# Homework Help: Differential equation substitution

1. Aug 20, 2010

### d.hatch75

(Apologies for not following the template for topic creation, but I wasn't sure how to adapt my problem to fit it). I'm following the derivation of the spherical harmonics in section 3.3 of Rae's "Quantum Mechanics", but have come across a step I can't quite understand. It seems like such a simple step, but for some reason I'm just not seeing it. For anyone with a copy of the book, it's the step between equations 3.30 and 3.31. For those without a copy, the initial equation is:

$$sin\theta \frac{d}{d\theta}(sin\theta \frac{d\Theta}{d\theta}) + (\lambda' sin^2\theta - m^2)\Theta = 0$$

By substituting:
$$v=cos\theta$$
and writing:
$$P(v)\equiv \Theta(\theta)$$
one should arrive at the following:

$$\frac{d}{dv}[(1-v^2)\frac{dP}{dv}] + [\lambda' - \frac{m^2}{1-v^2}]P = 0$$

whereas I'm getting:

$$-\frac{d}{dv}[\sqrt{1-v^2}\frac{dP}{dv}] + [\lambda' - \frac{m^2}{1-v^2}]P = 0$$

Am I missing a simple equivalence between what I'm getting and what I should be getting, or have I derived an incorrect result?

2. Aug 20, 2010

### LawlQuals

I suspect you did just miss one small detail. It appears you have made the translation: $$d\Theta /d\theta \rightarrow dP/dv$$, but this is not the case.

Beginning with the differential operator $$\frac{d}{d\theta}$$ itself, and noting that $$\theta = \theta (v)$$:

$$\frac{d}{d\theta} = \frac{d}{dv}\frac{dv}{d\theta} = -\sin\theta\frac{d}{dv}$$. If one proceeds as it appears you have done with $$\frac{d\Theta}{d\theta} = \frac{dP}{dv}$$, your term on the left would be:

$$\sin\theta\underbrace{\frac{d}{d\theta}}_{ = \, -\sin\theta d/dv}\underbrace{\sin\theta}_{= \, \sqrt{1 - v^2}}\underbrace{\frac{d\Theta}{d\theta}}_{= \, dP/dv} = -\sin^2\theta\frac{d}{dv}\left[\sqrt{1 - v^2}\frac{dP}{dv}\right]$$

After a subsequent division of the entire equation by $$\sin^2\theta$$, provided we tacitly enforce that $$\sin^2\theta \neq 0$$, the result you have written down may be retrieved. That is to say, by letting $$\sin^2\theta = 1 - v^2$$, and $$\frac{d\Theta}{d\theta} = \frac{dP}{dv}$$, the equation admits upon the aforementioned division:

$$-\frac{d}{dv}\left[\sqrt{1 - v^2}\frac{dP}{dv}\right] + \left[\lambda ' - \frac{m^2}{1 - v^2}\right]P = 0$$

Which is exactly the equation you have quoted . The improper step was on the treatment of $$dP/dv$$. The statement $$\frac{d\Theta}{d\theta} = \frac{dP}{dv}$$ is not true. To obtain the proper definition, chain rule must be engaged. Begin with the definition of $$P$$:

$$\Theta (\theta ) = P(v)$$, and differentiate both sides,

$$\frac{d\Theta}{d\theta} = \frac{dP}{d\theta} = \frac{dP}{dv}\frac{dv}{d\theta}$$

With this modification, the result in your textbook will follow directly.

By the way, just some friendly information: in LaTeX, there are specific commands for formatting elementary functions, for instance \sin\theta will format as $$\sin\theta$$ as opposed to $$sin\theta$$ should you have not used the command.

Last edited: Aug 20, 2010
3. Aug 21, 2010

### d.hatch75

The irony is that I thought twice about equating the two differentials but managed to convince myself that they were equal... Thank you ever so much, LawlQuals, I feel rather silly now!

I understand that this is more of a physics-related question and thus perhaps in the wrong forum, but I've already made this topic so it can be moved if mods would prefer it? I'm not too comfortable with dividing through by $$\sin^2\theta$$, unless of course there is a physical interpretation that justifies this move - is there? Or is this just clumsy algebra?

4. Aug 21, 2010

### LawlQuals

I also was uncomfortable with the division by $$\sin^2\theta$$, but regrettably I cannot provide a convincing reason why this step would be substantiated. Maybe there is a more proper way to obtain the same result that I cannot see. Hopefully a moderator may move this into a section where others may see it, now that a subsequent question has come up.

You are welcome and I am glad I could help!

5. Aug 23, 2010

### d.hatch75

I'm not so sure there is another way to obtain the same result; the (limited) workings in the book suggest a division by $$\sin^2\theta$$, so I can only assume there is a physical interpretation that justifies it. The only thing I can think of is that, going by the previous workings in the section, if one assumes $$\sin^2\theta = 0$$, and therefore that $$\sin\theta = 0$$, by the initial equation in my first post this requires that $$m = 0$$ (because $$\Theta = 0$$ gives a wavefunction of 0). But $$m = \sqrt{\frac{2m_{e}v}{\hbar^2}}$$ by its definition, so $$m = 0 \Rightarrow v = 0$$. Wait, hang on...yeah, I've lost my train of thought completely. In my head I came to the conclusion that $$\sin\theta = 0$$ implied a trivial constant spherical harmonic...

Anyone?

6. Aug 23, 2010

### diazona

I don't have a copy of Rae's book, but if the derivation there is anything like the derivation on Wikipedia, the original source of the differential equation is the Laplacian operator ∇², which includes factors of 1/(sin θ) in spherical coordinates. So the points where sin θ = 0 (i.e. the poles) have to be excluded from the problem from the very beginning, and there's no problem with dividing by sin θ.

When you go to actually solve the differential equation for P, you choose the boundary conditions that will allow the spherical harmonic to have a well-defined limit at the poles.* Then you can define the value of the spherical harmonic at each pole to be equal to its limit there, so that the function is continuous.

*Since the spherical harmonics take the form Θ(θ)e-imφ, hopefully you can see that in order to have a well-defined limit at e.g. θ=0, either m or Θ(0) has to be zero. That's why all the spherical harmonics with m≠0 have factors of sin θ.

7. Aug 24, 2010

### d.hatch75

Yeah, the source of the differential equation is indeed the Laplacian operator with factors of 1/(sin θ), and thus the poles have to be excluded. But is there a physical justification for excluding the poles? I won't pretend to know enough about it, but it seems to be just a mathematical necessity...

8. Aug 24, 2010

### diazona

No, there's no physical reason to exclude the poles. (If there were, the final spherical harmonics wouldn't even be defined at the poles - their domain would be the sphere with two holes in it.) That's because the spherical harmonics pop up in problems with spherical symmetry, and there are no physically special points that you could choose to be the poles.

The thing is, in the course of deriving the spherical harmonics, you're forced to use a mathematical operator (the Laplacian) that blows up at a couple of points on the sphere. So the derivation you do using that operator is not valid at the poles. It gives you functions defined only on the sphere with two holes. But since there's no physical reason to have those holes, you have to "patch" them - basically apply some other mathematical technique to assign the function a value at those points. That technique would be requiring continuity.