- #1
Salmone
- 101
- 13
The Hamiltonian of a particle of mass ##m## on the surface of a sphere of radius ##R## is ##H=\frac{L^2}{2mR^2}## where ##L## is the angular momentum operator. I want to solve the TISE ##\hat{H}\psi=E\psi## and in order to do that I rewrite ##L^2## in Schroedinger's representation in spherical coordinates so ##\frac{1}{sin(\theta)}\frac{\partial }{\partial \theta}(sin(\theta)\frac{\partial \psi}{\partial \theta})+\frac{1}{sin^2(\theta)}\frac{\partial^2 \psi}{\partial \phi^2}=-\frac{2mER^2}{\hbar^2}\psi=-k^2\psi## after some steps one can get ##\frac{sin(\theta)}{\Theta(\theta)}\frac{\partial }{\partial \theta}(sin(\theta)\frac{\partial \Theta(\theta)}{\partial \theta})+K^2sin^2(\theta)+\frac{1}{\Phi(\phi)}\frac{\partial^2 \Phi(\phi)}{\partial \phi^2}=0## having used ##\psi=\Theta(\theta)\Phi(\phi)##. Now the two parts dipend on different variables so they must be equal to a constant, let's call this constant ##m^2## and I obtain two equations:
##\frac{sin(\theta)}{\Theta(\theta)}\frac{\partial }{\partial \theta}(sin(\theta)\frac{\partial \Theta(\theta)}{\partial \theta})+K^2sin^2(\theta)=m^2## ##(1)##
##\frac{1}{\Phi(\phi)}\frac{\partial^2 \Phi(\phi)}{\partial \phi^2}=-m^2## ##(2)##
The equation ##(1)## could be rearranged and written as:
##\frac{\partial^2 \Theta(\theta)}{\partial \theta^2}+\frac{cos(\theta)}{sin(\theta)}\frac{\partial \Theta(\theta)}{\partial \theta}+[k^2-\frac{m^2}{sin^2(\theta)}]\Theta(\theta)=0##
which is, by choosing ##k^2=l(l+1)## the legendre associated differential equation according to https://mathworld.wolfram.com/AssociatedLegendreDifferentialEquation.html equation ##(9)##.
This leads to Energy eigenvalues ##E=\frac{\hbar^2l(l+1)}{2mR^2}##.
My questions:
1)
How do we know that the eigenvalues found are *all* the eigenvalues for this problem? If ##k^2## had a different value, for example ##k^2##=3, which is not a value obtainable with ##k^2=l(l+1)##, how do we know that the differential equation would have no solutions?
##\frac{\partial^2 \Theta(\theta)}{\partial \theta^2}+\frac{cos(\theta)}{sin(\theta)}\frac{\partial \Theta(\theta)}{\partial \theta}+[3-\frac{m^2}{sin^2(\theta)}]\Theta(\theta)=0## Does this TISE have solutions? How do we know that? Because if this TISE has solutions than I obtain the eigenvalue ##E=\frac{3\hbar^2}{2mR^2}## which is not in the list ##E=\frac{\hbar^2l(l+1)}{2mR^2}##.
2)
Usually in quantum mechanics problems, we find the eigenvalues by imposing that our eigenfunction is square-integrable, in this case what we do is putting ##k^2=l(l+1)##, does this represent the request that the eigenfunctions must be square-integrable?
##\frac{sin(\theta)}{\Theta(\theta)}\frac{\partial }{\partial \theta}(sin(\theta)\frac{\partial \Theta(\theta)}{\partial \theta})+K^2sin^2(\theta)=m^2## ##(1)##
##\frac{1}{\Phi(\phi)}\frac{\partial^2 \Phi(\phi)}{\partial \phi^2}=-m^2## ##(2)##
The equation ##(1)## could be rearranged and written as:
##\frac{\partial^2 \Theta(\theta)}{\partial \theta^2}+\frac{cos(\theta)}{sin(\theta)}\frac{\partial \Theta(\theta)}{\partial \theta}+[k^2-\frac{m^2}{sin^2(\theta)}]\Theta(\theta)=0##
which is, by choosing ##k^2=l(l+1)## the legendre associated differential equation according to https://mathworld.wolfram.com/AssociatedLegendreDifferentialEquation.html equation ##(9)##.
This leads to Energy eigenvalues ##E=\frac{\hbar^2l(l+1)}{2mR^2}##.
My questions:
1)
How do we know that the eigenvalues found are *all* the eigenvalues for this problem? If ##k^2## had a different value, for example ##k^2##=3, which is not a value obtainable with ##k^2=l(l+1)##, how do we know that the differential equation would have no solutions?
##\frac{\partial^2 \Theta(\theta)}{\partial \theta^2}+\frac{cos(\theta)}{sin(\theta)}\frac{\partial \Theta(\theta)}{\partial \theta}+[3-\frac{m^2}{sin^2(\theta)}]\Theta(\theta)=0## Does this TISE have solutions? How do we know that? Because if this TISE has solutions than I obtain the eigenvalue ##E=\frac{3\hbar^2}{2mR^2}## which is not in the list ##E=\frac{\hbar^2l(l+1)}{2mR^2}##.
2)
Usually in quantum mechanics problems, we find the eigenvalues by imposing that our eigenfunction is square-integrable, in this case what we do is putting ##k^2=l(l+1)##, does this represent the request that the eigenfunctions must be square-integrable?