Why Can Pl Be Taken Out of the Commutation in Quantum Mechanics?

AI Thread Summary
The discussion revolves around the commutation relations between momentum (P) and angular momentum (L) in quantum mechanics. A key point is the application of the identity [A, BC] = [A, B]C + B[A, C] to simplify the commutation, allowing Pl to be factored out. Participants clarify that Pi acting on Xk results in -i*h-bar deltaik, highlighting the significance of the delta function in this context. There is also a mention of the commutation of different momentum components, where [Pi, Pj] equals zero for i not equal to j. The conversation emphasizes the importance of understanding these commutation relations for solving quantum mechanics problems effectively.
Chronos000
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Homework Statement



the problem asks for the commutation between momentum P and angular momentum L.

My solutions give an intermediate step of:

ejkl [Pi, Xk *Pl]

ejkl[Pi,Xk]Pl

I don't understand why we can just take out a Pl from the commutation. its not at the end of both parts of the commutation and so will be acted on in a different order
 
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There is a very useful identity,
[A, BC] = [A, B]C + B[A, C],
I suggest you remember it forever :-)

You can easily prove it by writing out the commutator, if you wish.
 
thanks for pointing this out to me. but wouldn't I have two terms if I used that relation?

I have also been told that Pi acting on Xk results in -i*h-bar deltaik

I thought the delta was only used to denote a dot product whereas a differential was just shown as a differential?
 
Doesn't the different components of momentum commute, i.e. [Pi,Pj]=0, i != j ?

In that case the result follows by using the mentioned identity (if I correctly understand that when you write Pi and Pj you mean the different components of momentum P).
 
I can't believe I missed that... I still don't know where this delta comes from though
 
i figured it out, thanks anyway
 
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