Magnetic quantum number and integer orbital l

In spherical coordinates the z-component of r × p reduces to L_z = −iħ ∂/∂φ, acting on functions of the azimuth alone. The eigen-equation −iħ Φ′ = mħ Φ is first order, so Φₘ = C e^(imφ), and normalising on one turn, ∫|Φ|² dφ = 1 from 0 to 2π, gives C = 1/√(2π). Nothing so far restricts m: every real number solves the equation. The spectrum is decided by the domain, and the domain is decided by Hermiticity. Integrating by parts, ⟨f|L_z g⟩ − ⟨L_z f|g⟩ = −iħ [f*g] evaluated between 0 and 2π, which vanishes for every pair in the domain provided all of them share one twist, Φ(2π) = e^(iα) Φ(0). Each α gives a Hermitian L_z with eigenvalues (m + α/2π)ħ for integer m — still spaced by ħ, still mutually orthogonal. The periodic choice α = 0 is the one that returns the integers, and it is a choice you make when you declare that φ and φ + 2π label the same point. Keep that in view: the equation handed you e^(imφ) and a spacing of ħ; you supplied the integers.

The ladder of topic 9.3 runs m from −l to l in unit steps, 2l+1 rungs; with m integer, l is integer and the count is odd: 1, 3, 5, 7 for s, p, d, f. In a central potential [H₀, L±] = 0, so every rung shares the energy Eₙₗ and the spectrum never mentions m. Nor could it: the z-axis is yours to choose, and rotating it mixes the rungs by a unitary (2l+1)-square matrix without touching the energy. What m does fix is the phase. |Yₗ^m|² has no φ-dependence, so m never tilts an orbital; the factor e^(imφ) winds through m full cycles around z, and the probability current circulates in proportion to m — an electron in |2, 1, +1⟩ carries a real current and |2, 1, −1⟩ carries the opposite one. That circulation is what a magnetic field couples to. Until a field, or some other axis-picking perturbation, is switched on, mₗ is a label on a degenerate multiplet; the name 'magnetic' anticipates the experiment that turns it into an energy.

An electron circulating at angular frequency ω on a loop of radius r is a current I = −eω/2π enclosing area πr², so μ = IA = −eωr²/2 = −(e/2mₑ) L, since L = mₑ ω r². The ratio survives quantisation operator by operator: μ̂ = −(μ_B/ħ) L̂ with μ_B = eħ/2mₑ = 9.274 × 10⁻²⁴ J T⁻¹. For B = B ẑ the coupling is H_Z = −μ̂⋅B = (μ_B/ħ) B L_z, a multiple of L_z, so it commutes with H₀, L² and L_z, and |n, l, mₗ⟩ diagonalises the full Hamiltonian exactly with E = Eₙₗ + mₗ μ_B B. The sign matters: L along B means μ against B, so mₗ = +l is the highest rung, not the lowest. At 1 T the step is 57.9 μeV, 14.0 GHz or 0.467 cm⁻¹, some 10⁻⁵ of an optical spacing; the picture is perturbative, and the diamagnetic term e²B²(x² + y²)/8mₑ is smaller still. It also assumes S = 0, since with spin the moment is no longer proportional to L alone — which is why the etalon lab uses the cadmium singlet, and why it sees three lines whatever l is: upper and lower fans share one spacing, so hν = hν₀ + Δm μ_B B and Δm = 0, ±1 gives a single triplet.

The textbook line is short: ψ(x, y, z) is a function on R³, φ and φ + 2π name the same point, so e^(imφ) must be periodic, m is an integer, and with m running from −l to l so is l. Nothing in it is false, and it does deliver integer m for orbital motion. But watch where it leans. It is a boundary-condition choice among the Hermitian versions of −iħ d/dφ, and the rejected twists are not unphysical: on a ring threaded by flux Φ the mechanical angular momentum has eigenvalues (m − Φ/Φ₀)ħ, a twist realised in the laboratory, so the circle alone cannot decide what the sphere allows. It uses L_z only and never asks whether Lₓ and L_y act sensibly on the functions concerned. And what physics actually requires is that probabilities and expectation values be single-valued; a sign flip under one turn is invisible to |ψ|², which is precisely the freedom a spinor exercises. So single-valuedness explains integer m if you grant it. The strong statement is different: no half-integer tower is realised by functions on the sphere at all, and that has to be shown on the sphere.

In spherical coordinates L± = ħ e^(±iφ)(±∂_θ + i cot θ ∂_φ). A top state e^(ilφ) f(θ) needs L₊ on it to vanish, which reads f′ = l cot θ f, so f = sin^l θ for any real l. Take l = 1/2: Y = e^(iφ/2) √sin θ is normalisable, with ∫|Y|² dΩ = π², and L₊Y = 0 — everything the algebra asks of |1/2, 1/2⟩. Lower once: L₋Y = −ħ e^(−iφ/2) cos θ/√sin θ, whose squared norm π²ħ² is exactly the 2lħ² ‖Y‖² = ħ² π² that ‖L₋|l, l⟩‖² predicts. Lower again, where a two-state tower must return zero: L₋²Y = −ħ² e^(−3iφ/2) sin^(−3/2) θ. It is not zero, and it is not even in L²(S²), since ∫|⋅|² sin θ dθ = ∫ dθ/sin² θ diverges at both poles. Compare l = 1: e^(iφ) sin θ → −2ħ cos θ → −2ħ² e^(−iφ) sin θ → 0 exactly at the third step. The abstract proof did not lie; it assumed something. ‖L₋ψ‖² = ⟨ψ|L₊L₋|ψ⟩ needs L₊ to be the adjoint of L₋ on ψ, so the pole terms from integrating by parts in θ must vanish, and with a sin^(−1/2) θ factor they do not. Norm positivity, the step that terminated the ladder, is unavailable. The only towers that close inside L²(S²) are the integer ones: the spherical harmonics.

Work in u = cos θ, where ∫ dΩ becomes 2π ∫ du. On e^(imφ) f(u) the lowering operator is L₋ → ħ e^(i(m−1)φ) [√(1−u²) f′(u) − m u f/√(1−u²)]. Do not differentiate f itself on a grid: √(1−u²) has infinite slope at u = ±1, a polynomial interpolant gets that slope wrong at the end nodes, and the l = 1 ladder then refuses to close as well, a bug that imitates the theorem. Carry the pole factor by hand instead, f = (1−u²)^(m/2) g(u): the step becomes g → (1−u²) g′ − 2m u g, which maps polynomials to polynomials, so differentiating g spectrally on N Gauss–Legendre nodes from numpy.polynomial.legendre.leggauss(N) is exact, and after each step the norm is 2π Σ wᵢ (1−uᵢ²)^m |g(uᵢ)|². For l = 1, g runs 1, −2u, −2(1−u²); the norms come out 8.38, 16.8 ħ² and 33.5 ħ⁴, each ratio equal to the algebra's ħ²(l(l+1) − m(m−1)), and the fourth is round-off. For l = 1/2, g runs 1, −u, −1: the first norm is 9.87, the second creeps up to 9.87 ħ² (9.60 at N = 40, 9.80 at N = 160), slow but convergent as an integrable (1−u²)^(−1/2) end singularity allows, and the third never converges: it roughly doubles each time N doubles, because the integrand goes as (1−u²)^(−3/2) at the ends and every finer grid samples closer to the poles. Non-convergence is usually a bug; here it is the theorem. The same script checks the Zeeman fan: with the topic 9.4 matrices for l = 1, numpy.linalg.eigvalsh(E0*I + muB*B*Lz) returns E₀ + m μ_B B for m = −1, 0, 1 in ascending order, and replacing Lz by n⋅L for a tilted unit vector n returns the same three numbers.