How to derive [P(x), L(y)] = iℏP(z)

[rsaad]

Can someone please explain to me how do we get the following:

[P(x), L(y)]= i h(cut) P(z)

P(x) is the momentum operator with respect to x
and L(y) is the angular momentum operator with respect to y.

I have also attached the solution. I am stuck at the underlined part. I do not know how to proceed from there.

 

[dextercioby]

Px and Py commute, as per Born-Jordan commutation relations. Thus the term from with their commutator is 0 when you apply the general formula

[A, BC] = [A,B]C+B[A,C] with [A,B] =0

 

[rsaad]

How did you obtain the general formula that you have stated in your reply?

 

[rsaad]

it should be as follows:
[a,bc]=[a,b]c+b[a,c]

 

[andrien]

[x,p_x]=ih/2∏ is the usual commutation rule,if that is what you are asking.
EDIT:if you want to know how to get that underlined term then just write the commutator explicitly and see that p_z commutes with p_x.

 

[jfy4]

You got it right in post #4. Just work it out staring from $[A, BC]$ and write out the commutator, then in the middle add zero in a fancy way.

 

[rsaad]

Yes, I got the answer. Thank you all for your help =)