Commutation relations between Ladder operators and Spherical Harmonics

Click For Summary
SUMMARY

The discussion focuses on the commutation relations between the angular momentum operator ##L_z## and an operator ##A## in the context of spherical harmonics ##Y_{ll}##. Participants clarify that the commutation relation is given by $$[L_z, A] = \hbar A$$, which leads to the equation $$[L_z, A] Y_{ll} = \hbar A Y_{ll}$$. The eigenvalue associated with this relationship is ##\hbar (l+1)##, indicating that ##AY_{ll}## is also an eigenfunction of ##L_z##. The conversation concludes with the original poster successfully resolving their query with community assistance.

PREREQUISITES
  • Understanding of angular momentum operators in quantum mechanics
  • Familiarity with spherical harmonics and their properties
  • Knowledge of commutation relations and their implications
  • Basic grasp of eigenvalues and eigenfunctions in quantum systems
NEXT STEPS
  • Study the derivation of commutation relations in quantum mechanics
  • Explore the properties of spherical harmonics in detail
  • Learn about the application of ladder operators in quantum mechanics
  • Investigate the implications of eigenvalues and eigenfunctions in quantum systems
USEFUL FOR

Quantum physicists, students studying quantum mechanics, and anyone interested in the mathematical foundations of angular momentum in quantum systems.

PhysicsTruth
Messages
117
Reaction score
18
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?
 
Physics news on Phys.org
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##.
 
  • Like
  • Informative
Likes   Reactions: PhysicsTruth and DrClaude
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.
 
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?
 
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.
 
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.
 
  • Like
Likes   Reactions: PhysicsTruth and vanhees71
Thanks a lot everyone, I was able to solve it with all of yours help!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
26
Views
7K
  • · Replies 12 ·
Replies
12
Views
828
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
Replies
46
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K