Spherical Harmonics: Rigid Rotor, Laplace's Equation

In summary, the conversation discusses the solutions to the quantum rigid rotor and Laplace's equations, and how they are related. The spherical harmonic functions are the solutions to the angular portion of Laplace's equation, and also appear in the solution to the rigid rotor Schrödinger equation. The conversation also discusses the process of finding the general solution using a particular solution and the harmonic functions.
  • #1
Amok
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So, I was reading about the quantum http://en.wikipedia.org/wiki/Rigid_rotor" [Broken] 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[tex]\Psi[/tex]part 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



[tex]\Delta[/tex][tex]\Psi[/tex]=0 (Laplace's equation)

(-h-bar2/2[tex]\mu[/tex])[tex]\Delta[/tex][tex]\Psi[/tex]=E[tex]\Psi[/tex]

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.
 
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  • #2
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:

[tex]\nabla^2 \Psi = 0[/tex]

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

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

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

[tex]\nabla^2_\theta_, _\phi Y = CY[/tex]

Which is of the same form as the Schrödinger equation.
 
  • #3
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
[tex]-\frac{\hbar^2}{2\mu} \Delta\Psi = 0[/tex]
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 [itex]\Psi_0[/itex] satisfying [itex]-\frac{\hbar^2}{2\mu} \Delta\Psi_0 = E \Psi_0[/itex] 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:
[tex](-hbar^2 / 2\mu) \Delta\Psi = E \Psi[/tex]
already looks much better (it'd be perfect if I'd written \hbar instead of hbar).
 
  • #4
CompuChip said:
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
[tex]-\frac{\hbar^2}{2\mu} \Delta\Psi = 0[/tex]
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 [itex]\Psi_0[/itex] satisfying [itex]-\frac{\hbar^2}{2\mu} \Delta\Psi_0 = E \Psi_0[/itex] 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:
[tex](-hbar^2 / 2\mu) \Delta\Psi = E \Psi[/tex]
already looks much better (it'd be perfect if I'd written \hbar instead of hbar).

I think the equation is homogenous anyway:[tex](-hbar^2 / 2\mu) \Delta\Psi - E \Psi= 0[/tex]
phsopher said:
[tex]\nabla^2_\theta_, _\phi Y = CY[/tex]

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.
 
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1. What are spherical harmonics?

Spherical harmonics are a set of mathematical functions that describe the distribution of points on a sphere. They are commonly used in physics and engineering to represent the spatial variation of a physical quantity.

2. How are spherical harmonics related to the rigid rotor model?

In the rigid rotor model, the distribution of points on a rotating object can be described using spherical harmonics. This allows us to analyze the rotational motion of the object and calculate its energy levels.

3. What is Laplace's equation and how is it used in the context of spherical harmonics?

Laplace's equation is a partial differential equation that describes the behavior of a physical quantity in a region with no sources or sinks. In the context of spherical harmonics, it is used to solve for the eigenfunctions and eigenvalues of the rigid rotor model.

4. Can spherical harmonics be used to describe the shape of an object?

Yes, spherical harmonics can be used to represent the shape of an object by decomposing its surface into a series of spherical harmonics with different amplitudes and phases. This allows us to compare and analyze the shapes of different objects.

5. Are spherical harmonics only used in physics and engineering?

No, while spherical harmonics are commonly used in these fields, they have also found applications in other areas such as computer graphics, image processing, and geophysics. They are a versatile mathematical tool that can be applied in various fields.

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