## Spherical Harmonics and Fourier Series

Recently, I read a very interesting article on the origin of spherical harmonics. I like to summarize in here and add some personal comments.

Starting from Laplace equation

$\nabla^2 \phi(\vec{r}) = 0$

The Laplacian can be separated into radial and spherical part.

$\nabla^2 = \nabla_r^2 + \nabla_\Omega^2$

The solution is called harmonics, and it can be separated into radial part and angular part too,

$\phi(\vec{r}) = R(r) \Theta(\Omega)$

Since the Laplacian is coordinate-free, therefore, the solution is also coordinate free and is rotational invariant. We will come back to this point later.

A homogeneous function of degree n has property,

$f(t\vec{r}) = t^n f(\vec{r})$

In the case of homogeneous harmonics of degree n,

$\phi_n(\vec{r}) = r^n \Theta_n(\Omega)$

Here, the radial part is $R_n(r) = r^n$

Substitute this homogeneous harmonics into the Laplace equation, the $\nabla_r^2$ will produce a coefficient related to the order, and the radial part can be extracted.

$0 = f(r) ( \nabla_\Omega^2 - g(n) ) \Theta(\Omega)$

we have an eigenvalue problem for the angular part

$\nabla_\Omega^2 \Theta = g(n) \Theta$

The eigen function for 2-D Laplacian is the Fourier Series, and that for 3-D is the Spherical Harmonics. In this sense, Fourier Series is a “polar harmonics”.

In 3-D, the angular part of the Lapacian is proportional to the angular momentum operator, $-\hbar^2 \nabla_\Omega^2 = L^2$, where $\hbar$ is the reduced Planck constant, which has the dimension of angular momentum.

$L^2 Y_{lm}(\theta, \phi) = l(l+1) \hbar^2 Y_{lm}(\theta, \phi)$

Here, from the previous discussion, before we solve the equation, we know that the harmonic has maximum order of $l$. The $m$ is the degeneracy for same eigenvalue $l(l+1)$

As we mentioned before, the harmonics should be rotational invariant, such that any direction should be equal. However, when we look at the Spherical Harmonics, the poles are clearly two special points and the rotation around the “z-axis” has limited rotational symmetry with degree $l$. How come?

According to the article, the solution is not necessarily to be separated into $\theta, \phi$, such that

$\displaystyle Y_{lm}(\theta,\phi) = \sqrt{\frac{2l+1}{4\pi} \frac{(l-m)!}{(l+m)!}}P_{lm}(\cos\theta) e^{im\phi}$

I quote the original,

“It is not immediately obvious that we can separate variables and assume exponential functions in the φ direction. We are able to do this essentially because the lines of fixed θ are circles. We could also simply assume this form and show the construction succeeds. This organization is not forced, but separating the variables is so useful that there are no competitive options. A disadvantage of this organization is that it makes the poles into special points.”

The limited rotational symmetry with degree of $l$ is due to the limited “band-width” that restricted by the order of the homogeneous function. The relation between the band width and the order of the harmonics can be understood that the number of “sector” or “node” on the circle/sphere is proportional to the order, thus, the “resolution” is also limited by the order and thus the “band-width”.

Since the Platonic solid is coordinate-free that they are the most symmetry. In the next post, I will show the relation between Spherical Harmonics and Platonic solid. This is related to the section 3.2 in the article,

“One would like to have an uniform discretization for the sphere, with all portions equally represented. From such an uniform discretization we could construct a platonic solid. It is known, however, that there are only a few platonic solids, and the largest number of faces is 20 (icosohedron) and largest number of vertices is 20 (dodecahedron). If we want to discretize the sphere with many points, we cannot do it uniformly. Instead we set the goal of using the fewest points to resolve the Spherical Harmonics up to some degree. Since the Spherical Harmonics themselves are “fair” and “uniform”, this gives a good representation for functions on the sphere. “

As the Fourier Series and Spherical Harmonic are closely related, they should share many properties. For instant, they are orthonormal and form a basis. This leads to the Discrete Fourier Transform and also the “Spherical Transform”,

$\displaystyle f(\Omega) = \sum_{\alpha} a_\alpha \Theta_\alpha(\Omega)$

where $\alpha$ is the id of the basis. One can use the Parseval theorem,

$\displaystyle \int |f(\Omega)|^2 d\Omega = \sum_{\alpha} a_\alpha^2$

Also, the convolution using discrete Fourier transform can also be applied on the spherical harmonics.

Notice that, the Discrete Fourier Transform can “translate” to Continuous Fourier Transform. However, the order of the spherical harmonics is always discrete.

## on Diagonalization (reminder)

since i don’t have algebra book on my hand, so, it is just a reminder, very basic thing.

for any matrix $M$ , it can be diagonalized by it eigenvalue $\lambda_i$  and eigen vector $v_i$, given that it eigenvectors span all the space. thus, the transform represented by the matrix not contractive, which is to say, the dimension of the transform space is equal to the dimension of the origin space.

Let denote, D before Diagonal matrix, with it elements are eigenvalues.

$D_{ij} = \lambda_i \delta_{ij}$

P be the matrix that collect the eigenvectors:

$P_{i j} = \left( v_i \right)_j = \begin {pmatrix} v_1 & v_2 & ... & v_i \end {pmatrix}$

Thus, the matrix $M$ is :

$M = P \cdot D \cdot P^{-1}$

there are some special case. since any matrix can be rewritten by symmetric matrix $S$ and anti-symmetric matrix $A$. so we turn our focus on these 2 matrices.

For symmetric matrix $S$, the transpose of $P$ also work

$S =P \cdot D \cdot P^{-1} = (P^T)^{-1} \cdot D \cdot P^T$

which indicated that $P^T = P^{-1}$. it is because, for a symmetric matrix, $M = M^T$ ,  the eigenvalues are all different, then all eigenvector are all orthogonal, thus $P^T \cdot P = 1$.

For anti-symmetric matrix $A$

$A = P \cdot D \cdot P^{-1}$

since the interchange of row or column with corresponding exchange of eigenvalues in D still keep the formula working. Thus, the case $P = P^T$ never consider.

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For example, the Lorentz Transform

$L = \gamma \begin {pmatrix} \beta & 1 \\ 1 & \beta \end {pmatrix}$

which has eigenvalues:

$D = \gamma \begin {pmatrix} \beta-1 & 0 \\ 0 & \beta+1 \end {pmatrix}$

$P = \begin {pmatrix} -1 & 1 \\ 1 & 1 \end {pmatrix}$

the eigenvector are the light cone. because only light is preserved in the Lorentz Transform.

and it is interesting that

$L = P \cdot D \cdot P^{-1} = P^{-1} \cdot D \cdot P = P^T \cdot D \cdot (P^T)^{-1} = (P^T)^{-1} \cdot D \cdot P^T$

another example is the Rotation Matrix

$R = \begin {pmatrix} cos(\theta) & - sin(\theta) \\ sin(\theta) & cos(\theta) \end{pmatrix}$

$D = \begin {pmatrix} Exp( - i \theta) & 0 \\ 0 & Exp(i \theta) \end {pmatrix}$

$P = \begin {pmatrix} -i & i \\ 1 & 1 \end{pmatrix}$

the last example to give is the $J_x$ of the spin-½ angular momentum

$J_x = \begin {pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

$D = \begin {pmatrix} -1 & 0 \\ 0 & 1 \end {pmatrix}$

$P = \begin {pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}$