The theta(colatitude)-dependent portion of the Hamiltonian in physics provided by Atkins' Physical Chemistry 9th edition (p. 311):

$$\frac{\sin\theta}{\Theta}\frac{d}{d\theta}\sin\theta\frac{d}{d\theta}\Theta+\epsilon \sin^2\theta=m_l^2$$

has solutions based off the general legendre equation:


with $P^l_m(x)$ denoting a function P of x, in case the mathjax looks weird and this isn't clear.

My goal is to derive the physics form from the general legendre equation and thereby verify why their solutions match for $x=\cos\theta$.

I began with some basic algebra ($P^l_m(x)=P$ for simplicity):

  1. divide through by P


  1. add the negative term to both sides $$\frac{1}{P}\frac{d}{dx}\left[(1-x^2)\frac{d}{dx}P\right]+l(l+1)=\frac{m^2}{1-x^2}$$

  2. multiply through by $1-x^2$ $$\frac{1-x^2}{P}\frac{d}{dx}\left[(1-x^2)\frac{d}{dx}P\right]+l(l+1)(1-x^2)={m^2}$$

  1. Transform to spherical coordinates. Here's where I might have bent the rules a bit, and the purpose of this question is to verify I did the remaining steps correctly.

In fact, I'm not entirely sure a transformation to spherical coordinates is what's going on here, but the substitution $x \rightarrow \cos\theta$ has to have some motivating reason. I assumed the reason were related to spherical coordinates via the route below. If my math proves to be wrong, then I can probably conclude this assumption to also be wrong.

Beginning with

$$x=\rho \cos\theta \sin\phi$$

I assume $\rho$ is factored into the equation for the solutions to the wavefunction $\Psi$'s radial portion $R(\rho)$ and $\sin\phi$ into an azimuthal-dependent equation for the angular solutions $Y$ according to


where we're only concerned with the Legendre-related solutions $Y_m^l(\theta)$ that come from our Legendre equation.

So I ignored* $\rho$ and $\sin\phi$, effectively writing


This leads to the substitution

$$(1-x^2)=1-\cos^2\theta$$ $$1-\cos^2\theta=\sin^2\theta$$

$$(1-x^2) \rightarrow \sin^2\theta$$

yielding the legendre equation in the form


Lastly, I have to transform

$$\frac{d}{dx} \rightarrow \frac{d}{d\theta}$$

I did this as follows*:

$$x=\cos\theta$$ $$\frac{dx}{d\theta}=\sin\theta$$ $$\frac{d}{d\theta}=\sin\theta \frac{d}{dx}$$ $$\frac{1}{\sin\theta}\frac{d}{d\theta}=\frac{d}{dx}$$

Essentially, in the second equation above I multiplied through by $dx^{-1}$ like it had the following relation:


I feel like this might not be mathematically correct. For one thing, I "cheat" the $\sin\theta$ to the left of $\frac{d}{d\theta}$ because it's convenient. For another, I'm not sure I can move $dx$ around like a variable in that manner.

However, when I plug this in for $\frac{d}{dx}$ as the final manipulation to the legendre equation, I actually land at the "correct" result:

$$\frac{\sin\theta}{\Theta}\frac{d}{d\theta}\sin\theta\frac{d}{d\theta}\Theta+\epsilon \sin^2\theta=m_l^2$$

for $P=\Theta$ and $\epsilon=l(l+1)$

I know from separation of variables when solving ordinary differential equations, sometimes I can multiply a derivative term $dx$ like any other variable, but I was under the impression there were contexts where I cannot treat "$dx$" like a normal variable in algebraic manipulations. I wasn't sure if this were one of them.

* My questions pertain to the two asterisk-marked steps. First, I ignored variables $\rho$ and $\phi$ in $x=\rho \cos\theta \sin\phi$. Second, I multiplied $dx$ like a variable.

  • Am I allowed to do these two things, ie. does this all appear mathematically sound?

  • Is there a formal way to transform the derivative $\frac{d}{dx} \rightarrow \frac{d}{d\theta}$, or is this algebraic method perfectly acceptable?

  • $\begingroup$ Use \sin and \cos for better readability. Also, use parentheses for function arguments. Makes things clearer. $\endgroup$ – Adrian Keister Sep 26 '18 at 17:11
  • $\begingroup$ I changed the sin and cos. For the parentheses, are you referring to e.g. $\sin(\theta)$? $\endgroup$ – Blaisem Sep 26 '18 at 17:17
  • $\begingroup$ Exactly. People differ about this, but I think writing so that you can't be misunderstood is more important than writing so that you can be understood. $\endgroup$ – Adrian Keister Sep 26 '18 at 18:09

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