutator can be calculated using the position representation, and is equal to$${\displaystyle \left(\vec{L}\times\vec{L}\right)_{k}=\sum_{i,j Airton Rampim I have a question about this note: https://ocw.mit.edu/courses/physics...i-spring-2018/lecture-notes/MIT8_06S18ch2.pdf I don't understand the expression (2.2.15). The complete relation would be$$ \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} $$in gaussian unit. How can I prove that {πi, πj} = 0? In (2.2.15) you don't need the anticommutator, because the ##\epsilon_{ijk}## cancels this contribution identically. The commutator is most easily calculated in the position representation, where ##\hat{\vec{p}}=-\mathrm{i} \hbar \vec{\nabla}## or ##\hat{p}_i=-\mathrm{i} \partial_i##. Airton Rampim vanhees71 said: In (2.2.15) you don't need the anticommutator, because the ##\epsilon_{ijk}## cancels this contribution identically. The commutator is most easily calculated in the position representation, where ##\hat{\vec{p}}=-\mathrm{i} \hbar \vec{\nabla}## or ##\hat{p}_i=-\mathrm{i} \partial_i##. Sorry, but I can't see how the Levi-Civita tensor cancels the anticommutator. I calculated the commutator using the position representation, as you mentioned. What I can't figure out is how I relate the commutator with the vector product ##\vec{\pi}\times\vec{\pi}##. You have$$(\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],$$because ##\epsilon_{ijk}=-\epsilon_{jik}##. Airton Rampim vanhees71 said: You have$$(\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 :)

## 1. What is the Pauli equation?

The Pauli equation is a quantum mechanical equation that describes the behavior of a system of particles with half-integer spin, such as electrons. It is a relativistic extension of the Schrödinger equation and takes into account the spin of the particles.

## 2. Who developed the Pauli equation?

The Pauli equation was developed by the Austrian physicist Wolfgang Pauli in 1928. It was a significant contribution to the field of quantum mechanics and helped explain many experimental observations at the time.

## 3. What is the significance of the Pauli equation?

The Pauli equation is significant because it provides a more accurate description of the behavior of particles with spin, which was not accounted for in the original Schrödinger equation. It also helps explain the stability of atoms and the properties of materials, such as magnetism.

## 4. How is the Pauli equation different from the Schrödinger equation?

The Pauli equation differs from the Schrödinger equation in that it takes into account the spin of particles, while the Schrödinger equation does not. Additionally, the Pauli equation is a relativistic equation, meaning it accounts for the effects of special relativity, while the Schrödinger equation is non-relativistic.

## 5. What are the applications of the Pauli equation?

The Pauli equation has many applications in various fields, including quantum mechanics, atomic and molecular physics, and solid-state physics. It is used to study the behavior of electrons in atoms, molecules, and materials, and has also been applied in the development of technologies such as transistors and magnetic storage devices.

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