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Consider a particle on sphere. Its Hamiltonian in spherical polar coordinates is given by -
$-\frac{\hbar^2}{2mr^2}\Big(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\sin\theta\frac{\partial}{\partial\theta}+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\Big)=\frac{\hat L^2}{2mr^2}$

$\hat L_z=-i\hbar\frac{\partial}{\partial\phi}$
Commutator of both the operators is 0. So they have simultaneous eigen states.
$\hat L_z\psi_{l,m}=\hbar m\psi_{l,m}\;m\in\mathbb R\tag{1}$
$\hat L^2\psi_{l,m}=\hbar l(l+1)\psi_{l,m}\;l\in\mathbb R\tag{2}$

Solution to $(1)$ is given as $\psi_{lm}(\theta,\phi)=e^{im\phi}P_{l,m}(\theta)$

We require $\psi_{l,m}(\theta,\phi+2\pi)=\psi_{l,m}(\theta,\phi)$ which implies $m\in\mathbb Z$

But now if we consider $\psi_{l,m}(\theta,\phi+2\pi)=-\psi_{l,m}(\theta,\phi)$ ,then this implies $m=\frac{2n+1}{2}$ where $n\in\mathbb Z \tag{2}$

I feel that there is no harm in the above assumption because the probability density $|e^{im\phi}|^2$ remains same which is 1 (while considering only $\phi$ part) which is important.

But the solution of $\theta$ part is associated Legendre polynomials $\alpha\;\bigg(\frac{d}{dx}\bigg)^{|m|}P_l(x)\tag{3}$
But how can we solve the above equation if we consider the assumption in $(2)$ which yields fraction values of $m$?

In short, my question is that
i) if we consider particle on a ring (constant $\theta$), the $\psi_m=e^{im\phi}$. Can we use the assumption in $(2)$? Is there some violation in doing so?

ii) But if we consider particle on a sphere (varying $\theta$), then how can we solve the derivative in $(3)$ for fraction values of $m$?

Addendum.
In $(3)$, can the derivative $\bigg(\Big(\frac{d}{dx}\Big)^{|m|}\bigg)$ be defined for fractional values of $m$? For integer values of $m$, the derivative will be act on the function $m$ times.
If this is defined for fractional values of $m$ also then I think, assumption in $(2)$ can be valid?

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  • $\begingroup$ You need fractional derivatives. Feynman did some work on that. Cannot remember where I saw it. $\endgroup$ Sep 6, 2022 at 2:57
  • $\begingroup$ For fractional associated Legendre polynomials, this answer may be helpful. $\endgroup$
    – Luessiaw
    Sep 6, 2022 at 3:15
  • $\begingroup$ Mathematically you can define these objects, but physically they aren't relevant because the wavefunction needs to be single valued. $\endgroup$
    – Andrew
    Sep 6, 2022 at 3:24
  • $\begingroup$ @Andrew, I think probability density $|\psi|^2$ should be single valued. Because probability density os more physically relevant than the wavefunction? $\endgroup$
    – Manu
    Sep 6, 2022 at 3:48

1 Answer 1

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$l$ must be an integer, otherwise $P_{l}(x)$ is unnormalizable. So is $m$ since $m=-l,\cdots,l-1,l$.

Mathematically, the logic is that $m$ should be an integer, otherwise eigenfunctions of $L_z$ are multivalued. (We're talking about the orbital angular momentum. In real space a rotaion by $2\pi$ shouldn't change anything, namely, $\psi(\phi+2\pi)=\psi(\phi)$.) With $m$ an integer, the associated Legendre polynomials are normalizable iff $l$ is an integer no less than $|m|$.

When it comes to spin, things are different. A roration by $2\pi$ (in the spin space) leads to a negative sign for fermions.

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  • $\begingroup$ Yes, $l$ must be integer so that we get polynomial solution instead of power series which will diverge. The solution will be associated Legendre polynomials. I have one doubt in $(3)$ the polynomial is proportional to $\bigg(\frac{d}{dx}\bigg)^|m| P_l(x)$. Can this derivative be defined for the fractional m? $\endgroup$
    – Manu
    Sep 6, 2022 at 2:45
  • $\begingroup$ Sorry, I was wrong about the logic. See my updates. If $m$ is a fraction, the derivative can be defined, though a bit comlicated. But I'm not sure whether the polynomials are still solutions or not. $\endgroup$
    – Luessiaw
    Sep 6, 2022 at 3:02
  • $\begingroup$ Thanks for the reply. Can you tell why the rotation by $2\pi$ changes the sign in spin space. $\endgroup$
    – Manu
    Sep 6, 2022 at 10:01
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    $\begingroup$ Roughly speaking, fermions have half-integer spin, for example $\frac12$, which means half rotation will arouse an extral $\pi$ phase shift to the spin. I'm not confident to explain it simply. You can refer to abundant literature such as Littlejohn's lectures for details. $\endgroup$
    – Luessiaw
    Sep 6, 2022 at 12:40

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