University Physics V · Angular Momentum in Quantum Mechanics · 9.6
Magnetic Quantum Number & Integer Orbital l
Two jobs. First, make mₗ measurable: put an orbital in a field, Lz becomes the energy operator, and the 2l+1 sublevels fan out by μB B each. Second, settle why that count is always odd — not with the single-valuedness slogan, but by watching the l = 1/2 ladder fail to close on the sphere.
Build the model
Connect the measurement to the mechanism.
Separate the Laplacian and Lz is left as −iħ ∂/∂φ, whose eigenfunctions e(imφ) exist for every real m: the differential equation fixes nothing, and the spectrum is a statement about the domain. Demand periodicity and m is an integer; carry the ladder's m = −l, …, l along and l is an integer too, with 2l+1 states that a central Hamiltonian cannot tell apart. The label 'magnetic' is earned when a field picks an axis: for an electron μ = −(μB/ħ) L, so HZ = −μ⋅B = (μB/ħ) B Lz is a multiple of the operator whose eigenbasis you already hold, |n, l, mₗ⟩ stays exact, and the degeneracy opens into a fan of equally spaced rungs mₗ μB B, 57.9 μeV per tesla, highest at mₗ = +l.
What that picture costs is the honest reason the count is odd. Periodicity is a boundary-condition choice — a twisted domain also makes Lz Hermitian — and it says nothing about Lₓ and Ly. The argument that closes the case is the ladder itself: e(iφ/2) √sin θ is a normalisable top state that L₊ annihilates, but lowering it twice returns −ħ² e(−3iφ/2) sin(−3/2) θ where the algebra requires zero — a function outside L²(S²).
No half-integer tower fits on the sphere, so orbital l is integer because the representation does not exist there, not because a wavefunction 'must be single-valued'.
- Simple definition
- The magnetic quantum number mₗ labels the eigenvalue mₗ ħ of Lz = −iħ ∂/∂φ, one of the 2l+1 integers from −l to l, and it is 'magnetic' because a field along z makes Lz the Hamiltonian HZ = (μB/ħ) B Lz, placing each mₗ at its own energy mₗ μB B.
- Example
- A p orbital (l = 1) in B = 1.00 T has mₗ = −1, 0, +1 at −57.9, 0 and +57.9 μeV, since μB B = 5.788 × 10⁻⁵ eV; adjacent sublevels differ by μB B/h = 14.0 GHz, or 0.467 cm⁻¹, some 10⁻⁵ of an optical transition energy.
Fixes the whole φ-dependence of every Yₗm and shows that the differential equation leaves m free: the spectrum is decided by what you demand at φ = 2π.
φ the azimuth in radians; ħ = 1.055 × 10⁻³⁴ J s; ∫|Φₘ|² dφ = 1 over one turn — the equation itself is solved by every real m
Enough to make m integer once accepted, and with the ladder's m = −l … l it makes l integer too — but it never shows that the sphere cannot carry l = 1/2.
any twist Φ(2π) = e(iα) Φ(0) also makes −iħ d/dφ Hermitian, with spectrum (m + α/2π)ħ; α = 0 is a choice
The classical loop ratio μ/L = −e/2mₑ promoted to operators: no new physics enters, so μz has spectrum −mₗ μB and shares Lz's eigenstates.
the electron charge −e sets the minus sign; gₗ = 1 exactly for orbital motion, unlike gₛ ≈ 2 for spin
HZ is a multiple of Lz, so |n, l, mₗ⟩ stays an exact eigenstate and only its energy moves: a fan of 2l+1 equally spaced rungs, highest at mₗ = +l.
B in tesla; μB B = 57.9 μeV at 1 T, i.e. μB B/h = 14.0 GHz T⁻¹ and μB B/hc = 0.467 cm⁻¹ T⁻¹
The honest test of a proposed l: start here, lower 2l times, and demand that the (2l+1)-th lowering returns exactly zero.
L₋ = ħ e(−iφ)(−∂θ + i cot θ ∂φ); e(ilφ) sinl θ is normalisable on the sphere for any l > −1/2
The ladder never terminates and its functions leave L²(S²), so no orbital multiplet with half-integer l exists; for l = 1 the third lowering gives 0 exactly.
the second lowering should close a two-state tower; ∫|sin(−3/2) θ|² sin θ dθ = ∫ dθ/sin² θ diverges at both poles
Solve the Lz eigenproblem and notice what it leaves open
In spherical coordinates the z-component of r × p reduces to Lz = −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|Lz g⟩ − ⟨Lz 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 Lz 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.
Count the tower, and see why a field is needed to see m at all
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. |Yₗm|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.
Put the orbital in a field: Lz becomes the energy operator
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 = −(μB/ħ) L̂ with μB = eħ/2mₑ = 9.274 × 10⁻²⁴ J T⁻¹. For B = B ẑ the coupling is HZ = −μ̂⋅B = (μB/ħ) B Lz, a multiple of Lz, so it commutes with H₀, L² and Lz, and |n, l, mₗ⟩ diagonalises the full Hamiltonian exactly with E = Eₙₗ + mₗ μB 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 e²B²r²/8mₑ 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 μB B and Δm = 0, ±1 gives a single triplet.
Single-valuedness: what the usual argument cannot do
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 Lz only and never asks whether Lₓ and Ly 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.
Run the ladder for l = 1/2 and watch it fail to close
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 = sinl 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.
See the same failure numerically, then test any solver
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)]: one derivative and two multiplications per step. Sample f on N Gauss–Legendre nodes from numpy.polynomial.legendre.leggauss(N), differentiate spectrally, and after each step evaluate the norm 2π Σ wᵢ |f(uᵢ)|². For l = 1, starting from f = √(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, starting from (1−u²)¹⁄⁴, the first two norms are 9.87 and 9.87 ħ², again what the algebra says, and the third never converges: it grows without bound with N, 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 μB B for m = 1, 0, −1, and replacing Lz by n⋅L for a tilted unit vector n returns the same three numbers.
Change one variable at a time
Make the relationship visible.
Set l = 1 and step k from 0 to 3: sin θ, −2 cos θ, −2 sin θ, then a flat zero at k = 3 = 2l+1. Now set l = 1/2 and k = 2: the curve that must vanish is instead pinned to the frame at both poles, −1/sin³⁄² θ, and the marker sits below the tower's last rung.
m = l − k-1.5
CLOSING STEP 2l+12
k = 3 PREFACTOR 4l(l−1)-1.00
Θₖ AT θ = 90°-1.00 ħk
Live interpretationm = l − k: −1.5. CLOSING STEP 2l+1: 2. k = 3 PREFACTOR 4l(l−1): −1.00. Θₖ AT θ = 90°: −1.00 ħk
Catch the common trap
Explain before calculating.
A student proposes Y = e(iφ/2) √sin θ as the top state of an orbital tower with l = 1/2. It is normalisable on the sphere and L₊Y = 0. Which statement gives the real reason no such tower exists?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThe cadmium ¹D₂ level (L = 2, S = 0) is placed in B = 0.850 T. List the energies of its sublevels relative to the unsplit level, and give the spacing in μeV, in GHz and in cm⁻¹.
- S = 0, so the moment is purely orbital, μ = −(μB/ħ) L, and HZ = (μB/ħ) B Lz is diagonal in |L, ML⟩ with E = ML μB B.
- μB B = 5.788 × 10⁻⁵ eV T⁻¹ × 0.850 T = 4.92 × 10⁻⁵ eV = 49.2 μeV.
- L = 2 gives 2L+1 = 5 rungs, ML = −2, −1, 0, +1, +2: −98.4, −49.2, 0, +49.2, +98.4 μeV, the top rung being ML = +2 because μ is antiparallel to L.
- In frequency the step is μB B/h = 13.996 GHz T⁻¹ × 0.850 T = 11.9 GHz; in wavenumber μB B/hc = 0.4669 cm⁻¹ T⁻¹ × 0.850 T = 0.397 cm⁻¹.
AnswerFive equally spaced sublevels at ML μB B = 0, ±49.2, ±98.4 μeV; spacing 49.2 μeV = 11.9 GHz = 0.397 cm⁻¹, with ML = +2 highest.
MediumThe cadmium 643.8 nm line runs from ¹D₂ to ¹P₁, both with S = 0. In B = 1.20 T, count the allowed transitions, show how many distinct frequencies they produce, and find the separation of the components in GHz, in cm⁻¹ and in picometres of wavelength.
- Both terms are orbital-only, so each splits into ML μB B with the same spacing: μB B = 5.788 × 10⁻⁵ × 1.20 = 6.95 × 10⁻⁵ eV = 69.5 μeV. Upper rungs Mᵤ = −2 … +2, lower rungs Md = −1, 0, +1.
- A photon energy is hν = hν₀ + (Mᵤ − Md) μB B. Electric-dipole transitions need ΔM = Mᵤ − Md ∈ (−1, 0, +1): for each of the three lower rungs three upper rungs qualify, so 9 transitions are allowed.
- Because the upper and lower spacings are identical, hν depends only on ΔM, so the nine transitions collapse onto three frequencies, ν₀ and ν₀ ± μB B/h — the normal Zeeman triplet, with the same count for any L.
- Separation: μB B/h = 13.996 GHz T⁻¹ × 1.20 T = 16.8 GHz; in wavenumber 0.4669 cm⁻¹ T⁻¹ × 1.20 T = 0.560 cm⁻¹.
- In wavelength, |Δλ| = λ² Δν̃ = (6.438 × 10⁻⁵ cm)² × 0.560 cm⁻¹ = 2.32 × 10⁻⁹ cm = 23.2 pm, a fractional shift of 3.6 × 10⁻⁵ — which is why the lab resolves it with a Fabry–Perot etalon.
Answer9 allowed transitions on 3 frequencies ν₀, ν₀ ± 16.8 GHz; component spacing 0.560 cm⁻¹, or 23.2 pm at 643.8 nm.
HardWith L± = ħ e(±iφ)(±∂θ + i cot θ ∂φ), take Y = e(iφ/2) sin¹⁄² θ as a candidate |1/2, 1/2⟩ orbital state. Verify that L₊Y = 0 and that Y is normalisable, compute L₋Y and L₋²Y with their squared norms, and decide whether an orbital doublet with l = 1/2 exists on the sphere.
- L₊Y: ∂θ gives ½ cos θ sin(−1/2) θ · e(iφ/2), and i cot θ ∂φ gives i cot θ · (i/2) · e(iφ/2) sin¹⁄² θ = −½ cos θ sin(−1/2) θ · e(iφ/2); they cancel, so L₊Y = 0. Norm: ∫|Y|² dΩ = 2π ∫ sin θ · sin θ dθ over 0 to π = 2π · π/2 = π², finite.
- L₋Y = ħ e(−iφ) e(iφ/2) [−½ cos θ sin(−1/2) θ + i cot θ · (i/2) sin¹⁄² θ] = −ħ e(−iφ/2) cos θ sin(−1/2) θ. Its squared norm is ħ² · 2π ∫ cos² θ sin(−1) θ · sin θ dθ = ħ² · 2π · π/2 = π² ħ².
- Check against the algebra: ‖L₋|l, l⟩‖² = ħ²[l(l+1) − l(l−1)] ‖Y‖² = 2lħ² ‖Y‖² = ħ² π² for l = 1/2. Agreement — so far the function behaves exactly like |1/2, 1/2⟩, and L₋Y plays the part of |1/2, −1/2⟩.
- L₋ on Z = e(−iφ/2) cos θ sin(−1/2) θ: −∂θ Z = e(−iφ/2)[sin¹⁄² θ + ½ cos² θ sin(−3/2) θ], and i cot θ ∂φ Z = ½ cos² θ sin(−3/2) θ · e(−iφ/2); the sum is e(−iφ/2) sin(−3/2) θ (sin² θ + cos² θ) = e(−iφ/2) sin(−3/2) θ. Since L₋Y = −ħZ, L₋²Y = −ħ² e(−3iφ/2) sin(−3/2) θ.
- The algebra demands ‖L₋|1/2, −1/2⟩‖² = ħ²[3/4 − (−1/2)(−3/2)] ‖Z‖² = 0. Instead ∫|L₋²Y|² dΩ = ħ⁴ · 2π ∫ sin(−3) θ · sin θ dθ = 2πħ⁴ ∫ dθ/sin² θ, which diverges at both poles.
- Verdict: the ladder does not terminate, and its second step leaves L²(S²). The identity ‖L₋ψ‖² = ⟨ψ|L₊L₋|ψ⟩ that forces closure needs vanishing pole terms in the θ integration by parts, and sin(−1/2) θ factors deny it. No orbital doublet with l = 1/2 exists; for l = 1 the third lowering of e(iφ) sin θ is exactly zero.
AnswerL₊Y = 0, ‖Y‖² = π²; L₋Y = −ħ e(−iφ/2) cos θ/√sin θ with ‖L₋Y‖² = π²ħ² = 2lħ²‖Y‖²; L₋²Y = −ħ² e(−3iφ/2) sin(−3/2) θ ≠ 0 and non-normalisable — no l = 1/2 orbital doublet on the sphere.