
Advanced Quantum Mechanics course sequence · Interactive lesson
Spinor spherical harmonics and the action of σ⋅r̂
This is the step where a four-component equation in three dimensions becomes two ordinary differential equations in r. Quantum Mechanics II's coupled harmonics, one Pauli-matrix identity and a single phase are all it takes: the angles leave the central-field Dirac equation and only the number κ stays behind.
UAQM-07.03 · Lesson 3 of 9 in this unit. Position is not a completion score.Build the model
One Pauli identity and a sign read off on the z-axis turn every angle in the Dirac equation into the number κ.
The key idea
A spinor spherical harmonic χκᵐ is Quantum Mechanics II's coupled state |l, ½, j, m⟩ written as a two-component function of direction; a central-field Dirac state carries χκᵐ in its upper pair and χ(−κ)ᵐ in its lower, and σ⋅r̂ turns each into minus the other.
Example. For 1s, κ = −1 and the upper pair carries χ₋₁¹⁄² = ↑/√(4π). σ⋅r̂ turns it into −χ₁¹⁄² = (cos θ ↑ + sin θ e(iφ) ↓)/√(4π), a p(1/2) harmonic whose density is still 1/(4π) = 0.0796 sr⁻¹ in every direction.
The complete physical model and assumptions
In a central field the Dirac Hamiltonian cα⋅p + βmc² + V(r) is block-diagonal apart from cσ⋅p, the one term that joins the upper pair of components to the lower, so the angular content of σ⋅p decides how the four components are built. σ⋅p is a pseudoscalar: it commutes with J and reverses parity. Whatever harmonic the upper pair carries, the lower pair must carry the single partner with the same j and m and the opposite orbital parity. Quantum Mechanics II built both: the coupled state |l, ½, j, m⟩, labelled here χκᵐ, and its partner with l′ = 2j − l, which is χ(−κ)ᵐ because the β inside K reverses the sign of σ⋅L + ħ on the lower pair. Now write σ⋅p as σ⋅r̂ times the sum of a radial derivative and iσ⋅L/r. On each harmonic σ⋅L is a number, −(κ + 1)ħ; σ⋅r̂ is a unitary pseudoscalar that swaps the partners with one fixed phase, −1 in the Condon-Shortley convention, found by looking along the z-axis. The result, σ⋅p[(G/r)χκᵐ] = (iħ/r)(dG/dr + κG/r)χ(−κ)ᵐ, removes every angle: the Dirac equation becomes two radial equations in which the angular motion survives only as κ. The reduction needs V to depend on r alone. A uniform magnetic field, a deformed nucleus or a tensor force mixes different κ. The −1 itself is a convention: rephase the harmonics and it flips, F flips with it, and nothing measurable changes.
1The lower pair carries −κ, and with it the other l
Units are SI with ħ kept, and the metric signature is (+, −, −, −) wherever a four-vector appears; none is needed here. Quoted from the previous topic: in a central field H commutes with J², Jz, the parity operator βP and K = β(Σ⋅L + ħ), whose eigenvalue −κħ has κ = −(j + ½) when l = j − ½ and κ = +(j + ½) when l = j + ½. Write ψ as an upper pair over a lower pair. Since β = diag(1, −1) and Σ = diag(σ, σ), K acts as +(σ⋅L + ħ) on the upper pair and as −(σ⋅L + ħ) on the lower. One eigenvalue −κħ for the whole spinor therefore needs σ⋅L + ħ to give −κħ above and +κħ below: the lower pair is the harmonic labelled −κ. Same |κ|, so the same j; opposite sign, so the other orbital value, l′ = 2j − l. For 2P(3/2), κ = −2, l = 1 and l′ = 2. Parity agrees: βP gives the upper pair (−1)ˡ and the lower −(−1)(l′) = (−1)ˡ, one parity for the state, read from the upper l.
2Writing χκᵐ out with Clebsch-Gordan coefficients
χκᵐ is the coupled state |l, ½, j, m⟩ of Quantum Mechanics II, quoted with Condon-Shortley phases and l coupled before s, read as a two-component function of direction. For j = l + ½, κ < 0, it is √((l + m + ½)/(2l + 1)) Yₗ(m−1/2) ↑ + √((l − m + ½)/(2l + 1)) Yₗ(m+1/2) ↓. For j = l − ½, κ > 0, the coefficients become −√((l − m + ½)/(2l + 1)) and +√((l + m + ½)/(2l + 1)). So χ₋₁¹⁄² = Y₀⁰ ↑ = ↑/√(4π) is the s(1/2) harmonic, and χ₁¹⁄² = −√(1/3) Y₁⁰ ↑ + √(2/3) Y₁¹ ↓ = −(cos θ ↑ + sin θ e(iφ) ↓)/√(4π) is p(1/2). Each is dimensionless and orthonormal over the sphere, so |χκᵐ|² is a density per steradian. On each, σ⋅L = 2L⋅S/ħ takes the value [j(j + 1) − l(l + 1) − ¾]ħ = −(κ + 1)ħ: 0 for s(1/2), +ħ for p(3/2), −2ħ for p(1/2).
3What symmetry alone says about σ⋅r̂
σ⋅r̂ = σₓ sin θ cos φ + σy sin θ sin φ + σz cos θ depends on direction only. It is Hermitian, and (σ⋅a)(σ⋅b) = a⋅b + iσ⋅(a×b) gives (σ⋅r̂)² = r̂⋅r̂ = 1, so it is unitary as well. It is the dot product of two vectors, r̂ turned by L and σ turned by S, so it commutes with J = L + (ħ/2)σ and cannot change j or m. Under parity r̂ reverses and σ, an axial vector, does not: σ⋅r̂ is a pseudoscalar and reverses orbital parity. Of the harmonics with given j and m exactly one has the other parity, so σ⋅r̂χκᵐ = cχ(−κ)ᵐ. The same follows algebraically from the identity (σ⋅L + ħ)(σ⋅r̂) = −(σ⋅r̂)(σ⋅L + ħ), derived from r̂⋅L = 0 and L×r̂ = −r̂×L + 2iħr̂: σ⋅r̂ turns the eigenvalue −κħ into +κħ. Hermiticity and (σ⋅r̂)² = 1 force |c| = 1, and because σ⋅r̂ is a rotational scalar, c is the same for every m. A single evaluation fixes it.
4Fixing the phase on the z-axis
Look along +z, where θ = 0. Every Yₗ(mₗ) vanishes there except Yₗ⁰ = √((2l + 1)/(4π)), so for m = ½ only the spin-up term with mₗ = 0 survives. For κ < 0, with l = |κ| − 1, the coefficient √((l + 1)/(2l + 1)) gives χκ¹⁄²(ẑ) = +√(|κ|/(4π)) ↑; for κ > 0, with l = |κ|, the coefficient −√(l/(2l + 1)) gives −√(|κ|/(4π)) ↑. Partners of equal |κ| are equal and opposite on the axis. There σ⋅r̂ = σz and σz ↑ = ↑, so σ⋅r̂ leaves χκ¹⁄²(ẑ) unchanged, which is −χ(−κ)¹⁄²(ẑ): c = −1, derived. For 1s, 0.2821 ↑ against χ₁¹⁄²(ẑ) = −0.2821 ↑. One c serves every m and direction, so σ⋅r̂χκᵐ = −χ(−κ)ᵐ everywhere, and |χ(−κ)ᵐ|² = |χκᵐ|² point by point because σ⋅r̂ is unitary in each direction. The −1 is a convention: coupling s before l flips it, F changes sign with it, and no observable moves.
5Reducing σ⋅p to the number κ
Put (σ⋅r̂)² = 1 in front of σ⋅p and multiply out the right-hand pair: σ⋅p = (σ⋅r̂)[r̂⋅p + iσ⋅(r̂×p)]. Here r̂⋅p = −iħ∂/∂r, and r̂×p = L/r with 1/r on the left, so σ⋅p = (σ⋅r̂)[−iħ∂/∂r + iσ⋅L/r], derived. Act on (G/r)χκᵐ. The derivative gives −iħ(G′/r − G/r²)χκᵐ and the angular term gives −iħ(κ + 1)(G/r²)χκᵐ, so the bracket is −(iħ/r)(G′ + κG/r)χκᵐ: the G/r² from differentiating G/r has cancelled the 1 in κ + 1. Then σ⋅r̂ supplies −χ(−κ)ᵐ, and σ⋅p[(G/r)χκᵐ] = (iħ/r)(dG/dr + κG/r)χ(−κ)ᵐ, derived. The lower pair follows from κ → −κ: σ⋅p[(F/r)χ(−κ)ᵐ] = (iħ/r)(dF/dr − κF/r)χκᵐ. With ψ = (1/r)(Gχκᵐ, iFχ(−κ)ᵐ), every term of the Dirac equation now carries one of the two harmonics, which divide out; the next topic writes the two radial equations that remain.
6Checks: dimensions, p², a free wave and the 2p doublet
G is in m(−1/2), so (G/r)χκᵐ is in m(−3/2) like ψ, and (ħ/r)(dG/dr) and (ħ/r)(κG/r) both carry kg m s⁻¹ times m(−3/2), as σ⋅p acting on ψ must. Applied twice, the result must give p², since (σ⋅p)² = p²: it gives −(ħ²/r)[G″ − κ(κ + 1)G/r²]χκᵐ, and κ(κ + 1) = l(l + 1) for either sign, 2 for κ = −2 and for κ = +1, which is Quantum Mechanics II's radial kinetic operator. For a free spherical wave, G equal to r times the spherical Bessel function of order l at kr, the Bessel recurrences turn dG/dr + κG/r into −kr (κ < 0) or +kr (κ > 0) times the Bessel function of order l′, as a free particle of orbital l′ requires. The measured check runs through σ⋅L = −(κ + 1)ħ: Unit 6's spin-orbit term, quoted, weights 2P(3/2) by +1 and 2P(1/2) by −2, a splitting of mc²α⁴/32 = 4.53 × 10⁻⁵ eV, or 10.95 GHz. Hydrogen's measured interval is 10.97 GHz; the 0.2% rest is mostly the Lamb shifts.
Key equations
Read the units, assumptions and limits before substituting.
Upper and lower angular functions
ψ = (1/r) (G(r) χκᵐ, i F(r) χ(−κ)ᵐ)ᵀ
Derived here: the lower pair carries −κ, so the same j and m, orbital l′ = 2j − l and opposite orbital parity, because the β inside K = β(Σ⋅L + ħ) reverses the sign of σ⋅L + ħ on it.
G and F in m(−1/2); χ dimensionless, normalised over solid angle; κ = −(j + ½) for l = j − ½, +(j + ½) for l = j + ½, quoted.σ⋅r̂ reverses the sign of κ
(σ⋅L + ħ)(σ⋅r̂) = −(σ⋅r̂)(σ⋅L + ħ)
Applied to χκᵐ it shows that σ⋅r̂χκᵐ has eigenvalue +κħ under σ⋅L + ħ, and, since σ⋅r̂ also commutes with J, it can only be a multiple of χ(−κ)ᵐ.
Exact operator identity; σ dimensionless, L and ħ in J s. Derived from r̂⋅L = 0 and L×r̂ = −r̂×L + 2iħr̂.Action of σ⋅r̂ on a spinor harmonic
σ⋅r̂ χκᵐ = −χ(−κ)ᵐ
Derived on the z-axis and valid for every κ, m and direction. Since (σ⋅r̂)² = 1, the same operator takes χ(−κ)ᵐ back to −χκᵐ.
σ⋅r̂ = σₓ sin θ cos φ + σy sin θ sin φ + σz cos θ; Condon-Shortley phases, l coupled before s.σ⋅p split into radial and angular parts
σ⋅p = (σ⋅r̂)(−iħ ∂/∂r + iσ⋅L/r)
Derived from (σ⋅r̂)² = 1. The derivative acts only on the radial function and σ⋅L only on the harmonic, where it is the number −(κ + 1)ħ.
p and L/r in kg m s⁻¹; uses r̂⋅p = −iħ ∂/∂r and r̂×p = L/r, with 1/r standing to the left of L.σ⋅p on a central-field spinor
σ⋅p [(G/r) χκᵐ] = (iħ/r)(dG/dr + κG/r) χ(−κ)ᵐ
Derived here, and every angle has gone. Used in both rows of the Dirac equation, it leaves the two coupled radial equations for G and F that the next topic writes down.
G in m(−1/2), ħ in J s, both sides in kg m s⁻¹ m(−3/2). For the lower pair put −κ: (iħ/r)(dF/dr − κF/r)χκᵐ.One angular density for both pairs
|χ(−κ)ᵐ(θ, φ)|² = |χκᵐ(θ, φ)|²
The 1s lower pair is a p(1/2) harmonic yet spreads as 1/(4π) = 0.0796 sr⁻¹ in every direction: l′ shows in its spin, not in its density.
Both in sr⁻¹. Derived: it holds direction by direction, because σ⋅r̂ is a unitary 2 × 2 matrix at each point.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start at κ = −1: the upper spin points up everywhere while the lower one turns through 2θ. Flip the sign and the two swap. Raise |κ|: the faint spin-up curves show l and l′ nodes, yet the two totals stay exact mirror images.
Live results
- κ
- -1
- UPPER l
- 0
- LOWER l′ = 2j − l
- 1
- 4π|χ|² AT θ, BOTH
- 1.000
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. |κ| = j + ½: 1; Sign of κ: −1; Polar angle θ at φ = 0: 60 °
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
A first guess is that the small component is the large one made small: the same angular function times a radial factor of order v/c, so that 1s is an s-wave in all four components. It cannot be, because σ⋅p, the operator that builds the lower pair from the upper, is a pseudoscalar: it keeps j and m and reverses orbital parity. The 1s lower pair is iF/r times χ₁¹⁄² = −(cos θ ↑ + sin θ e(iφ) ↓)/√(4π), a p(1/2) harmonic with l′ = 1. Two facts disguise it. Its density is 1/(4π) in every direction, exactly as for s(1/2), because σ⋅r̂ is unitary at each point. And the state is still even, because the parity operator βP adds the sign the odd l′ would otherwise remove. The spin gives it away: along a meridian the lower pair's spin turns through 2θ while the upper pair's points up everywhere.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Fix the phase of σ⋅r̂ for the d(5/2) and f(5/2) harmonics, κ = −3 and κ = +3, by evaluating both m = ½ harmonics on the +z axis from their Clebsch-Gordan coefficients and then applying σ⋅r̂ there. Give the angular density of each on the axis.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- On the +z axis only Yₗ⁰ survives, with Yₗ⁰(ẑ) = √((2l + 1)/(4π)), so each m = ½ harmonic is the spin-up coefficient for mₗ = 0 times that value, times ↑.
- κ = −3 is d(5/2): l = 2, j = l + ½, coefficient √((l + m + ½)/(2l + 1)) = √(3/5) = 0.7746, and Y₂⁰(ẑ) = √(5/(4π)) = 0.6308, so χ₋₃¹⁄²(ẑ) = 0.4886 ↑ = +√(3/(4π)) ↑.
- κ = +3 is f(5/2): l = 3, j = l − ½, coefficient −√((l − m + ½)/(2l + 1)) = −√(3/7) = −0.6547, and Y₃⁰(ẑ) = √(7/(4π)) = 0.7464, so χ₃¹⁄²(ẑ) = −0.4886 ↑.
- On the axis σ⋅r̂ = σz and σz ↑ = ↑, so σ⋅r̂χ₋₃¹⁄²(ẑ) = 0.4886 ↑ = −χ₃¹⁄²(ẑ). The phase is −1, the same as for 1s.
- Densities: |χ|² = 0.4886² = 3/(4π) = 0.2387 sr⁻¹ for both, which is (j + ½)/(4π) with j = 5/2.
Answer. σ⋅r̂χ₋₃¹⁄² = −χ₃¹⁄², a phase of −1; both harmonics have density 0.239 sr⁻¹ on the axis.
MediumExample 2
At θ = 60°, φ = 0, build χ₋₂¹⁄² (p(3/2)) and χ₂¹⁄² (d(3/2)) from their Clebsch-Gordan coefficients, apply the matrix σ⋅r̂ to the first, and confirm that σ⋅r̂χ₋₂¹⁄² = −χ₂¹⁄² away from the axis. Compare the two angular densities and their spin-up shares.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- At φ = 0 every Yₗ(mₗ) is real. At θ = 60°: Y₁⁰ = 0.24430, Y₁¹ = −0.29921, Y₂⁰ = −0.07885 and Y₂¹ = −0.33452, in sr(−1/2).
- p(3/2), m = ½, j = l + ½: coefficients √(2/3) = 0.81650 on Y₁⁰ ↑ and √(1/3) = 0.57735 on Y₁¹ ↓, so χ₋₂¹⁄² has (↑, ↓) components (0.19947, −0.17275).
- d(3/2), m = ½, j = l − ½: coefficients −√(2/5) = −0.63246 on Y₂⁰ ↑ and √(3/5) = 0.77460 on Y₂¹ ↓, so χ₂¹⁄² = (0.04987, −0.25912).
- At φ = 0, σ⋅r̂ is the real matrix with rows (cos θ, sin θ) and (sin θ, −cos θ), here (0.50000, 0.86603) and (0.86603, −0.50000). Acting on (0.19947, −0.17275) it gives (−0.04987, 0.25912), which is −χ₂¹⁄²: the phase found on the axis holds at 60°.
- Densities: 0.19947² + 0.17275² = 0.06963 sr⁻¹ and 0.04987² + 0.25912² = 0.06963 sr⁻¹, equal as unitarity requires. The spin-up shares differ, 57% for p(3/2) against 3.6% for d(3/2): two different functions with one density.
Answer. σ⋅r̂χ₋₂¹⁄² = (−0.04987, 0.25912) = −χ₂¹⁄² at θ = 60°; both densities are 0.0696 sr⁻¹.
HardExample 3
In hydrogen, take the non-relativistic 2p radial shape for the upper pair, G(r) = N r² e(−r/2a₀). Unit 6's leading-order elimination sets the lower pair to σ⋅p/2mc times the upper. Use this topic's reduction to find F for 2P(3/2) and for 2P(1/2), evaluate F/G at r = 2a₀, and find the probability carried by the lower pair in each, checking it against ⟨T⟩/2mc².
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Quoted from Unit 6: lower pair ≈ (σ⋅p/2mc) × upper pair. With this topic's result the lower pair is (1/2mc)(iħ/r)(dG/dr + κG/r)χ(−κ)ᵐ, which must equal i(F/r)χ(−κ)ᵐ. The harmonics divide out: F = (ħ/2mc)(dG/dr + κG/r), with ħ/2mc = αa₀/2 = 1.931 × 10⁻¹³ m.
- dG/dr = N(2r − r²/2a₀)e(−r/2a₀). For 2P(3/2), κ = −2: dG/dr − 2G/r = −N(r²/2a₀)e(−r/2a₀) = −G/2a₀, so F = −(αa₀/2)(G/2a₀) = −(α/4)G. F/G = −1.824 × 10⁻³ at every r, including 2a₀.
- For 2P(1/2), κ = +1: dG/dr + G/r = N(3r − r²/2a₀)e(−r/2a₀), so F = (αa₀/2)(3/r − 1/2a₀)G. At r = 2a₀, F/G = (α/2)(3/2 − 1/2) = α/2 = 3.649 × 10⁻³. Near the nucleus F grows as r, so F/r stays finite: an s-wave lower pair, as l′ = 0 demands.
- Probabilities, with x = r/a₀: ∫G² dr is proportional to ∫x⁴e(−x) dx = 24. For 2P(3/2), ∫F² dr/∫G² dr = (α/4)² = α²/16. For 2P(1/2), (α/2)² ∫(3x − x²/2)² e(−x) dx/24 = (α²/4)(18 − 18 + 6)/24 = α²/16 as well: 3.328 × 10⁻⁶ for both.
- Check: because (σ⋅p)² = p², the elimination predicts a ratio ⟨p²⟩/4m²c² = ⟨T⟩/2mc². For n = 2, ⟨T⟩ = 13.606 eV/4 = 3.401 eV and 2mc² = 1.022 MeV, giving 3.328 × 10⁻⁶. Dimensions: F/G = α/4 is a pure number, as a ratio of two radial functions in m(−1/2) must be.
Answer. F = −(α/4)G for 2P(3/2), so F/G = −1.82 × 10⁻³ everywhere; F = (αa₀/2)(3/r − 1/2a₀)G for 2P(1/2), so F/G = 3.65 × 10⁻³ at 2a₀. Each lower pair holds 3.33 × 10⁻⁶ of the probability.
Quick check
Select an answer and explain it before checking.
Review & continue
A spinor spherical harmonic χκᵐ is Quantum Mechanics II's coupled state |l, ½, j, m⟩ written as a two-component function of direction; a central-field Dirac state carries χκᵐ in its upper pair and χ(−κ)ᵐ in its lower, and σ⋅r̂ turns each into minus the other.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.