# Quantization of particle on a sphere

While studying particle on a sphere, I read that boundary condition on $$\theta$$ and $$\phi$$ results in quantization. I understand boundary condition for $$\phi$$ part which varies from 0 to 2$$\pi$$. As $$\psi(0)$$ should be equal to $$\psi(2\pi)$$ otherwise in standing wave the wave function can have more than one value and can interfere destructively.

But I am not able to figure out boundary condition for $$\theta$$ part which varies from 0 to $$\pi$$. In books,the result is directly quoted as "Boundary condition in $$\theta$$ part resulted in quantization which restricts the quantization in $$\phi$$ part". Quantization in $$\theta$$ part is called as azimuthal quantum number$$(l)$$ and $$\phi$$ part is called magnetic quantum number($$m$$) which varies from $$-l$$ to $$l$$.

Please tell what would be the boundary condition for $$\theta$$ part. I am not able to figure out and visualize it.

The idea of boundary conditions is nice and intuitive, but this is one case in which intuition cannot really help us.

In general when you are solving the Schrödinger equation in spherical coordinates, you can separate the radial and angular parts (unless you have some pathological potential energy term, I think). See Griffiths' Introduction to Quantum Mechanics (2nd ed.), Chpt 4 for details. It doesn't matter whether you're solving the H atom, a particle on a sphere, or indeed a free particle: in all cases the potential energy has no angular dependence in the potential energy, so the angular component is the same problem.

When you solve the angular bit, you can (again) separate out the two angles, so you end up with two differential equations: one in $$\theta$$ and one in $$\phi$$. The $$\phi$$ bit is easy enough:

$$\frac{1}{g}\frac{\mathrm d^2 g}{\mathrm d\phi^2} = -m^2$$

where $$m$$ is unconstrained for now, and $$g$$ is a function of only $$\phi$$. This is easy to solve and applying the "common-sense" boundary condition that $$g(\phi) = g(\phi + 2\pi)$$, you can find that $$g = \exp(\mathrm i m\phi)$$ with $$m$$ an integer. So far so good.

The $$\theta$$ bit is substantially more difficult, but this differential equation is so important that it has been studied for centuries and has its own name, the Legendre equation:

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

where $$f$$ is again a function of only $$x$$, and we have made the substitution $$x = \cos\theta$$. In our case, it turns out that unless $$(l, m)$$ obey certain constraints (i.e. the now-familiar $$l = 0, 1, 2\ldots$$ and $$m = -l, \ldots, +l$$), then the solutions $$f(x) = f(\cos\theta)$$ go to $$\pm\infty$$ at $$x = \pm 1$$, i.e. $$\theta = 0$$ or $$\theta = \pi$$. That's not acceptable for a well-behaved wavefunction, so we have to reject all the other cases.

So, you could say that the boundary condition on $$\theta$$ is that the wavefunction must be welll-behaved (i.e. not diverge to $$\pm\infty$$) at $$\theta = 0$$ and $$\theta = \pi$$. But that is not something we could say without delving into the maths. A priori, we could certainly say that it is important for the wavefunction to be well-behaved at all points. But we cannot pick out the two particularly crucial points $$\theta = 0$$ and $$\theta = \pi$$ just using common sense, or "by observation".

More details can be found in Griffiths (there is a fair bit of mathematical detail in there), or any other book that covers differential equations, really.

Bonus: if you multiply the solutions $$f(\theta)$$ with $$g(\phi)$$, then normalise the entire thing, you get what are known as the spherical harmonics, which are also extremely important functions in their own right; that's where the shapes of the atomic orbitals come from.