I Solving a Particle on the Surface of a Sphere: Obtaining Eigenvalues

[Salmone]

The Hamiltonian of a particle of mass $m$ on the surface of a sphere of radius $R$ is $H=\frac{L^2}{2mR^2}$ where $L$ is the angular momentum operator. I want to solve the TISE $\hat{H}\psi=E\psi$ and in order to do that I rewrite $L^2$ in Schroedinger's representation in spherical coordinates so $\frac{1}{sin(\theta)}\frac{\partial }{\partial \theta}(sin(\theta)\frac{\partial \psi}{\partial \theta})+\frac{1}{sin^2(\theta)}\frac{\partial^2 \psi}{\partial \phi^2}=-\frac{2mER^2}{\hbar^2}\psi=-k^2\psi$ after some steps one can get $\frac{sin(\theta)}{\Theta(\theta)}\frac{\partial }{\partial \theta}(sin(\theta)\frac{\partial \Theta(\theta)}{\partial \theta})+K^2sin^2(\theta)+\frac{1}{\Phi(\phi)}\frac{\partial^2 \Phi(\phi)}{\partial \phi^2}=0$ having used $\psi=\Theta(\theta)\Phi(\phi)$. Now the two parts dipend on different variables so they must be equal to a constant, let's call this constant $m^2$ and I obtain two equations:

$\frac{sin(\theta)}{\Theta(\theta)}\frac{\partial }{\partial \theta}(sin(\theta)\frac{\partial \Theta(\theta)}{\partial \theta})+K^2sin^2(\theta)=m^2$ $(1)$

$\frac{1}{\Phi(\phi)}\frac{\partial^2 \Phi(\phi)}{\partial \phi^2}=-m^2$ $(2)$

The equation $(1)$ could be rearranged and written as:
$\frac{\partial^2 \Theta(\theta)}{\partial \theta^2}+\frac{cos(\theta)}{sin(\theta)}\frac{\partial \Theta(\theta)}{\partial \theta}+[k^2-\frac{m^2}{sin^2(\theta)}]\Theta(\theta)=0$
which is, by choosing $k^2=l(l+1)$ the legendre associated differential equation according to https://mathworld.wolfram.com/AssociatedLegendreDifferentialEquation.html equation $(9)$.

This leads to Energy eigenvalues $E=\frac{\hbar^2l(l+1)}{2mR^2}$.

My questions:

1)
How do we know that the eigenvalues found are *all* the eigenvalues for this problem? If $k^2$ had a different value, for example $k^2$=3, which is not a value obtainable with $k^2=l(l+1)$, how do we know that the differential equation would have no solutions?

$\frac{\partial^2 \Theta(\theta)}{\partial \theta^2}+\frac{cos(\theta)}{sin(\theta)}\frac{\partial \Theta(\theta)}{\partial \theta}+[3-\frac{m^2}{sin^2(\theta)}]\Theta(\theta)=0$ Does this TISE have solutions? How do we know that? Because if this TISE has solutions than I obtain the eigenvalue $E=\frac{3\hbar^2}{2mR^2}$ which is not in the list $E=\frac{\hbar^2l(l+1)}{2mR^2}$.

2)
Usually in quantum mechanics problems, we find the eigenvalues by imposing that our eigenfunction is square-integrable, in this case what we do is putting $k^2=l(l+1)$, does this represent the request that the eigenfunctions must be square-integrable?

 

[Orodruin]

1) This follows from the Sturm-Liouville theorem. It is not that we are assuming $k^2 = \ell(\ell+1)$, it is the Sturm-Liouville operator involved that has those eigenvalues. Any other $k^2$ would not have a corresponding solution just as there are no vectors $x$ such that $Ax = \lambda x$ for a matrix $A$ unless $\lambda$ is an eigenvalue of $A$.

2) No, it is from the requirement that the function must satisfy certain boundary conditions. Compare to fitting the sines $\sin(kx)$ that solves the particle-in-a-box problem with $0<x<L$. In order to satisfy $\sin(kL)=0$, only a discrete set of values $k = \pi n/L$ ($n\in \mathbb Z$) are allowed.

 

[Salmone]

@Orodruin Thanks for your answers, about anwer no.2: so in this case $k^2=l(l+1)$ is a boundary condition for the eigenfunctions?

 

[Orodruin]

@Orodruin Thanks for your answers, about anwer no.2: so in this case $k^2=l(l+1)$ is a boundary condition for the eigenfunctions?

Not per se. You should consider it as a result for the eigenvalues of the Sturm-Liouville problem (which includes appropriate boundary conditions).

In the case of the Legendre polynomials, those boundary conditions are essentially that the function is regular at the poles. One way of seeing this is considering the Taylor expansion around zero of the solution to the Legendre differential equation. The differential equation will lead to a recursion relation for the expansion coefficients and you can show that the Taylor series diverges at $\pm 1$ unless it terminates - which will only happen for eigenvalues $\ell(\ell+1)$. Hence, Legendre polynomials.

Just as for the sine and cosine solutions for the particle in a box, the Legendre differential equation is of order two and therefore has two linearly independent solutions. For the eigenvalues of the Sturm-Liouville problem the second solution (often denoted $Q_\ell$) is not regular at $\pm 1$.

If you can get access to my book this is discussed in some detail in section 5.5.2 which treats finding the eigenfunctions of the Laplace operator in spherical coordinates.

 

[vanhees71]

I think the modern treatment is much more clear (and incomplete in almost all standard QM textbooks; the only exceptions I'm aware of is Ballentine and the German textbook by Münster).

In my QM lecture I do it in the following way:

(1) Start with the representation-free treatment, based solely on the commutation relations of (Cartesian) angular-momentum components,
$$[\hat{J}_a,\hat{J}_b]=\mathrm{i} \epsilon_{abc} \hat{J}_c.$$
Then you can look for the common eigenvectors of $\hat{\vec{J}}^2$ and $\hat{J}_3$ using the "ladder operators" $\hat{J}_{\pm}=\hat{J}_1 \pm \mathrm{i} \hat{J}_2$ with the standard result that the eigenvalues of $\hat{\vec{J}}^2$ are $j(j+1)$ with $j \in \{0,1/2,1,3/2,\ldots \}$ and those for $\hat{J}_3$ being $m \in \{-j,-j+1,\ldots,j-1,j \}$. You make heavy use of the properties of self-adjoint operators.

(2) Then consider the special case of orbital angular momentum, which is given in terms of position and momentum operators (within the Heisenberg algebra),
$$\hat{\vec{L}}=\hat{\vec{x}} \times \hat{\vec{p}}.$$
This enables you to consider the 2D isotropic harmonic oscillator in the (12)-plane with the annihilation operators for "phonons" $\hat{a}_1$, $\hat{a}_2$ and, in addition, $\hat{b}=1/\sqrt{2} (\hat{a}_1 + \mathrm{i} \hat{a}_2)$. Then you can solve the common eigenvalue problem for $\hat{N}=\hat{N}_1+\hat{N}_2$ and $\hat{N}_b=\hat{b}^{\dagger} \hat{b}$, for which $\hat{L}_3=\hat{N}-2 \hat{N}_b$, which shows that the eigenvalues of $\hat{L}_3$ take only (all) integer values, i.e., there are only integer (true) representations for the rotation algebra in case of orbital angular momentum.

(3) Only now you go into the "position representation" in spherical coordinates. The orbital-angular momentum eigenfunctions then live in $\mathrm{L}^2(S)$, where $S$ is the unit sphere in $\mathbb{R}^3$, parametrized with the angles $(\vartheta,\varphi)$. The scalar product is
$$\langle u_1 | u_2 \rangle=\int_0^{\pi} \mathrm{d} \vartheta \sin \vartheta \int_0^{2 \pi} \mathrm{d} \varphi u_1^*(\vartheta,\varphi) u_2(\vartheta,\varphi).$$
The eigenfunctions are then easily constructed starting with $\text{Y}_{\ell \ell}$ from $\hat{L}_+ \text{Y}_{\ell \ell}=0$ and then using $\hat{L}_-$ to get all the $\text{Y}_{\ell m}$ for fixed $\ell \in \mathbb{N}_0$.

It's also an amusing excercise to show, why this construction fails when assuming falsely that $\ell=1/2$ (or any other half-integer value) were a solution. It's, because you get apparent "eigenfunctions" which lie not all in the domain of the $\hat{\vec{L}}$ as self-adjoint operators. Particularly you can start with $\text{Y}_{1/2,1/2}$ defined by $\hat{L}_+ \text{Y}_{1/2,1/2}=0$ and then repeatly apply $\hat{L}_-$. You get also a function $\text{Y}_{1/2,-1/2}$ but then applying $\hat{L}_-$ once more you don't get 0 but a function which is no longer in $\mathrm{L}^2(S)$.

The way using partial differential equations is quite tedious but of course leads to the same conclusions:

W. Pauli, Helv. Acta. Physica 12, 147 (1939) (in German)