Commutators of Angular momentum operator

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The discussion centers on the commutation relations of angular momentum operators, specifically addressing the calculation of [Lz, px]. Participants question why [x, px] is not zero, despite the expectation that they commute, and why py is placed on the left in certain steps of the calculation. There is also confusion regarding the multiplication by i(hbar) instead of i/(hbar) when transitioning from derivatives to momentum operators. The importance of the order of operators in quantum mechanics is emphasized, but it is noted that the specific placement of py does not affect the outcome due to the commutation relation being a constant. Overall, the conversation highlights the nuances of operator algebra in quantum mechanics.
jaobyccdee
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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"?
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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