A Cartoon Guide to Spherical Harmonics

Calvin W. Johnson

What are spherical harmonics, anyway? Spherical harmonics are useful functions. Essentially, if you have the ∇2 operator, spherical harmonics are the of the angular part. This means if you have a differential equation and the rest of the equation apart 2 from ∇ depends only on r, you can always write the solution in the form R(r)Ylm(θ,φ). Always. This makes the differential equations much simpler and your life much easier. I refer the interested reader to his or her favorite textbook on or mathematical methods for details. Here I will give you a very quick introduction.

I. NOT SO VERY CLEVER AFTER ALL

Spherical harmonics arise in different contexts, most of which are solving partial differential equations. Consider Schr¨odinger’s time-independent equation in three dimensions, with a spherically symmetric potential:

2 2 2 ¯h ∂2 ¯h ∂2 ¯h ∂2 − − − + V x2 + y2 + z2 ψ(x,y,z)= Eψ(x,y,z). (1) m ∂x2 m ∂y2 m ∂z2  2 2 2 p  This is a 3-dimension differential equation. Despite the fact that we have had differential equations around for three hundred years, we really only know how to solve a one-dimensional differential equation. Even with powerful computers a two-dimensional differential equation is very difficult. Fortunately, we can cheat. We can rewrite the above equation as a set of one-dimensional differential equations, all of which we can solve. This is known as . Cases which do not allow separation of variables are very, very hard to solve.

II. SEPARATION OF VARIABLES

The first step is to choose the right variables. We usually rely upon someone much smarter than us (Euler, Gauss, Bessel) to do this. For the above case, we choose spherical coordinates (r,θ,φ): x = r sin θ cos φ, y = r sin θsinφ, and z = r cos θ . In spherical coordinates the Sch¨odinger equation becomes

2 ¯h ∂2 2 ∂ [− + + V (r) 2m ∂r2 r ∂r  2 ¯h 1 ∂ ∂ − sin θ 2m r2 sin θ ∂θ ∂θ 2 ¯h 1 ∂2 − 2 ]ψ(r,θ,φ)= Eψ(r,θ,φ) 2m r2 sin θ ∂φ2 Now we make an ansatz, or guess, for the form of ψ. We hope it can take a very simple form:

ψ(r,θ,φ)= R(r)Θ(θ)Φ(φ). (2)

If such a guess can be used to solve exactly the Schr¨odinger equation, then we can say we have used separation of variables.

III. SOLVING PIECEWISE

We begin with trying to tackle Φ(φ). Just consider the smaller eigenvalue equation

∂2 Φ(φ)= λΦ(φ). (3) ∂φ2 2

∂2 2 We solve this differential equation by guessing: let Φ(φ) = exp(imφ), then ∂φ2 Φ(φ)= −m Φ(φ). Furthermore, we have a boundary condition, that we want Φ(φ)=Φ(φ + 2π), which means m must be an integer (any integer). We put this back into the Schr¨odinger equation and get:

2 ¯h ∂2 2 ∂ [− + + V (r) 2m ∂r2 r ∂r  2 ¯h 1 ∂ ∂ − sin θ 2m r2 sin θ ∂θ ∂θ 2 ¯h 1 2 imφ imφ + 2 m ]R(r)Θ(θ)e = ER(r)Θ(θ)e 2m r2 sin θ Notice that now the Schr¨odinger equation has no dependence on φ any longer; we have already solved completely the φ dependence. So we can really drop Φ(φ) = eimφ , or at least put it aside until we need it later, and we now have reduced a 3-D problem to a 2-D problem. Hurray! Now we do this trick again for the θ dependence:

1 ∂ ∂ 1 2 ′ sin θ + 2 m Θ(θ)= λ Θ(θ) (4) sin θ ∂θ ∂θ sin θ 

This is not as easy as for φ, but it turns out this was solved by a guy named Legendre a long time ago (actually, by l one of his associates, hence the solution, Θ(θ)= Pm(cos θ) are called associated ). The solution is

1 ∂ ∂ 1 2 l l sin θ + 2 m P (cos θ)= −l(l + 1)P (cos θ) (5) sin θ ∂θ ∂θ sin θ  m m

Again, boundary conditions require that l be an nonnegative integer, and furthermore that l ≥|m|.

IV. THE FINAL ANSWER

l imφ Spherical harmonics are of the form Ylm(θ,φ) = NlmPm(cos θ)e . The normalization Nlm is determined by the requirement that the Ylm’s be orthonormal. You can look up the Ylm’s and their useful properties. They tell us that in a system with spherical symmetry, all of the angular properties are universal and independent of the radial functions. All answers for Schrodinger’s equation with a spherically symmetric potential must be of the form R(r)Ylm(θ,φ).