Questions: Spherical Harmonics in Electrostatics

Question 1

A conducting sphere of radius R is placed in a uniform external electric field E₀ pointing in the z-direction. Which term in the spherical harmonic expansion dominates the boundary condition at large r, and why?

• A. The ℓ=0 term, because uniform fields have no angular variation
• B. The ℓ=1 term, because E₀z = E₀r cos θ = E₀r P₁(cos θ)
• C. The ℓ=2 term, because the sphere introduces quadrupole distortions
• D. All ℓ terms contribute equally to a uniform field

Answer: (B)

The external uniform field has potential V = −E₀z = −E₀r cos θ. Recognizing that cos θ = P₁(cos θ) (the ℓ=1 Legendre polynomial), this boundary condition at large r is entirely captured by the ℓ=1 term with radial dependence r¹. This is why the general solution only needs ℓ=1 terms: the far-field condition pins all higher-ℓ coefficients to zero, and orthogonality then determines the induced dipole coefficient (the r⁻² term) from the boundary condition on the sphere surface.

Question 2

What property of spherical harmonics transforms the problem of finding the potential with an arbitrary spherical boundary condition from a differential equation problem into an algebraic one?

• A. They satisfy Laplace's equation individually, so any linear combination also satisfies it
• B. Their orthonormality over the sphere allows any boundary condition to be expanded uniquely, with coefficients extracted by integration
• C. They are real-valued, simplifying the mathematics for physical problems
• D. They form a finite set for any given problem, making computations exact

Answer: (B)

Orthonormality plus completeness is the key combination. Because ∫Y_ℓᵐ* Y_ℓ'ᵐ' dΩ = δ_ℓℓ' δ_mm', any function on the sphere can be written as a sum of harmonics with unique coefficients, and each coefficient can be isolated by multiplying both sides by a specific harmonic and integrating — all other terms drop out. This turns 'match the boundary condition' into 'compute an integral.' Option A is true but doesn't explain the systematic solution; option D is false — there are infinitely many harmonics.

Question 3

The same spherical harmonics that solve electrostatic boundary value problems also appear as the angular part of electron wavefunctions in hydrogen — the shapes called s, p, d, f orbitals.

• T. True
• F. False

Answer: True

Atomic orbital shapes are exactly |Y_ℓᵐ(θ,φ)|² plotted on the surface of a sphere. The reason is that both problems — electrostatics in spherical coordinates and quantum mechanics of a spherically symmetric potential — require solving an equation with the spherical Laplacian, which separates into the same angular equation. Spherical harmonics appear wherever Laplace's equation or the angular part of the Schrödinger equation is solved on a sphere, regardless of the physical context. This is why the misconception that they are specific to electrostatics is important to correct.

Question 4

When solving Laplace's equation outside a sphere (r > R), the radial terms of the form rˡ are retained because they remain finite at large r.

• T. True
• F. False

Answer: False

For ℓ ≥ 1, rˡ diverges as r → ∞, so it cannot appear in the exterior solution. The exterior solution uses r^(−ℓ−1) terms, which decay to zero at infinity as required. Only the ℓ=0 term r⁰ = 1 (a constant) is acceptable at large r from the rˡ family. The rˡ terms are used in the *interior* solution (r < R) where r → 0 would make r^(−ℓ−1) diverge. The boundary conditions at r = 0 and r = ∞ determine which radial solutions to keep.

Question 5

Explain why the orthogonality of spherical harmonics allows you to extract individual expansion coefficients from a boundary condition, rather than having to solve for all coefficients simultaneously.

Model answer: If you expand the boundary condition f(θ,φ) = Σ A_ℓᵐ Y_ℓᵐ(θ,φ) and multiply both sides by a specific Y_ℓ'ᵐ'* and integrate over the sphere, the orthonormality relation ∫Y_ℓᵐ* Y_ℓ'ᵐ' dΩ = δ_ℓℓ'δ_mm' causes every term in the sum except the one with ℓ=ℓ', m=m' to vanish. This leaves A_ℓ'ᵐ' = ∫f(θ,φ) Y_ℓ'ᵐ'* dΩ — each coefficient is determined by a single integral, independently of all others.

This is the spherical analog of extracting Fourier coefficients: orthogonality makes all cross-terms vanish, decoupling what would otherwise be an infinite system of simultaneous equations into an infinite set of independent one-variable integrals. Without orthogonality, determining one coefficient would require knowing all others. The completeness property guarantees that any smooth boundary function can be expressed this way, making the expansion exact rather than approximate.