# Eigenfunctions of the angular momentum operator

1. Oct 17, 2015

### klpskp

Hi everyone,

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

$$L_zY_{lm}=mħY_{lm}$$ and $$L^2Y_{lm}=l(l+1)ħ^2Y_{lm}$$

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

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

So where is the mistake? I did not find this solution anywhere.
Thank you for your help :)

2. Oct 17, 2015

### fzero

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.