Commutators of Angular momentum operator

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SUMMARY

The discussion centers on the commutation relations of angular momentum operators, specifically the calculation of the commutator [Lz, px]. The participants clarify that [x, px] is not zero due to the nature of quantum mechanical operators, despite their commutation. The placement of the operator py in the calculation is justified by the rules governing operator order, and the multiplication by iħ is confirmed as the correct approach in quantum mechanics. Misunderstandings regarding these principles highlight the importance of foundational knowledge in quantum mechanics.

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  • Quantum Mechanics fundamentals
  • Understanding of angular momentum operators
  • Commutation relations in quantum mechanics
  • Operator algebra in quantum mechanics
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The letters next to p and L should be subscripts.
[Lz, px] = [xpy − ypx, px] = [xpy, px] − [ypx, px] = py[x, px] −0 = i(hbar)py

1.In this calculation, why is [x, px] not 0 even they commute?

2.Why is py put on the left instead of the right in the second last step? i thought it should be put on the right bec it's on the right of x in the third step, and we have to keep the orders for operators.

3.With L=rxp, why are we multiplying i(hbar) instead of multiplying by i/ (hbar), coz at the beginning, we change all the d/dx or d/dy or d/dz to px, py, pz, why aren't we multiplying i/(hbar) to compensate what we change for convenient calculation?
 
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Why do you think [x,p_x]=0? That should be one of the first things you learned about QM.

Your p_y should be to the right, yes, but inevitably it doesn't matter considering [x,p_x] is equal to a constant.

As for why you multiply by i\hbar, what do you mean "at the beginning"?
 

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