Spherical Harmonics: Rigid Rotor, Laplace's Equation

[Amok]

So, I was reading about the quantum http://en.wikipedia.org/wiki/Rigid_rotor" and apparently its solutions are the so-called spherical harmonic functions, which are the solution to the angular portion of Laplace's equation. The way I see it, the rigid rotor Schroedinger equation is not the angular part of Laplace's equation (because of that E\Psipart of the rigid rotor equation). Am I right? Am I missing something? Am I completely wrong? Maybe these functions are solutions to both equations?

Homework Equations

\Delta\Psi=0 (Laplace's equation)

(-h-bar²/2\mu)\Delta\Psi=E\Psi

Homework Statement

Errrr... I'm have a hard time writing these equations here, it's just worth mentioning that it's not 2 to the power of mu, but 2 times mu.

 

[phsopher]

If you actually solve the Laplace's equation you'll realize that the angular part is of the same form as the Shrödinger's equation. Start with the Laplace equation:

\nabla^2 \Psi = 0

Assume the solution is of the form \Psi(r,\theta,\phi) = R(r) Y(\theta,\phi) so that you can separate radius and angular dependence. You'll get:

\frac{1}{R}\nabla_r^2R + \frac{1}{Y}\nabla^2_\theta_, _\phi Y = 0

Now the first term only depends on r while the second term depends only on the angles which means that both must be constant (denote C) with opposite signs. That means that you can solve them separately and the angular part becomes

\nabla^2_\theta_, _\phi Y = CY

Which is of the same form as the Schrödinger equation.

 

[CompuChip]

It's been a while for me, so please correct me if I am wrong here.
If you want to solve an equation such as the second, it is common to divide the problem in finding a particular solution and the general solution to the homogeneous equation. In this case, the solutions to the homogeneous equation
-\frac{\hbar^2}{2\mu} \Delta\Psi = 0
are the spherical harmonic functions (or a multiple thereof) because the homogeneous equation simply is Laplace's equation. After you have found a particular solution \Psi_0 satisfying -\frac{\hbar^2}{2\mu} \Delta\Psi_0 = E \Psi_0 you can construct the general solution as the sum of this particular solution and the general solution, which is in turn a linear combination of spherical harmonics.

As for the equations, it would help if you all put it in just one pair of tags:
(-hbar^2 / 2\mu) \Delta\Psi = E \Psi
already looks much better (it'd be perfect if I'd written \hbar instead of hbar).

 

[Amok]

It's been a while for me, so please correct me if I am wrong here.
If you want to solve an equation such as the second, it is common to divide the problem in finding a particular solution and the general solution to the homogeneous equation. In this case, the solutions to the homogeneous equation
-\frac{\hbar^2}{2\mu} \Delta\Psi = 0
are the spherical harmonic functions (or a multiple thereof) because the homogeneous equation simply is Laplace's equation. After you have found a particular solution \Psi_0 satisfying -\frac{\hbar^2}{2\mu} \Delta\Psi_0 = E \Psi_0 you can construct the general solution as the sum of this particular solution and the general solution, which is in turn a linear combination of spherical harmonics.

As for the equations, it would help if you all put it in just one pair of tags:
(-hbar^2 / 2\mu) \Delta\Psi = E \Psi
already looks much better (it'd be perfect if I'd written \hbar instead of hbar).

I think the equation is homogenous anyway:(-hbar^2 / 2\mu) \Delta\Psi - E \Psi= 0

\nabla^2_\theta_, _\phi Y = CY

Which is of the same form as the Schrödinger equation.

Ok, so C is a constant with respect to the radius (and everything else actually), which means I had forgotten this step in the separation of variables. Thanks a lot.