Eigenfunctions of the angular momentum operator

Click For Summary
SUMMARY

The discussion focuses on the Eigenfunctions of the angular momentum operator, specifically the function Y_{lm}(\theta,\phi) that satisfies the equations L_zY_{lm}=mħY_{lm} and L^2Y_{lm}=l(l+1)ħ^2Y_{lm}. The standard solution for the case m=l=0 is Y_{00}(\theta,\phi)=\frac{1}{\sqrt{4\pi}}, but an alternative solution Y_{00}(\theta,\phi)=A\ln(\cot\theta+\csc\theta) is proposed. However, this alternative function is deemed unsuitable as a wavefunction due to its singularity at the poles of the sphere (θ = 0, π), violating quantum mechanics requirements for wavefunctions.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly angular momentum operators.
  • Familiarity with spherical harmonics and their properties.
  • Knowledge of differential equations and their solutions in the context of quantum mechanics.
  • Basic grasp of normalization conditions for wavefunctions.
NEXT STEPS
  • Study the properties of spherical harmonics in quantum mechanics.
  • Learn about the normalization of wavefunctions in quantum systems.
  • Explore the implications of singularities in wavefunctions and their physical interpretations.
  • Investigate the mathematical derivation of angular momentum operators in quantum mechanics.
USEFUL FOR

Quantum mechanics students, physicists specializing in angular momentum, and researchers exploring wavefunction properties in quantum systems.

klpskp
Messages
8
Reaction score
0
Hi everyone,

I tried to find the Eigenstate of the angular momentum operator myself, more specifically I tried to find a Function [itex]Y_{lm}(\theta,\phi)[/itex] with

[tex]L_zY_{lm}=mħY_{lm}[/tex] and [tex]L^2Y_{lm}=l(l+1)ħ^2Y_{lm}[/tex]

where [tex]L_z=-iħ\frac{\partial}{\partial \phi}[/tex]
and [tex]L^2=-ħ^2(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta\frac{\partial}{\partial\theta})+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial^2\phi})[/tex]

These representations can be found here.
Now let's look at the simple case of [itex]m=l=0[/itex]. The standard solution here is [itex]Y_{00}(\theta,\phi)=\frac{1}{\sqrt{4\pi}}[/itex]. However, it seems like the function [itex]Y_{00}(\theta,\phi)=A\ln(\cot\theta+\csc\theta)[/itex] is as well a solution to the differential equations above, since [itex]\frac{\partial Y}{\partial \phi}=0=mħY[/itex] and [itex]\frac{\partial Y}{\partial \theta}=-A\csc\theta[/itex] and therefore [tex]L^2Y=-ħ^2(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}(-A \sin\theta\csc\theta))=0=l(l+1)ħ^2Y[/tex]
Since [tex]\int_0^\pi YY^*\sin\theta d\theta[/tex] converges (you can look at the graph here) it is possible to find an [itex]A[/itex] that normalizes [itex]Y_{00}[/itex].

So where is the mistake? I did not find this solution anywhere.
Thank you for your help :)
 
Physics news on Phys.org
That function is singular at the poles of the sphere ##\theta = 0,\pi##. To do QM, generally we require the wavefunction and at least the 1st derivative to exist everywhere (2nd is good too), so we would not admit this as a possible wavefunction.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K