Shankar (1994)

Shankar [47] is a widely used graduate textbook on quantum mechanics. Shankar uses the standard (physicists') spherical coordinates, and writes the spherical harmonics as $Y_ℓ^m$.

Angular-momentum operators

Just below Eq. (12.5.27), Shankar gives the angular-momentum operators in spherical coordinates as

\[\begin{aligned} L_x &= i \hbar \left( \sin ϕ \frac{\partial} {\partial θ} + \cos ϕ \cot θ \frac{\partial} {\partial ϕ} \right), \\ L_y &= i \hbar \left( -\cos ϕ \frac{\partial} {\partial θ} + \sin ϕ \cot θ \frac{\partial} {\partial ϕ} \right), \\ L_z &= -i \hbar \frac{\partial} {\partial ϕ}, \end{aligned}\]

which agree (up to the factor of $\hbar$) with our expressions.

Spherical harmonics

Equation (12.5.35) gives the spherical harmonics for $m \geq 0$ as

\[Y_{ℓ}^{m}(θ, ϕ) = (-1)^ℓ \left[ \frac{(2ℓ+1)!}{4π} \right]^{1/2} \frac{1}{2^ℓ ℓ!} \left[ \frac{(ℓ+m)!}{(2ℓ)!(ℓ-m)!} \right]^{1/2} e^{i m ϕ} (\sin θ)^{-m} \frac{d^{ℓ-m}}{d(\cos θ)^{ℓ-m}} (\sin θ)^{2ℓ},\]

with Eq. (12.5.40) extending this to negative $m$,

\[Y_{ℓ}^{-m}(θ, ϕ) = (-1)^m \left( Y_{ℓ}^{m}(θ, ϕ) \right)^\ast.\]

Equation (12.5.39) lists the first few explicitly:

\[\begin{aligned} Y_0^0 &= \frac{1}{\sqrt{4π}}, & Y_1^{\pm 1} &= \mp\sqrt{\frac{3}{8π}} \sin θ\, e^{\pm iϕ}, & Y_1^0 &= \sqrt{\frac{3}{4π}} \cos θ, \\ Y_2^{\pm 2} &= \sqrt{\frac{15}{32π}} \sin^2 θ\, e^{\pm 2iϕ}, & Y_2^{\pm 1} &= \mp\sqrt{\frac{15}{8π}} \sin θ \cos θ\, e^{\pm iϕ}, & Y_2^0 &= \sqrt{\frac{5}{16π}} \left(3\cos^2 θ - 1\right). \end{aligned}\]

This is the Condon–Shortley form, so we expect agreement with ours.

Rotation operator

In Exercise 12.5.7, the rotation operator is defined by

\[U\left[ R(α, β, γ) \right] = e^{-i α J_z/\hbar} e^{-i β J_y/\hbar} e^{-i γ J_z/\hbar}.\]

Shankar never actually uses notation like $D^{(j)}_{m', m}$, but he does talk about $\langle j, m' | D^{(j)}\left[ R(α, β, γ) \right] | j, m \rangle$, the matrix elements of $U$ restricted to the states $|j, m\rangle$ for a given $j$. This is precisely our definition of $𝔇$, so we expect agreement. To test it without relying on any further formula from Shankar, we construct the matrices of $J_z$ and $J_y = (J_+ - J_-)/2i$ in the $|j, m\rangle$ basis from the standard ladder relations (which Shankar derives in Sec. 12.5), exponentiate them numerically, and take the matrix elements of $U$.

Implementing formulas

We begin by writing code that implements the formulas from Shankar. We encapsulate the formulas in a module so that we can test them against the SphericalFunctions package.

module Shankar

We'll use some predefined utilities to make the code look more like the equations, and the matrix exponential from LinearAlgebra for the rotation operator.

import ..ConventionsUtilities: 𝒾, ❗, dʲsin²ᵏθdcosθʲimport LinearAlgebra: exp, Diagonal

Equation (12.5.35) for $m \geq 0$, and Eq. (12.5.40) for $m < 0$. We capture the floating-point type T to ensure that we don't lose precision when converting π and the factorials to floating-point numbers.

function Y(ℓ, m, θ::T, ϕ::T) where {T<:Real}    if m < 0        return (-1)^m * conj(Y(ℓ, -m, θ, ϕ))    end    (-1)^ℓ * √T((2ℓ+1)❗ / (4big(π))) * (1 / T(2^ℓ * (ℓ)❗)) * √T((ℓ+m)❗ / ((2ℓ)❗ * (ℓ-m)❗)) *    exp(𝒾 * m * ϕ) * sin(θ)^(-m) * dʲsin²ᵏθdcosθʲ(j=ℓ-m, k=ℓ, θ=θ)end

The explicit formulas of Eq. (12.5.39):

Y₀⁰(θ, ϕ) = 1 / √(4π)Y₁⁻¹(θ, ϕ) = +√(3/(8π)) * sin(θ) * exp(-𝒾*ϕ)Y₁⁰(θ, ϕ) = √(3/(4π)) * cos(θ)Y₁⁺¹(θ, ϕ) = -√(3/(8π)) * sin(θ) * exp(+𝒾*ϕ)Y₂⁻²(θ, ϕ) = √(15/(32π)) * sin(θ)^2 * exp(-2𝒾*ϕ)Y₂⁻¹(θ, ϕ) = +√(15/(8π)) * sin(θ) * cos(θ) * exp(-𝒾*ϕ)Y₂⁰(θ, ϕ) = √(5/(16π)) * (3cos(θ)^2 - 1)Y₂⁺¹(θ, ϕ) = -√(15/(8π)) * sin(θ) * cos(θ) * exp(+𝒾*ϕ)Y₂⁺²(θ, ϕ) = √(15/(32π)) * sin(θ)^2 * exp(+2𝒾*ϕ)

The matrices of $J_z$, $J_\pm$, and $J_y$ in the $|j, m\rangle$ basis (with $\hbar = 1$), ordered so that row and column k correspond to $m = -j + k - 1$:

function Jz(j)    Diagonal([m for m ∈ -j:j])endfunction J₊(j)    M = zeros(2j+1, 2j+1)    for (k, m) ∈ enumerate(-j:j-1)        M[k+1, k] = √((j-m) * (j+m+1))  # ⟨j, m+1| J₊ |j, m⟩    end    MendJ₋(j) = J₊(j)'Jy(j) = (J₊(j) - J₋(j)) / 2𝒾

The rotation operator of Exercise 12.5.7, and its matrix element $\langle j, m' | U | j, m \rangle$:

function U(j, α, β, γ)    exp(-𝒾 * α * Matrix(Jz(j))) * exp(-𝒾 * β * Matrix(Jy(j))) * exp(-𝒾 * γ * Matrix(Jz(j)))endfunction D(j, m′, m, α, β, γ)    U(j, α, β, γ)[m′ + j + 1, m + j + 1]end

end  # module Shankar

Tests

We can now test the functions against the equivalent functions from the SphericalFunctions package. We will need to test approximate floating-point equality, so we set absolute and relative tolerances (respectively) in terms of the machine epsilon:

ϵₐ = 100eps()ϵᵣ = 1000eps()

We only test up to

ℓₘₐₓ = 4

because the formulas are slow, and this will be sufficient to sort out any sign or normalization differences, which are the most likely source of error. For the same reason we use a modest grid of Euler angles.

αβγs = αβγrange(rng, Float64, 5)

First, the explicit formulas of Eq. (12.5.39) agree with the general formula (12.5.35). The general formula has a factor of $1/\sin^m θ$, so we avoid the poles.

for (θ, ϕ) ∈ θϕrange(rng; avoid_poles=ϵₐ/40)    @test Shankar.Y₀⁰(θ, ϕ) ≈ Shankar.Y(0, 0, θ, ϕ) atol=ϵₐ rtol=ϵᵣ    @test Shankar.Y₁⁻¹(θ, ϕ) ≈ Shankar.Y(1, -1, θ, ϕ) atol=ϵₐ rtol=ϵᵣ    @test Shankar.Y₁⁰(θ, ϕ) ≈ Shankar.Y(1, 0, θ, ϕ) atol=ϵₐ rtol=ϵᵣ    @test Shankar.Y₁⁺¹(θ, ϕ) ≈ Shankar.Y(1, 1, θ, ϕ) atol=ϵₐ rtol=ϵᵣ    @test Shankar.Y₂⁻²(θ, ϕ) ≈ Shankar.Y(2, -2, θ, ϕ) atol=ϵₐ rtol=ϵᵣ    @test Shankar.Y₂⁻¹(θ, ϕ) ≈ Shankar.Y(2, -1, θ, ϕ) atol=ϵₐ rtol=ϵᵣ    @test Shankar.Y₂⁰(θ, ϕ) ≈ Shankar.Y(2, 0, θ, ϕ) atol=ϵₐ rtol=ϵᵣ    @test Shankar.Y₂⁺¹(θ, ϕ) ≈ Shankar.Y(2, 1, θ, ϕ) atol=ϵₐ rtol=ϵᵣ    @test Shankar.Y₂⁺²(θ, ϕ) ≈ Shankar.Y(2, 2, θ, ϕ) atol=ϵₐ rtol=ϵᵣend

Next, the spherical harmonics agree with ours:

for (θ, ϕ) ∈ θϕrange(rng; avoid_poles=ϵₐ/40)    for (ℓ, Yˡ) ∈ SphericalFunctions.YlmCalculator(θ, ϕ, ℓₘₐₓ), m ∈ -ℓ:ℓ        @test Shankar.Y(ℓ, m, θ, ϕ) ≈ Yˡ[m] atol=ϵₐ rtol=ϵᵣ    endend

Finally, the matrix elements of Shankar's rotation operator agree with our $𝔇$:

for (α, β, γ) ∈ αβγs    for (j, 𝔇ʲ) ∈ SphericalFunctions.DCalculator(α, β, γ, ℓₘₐₓ), m′ ∈ -j:j, m ∈ -j:j        @test Shankar.D(j, m′, m, α, β, γ) ≈ 𝔇ʲ[m′, m] atol=ϵₐ rtol=ϵᵣ    endend

These successful tests show that Shankar's spherical harmonics and rotation operator agree with the corresponding functions defined by the SphericalFunctions package.

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