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Quantum Mechanics}Commutators

Write it out and recall the commutator identities [B,AC]=A[B,C]+[B,A]C and [A,B]=-[B,A].

Thus -[L_z,L_xL_y]=-L_x[L_z,L_y]-[L_Z,L_x]L_y=i\hbar (L_x^2-L_y^2), using the identities supplied by ETS.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
QM320
2008-11-05 20:50:33
You can also do it by brute force:

[L_x L_y,L_z] = L_x L_y L_z - L_z L_y L_x

Using the given commutation relations, this is equal to

L_x(i\hbar L_x+L_z L_y)-(i\hbar L_y+L_x L_z) L_y

Rearranging and canceling gives choice (D).

Note: This was supposed to be an alternate solution, so I've reposted it as such. If someone could tell me how to delete my original (duplicate) comment, below, that would be helpful. :)
Alternate Solution - Unverified
Comments
QM320
2008-11-05 20:50:33
You can also do it by brute force:

[L_x L_y,L_z] = L_x L_y L_z - L_z L_y L_x

Using the given commutation relations, this is equal to

L_x(i\hbar L_x+L_z L_y)-(i\hbar L_y+L_x L_z) L_y

Rearranging and canceling gives choice (D).

Note: This was supposed to be an alternate solution, so I've reposted it as such. If someone could tell me how to delete my original (duplicate) comment, below, that would be helpful. :)
thisguy
2009-10-23 15:16:39
There's a small typo. [LxLy,Lz] = LxLyLz - LzLxLy. Not LxLyLz - LzLyLx as you wrote.
hybridusmanus
2010-06-15 18:06:16
Why can't re-associate the commutations after expansion? rnrni.e. rn[L_{x}L_{y},L_{z}] = rnL_{x}L_{y}L_{z} - L_{z}L_{x}L_{y} =rnL_{x}(L_{y}L_{z}) - (L_{z}L_{x})L_{y} = rnL_{x}ihL_{x} - ihL_{y}L_{y} = rnih(L^2_{x} - L^2_{y})rnrndoes this not always work?
walczyk
2011-04-07 08:51:16
Just wanted to clarify what the typo was and why it is significant. We cannot commute these orbital momentum operators willy nilly, without adding extra terms and whatnot so its a delicate procedure you could say. So beginning we have [L_xL_y,L_z] = L_xL_yL_z-L_zL_xL_y. Let's follow the original approach and try to substitute something for L_yL_z and L_zL_x. Quickly write out [L_y,L_z] = L_yL_z - L_zL_y = i\hbar L_x, rearrange to get L_yL_z=i\hbar L_x+L_zL_y. You can keep going with it if you'd like.

I'm curious what happens if you try to do something different.. [L_x,L_y]=L_xL_y-L_yL_x=i\hbar L_z, so we've got L_xL_y=i\hbar L_z+L_yL_x.
Alternate Solution - Unverified
QM320
2008-11-05 20:49:13
You can also do it by brute force:

[L_x L_y,L_z] = L_x L_y L_z - L_z L_y L_x

Using the given commutation relations, this is equal to

L_x(i\hbar L_x+L_z L_y)-(i\hbar L_y+L_x L_z) L_y

Rearranging and canceling gives choice (D).
NEC
Anastomosis
2008-04-10 14:20:54
A quick way to do this:
As the commutator is [L_xL_y, L_z],
Use the right hand rule to point first in the +X+Y (diagonal) direction, and then curl up to the Z direction. Your thumb will be pointing in the +X-Y direction, so thus choice (D)
NEC
bootstrap
2007-04-06 20:15:33
The question have it setup as [AB,C] which is [AB,C] = A[B,C] + [A,C]B. Its very similar to what is posted.. Typo Alert!

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You can also do it by brute force: [L_x L_y,L_z] = L_x L_y L_z - L_z L_y L_x Using the given commutation relations, this is equal to L_x(i\hbar L_x+L_z L_y)-(i\hbar L_y+L_x L_z) L_y Rearranging and canceling gives choice (D).

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