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I found the operators to be $J_z, J^2, L^2, S^2$, but how do I prove that they commute?

My attempt:

  1. For $L^2$, we know that $[\vec{L},L^2]=0$, so $[\vec{L} \cdot \vec{S}, L^2]=0$. But I don't understand why.

  2. For $S^2$, we know that $[\vec{S},S^2]=0$, so $[\vec{L} \cdot \vec{S}, S^2]=0$. But I'm not sure why.

  3. For $J^2=L^2+S^2+2\vec{L} \cdot \vec{S}$. We also know that $[\vec{L} \cdot \vec{S},\vec{S} \cdot \vec{L}] = 0$, so $[\vec{L} \cdot \vec{S}, J^2]=0$

  4. For $J_z$ ???

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  • $\begingroup$ @AlternativeFacts this is exactly what I was reading. I'm confused because he leaves out a lot of steps. $\endgroup$
    – loltospoon
    Feb 21, 2017 at 23:41
  • $\begingroup$ OK, I looked through my books and it would be a slog of an answer. What I can do is upvote your question, and suggest you post a question with the steps you do know included from the sources you find and then ask for the intermediate steps, .....as it's 2 am where I am, sorry. The more you show you are trying, the more chance of an answer. $\endgroup$
    – user146020
    Feb 22, 2017 at 0:08
  • $\begingroup$ Related: Google HW_3_solutions_sebastian-1.pdf $\endgroup$
    – user146020
    Feb 22, 2017 at 0:16
  • 1
    $\begingroup$ True, thank you. I appreciate your help. Get some sleep! It's important!! $\endgroup$
    – loltospoon
    Feb 22, 2017 at 0:18

1 Answer 1

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Since $S^2$ and $L^2$ are just multiples of the identity, they commute with the interaction. This gives 1 and 2.

Recall that $\vec J:=\vec L+\vec S$, so $$J^2=L^2+S^2+\vec L\cdot \vec S+\vec S\cdot\vec L.$$ Now, since $\vec L$ and $\vec S$ act on difference spaces$^\dagger$, they commute, so we have $$\vec L\cdot\vec S=\frac{J^2-S^2-L^2}{2}.$$ Since the spin is just $\frac{1}{2}$, and the orbital angular momentum is a fixed integer $\ell$, we have $$\vec L\cdot\vec S=\frac{J^2-\frac{3}{4}\hbar^2-\ell(\ell+1)\hbar^2}{2}.$$ This shows 3 and 4, since $J^2$ and $J_z$ commute with $J^2$.


$^\dagger$ The Hilbert space is $L^2(\Bbb R^3)\otimes V$, where $V$ carries a representation of $\mathrm{SU}(2)$. $\vec L$ acts on the first factor, $\vec S$ on the second.

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  • $\begingroup$ "then commute," or "they commute"? $\endgroup$
    – loltospoon
    Feb 22, 2017 at 0:57
  • $\begingroup$ @loltospoon They. $\endgroup$
    – Ryan Unger
    Feb 22, 2017 at 1:13

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