Spherical Harmonics: Arriving at Equation

[davidge]

How does one arrive at the equation

$$\bigg( (1-z^2) \frac{d^2}{dz^2} - 2z \frac{d}{dz} + l(l+1) - \frac{m^2}{1-z^2} \bigg) P(z) = 0$$
Solving this equation for $P(z)$ is one step in deriving the spherical harmonics "$Y^{m}{}_{l}(\theta, \phi)$".

The problem is that the book I'm following doesn't show how to arrive at the above equation. It shows how to arrive at it only for the special case $m=0$.

I've tried googling "Associate Legendre's equation" and "Legendre's general equation derivation" but it seems there's no such derivation on web.

 

[blue_leaf77]

Are you trying to solve the hydrogen atom eigenfunction?
As a matter of fact, that differential equation is the defining equation for the associated Legendre polynomial. Therefore, unless you encounter it in a different context, there is no way to derive it. It is simply defined such that the associated Legendre polynomial is the solution of that equation.

 

[vanhees71]

The standard way to derive the equation is to apply the separation ansatz
$$\psi(r,\vartheta,\varphi)=R(r) \Theta(\vartheta) \Phi(\varphi)$$
for the eigenfunctions of the Laplacian in spherical coordinates.
$$\Delta \psi=\frac{1}{r} \partial_r^2 (r \psi) + \frac{1}{r^2 \sin \vartheta} \partial_{\vartheta}(\sin \vartheta \partial_{\vartheta} \psi) + \frac{1}{r^2 \sin^2 \vartheta} \partial_{\varphi}^2 \psi.$$

 

[davidge]

Are you trying to solve the hydrogen atom eigenfunction?

I'm trying to derive the spherical harmonics functions, often denoted by $Y^{m}{}_{l}(\theta, \phi)$. It's part of the hydrogen atom problem.

As a matter of fact, that differential equation is the defining equation for the associated Legendre polynomial. Therefore, unless you encounter it in a different context, there is no way to derive it. It is simply defined such that the associated Legendre polynomial is the solution of that equation.

Maybe I should have put my question in the following way

The book I'm reading shows how to arrive at the equation $$\bigg( (1-z^2) \frac{d^2}{dz^2} - 2z \frac{d}{dz} + A - \frac{m^2}{1-z^2} \bigg) P(z) = 0$$
where one needs to find what $A$ is. The book shows how to find $A$ for the case $m=0$; it's $A = l(l+1)$. Then it states that $A = l(l+1)$ even for $m \neq 0$, but does not proves that, as it proves for the former case.

@vanhees71 By doing this way, can one show that $A = l(l+1)$ is the correct term?

 

[PeterDonis]

the book I'm following

Which book?

 

[vanhees71]

If I think about it, an even simpler way is to look for harmonic functions, i.e., the Laplace equation
$$\Delta \psi=0.$$
You make the above separation ansatz. Plugging it into the Laplace Equation leads to
$$\frac{\Theta \Phi}{r} (r R)''+\frac{R \Phi}{r^2 \sin \vartheta}(\sin \vartheta \Theta)'+\frac{R \Theta}{r^2 \sin^2 \vartheta} \Phi''=0.$$
Multiply the entire expression with $r^2/(R \Theta \Phi)$, and you get
$$\frac{r}{R} (r R)''=-\frac{1}{\Theta \sin \vartheta}(\sin \vartheta \Theta)'-\frac{1}{\Phi \sin^2 \vartheta} \Phi''.$$
Sine the left-hand side depends on $r$ only but the right-hand side on $\vartheta$ and $\varphi$ the entire expression must be some constant $A$:
$$r (r R)''=A R.$$
The equation can obviously solved by the ansatz $R=c r^{\lambda}$:
$$r (r R)''=c r (r^{\lambda+1})''=c r (\lambda+1) \lambda r^{\lambda-1}=c A r^{\lambda} \; \Rightarrow\; \lambda(\lambda+1)=A.$$
Then you repeat the argument with the right-hand side
$$-\frac{1}{\Theta \sin \vartheta}(\sin \vartheta \Theta)'-\frac{1}{\Phi \sin^2 \vartheta} \Phi''=\lambda(\lambda+1)$$
or, a bit rewritten,
$$-\frac{\sin \vartheta }{\Theta}(\sin \vartheta \Theta')' - \lambda(\lambda+1) \sin^2 \vartheta = \frac{1}{\Phi} \Phi''=-m^2,$$
where again $m=\text{const}$ since the left-hand side depends on $\vartheta$ only and the right-hand side on $\varphi$ only. You then find
$$\Phi(\varphi)=\exp(\mathrm{i} m \varphi),$$
and since the solution should be a $2 \pi$-periodic function in $\varphi$ you get $m \in \mathbb{N}$.

Finally you are left with [Typos corrected]
$$\sin \vartheta (\sin \vartheta \Theta')'+ \left [\lambda(\lambda+1) \sin^2 \vartheta - m^2 \right ]\Theta=0.$$
Now substitute
$$z=\cos \vartheta, \quad \Theta(\vartheta)=P(z).$$
From this you get immediately your formula.

 

[davidge]

Which book?

McIntyre - Quantum Mechanics - A Paradigms Approach

@vanhees71 Perfect. Thank you.

 

[davidge]

Is there a typo here?

or, a bit rewritten,
$$-\frac{\sin \vartheta }{\Theta}(\sin \vartheta \Theta')' - \lambda(\lambda+1) \sin^2 \vartheta = \frac{1}{\Phi \sin^2 \vartheta} \Phi''=-m^2,$$

$$ \Longrightarrow -\frac{\sin \vartheta }{\Theta}(\sin \vartheta \Theta)' - \lambda(\lambda+1) \sin^2 \vartheta = \frac{\Phi''}{\Phi}=-m^2$$

And, also, here?

Finally you are left with
$$\sin \vartheta (\sin \vartheta \Theta')'+ \left [\lambda(\lambda-1) \sin^2 \vartheta + m^2 \right ]\Theta=0.$$

$$ \Longrightarrow \sin \vartheta (\sin \vartheta \Theta)'+ \left [\lambda(\lambda+1) \sin^2 \vartheta - m^2 \right ]\Theta=0.$$

 

[vanhees71]

I've corrected the typos in #6. I hope, now all signs are correct ;-).

 

[davidge]

I've corrected the typos in #6. I hope, now all signs are correct ;-).

You are still keeping the derivative symbol on $\Theta$. Is this so?

Also, you have actually forgotten the first one that I have pointed out.

 

[vanhees71]

Argh. Yes, I corrected it, but the $\Theta'$ must stay. You have a 2nd-order differential operator (it's more or less $\hat{\vec{L}}^2$, and it's much more understandable what's coming out using the semi-algebraic way in my QM manuscript).