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

[Chronos000]

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

 

[CompuChip]

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.

 

[Chronos000]

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?

 

[kloptok]

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).

 

[Chronos000]

I can't believe I missed that... I still don't know where this delta comes from though

 

[Chronos000]

i figured it out, thanks anyway