I Particle on a sphere problem in quantum mechanics and its solution

[Kashmir]

To solve a particle on a sphere problem in quantum mechanics we get the below equation :$\left[\frac{1}{\sin \theta} \frac{d}{d \theta}\left(\sin \theta \frac{d}{d \theta}\right)-\frac{m^{2}}{\sin ^{2} \theta}\right] \Theta(\theta)=-A \Theta(\theta)$

To solve this differential equation, we start with a change of independent variable $z=\cos \theta$, where $z$ is the rectangular coordinate for the particle, assuming a unit sphere. We also introduce a new function
$P(z)=\Theta(\theta) .$
Using this substitution we've this equation
$\left(\left(1-z^{2}\right) \frac{d^{2}}{d z^{2}}-2 z \frac{d}{d z}+A-\frac{m^{2}}{\left(1-z^{2}\right)}\right) P(z)=0$ whose solutions are the associated Legendre polynomials :

$P_{\ell}^{-m}(z)=P_{\ell}^{m}(z)$

To get back the angular solution I should use $P(z)=\Theta(\theta)$ and also normalise it, giving me

$\Theta_{\ell}^{m}(\theta)= \frac{(2 \ell+1)}{2} \frac{(\ell-m) !}{(\ell+m) !} P_{\ell}^{m}(\cos \theta)$

But my book writes the solution with an additional $(-1)^{m}$ term as
$\Theta_{\ell}^{m}(\theta)= (-1)^{m}\frac{(2 \ell+1)}{2} \frac{(\ell-m) !}{(\ell+m) !} P_{\ell}^{m}(\cos \theta)$

Why is that so?

 

[dextercioby]

What is your book? Do you know about parity transformations?

 

[Kashmir]

What is your book? Do you know about parity transformations?

David McIntyre, Quantum mechanics.

I know little bit about parity operator.

 

[dextercioby]

Look up on the internet the "Condon-Shortley phase factor".

 

[Kashmir]

Look up on the internet the "Condon-Shortley phase factor".

So the factor is there to simplify calculations.

Similarly defining for negative $m$:
$\Theta_{\ell}^{-m}(\theta)=(-1)^{m} \Theta_{\ell}^{m}(\theta), \quad m \geq 0 .$ is also to simplify calculations?

 

[dextercioby]

I think that is symmetry (parity)-related.

 

[PeroK]

$\Theta_{\ell}^{m}(\theta)= (-1)^{m}\frac{(2 \ell+1)}{2} \frac{(\ell-m) !}{(\ell+m) !} P_{\ell}^{m}(\cos \theta)$

Why is that so?

What happens if you apply the raising operator $L_+$ to $\Theta_{1}^{0}(\theta)$? Do you need a factor of $(-1)^m$ to satisfy $L_+\Theta_{1}^{0} = \Theta_{1}^{1}$?

 

[Kashmir]

What happens if you apply the raising operator $L_+$ to $\Theta_{1}^{0}(\theta)$? Do you need a factor of $(-1)^m$ to satisfy $L_+\Theta_{1}^{0} = \Theta_{1}^{1}$?

Hello sir.
I've not done the ladder operators for angular momentum. I'll do them and come back.
Thank you :)