Commutation relations between Ladder operators and Spherical Harmonics

In summary: Your starting point is the equality$$[L_z,A] = \hbar A$$which you apply to the spherical harmonic ##Y_{ll}##,$$[L_z,A] Y_{ll} = \hbar A Y_{ll}$$You expand the commutator, but the result should still be an equality.Thanks a lot everyone, I was able to solve it with all of yours help!
  • #1
PhysicsTruth
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Homework Statement
Consider an operator A such that it satisfies the following commutation relations-

##[L_+,A] = 0##
##[L_z,A] = \hbar A##
Using these, find ##L_z(AY_{ll})## and ##L^2(AY_{ll})## , where ##AY_{ll}## is an eigenfunction of ##L_z## and ##L^2##.

Also, deduce ##AY_{ll}##.
Relevant Equations
##L_+ = L_x +iL_y##
##L_z(Y_{ll}) = l\hbar (Y_{ll})##
I've tried figuring out commutation relations between ##L_+## and various other operators and ##L^2## could've been A, but ##L_z, L^2## commute. Can someone help me out in figuring how to actually proceed from here?
 
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  • #2
Just figure out what you get when applying the given commutators to ##\mathrm{Y}_{ll}## and use the information that ##\hat{A} Y_{ll}## is also an eigenfunction of ##\hat{L}_z## and ##\hat{L}^2##.
 
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  • #3
vanhees71 said:
Just figure out what you get when applying the given commutators to ##\mathrm{Y}_{ll}## and use the information that ##\hat{A} Y_{ll}## is also an eigenfunction of ##\hat{L}_z## and ##\hat{L}^2##.
If I use the commutator between ##L_z, A##, I get

##(l\hbar +\hbar)A(Y_{ll})## for the 1st part. But I don't know how to figure out ##A(Y_{ll})## from the given information.
 
  • #4
PhysicsTruth said:
If I use the commutator between ##L_z, A##, I get

##(l\hbar +\hbar)A(Y_{ll})## for the 1st part. But I don't know how to figure out ##A(Y_{ll})## from the given information.
You should get an equality here. What is it?
 
  • #5
DrClaude said:
You should get an equality here. What is it?
I'm really sorry but I'm not being able to get you. Equality in which sense? Like it's not an equation, I'm just trying to find out ##L_z(AY_{ll})##. The eigenvalue is ##\hbar (l+1)##, but I also need to deduce what ##AY_{ll}## is.

Once again, I'm really sorry for not being able to follow you.
 
  • #6
PhysicsTruth said:
I'm really sorry but I'm not being able to get you. Equality in which sense? Like it's not an equation, I'm just trying to find out ##L_z(AY_{ll})##. The eigenvalue is ##\hbar (l+1)##, but I also need to deduce what ##AY_{ll}## is.

Once again, I'm really sorry for not being able to follow you.
Your starting point is the equality
$$
[L_z,A] = \hbar A
$$
which you apply to the spherical harmonic ##Y_{ll}##,
$$
[L_z,A] Y_{ll} = \hbar A Y_{ll}
$$
You expand the commutator, but the result should still be an equality.
 
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  • #7
Thanks a lot everyone, I was able to solve it with all of yours help!
 
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