University Quantum Mechanics II · Rotations and the Addition of Angular Momentum · 5.7
Orbit plus spin: the coupled states |l, ½, j, mⱼ⟩
Every fine-structure and Zeeman calculation later in this course opens with a state such as 2p₃/₂, mⱼ = ½. Here you build all of them, for any l, from one 2 × 2 diagonalisation, so you can write the square roots yourself, check them in three lines, and test them against g-factors a spectroscopist has measured.
Build the model
Connect the measurement to the mechanism.
An electron with orbital quantum number l and spin ½ has 2(2l + 1) states, and Quantum Mechanics I labelled them |l, mₗ⟩|mₛ⟩, the uncoupled basis. That basis is right only while nothing couples orbit to spin. The spin-orbit term ξ(r)L⋅S does: L⋅S commutes with L², S², J² and Jz but not with Lz or Sz, so mₗ and mₛ stop being good labels while mⱼ = mₗ + mₛ survives.
Written as LzSz + (L₊S₋ + L₋S₊)/2, the operator can only trade one unit of ħ between orbit and spin, so it never leaves the pair |mⱼ − ½, ↑⟩, |mⱼ + ½, ↓⟩. The whole problem falls apart into 2 × 2 blocks, plus a 1 × 1 block at each end, mⱼ = ±(l + ½). Diagonalising one block derives everything here: eigenvalues lħ²/2 and −(l + 1)ħ²/2 that do not depend on mⱼ, which is why j = l + ½ and j = l − ½ appear as degenerate multiplets of 2l + 2 and 2l states, and eigenvectors whose amplitudes √[(l ± mⱼ + ½)/(2l + 1)] are the Clebsch-Gordan coefficients.
No α and no radial integral enters; how far apart the two levels sit needs ξ(r), which Unit 8 derives from the field a moving electron sees, with the Thomas factor quoted from relativity. The coupled basis stops being the right one when something picks out an axis more strongly than L⋅S does: a magnetic field with μB B comparable to the fine-structure gap, about 0.8 T for hydrogen 2p but 37 T for sodium 3p, hands the job back to the uncoupled states.
- Simple definition
- The coupled states |l, ½, j, mⱼ⟩ are the joint eigenstates of L², S², J² and Jz for one electron: each is a fixed superposition of the two uncoupled kets |mⱼ − ½, ↑⟩ and |mⱼ + ½, ↓⟩, with j = l + ½ or j = l − ½.
- Example
- For a 2p electron, |j = 3/2, mⱼ = ½⟩ = √(2/3)|mₗ = 0, ↑⟩ + √(1/3)|mₗ = 1, ↓⟩, so a measurement of Sz returns +ħ/2 with probability 2/3 = 0.667 and −ħ/2 with probability 1/3.
Each ladder term moves one ħ between orbit and spin, so mⱼ = mₗ + mₛ never changes: L⋅S is block-diagonal in mⱼ, with blocks of size two or one.
L and S in J s, so L⋅S is in J² s², natural unit ħ² with ħ = 1.055 × 10⁻³⁴ J s. L± = Lₓ ± iLy, and S± likewise.
Derived from LzSz and one ladder element, ħ√[l(l + 1) − (mⱼ + ½)(mⱼ − ½)]. Its trace, −ħ²/2, and determinant, −l(l + 1)ħ⁴/4, do not depend on mⱼ.
Basis |mⱼ − ½, ↑⟩, |mⱼ + ½, ↓⟩, kets |mₗ, mₛ⟩; entries in J² s². At |mⱼ| = l + ½ only one ket exists and the block is 1 × 1.
Derived as the roots of λ² + λ − l(l + 1) = 0, with λ in units of ħ²/2, and again from L⋅S = (J² − L² − S²)/2. Weighted by 2l + 2 and 2l states they sum to zero.
In J² s². For p: +ħ²/2 = 5.56 × 10⁻⁶⁹ J² s² and −ħ² = −1.11 × 10⁻⁶⁸ J² s². An s level has only the first, and it is zero.
Derived as the λ = l eigenvector of the block. At mⱼ = ±(l + ½) one coefficient vanishes, leaving the plain products |l, ↑⟩ and |−l, ↓⟩; at every other mⱼ the spin is entangled with mₗ.
Coefficients are dimensionless; their squares add to (2l + 1)/(2l + 1) = 1. Valid for every mⱼ from −(l + ½) to l + ½.
Orthogonality fixes it up to sign; the Condon-Shortley convention of every published table makes the |mₗ = l, ↓⟩ term of the top state positive, so the minus sits on spin up.
Dimensionless coefficients; exists only for l ≥ 1 and |mⱼ| ≤ l − ½, which gives 2l states.
Derived from the squared coefficients. Then ⟨Lz + 2Sz⟩ = mⱼħ[1 ± 1/(2l + 1)]: 4/3 and 2/3 for p, against 1.3342 and 0.6658 measured in sodium.
⟨Sz⟩ in J s. For 2p₃/₂, mⱼ = 3/2 it is ħ/2, fully up; for 2p₁/₂, mⱼ = ½ it is −ħ/6.
Which labels survive spin-orbit coupling
Quantum Mechanics I gave one electron the uncoupled basis |l, mₗ⟩|mₛ⟩: 2(2l + 1) states, six for a p electron and ten for d. The spin-orbit term that Unit 8 derives has the form ξ(r)L⋅S, so ask which labels it respects. The commutators are [Lz, L⋅S] = iħ(LySₓ − LₓSy) and [Sz, L⋅S] = iħ(LₓSy − LySₓ). Neither vanishes, so mₗ and mₛ are no longer good quantum numbers, but the two add to zero: Jz = Lz + Sz commutes with L⋅S, and by the same argument so do Jₓ, Jy and therefore J². L² and S² commute with everything in sight. The good labels are l, s = ½, j and mⱼ. Counting m-values as topic 05.05 did, the largest mⱼ, l + ½, occurs once and every lower one twice down to −(l − ½), so j = l + ½ carries 2l + 2 states and j = l − ½ carries 2l: 4 + 2 = 6 for p and 6 + 4 = 10 for d. An s electron has j = ½ only, because l − ½ would be negative.
L⋅S only trades one unit between orbit and spin
Rewrite the dot product with ladder operators: LₓSₓ + LySy = (L₊S₋ + L₋S₊)/2, because the cross terms cancel. So L⋅S = LzSz + (L₊S₋ + L₋S₊)/2. The first term is diagonal in |mₗ, mₛ⟩. The second raises mₗ while lowering mₛ, or the reverse, and a spin-½ can only be lowered from ↑ or raised from ↓. Starting from |mⱼ − ½, ↑⟩, the only other ket L⋅S can reach is |mⱼ + ½, ↓⟩, and from there only the first again. The 2(2l + 1)-dimensional problem therefore splits into independent blocks of fixed mⱼ. At the ends the pair is broken: at mⱼ = l + ½ the ket |l + 1, ↓⟩ does not exist, so |l, ↑⟩ sits alone in a 1 × 1 block and is an eigenvector without any work. Both ladder terms annihilate it, leaving LzSz = (lħ)(ħ/2) = lħ²/2. That stretched state hands you one eigenvalue free; the 2 × 2 blocks must reproduce it and supply the other.
Diagonalising the 2 × 2 block
Call the kets |a⟩ = |mⱼ − ½, ↑⟩ and |b⟩ = |mⱼ + ½, ↓⟩. The diagonal entries are LzSz: (mⱼ − ½)ħ²/2 and −(mⱼ + ½)ħ²/2. The off-diagonal entry is ⟨a|L₋S₊|b⟩/2: S₊ turns ↓ into ħ↑, and the Quantum Mechanics I ladder element gives L₋ a factor ħ√[l(l + 1) − (mⱼ + ½)(mⱼ − ½)] = ħ√[(l + ½)² − mⱼ²]. In units of ħ²/2 the block has trace −1 and determinant −(mⱼ² − ¼) − (l + ½)² + mⱼ² = −l(l + 1), so λ² + λ − l(l + 1) = 0 and λ = l or −(l + 1). The eigenvalues lħ²/2 and −(l + 1)ħ²/2 are the same in every block, because L⋅S commutes with J₋, which walks each eigenstate down its ladder. The identity L⋅S = (J² − L² − S²)/2 confirms both: [(l + ½)(l + 3/2) − l(l + 1) − ¾]/2 = l/2 and [(l − ½)(l + ½) − l(l + 1) − ¾]/2 = −(l + 1)/2. For p that is +ħ²/2 and −ħ², and 4 × ½ + 2 × (−1) = 0: L⋅S is traceless, so its degeneracy-weighted shifts cancel.
Reading off the coefficients and fixing the sign
For λ = l the first row of the block reads (mⱼ − ½ − l)α + √[(l + ½ − mⱼ)(l + ½ + mⱼ)] β = 0, so β/α = √[(l − mⱼ + ½)/(l + mⱼ + ½)]. Take α ∝ √(l + mⱼ + ½) and β ∝ √(l − mⱼ + ½); their squares add to 2l + 1, which sets the normalisation. The j = l − ½ state is the orthogonal vector, which orthogonality fixes only up to an overall sign. The Condon-Shortley convention, used by every published table, makes the coefficient of the ket with mₗ = l positive in the top state |l − ½, l − ½⟩; that ket is |l, ↓⟩, so the minus sign lands on the spin-up term. Now check against the 1 × ½ table that topic 05.06 built by lowering. With l = 1, mⱼ = ½ the formulas give |3/2, ½⟩ = √(2/3)|0, ↑⟩ + √(1/3)|1, ↓⟩ and |½, ½⟩ = −√(1/3)|0, ↑⟩ + √(2/3)|1, ↓⟩. Independently, J₋ = L₋ + S₋ acting on |3/2, 3/2⟩ = |1, ↑⟩ gives ħ√2|0, ↑⟩ + ħ|1, ↓⟩, of norm ħ√3, which is the first line again.
What 2p₃/₂ and 2p₁/₂ name, and what is measured
The spectroscopic label n l with subscript j names a whole multiplet: 2p₃/₂ is the four states j = 3/2, mⱼ = ±3/2, ±½, and 2p₁/₂ the two with j = ½. Only the stretched pair mⱼ = ±3/2 are products of an orbital state and a spin state; the other four are superpositions with the weights above. An s level has j = ½ alone, and its two states are simply |0, ↑⟩ and |0, ↓⟩, on which L⋅S gives zero. The 4 : 2 count is visible in sodium's yellow doublet: the D₂ line at 589.0 nm reaches 3p₃/₂ and D₁ at 589.6 nm reaches 3p₁/₂, with tabulated oscillator strengths 0.641 and 0.320, a ratio of 2.00, because each of the six 3p states couples equally to the ground level, as topics 05.08 and 05.09 show. The size of the gap is not this lesson's to derive. For hydrogen 2p, Unit 8 finds α²(13.598 eV)/16 = 4.526 × 10⁻⁵ eV, or 10.943 GHz (13.598 eV is the Rydberg energy with the reduced mass), of order α² times the Bohr energy, with 2p₃/₂ on top as the positive eigenvalue and a positive ξ require. The measured interval is 10.969 GHz; the 0.2% remainder is mostly quantum electrodynamics, quoted.
The spin inside a coupled state, and when the basis fails
Squaring the coefficients gives something measurable. In |l + ½, mⱼ⟩ the probability of spin up is (l + mⱼ + ½)/(2l + 1), so ⟨Sz⟩ = (ħ/2)[(l + mⱼ + ½) − (l − mⱼ + ½)]/(2l + 1) = mⱼħ/(2l + 1), and in |l − ½, mⱼ⟩ the sign flips. Then ⟨Lz + 2Sz⟩ = mⱼħ + ⟨Sz⟩ = mⱼħ[1 ± 1/(2l + 1)], which Unit 9 turns into the weak-field Zeeman shift: g = 4/3 for p₃/₂, 2/3 for p₁/₂ and 2 for s₁/₂. Sodium's measured 3p values are 1.3342 and 0.6658; the differences of about 0.1% come from gₛ = 2.0023 rather than 2, the factor 1 + α/2π quoted from quantum electrodynamics. At large l both weights near mⱼ = 0 tend to ½, so ⟨Sz⟩ shrinks while L and S stay locked together. The coupled basis diagonalises L⋅S only. Add a field B along z and the term (μB B/ħ)(Lz + 2Sz) is diagonal in the uncoupled basis instead; which basis wins is set by μB B against the fine-structure gap, and the crossover is 0.78 T for hydrogen 2p.
Change one variable at a time
Make the relationship visible.
At l = 1, step mⱼ from −1.5 to 1.5: the solid arrow swings from the ↓ axis to the ↑ axis and the circles slide, but the bars never move. Then hold mⱼ = 0.5 and raise l: both amplitudes drift towards 0.707 as ⟨Sz⟩ = mⱼħ/(2l + 1) shrinks.
SPIN-UP AMPLITUDE, j = l + ½0.816
SPIN-DOWN AMPLITUDE, j = l + ½0.577
⟨Sz⟩ IN j = l + ½0.167 ħ
MIXING ANGLE θ35.3 °
Live interpretationSPIN-UP AMPLITUDE, j = l + ½: 0.816. SPIN-DOWN AMPLITUDE, j = l + ½: 0.577. ⟨Sz⟩ IN j = l + ½: 0.167 ħ. MIXING ANGLE θ: 35.3 °
Catch the common trap
Explain before calculating.
A 3d electron (l = 2) is prepared in the coupled state j = 5/2, mⱼ = ½. What is the probability that a measurement of Sz returns +ħ/2?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron occupies a 4f level (l = 3). List the allowed j with the number of states in each and confirm the total, give the two eigenvalues of L⋅S in units of ħ² and in J² s², and check that the eigenvalues weighted by their degeneracies sum to zero.
- Uncoupled count: (2l + 1) × 2 = 7 × 2 = 14 states.
- Coupling with s = ½ allows only j = l ± ½ = 7/2 and 5/2, with 2j + 1 = 8 and 6 states: 8 + 6 = 14, matching. The labels are 4f₇/₂ and 4f₅/₂.
- Eigenvalues: lħ²/2 = 1.5ħ² for j = 7/2 and −(l + 1)ħ²/2 = −2ħ² for j = 5/2. Cross-check with (J² − L² − S²)/2: [(7/2)(9/2) − 12 − ¾]/2 = (15.75 − 12.75)/2 = 1.5.
- With ħ² = (1.0546 × 10⁻³⁴ J s)² = 1.1121 × 10⁻⁶⁸ J² s², the eigenvalues are 1.668 × 10⁻⁶⁸ J² s² and −2.224 × 10⁻⁶⁸ J² s².
- Weighted sum: 8 × 1.5 + 6 × (−2) = 12 − 12 = 0, as the tracelessness of L⋅S requires.
Answerj = 7/2 (8 states) and j = 5/2 (6 states), 14 in all; L⋅S = +1.5ħ² = 1.67 × 10⁻⁶⁸ J² s² and −2ħ² = −2.22 × 10⁻⁶⁸ J² s², with weighted sum zero.
MediumBuild the two coupled states of a d electron (l = 2) with mⱼ = −½ from the uncoupled kets, confirm they are normalised and orthogonal, and verify the j = 5/2 state exactly by acting on it with the 2 × 2 block of L⋅S.
- At mⱼ = −½ the pair is |mₗ = −1, ↑⟩ and |mₗ = 0, ↓⟩. With 2l + 1 = 5: l + mⱼ + ½ = 2 and l − mⱼ + ½ = 3.
- |5/2, −½⟩ = √(2/5)|−1, ↑⟩ + √(3/5)|0, ↓⟩ = 0.632|−1, ↑⟩ + 0.775|0, ↓⟩, and |3/2, −½⟩ = −√(3/5)|−1, ↑⟩ + √(2/5)|0, ↓⟩.
- Norms: 2/5 + 3/5 = 1 for both. Overlap: √(2/5)(−√(3/5)) + √(3/5)√(2/5) = 0.
- The block in units of ħ²/2 has diagonal mⱼ − ½ = −1 and −(mⱼ + ½) = 0, and off-diagonal √[(l + ½)² − mⱼ²] = √(6.25 − 0.25) = √6 = 2.449.
- Acting on (√(2/5), √(3/5)): the first row gives −√(2/5) + √6√(3/5) = −√(2/5) + 3√(2/5) = 2√(2/5), and the second gives √6√(2/5) = √(12/5) = 2√(3/5). Both are twice the input, so the eigenvalue is 2 × ħ²/2 = ħ² = lħ²/2 for l = 2.
Answer|5/2, −½⟩ = √(2/5)|−1, ↑⟩ + √(3/5)|0, ↓⟩ and |3/2, −½⟩ = −√(3/5)|−1, ↑⟩ + √(2/5)|0, ↓⟩; L⋅S eigenvalues ħ² and −1.5ħ².
HardSodium's valence electron is in 3p₁/₂, mⱼ = +½. From the coupled-state coefficients find the probability of spin up, ⟨Sz⟩, ⟨Lz⟩ and ⟨Lz + 2Sz⟩. Unit 9 shows the weak-field shift is (μB B/ħ)⟨Lz + 2Sz⟩: use it to find g, the mⱼ = ±½ splitting in 0.250 T in GHz, and whether 0.250 T is weak against the 17.2 cm⁻¹ fine-structure gap.
- With l = 1, mⱼ = ½: |½, ½⟩ = −√(1/3)|0, ↑⟩ + √(2/3)|1, ↓⟩. Spin up has probability 1/3 and spin down 2/3.
- ⟨Sz⟩ = (ħ/2)(1/3 − 2/3) = −ħ/6, matching −mⱼħ/(2l + 1). ⟨Lz⟩ = (1/3)(0) + (2/3)(ħ) = 2ħ/3, and ⟨Lz⟩ + ⟨Sz⟩ = ħ/2 = mⱼħ, as it must.
- ⟨Lz + 2Sz⟩ = 2ħ/3 − ħ/3 = ħ/3 = g mⱼħ, so g = 2/3.
- The mⱼ = ±½ states shift by ±g μB B/2, so they split by g μB B. With μB/h = 13.996 GHz T⁻¹: (2/3)(13.996 GHz T⁻¹)(0.250 T) = 2.333 GHz; the measured g = 0.6658 gives 2.330 GHz.
- Weak-field test: μB B = (5.788 × 10⁻⁵ eV T⁻¹)(0.250 T) = 1.45 × 10⁻⁵ eV, while 17.2 cm⁻¹ is 2.13 × 10⁻³ eV, a ratio of 0.0068. Mixing with 3p₃/₂ is negligible and the coupled basis stays good.
AnswerP(↑) = 1/3, ⟨Sz⟩ = −ħ/6, ⟨Lz⟩ = 2ħ/3, ⟨Lz + 2Sz⟩ = ħ/3, so g = 2/3; the splitting is 2.33 GHz at 0.250 T, and μB B is 0.0068 of the gap, so the field is weak.