# Wavefunction with two different values at same point

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 $$m$$?

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?

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

$$l$$ 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.

• 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?
– Manu
Sep 6, 2022 at 2:45
• 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. Sep 6, 2022 at 3:02
• Thanks for the reply. Can you tell why the rotation by $2\pi$ changes the sign in spin space.
– Manu
Sep 6, 2022 at 10:01
• 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. Sep 6, 2022 at 12:40