Spherical Harmonics

Plots of the real parts of the first few spherical harmonics, where distance from origin gives the value of the spherical harmonic as a function of the spherical angles $\phi$ and $\theta$. Blue represents positive values and yellow represents negative values [1].

Spherical harmonics are a set of functions used to represent functions on the surface of the sphere $S^2$. They are a higher-dimensional analogy of Fourier series, which form a complete basis for the set of periodic functions of a single variable $($functions on the circle $S^1).$ Spherical harmonics are defined as the eigenfunctions of the angular part of the Laplacian in three dimensions. As a result, they are extremely convenient in representing solutions to partial differential equations in which the Laplacian appears. Since the Laplacian appears frequently in physical equations (e.g. the heat equation, Schrödinger equation, wave equation, Poisson equation, and Laplace equation) ubiquitous in gravity, electromagnetism/radiation, and quantum mechanics, the spherical harmonics are particularly important for representing physical quantities of interest in these domains, most notably the orbitals of the hydrogen atom in quantum mechanics.

The spherical harmonics are constructed to be the eigenfunctions of the angular part of the Laplacian in three dimensions, also called the Laplacian on the sphere. This construction is analogous to the case of the usual trigonometric functions $\sin (m \phi)$ and $\cos (m \phi)$ which form a complete basis for periodic functions of a single variable (the Fourier series) and are eigenfunctions of the angular Laplacian in two dimensions, $\nabla^2_{\phi} = \frac{\partial^2}{\partial \phi^2}$, with eigenvalue $-m^2$.

In Cartesian coordinates, the three-dimensional Laplacian is typically defined as

$\nabla^2 = \frac{\partial}{\partial x^2} + \frac{\partial}{\partial y^2} + \frac{\partial}{\partial z^2}.$

In spherical coordinates $(x = r\sin \theta \cos \phi, y=r\sin \theta \sin \phi, z = r\cos \theta),$ it takes the form

$\nabla^2 = \frac{1}{r^2 \sin \theta} \left(\frac{\partial}{\partial r} r^2 \sin \theta \frac{\partial}{\partial r} + \frac{\partial}{\partial \theta} \sin \theta \frac{\partial}{\partial \theta} + \frac{\partial}{\partial \phi} \csc \theta \frac{\partial}{\partial \phi} \right).$

The Laplace equation $\nabla^2 f = 0$ can be solved via separation of variables. Make the ansatz $f(r,\theta, \phi) = R(r) Y(\theta, \phi)$ to separate the radial and angular parts of the solution. In spherical coordinates, one obtains the two eigenvalue equations for $R(r)$ and $Y(\theta, \phi)$:

$\begin{aligned} \frac{\partial}{\partial r} \left(r^2 \frac{\partial R(r)}{\partial r} \right) &= \ell (\ell+1) R(r) \\ \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial Y(\theta, \phi)}{\partial \theta} \right) + \frac{1}{\sin^2 \theta} \frac{d^2 Y(\theta, \phi)}{d\phi^2} &= -\ell (\ell+1) Y(\theta, \phi), \end{aligned}$

where $\ell(\ell+1)$ is some constant called the separation constant, written in what will ultimately be the most convenient form. Multiplying the top equation by $Y(\theta, \phi)$ on both sides, the bottom equation by $R(r)$ on both sides, and adding the two would recover the original three-dimensional Laplace equation in spherical coordinates; the separation constant is obtained by recognizing that the original Laplace equation describes two eigenvalue equations of opposite signs.

The angular equation above can also be solved by separation of variables. Note that the $\phi$ dependence is the same as in the case of the two-dimensional angular Laplacian; the solutions there were simply the trigonometric functions $\sin (m\phi)$ and $\cos (m\phi)$ or $e^{\pm im\phi}$. Therefore, make the ansatz $Y(\theta, \phi) = \Theta (\theta) e^{i m\phi}$ for some second separation constant $m$ which can take negative values. This gives the equation for $\Theta (\theta)$:

$\sin \theta\frac{\partial}{\partial \theta} \left(\sin \theta \frac{\partial \Theta (\theta)}{\partial \theta} \right) = m^2 \Theta (\theta) - \ell (\ell+1) \sin^2 \theta\, \Theta (\theta).$

The solutions for $\Theta (\theta)$ can be found by putting the equation into a canonical form, the solutions of which are given in terms of the Legendre polynomials:

$P^m_{\ell} (x) = \frac{(-1)^m}{2^{\ell} \ell!} (1-x^2)^{m/2} \frac{d^{\ell + m}}{dx^{\ell + m}} (x^2 - 1)^{\ell}.$

Notably, this formula is only well-defined and nonzero for $\ell \geq 0$ and $m$ integers such that $|m| \leq \ell$. Formally, these conditions on $m$ and $\ell$ can be derived by demanding that solutions be periodic in $\theta$ and $\phi$.

The general solutions for each linearly independent $Y(\theta, \phi)$ are the spherical harmonics, with a normalization constant multiplying the solution as described so far to make independent spherical harmonics orthonormal:

$Y^m_{\ell} (\theta, \phi) = \sqrt{\frac{2\ell + 1}{4\pi} \frac{(\ell - m)!}{(\ell + m)!}} P^m_{\ell} (\cos \theta) e^{im\phi}.$

Every spherical harmonic is labeled by the integers $\ell$ and $m$, the order and degree of a solution, respectively. Note that the normalization factor of $(-1)^m$ here included in the definition of the Legendre polynomials is sometimes included in the definition of the spherical harmonics instead or entirely omitted.

Some of the low-lying spherical harmonics are enumerated in the table below, as derived from the above formula:

$\begin{array}{ccl} \\ \hspace{15mm} \ell & \hspace{15mm} m&\hspace{15mm} Y^m_{\ell} (\theta, \phi) \\ \hline \hspace{15mm} 0&\hspace{15mm} 0&\hspace{15mm} \sqrt{\frac{1}{4\pi}} \\ \hspace{15mm} 1&\hspace{15mm} -1&\hspace{15mm} \sqrt{\frac{3}{8\pi}} \sin \theta e^{-i \phi}\\ \hspace{15mm} 1&\hspace{15mm} 0&\hspace{15mm} \sqrt{\frac{3}{4\pi}} \cos \theta\\ \hspace{15mm} 1&\hspace{15mm} 1&\hspace{15mm} -\sqrt{\frac{3}{8\pi}} \sin \theta e^{i \phi} \\ \hspace{15mm} 2&\hspace{15mm} -2&\hspace{15mm} \sqrt{\frac{15}{32\pi}} \sin^2 \theta e^{-2i\phi} \\ \hspace{15mm} 2&\hspace{15mm} -1&\hspace{15mm} \sqrt{\frac{15}{8\pi}} \sin \theta \cos \theta e^{-i \phi}\\ \hspace{15mm} 2&\hspace{15mm} 0&\hspace{15mm} \sqrt{\frac{5}{16\pi}} (3\cos^2 \theta -1 )\\ \hspace{15mm} 2&\hspace{15mm} 1&\hspace{15mm} -\sqrt{\frac{15}{8\pi}} \sin \theta \cos \theta e^{i \phi} \\ \hspace{15mm} 2&\hspace{15mm} 2&\hspace{15mm} \sqrt{\frac{15}{32\pi}} \sin^2 \theta e^{2i\phi} \end{array}$

Note that $Y^0_0 (\theta, \phi)$ is spherically symmetric while the $Y^m_1 (\theta, \phi)$ are axially symmetric. Physically, $Y^0_0 (\theta, \phi)$ represents the overall average or monopole moment of a function on the sphere, while the $Y^m_1 (\theta, \phi)$ represent the dipole moments of this function. The generalization to higher $\ell$ is similar.

Consider the real function on the sphere given by $f(\theta, \phi) = 1 + \sin \theta\cos \phi$. Write $f$ as a linear combination of spherical harmonics.

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The overall shift of $1$ comes from the lowest-lying harmonic $Y^0_0 (\theta, \phi)$. At the $\ell = 1$ level, both $m= \pm 1$ have a $\sin \theta$ factor; their difference will give $e^{i\phi} + e^{-i\phi}$ giving a factor of $\cos \phi$ as desired. So $f$ can be written as

$f(\theta, \phi) = \sqrt{4\pi} Y^0_0 (\theta, \phi ) + \frac12 \sqrt{\frac{8\pi}{3}} \left(Y^{-1}_{1}(\theta, \phi) - Y^{1}_1 (\theta, \phi)\right).$

Visually, this corresponds to the decomposition below: The function $f(\theta, \phi)$ decomposed into the sum of spherical harmonics given above. Color represents the phase of the spherical harmonic.

Hydrogen Atom

One of the most well-known applications of spherical harmonics is to the solution of the Schrödinger equation for the wavefunction of the electron in a hydrogen atom in quantum mechanics. The Schrödinger equation for hydrogen reads in S.I. units as follows:

$-\frac{\hbar^2}{2m} \nabla^2 \psi - \frac{e^2}{4\pi \epsilon_0 r}\psi = E\psi,$

with $\hbar$ Planck's constant, $m$ the electron mass, and $E$ the energy of any particular state of the electron. Since the electric potential energy $U(r) = - \frac{e^2}{4\pi \epsilon_0 r}$ is spherically symmetric, the separation of variables procedure used above still works and the potential only modifies the radial solution $R(r)$. The electron wavefunction in the hydrogen atom is still written $\psi (r,\theta \phi) = R_{n\ell} (r) Y^m_{\ell} (\theta, \phi)$, where the index $n$ corresponds to the energy $E_n$ of the electron obtained by solving the new radial equation.

At each fixed energy, the solutions to the hydrogen atom are degenerate: one can modify the $Y^m_{\ell} (\theta, \phi)$ in any solution for the electron wavefunction without changing the energy of the electron (provided that the spin of the electron is ignored). For each fixed $n$ and $\ell$ there are $2\ell + 1$ solutions corresponding to the $2\ell + 1$ choices of $m$ at fixed $\ell.$ The $Y^m_{\ell} (\theta, \phi)$ thus correspond to the different possible electron orbitals; they label the unique states of the electron in hydrogen at a single fixed energy.

This correspondence can be made more precise by considering the angular momentum of the electron. In quantum mechanics, the total angular momentum operator is defined as the Laplacian on the sphere up to a constant:

$\hat{L}^2 = -\hbar^2 \left(\frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \left(\sin \theta \frac{\partial}{\partial \theta} \right) + \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} \right),$

and similarly the operator for the angular momentum about the $z$-axis is

$\hat{L}_z = -i\hbar \frac{\partial}{\partial \phi}.$

The spherical harmonics are eigenfunctions of both of these operators, which follows from the construction of the spherical harmonics above: the solutions for $Y^m_{\ell} (\theta, \phi)$ and its $\phi$ dependence were both eigenvalue equations corresponding to these operators (or their squares). The (total and axial) angular momentum of an electron in the orbital corresponding to the spherical harmonic $Y^m_{\ell} (\theta, \phi)$ is therefore

$L^2 = \hbar^2 \ell (\ell + 1), \quad L_z = \hbar m.$

One concludes that the spherical harmonics in the solution for the electron wavefunction in the hydrogen atom identify the angular momentum of the electron. In quantum mechanics the constants $\ell$ and $m$ are called the azimuthal quantum number and magnetic quantum number due to their association with rotation and how the energy of an electron in a nonzero $m$ state changes in a magnetic field.

Classical Electrodynamics

As stated, spherical harmonics routinely arise in physical settings due to the prevalence of the Laplacian in many physical equations. The simplest of these is the Laplace equation from classical electrodynamics, which appears as Gauss's law:

$\nabla^2 V = -\frac{\rho}{\epsilon_0}.$

The general solution for the electric potential $V$ can be expanded in a basis of spherical harmonics as

$V(r,\theta, \phi ) = \sum_{\ell = 0}^{\infty} \sum_{m=-\ell }^{\ell } \left( A_{m}^{\ell} r^{\ell} + \frac{B_{m}^{\ell}}{r^{\ell +1}}\right) Y_{\ell}^m (\theta, \phi) ,$

where the $A_{m}^{\ell}$ and $B_{m}^{\ell}$ are some set of coefficients depending on the boundary conditions. Note that the first term inside the sums is essentially just a Laurent series in $r$ describing every possible power of $r$ up to order $\ell$.

A conducting sphere of radius $R$ with a layer of charge $Q$ distributed on its surface has the electric potential on the surface of the sphere given by

$V = \frac{1}{4\pi \epsilon_0} \frac{Q}{R} \sin \theta \cos \theta \cos (\phi).$

Find the potential in terms of spherical harmonics in all of space $(r<R$ and $r>R).$

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From the table of low-lying spherical harmonics, the angular part $\sin \theta \cos \theta \cos (\phi)$ of the potential can be generated from the correct linear combination of $Y^{-1}_2 (\theta, \phi)$ and $Y^1_2 (\theta, \phi)$. The exact combination including the correct coefficient is

$V = \frac{1}{4\pi \epsilon_0} \frac{Q}{R} \sin \theta \cos \theta \cos (\phi) = \frac{1}{4\pi \epsilon_0} \frac{Q}{R} \sqrt{\frac{2\pi}{15}} \big(Y^{-1}_2 (\theta, \phi) - Y^1_2 (\theta, \phi) \big).$

As written above, the general solution to Laplace's equation in all of space is

$V(r,\theta, \phi ) = \sum_{\ell = 0}^{\infty} \sum_{m=-\ell }^{\ell } \left( A_{m}^{\ell} r^{\ell} + \frac{B_{m}^{\ell}}{r^{\ell +1}}\right) Y_{\ell}^m (\theta, \phi).$

The full solution may only include a combination of $Y^{-1}_2$ and $Y^1_2$ in the angular part because the angular dependence is completely independent of the radial dependence. The angular dependence at $r=R$ solved for above in terms of spherical harmonics is therefore the angular dependence everywhere.

To specify the full solution, the coefficients $A_m^{\ell}$ and $B_m^{\ell}$ must be found. When $r>R$, all $A_m^{\ell} = 0$ since in this case the potential will otherwise diverge as $r \to \infty$, where the potential ought to vanish (or at the very least be finite, depending on where the zero of potential is set in this case). So the solution can thus far be written in the form

$V(r,\theta, \phi ) = \frac{B_{-1}^2 Y_{2}^{-1} (\theta, \phi) + B_{1}^2 Y_2^1 (\theta, \phi)}{r^3}.$

The problem for $r>R$ is thus reduced to finding only the two coefficients $B_{-1}^2$ and $B_1^2$. These can be found by demanding continuity of the potential at $r=R$. From the solution on $r=R$ in terms of spherical harmonics, these coefficients can be read off:

$B_{-1}^2 = \frac{1}{4\pi \epsilon_0} QR^2 \sqrt{\frac{2\pi}{15}} = -B_1^2.$

The full solution for $r>R$ is therefore

$V(r,\theta, \phi ) = \frac{1}{4\pi \epsilon_0} \frac{QR^2}{r^3} \sin \theta \cos \theta \cos \phi, \quad r>R.$

A similar analysis obtains the solution for $r<R$. In this case, the coefficients $B_m^{\ell}$ must all vanish or the potential diverges as $r \to 0$, and the only nonzero coefficients are $A_{-1}^2$ and $A_1^2$ due to the angular dependence. So the solution takes the form

$V(r,\theta, \phi ) = \big(A_{-1}^2 Y_{2}^{-1} (\theta, \phi) + A_{1}^2 Y_2^1 (\theta, \phi)\big)r^2$

and again requiring continuity at $r=R$ yields the solution for $r<R$:

$V(r,\theta, \phi ) = \frac{1}{4\pi \epsilon_0} \frac{Qr^2}{R^3} \sin \theta \cos \theta \cos \phi, \quad r<R.\ _\square$

Black Hole Physics

Spherical harmonics are also generically useful in expanding solutions in physical settings with spherical symmetry. One interesting example of spherical symmetry where the expansion in spherical harmonics is useful is in the case of the Schwarzschild black hole. Perturbations of a massless complex scalar field $\Phi$ outside a Schwarzschild black hole of mass $M$ satisfy a version of Laplace's equation generalized for curved spacetime:

$\nabla^2 \Phi \sim \nabla_{\mu} \nabla^{\mu} \Phi = \left(-\frac{1}{r^2} \partial_r \big((r^2-2Mr) \partial_r\big) + \nabla_{\theta, \phi}^2 - \frac{r^4}{r^2-2Mr} \partial_t^2 \right) \Phi = 0,$

where $\nabla_{\theta, \phi}^2$ denotes the Laplacian on the sphere. These perturbations correspond to dissipative waves caused by probing a black hole, like the dissipative waves caused by dropping a pebble into water. Due to the spherical symmetry of the black hole and the presence of the Laplacian on the sphere, the general solution for perturbations can be written as a Fourier transform:

$\Phi(t,r, \theta, \phi) = \int d\omega e^{-i\omega t} \sum_{\ell ,m} \frac{\Psi (r)}{r} Y_{\ell m} (\theta, \phi).$

This decomposition is typically performed as part of an analysis of the modes $\omega$ describing the evolution of the perturbation $\Phi$, called quasinormal modes [3]. The ability to expand in the basis of spherical harmonics is essential in permitting the separation of the radial dependence which ultimately constrains the modes $\omega$.

[1] Image from https://en.wikipedia.org/wiki/Spherical_harmonics#/media/File:Spherical_Harmonics.png under Creative Commons licensing for reuse and modification.

[2] Griffiths, David J. Introduction to Quantum Mechanics. Second Edition. Pearson: Upper Saddle River, NJ, 2006.

[3] E. Berti, V. Cardoso, and A.O. Starinets. Quasinormal modes of black holes and black branes. http://arxiv.org/pdf/0905.2975v2.pdf.