The operator K = β(Σ⋅L + ħ) and the quantum number κ

Work in SI units with the Dirac representation, β = diag(1, 1, −1, −1) and α carrying σ off the diagonal; the metric is (+, −, −, −), though nothing here needs a four-vector. For H = cα⋅p + βmc² + V(r), topic 7.01 showed that J = L + ħΣ/2 and the Dirac parity γ⁰P commute with H while L² and S⋅L do not. α⋅p is odd, so it moves amplitude between the upper and lower pairs, and as a vector operator it changes l by one. A state of definite j and parity therefore has two orbital labels, l above and l′ = l ± 1 below, and l names neither the state nor its energy. J² and parity still sort the states; what is wanted is a single conserved operator whose eigenvalue does that sorting in one number and can stand in the radial equations. Dirac's answer in 1928 was K = β(Σ⋅L + ħ). It is a rotational scalar, so [K, J] = 0, and it is unchanged by inversion, because Σ and L are axial vectors and β commutes with γ⁰ = β.

Write Λ = Σ⋅L + ħ, so K = βΛ; β is block-diagonal and commutes with Σ and L, so K = Λβ too. The kinetic term is odd: βα⋅p = −α⋅pβ. In the Dirac representation α = γ⁵Σ, with γ⁵ the off-diagonal identity blocks, which commute with Σ, so Λ anticommutes with α⋅p if σ⋅L + ħ anticommutes with σ⋅p. The Pauli identity (σ⋅a)(σ⋅b) = a⋅b + iσ⋅(a × b) gives (σ⋅L)(σ⋅p) + (σ⋅p)(σ⋅L) = L⋅p + p⋅L + iσ⋅(L × p + p × L). Here L⋅p = p⋅L = 0 and L × p + p × L = 2iħp, so that anticommutator is −2ħσ⋅p, and the +ħ adds +2ħσ⋅p: the sum vanishes, which is why the ħ is there. Then Kα⋅p = βΛα⋅p = −βα⋅pΛ = α⋅pβΛ = α⋅pK. β commutes with K, and V(r) commutes with L, Σ and β, so [K, H] = 0, derived, for any central V. Dimensions: K is in J s, like L. Free-field limit: with V = 0 nothing changes, so free spherical waves carry κ as well.

Since β² = 1 and β commutes with Λ, K² = Λ² = (Σ⋅L)² + 2ħΣ⋅L + ħ². The Pauli identity again, with L × L = iħL, gives (Σ⋅L)² = L² + iΣ⋅(iħL) = L² − ħΣ⋅L, so K² = L² + ħΣ⋅L + ħ². Expanding J² = (L + ħΣ/2)² = L² + ħΣ⋅L + 3ħ²/4, because Σ⋅Σ = 3, gives K² = J² + ħ²/4, derived; both sides are in J² s². On a state of total angular momentum j the eigenvalue is ħ²[j(j + 1) + ¼] = (j + ½)²ħ², a perfect square, so K = ±(j + ½)ħ. The convention writes the eigenvalue as −κħ, and κ runs over ±1, ±2, ±3 and so on; κ = 0 would need j = −½, which no state has. For j = ½ the choices are K = ±ħ, for j = 3/2 they are ±2ħ. K² cannot say which sign a given state takes. That needs K itself, evaluated where it is simplest.

Where β = +1, K = σ⋅L + ħ = 2S⋅L/ħ + ħ, with S = ħσ/2: Quantum Mechanics II's spin-orbit operator, doubled, divided by ħ and shifted by ħ. In QM-II's coupled state |l, ½, j, mⱼ⟩, 2S⋅L/ħ = ħ[j(j + 1) − l(l + 1) − ¾], which is lħ for j = l + ½ and −(l + 1)ħ for j = l − ½. So K = (l + 1)ħ = (j + ½)ħ on the first branch and −lħ = −(j + ½)ħ on the second, and with Kψ = −κħψ, κ = −(l + 1) for j = l + ½ and κ = +l for j = l − ½, derived. s₁/₂ is −1, p₁/₂ is +1, p₃/₂ is −2, d₃/₂ is +2 and d₅/₂ is −3. Read backwards, S⋅L = −(κ + 1)ħ²/2 on the upper pair: 0 for s₁/₂, −ħ² for p₁/₂, +ħ²/2 for p₃/₂, the values QM-II found. Non-relativistic limit: the upper pair becomes the Pauli spinor, so κ is QM-II's (l, j) relabelled and nothing about a slow electron changes.

Where β = −1, K = −(σ⋅L + ħ), so to give the same eigenvalue −κħ the lower pair needs σ⋅L + ħ = +κħ: the coupled state of the same j on the other branch, with orbital label l′ = 2j − l. For 1s₁/₂, κ = −1 and l′ = 1, so the small component of the ground state is a p wave. Compactly, l(l + 1) = κ(κ + 1) and l′(l′ + 1) = κ(κ − 1): the lower pair of κ has the angular form of the upper pair of −κ, which topic 7.03 builds explicitly. The Dirac parity γ⁰P gives the upper pair (−1)^l and the lower −(−1)^(l′), which equals (−1)^l because l′ = l ± 1, so the state has parity (−1)^l = sgn(κ)(−1)^κ. At fixed j the two signs of κ are the two parities, and |κ| is j + ½: κ carries j and parity together. l alone cannot do this, because l = 1 belongs both to κ = +1, p₁/₂, and to κ = −2, p₃/₂.

This course's Kψ = −κħψ is the convention of Rose's Relativistic Electron Theory, Sakurai and Greiner, and gives the same κ as Johnson's Atomic Structure Theory, which reaches it through a different but equivalent operator: κ = −1 for s₁/₂, which is also how relativistic atomic-structure codes such as GRASP label orbitals. Papers on the hidden symmetries of the Dirac-Coulomb problem, such as Khelashvili and Khachidze's or Al-Hashimi, Shalaby and Wiese's, label states by the eigenvalue of K/ħ itself, making s₁/₂ +1 and p₁/₂ −1, and every sign-dependent formula, the radial equations included, flips with it. Test any borrowed formula on 1s, where it must give l = 0. Hydrogen's measured n = 2 levels back the label. 2P₁/₂ and 2P₃/₂ share l = 1 but differ in |κ|, and lie 10.969 GHz apart. 2S₁/₂ and 2P₁/₂ differ in l and in the sign of κ but share |κ| = 1, and lie only 1.058 GHz apart, a Lamb shift Unit 13 computes. The Dirac energy depends on n and |κ| alone, which topic 7.06 derives and this lesson only quotes.