Matching the general legendrian equation to the physics form used in the hamiltonian

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:

$$\frac{d}{dx}\left[(1-x^2)\frac{d}{dx}P^l_m(x)\right]+\left[l(l+1)-\frac{m^2}{1-x^2}\right]P^l_m(x)=0$$

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

$$\frac{1}{P}\frac{d}{dx}\left[(1-x^2)\frac{d}{dx}P\right]+\left[l(l+1)-\frac{m^2}{1-x^2}\right]=0$$

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

$$\Psi=R(\rho)\left[Y_m(\phi)Y_m^l(\theta)\right]$$

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

$$x=\cos\theta$$

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

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

yielding the legendre equation in the form

$$\frac{\sin^2\theta}{P}\frac{d}{dx}\left[(\sin^2\theta)\frac{d}{dx}P\right]+l(l+1)\sin^2\theta={m^2}$$

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:

$$dx^{-1}=\frac{d}{dx}$$

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?

• Use \sin and \cos for better readability. Also, use parentheses for function arguments. Makes things clearer. – Adrian Keister Sep 26 '18 at 17:11
• I changed the sin and cos. For the parentheses, are you referring to e.g. $\sin(\theta)$? – Blaisem Sep 26 '18 at 17:17
• 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. – Adrian Keister Sep 26 '18 at 18:09