Eigenvalues of spherical harmonics in $d$ dimensions

I'm working on the Schrodinger equation for a hydrogen atom in a $d$-dimensional space, so I'm interested in the possible eigenvalues of the angular momentum part of the $d$-dimensional Laplace operator.

Since the angular momentum part corresponds to the quadratic casimir operator of the special orthogonal group in $d$ dimensions one can calculate the eigenvalues of the casimir operator and gets $$\sum_{n = 1}^{\lfloor d/2 \rfloor} \lambda_n (\lambda_n + d - 2n),$$ where $\lambda_n$ is a positive integer. My problem is that the eigenvalues of the spherical harmonics (related to the angular momentum part) in higher dimensions are $$\lambda (\lambda + d - 2),$$ so not all possible eigenvalues of the casimir operator appear in the spherical harmonics.

I found that the spherical harmonics correspond to a "totally symmetric tensor representation", which would explain this fact, but what is the relation between the spherical harmonics and the symmetric tensor representations? Or what is the reason that not all possible eigenvalues for the quadratic Casimir operator appear in the spherical harmonics?

• – Cosmas Zachos Jun 24 '16 at 22:33
• Side note: you shouldn't do a 1/r^2 potential so you don't copy the same kind of Schrodinger equation. It's a mistake I've seen before. – AHusain Jun 25 '16 at 11:22

One reason there are more possible eigenvalues of the Casimir operator of the rotations than appear in the spherical harmonics is that the spherical harmonics are proper representations of $\mathrm{SO}(n)$ while the possible values for the Casimir operator classify the possible irreducible representations of $\mathfrak{so}(n)$.
By general Lie theoretic arguments, these lift to representations of the universal cover $\mathrm{Spin}(n)\overset{2:1}{\to}\mathrm{SO}(n)$, but not every representation of the cover descends to a representation of $\mathrm{SO}(n)$.
In 3D, this is the only reason for eigenvalues that do not appear on the spherical harmonics. In general, the spherical harmonics only correspond to the $\mathrm{SO}(n)$ representations of the traceless symmetric tensors. How to see this depends on your definition of the spherical harmonics of degree $\ell$.
The convenient way is to define the spherical harmonics of degree $\ell$ in $d$ dimensions as the restriction of harmonic homogeneous polynomials of degree $\ell$ to the unit sphere, that is polynomials $P_\ell$ in $x_1,\dots,x_d$ such that $P_\ell(\lambda x_1,\dots,\lambda x_d) = \lambda^\ell P_\ell(x_1,\dots,x_d)$ and $\Delta P_\ell = 0$. Such a polynomial can be written as $$P_\ell = \sum_{i_1,\dots,i_\ell, \sum_k i_k = \ell} p_{i_1\dots i_d} x_1^{i_1}\dots x_d^{i_d}$$
Now the monomes $\prod_{k} x_i^{i_k}$ with $\sum_k i_k = \ell$ correspond exactly to symmetric $\ell$-tensors in $k$ dimensions, see this answer of mine, so each $P_\ell$ as a whole also corresponds to a symmetric tensor.
That this tensor is traceless follows from $\Delta P_\ell = 0$ by a straightforward computation.