University Quantum Mechanics II · Rotations and the Addition of Angular Momentum · 5.3
Total angular momentum J = J₁ + J₂ and its two bases
Every system with two sources of angular momentum poses the same question: label the parts, or label the whole? This lesson proves the total is a true angular momentum, finds exactly which labels can be sharp at once, and shows how the Hamiltonian, not taste, picks the basis to work in.
Build the model
Connect the measurement to the mechanism.
An orbit and a spin, or an electron and a nucleus, share the product space of Unit 4, and their sum J = J₁ + J₂ is a genuine angular momentum because components acting on different factors commute. Quantum Mechanics I's ladder argument therefore applies to J unchanged: J² has eigenvalues j(j + 1)ħ² and Jz has Mħ. What the sum does not do is respect the parts.
Expanding J² produces J₁₊J₂₋ + J₁₋J₂₊, which hands one unit of m from one factor to the other; it conserves M = m₁ + m₂ but not m₁ or m₂, and [J², J₁z] = 2iħ(J₁ × J₂)z. So j and m₁ cannot share a basis, and the space carries two complete labellings: the uncoupled |m₁ m₂⟩, sharp in each part's z-component, and the coupled |j M⟩, sharp in the total. They are orthonormal bases of the same (2j₁ + 1)(2j₂ + 1) states, joined by a unitary matrix that is block-diagonal in M because Jz is diagonal in both.
Neither basis is more fundamental; the Hamiltonian chooses. A coupling a J₁⋅J₂/ħ², the form spin–orbit and hyperfine terms take, is diagonal in the coupled basis; a field acting on one part alone is diagonal in the uncoupled one. The picture stops helping when the two are comparable, where only the M blocks survive and must be diagonalised, and it presumes j₁ and j₂ stay fixed, which fails once H mixes different l, as an electric field does.
- Simple definition
- The total angular momentum J = J₁ + J₂ of a two-part system is itself an angular momentum, and its states can be labelled by m₁ and m₂ (the uncoupled basis) or by j and M (the coupled basis): two orthonormal bases of one space, needed because J² and J₁z do not commute.
- Example
- A spin-3/2 nucleus and a spin-½ electron have 4 × 2 = 8 states: uncoupled, the eight pairs (m₁, m₂); coupled, five j = 2 and three j = 1 states. In ⁸⁷Rb the coupling a J₁⋅J₂/ħ² splits j = 2 from j = 1 by a measured 6.835 GHz.
The cross commutators [J₁ᵢ, J₂ⱼ] vanish, so J inherits the algebra and every ladder result of Quantum Mechanics I: J² gives j(j + 1)ħ² and Jz gives Mħ.
J₁ acts on the first factor and J₂ on the second; each component in J s, with ħ = 1.055 × 10⁻³⁴ J s
The ladder pair moves one ħ of z-angular momentum between the parts, keeping M = m₁ + m₂ but changing m₁ and m₂, so J² is not diagonal on product states.
J₁± = J₁x ± iJ₁y, and likewise for J₂; every term in J² s²
Nonzero whenever both parts carry angular momentum, so J² and J₁z share no basis; only the two stretched states are sharp in both.
In J³ s³. [J², J₂z] is its negative, so J² still commutes with Jz = J₁z + J₂z.
Jz is diagonal in both bases, so the matrix never mixes different M: each block is as large as the number of ways to split M between the parts.
Coefficients dimensionless; with the Condon–Shortley phase they are real, an orthogonal matrix of order (2j₁ + 1)(2j₂ + 1)
Any H = a J₁⋅J₂/ħ² is diagonal here with (2j + 1)-fold levels: 3a/4 for j = 2 and −5a/4 for j = 1 when j₁ = 3/2 and j₂ = ½.
ħ² = 1.112 × 10⁻⁶⁸ J² s²; the eigenvalue does not depend on M
Intervals in the ratio of their upper j test the J₁⋅J₂ form: Cs 6²P₃/₂ gives 251.09, 201.29 and 151.22 MHz, within 0.4% of 5 : 4 : 3.
a in J, often quoted as a/h in Hz; j is the upper level of the adjacent pair
The sum inherits the algebra
On Unit 4's product space J₁ stands for J₁ ⊗ 1 and J₂ for 1 ⊗ J₂, so every component of one commutes with every component of the other. Then [Jₓ, Jy] = [J₁x, J₁y] + [J₂x, J₂y] + [J₁x, J₂y] + [J₂x, J₁y] = iħJ₁z + iħJ₂z + 0 + 0 = iħJz, and the other two relations follow by cycling x, y, z. Quantum Mechanics I's ladder argument used nothing but these commutators and Hermiticity, so it applies to J unchanged: J² has eigenvalues j(j + 1)ħ², Jz has Mħ with M from −j to j, and J±|j M⟩ = ħ√[(j ∓ M)(j ± M + 1)] |j, M ± 1⟩. The product states already have one new label sharp, since Jz|m₁ m₂⟩ = (m₁ + m₂)ħ|m₁ m₂⟩. For j₁ = 3/2 and j₂ = ½ the 4 × 2 = 8 product states have M = 2, 1, 0, −1, −2, with 1, 2, 2, 2 and 1 states. Which values of j occur is not decided by the algebra; the lesson on the allowed values of j counts them.
Expanding J² exposes the ladder terms
Because J₁ and J₂ commute, J² = J₁² + J₂² + 2J₁⋅J₂ with no ordering to worry about. Write the transverse part with ladder operators: J₁₊J₂₋ + J₁₋J₂₊ = 2(J₁xJ₂x + J₁yJ₂y), the imaginary cross terms cancelling, so J₁⋅J₂ = J₁zJ₂z + ½(J₁₊J₂₋ + J₁₋J₂₊). On a product state the first term is diagonal, m₁m₂ħ². The other two are not: J₁₊J₂₋ takes |m₁ m₂⟩ to |m₁ + 1, m₂ − 1⟩, handing one ħ of z-angular momentum from the second part to the first. M is untouched, but m₁ and m₂ each change. For 3/2 ⊗ ½ in the M = 1 block, ordered |½, ½⟩ then |3/2, −½⟩, the ladder elements give J²/ħ² as the matrix with rows (5, √3) and (√3, 3): diagonal entries 15/4 + 3/4 + 2m₁m₂, off-diagonal √(1 × 3) × √(1 × 1). Its eigenvalues are 4 ± 2, that is 6 and 2, which are j(j + 1) for j = 2 and j = 1, and neither product state is an eigenvector.
Which operators can be sharp together
J₁² commutes with every component of J₁ and trivially with J₂, so it commutes with J² and with Jz; so does J₂². Jz commutes with J² because J is an angular momentum. The failure is J₁z: [J², J₁z] = 2[J₁⋅J₂, J₁z] = 2([J₁x, J₁z]J₂x + [J₁y, J₁z]J₂y) = 2iħ(J₁xJ₂y − J₁yJ₂x) = 2iħ(J₁ × J₂)z. Its matrix element ⟨3/2, −½|[J², J₁z]|½, ½⟩ = (½ − 3/2)ħ × √3ħ² = −√3ħ³ is not zero. [J², J₂z] is the same with the opposite sign, which is why the sum commutes. So there are two complete sets of commuting observables: the uncoupled (J₁², J₂², J₁z, J₂z), labelling |j₁ j₂; m₁ m₂⟩, and the coupled (J₁², J₂², J², Jz), labelling |j₁ j₂; j M⟩. Each has four labels that fix a state uniquely, and no commuting set holds both J² and J₁z. In the vector model J₁ and J₂ precess about their fixed sum: Jz stays put while J₁z and J₂z swing.
One space, two bases, a block-diagonal unitary
Both sets span the same (2j₁ + 1)(2j₂ + 1) states, so each coupled state is a combination of uncoupled ones, |j M⟩ = Σ ⟨m₁ m₂|j M⟩|m₁ m₂⟩, and because both bases are orthonormal the coefficients form a unitary matrix. Hermiticity of Jz makes it sparse: the two states have Jz eigenvalues Mħ and (m₁ + m₂)ħ, so their overlap vanishes unless M = m₁ + m₂. For 3/2 ⊗ ½ the blocks have sizes 1, 2, 2, 2, 1, and the 1 × 1 blocks are the stretched states |3/2, ½⟩ and |−3/2, −½⟩, members of both bases. In the M = 1 block the eigenvectors of the J² matrix are |j = 2, M = 1⟩ = (√3/2)|½, ½⟩ + ½|3/2, −½⟩ and |j = 1, M = 1⟩ = −½|½, ½⟩ + (√3/2)|3/2, −½⟩, signs fixed by the Condon–Shortley rule. Squared, they are probabilities: in |j = 2, M = 1⟩, J₁z gives ħ/2 with probability ¾ and 3ħ/2 with probability ¼. These entries are Clebsch–Gordan coefficients, which a later lesson builds by lowering, with no matrix to diagonalise.
A J₁⋅J₂ coupling picks the coupled basis
A Hamiltonian that commutes with a complete set is diagonal in its basis. A coupling a J₁⋅J₂/ħ² = (a/2ħ²)(J² − J₁² − J₂²) commutes with the coupled set, so its levels are (a/2)[j(j + 1) − j₁(j₁ + 1) − j₂(j₂ + 1)], each (2j + 1)-fold: 3a/4 for j = 2 and −5a/4 for j = 1 when j₁ = 3/2, j₂ = ½. The trace checks it: in the uncoupled basis the diagonal of J₁⋅J₂ is m₁m₂ħ², summing to zero, and a trace does not depend on basis, so 5 × 3/4 + 3 × (−5/4) = 0. Adjacent levels differ by a j, the Landé interval rule. That spin–orbit and hyperfine couplings take this form, in hydrogen of order α² and α² mₑ/mₚ times the Bohr energies, is quoted here and derived in Units 8 and 9; a comes from measurement. The ⁸⁷Rb ground state splits by a measured 6.834683 GHz, so a/h = 3.417 GHz. Cs 6²P₃/₂, with j₁ = 7/2 and j₂ = 3/2, has measured intervals 251.09, 201.29 and 151.22 MHz; divided by 5, 4 and 3 they give 50.22, 50.32 and 50.41 MHz, equal to within 0.4%; the small residual is a nuclear electric-quadrupole term, quoted from the hyperfine tables, which a pure J₁⋅J₂ form leaves out.
A term on one part picks the uncoupled basis
A term b J₂z/ħ acting on one part alone, such as a field along z acting on an electron's spin moment (the nucleus's own Zeeman term, about 2000 times smaller per unit of m₁ in ⁸⁷Rb, is dropped), commutes with J₁z and J₂z but not with J², so the uncoupled basis diagonalises it with energies b m₂. Put both terms in, H = a J₁⋅J₂/ħ² + b J₂z/ħ, and neither basis works. Both terms commute with Jz, though, so H is block-diagonal in M and each block can be solved exactly. For M = 1 the uncoupled diagonal entries a/4 + b/2 and −3a/4 − b/2, with off-diagonal (√3/2)a, give E± = −a/4 ± ½√(4a² + 2ab + b²). At b = 0 these are the coupled levels 3a/4 and −5a/4; for b ≫ a they approach b m₂ + a m₁m₂, the uncoupled diagonal entries; at every b, E₊ + E₋ = −a/2, the block's trace. For ⁸⁷Rb, where b = gₛ μB B with gₛ = 2.0023 quoted from the Dirac equation and quantum electrodynamics as in Unit 3, b = a at B = 0.122 T: at 0.024 T the upper M = 0 level is still 99.8% j = 2, and at 1 T it is 98.6% m₂ = +½.
Change one variable at a time
Make the relationship visible.
At b/a = 0 the upper level is pure j = 2, yet its probability of m₂ = +½ is only {(2+m)/4|2}. Drag b/a to 5: the solid curves peel onto the dashed uncoupled lines and the bars trade places. Change M to start from ¾, ½ or ¼; E₊ + E₋ stays −a/2 throughout.
UPPER LEVEL E₊/a0.868
⟨J²⟩/ħ² IN UPPER LEVEL5.79
P(j = 2), UPPER LEVEL0.947
P(m₂ = +½), UPPER LEVEL0.724
Live interpretationUPPER LEVEL E₊/a: 0.868. ⟨J²⟩/ħ² IN UPPER LEVEL: 5.79. P(j = 2), UPPER LEVEL: 0.947. P(m₂ = +½), UPPER LEVEL: 0.724
Catch the common trap
Explain before calculating.
Two angular momenta with j₁ = 2 and j₂ = 1 have H = a J₁⋅J₂/ħ² + b Jz/ħ, with a and b of similar size. In which basis is H diagonal?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTwo angular momenta with j₁ = 2 and j₂ = 1 are coupled by H = a J₁⋅J₂/ħ² with a/h = 40 MHz, and the allowed totals are j = 3, 2 and 1. Check the state count, find each level's J₁⋅J₂ eigenvalue and energy, and test the result with the trace and the interval rule.
- Uncoupled count: (2j₁ + 1)(2j₂ + 1) = 5 × 3 = 15. Coupled count: 7 + 5 + 3 = 15 for j = 3, 2, 1. The two bases span the same space, as they must.
- J₁⋅J₂ = ½(J² − J₁² − J₂²), and j₁(j₁ + 1) + j₂(j₂ + 1) = 6 + 2 = 8, so the eigenvalue is ½[j(j + 1) − 8]ħ²: 2ħ² for j = 3, −ħ² for j = 2 and −3ħ² for j = 1.
- Energies are a times the eigenvalue over ħ²: E/h = +80 MHz, −40 MHz and −120 MHz for j = 3, 2 and 1.
- Trace check: 7 × 2 + 5 × (−1) + 3 × (−3) = 14 − 5 − 9 = 0, matching Σ m₁m₂ = (Σ m₁)(Σ m₂) = 0 in the uncoupled basis.
- Intervals: E(3) − E(2) = 3a gives 120 MHz and E(2) − E(1) = 2a gives 80 MHz, the ratio 3 : 2 that the interval rule predicts.
Answer15 states in each basis; J₁⋅J₂ = 2ħ², −ħ² and −3ħ² for j = 3, 2, 1, so E/h = +80, −40 and −120 MHz, with a (2j + 1)-weighted sum of zero and intervals of 120 MHz and 80 MHz.
MediumTwo angular momenta have j₁ = j₂ = 1. Build the matrix of J² in the M = 1 block of the uncoupled basis, diagonalise it to obtain the coupled states, and find the possible results of measuring J² on the product state m₁ = 1, m₂ = 0. Show also that J² and J₁z do not commute in this block.
- Write uncoupled states |m₁, m₂⟩. M = 1 is reached by (1, 0) and (0, 1), so the block is 2 × 2; order it |1, 0⟩, |0, 1⟩.
- Diagonal entries of J² = J₁² + J₂² + 2J₁zJ₂z + J₁₊J₂₋ + J₁₋J₂₊ are (2 + 2 + 2m₁m₂)ħ² = 4ħ² for both, since m₁m₂ = 0.
- Off-diagonal: J₁₊J₂₋|0, 1⟩ = ħ√(1 × 2) × ħ√(2 × 1)|1, 0⟩ = 2ħ²|1, 0⟩. So J²/ħ² has rows (4, 2) and (2, 4).
- Eigenvalues 4 ± 2: 6 = j(j + 1) for j = 2 and 2 for j = 1. Eigenvectors |j = 2, M = 1⟩ = (|1, 0⟩ + |0, 1⟩)/√2 and |j = 1, M = 1⟩ = (|1, 0⟩ − |0, 1⟩)/√2, the Condon–Shortley rule putting a plus sign on |1, 0⟩, the term with m₁ = j₁, in the top state of j = 1.
- Invert: |1, 0⟩ = (|j = 2, M = 1⟩ + |j = 1, M = 1⟩)/√2, so J² gives 6ħ² or 2ħ² with probability 0.50 each, and ⟨J²⟩ = 0.50 × 6ħ² + 0.50 × 2ħ² = 4ħ², equal to the diagonal element, as it must be.
- Commutator: ⟨0, 1|[J², J₁z]|1, 0⟩ = (1 − 0)ħ × 2ħ² = 2ħ³, not zero, so no basis of this block diagonalises both J² and J₁z.
Answer|j = 2, M = 1⟩ = (|1, 0⟩ + |0, 1⟩)/√2 with J² = 6ħ², and |j = 1, M = 1⟩ = (|1, 0⟩ − |0, 1⟩)/√2 with 2ħ²; on |1, 0⟩ each result has probability 0.50 and ⟨J²⟩ = 4ħ².
HardA spin-3/2 J₁ and a spin-½ J₂ are coupled by H = a J₁⋅J₂/ħ² with a/h = 3.417 GHz, the value measured for the ⁸⁷Rb ground state. At t = 0 the system is in the uncoupled state m₁ = ½, m₂ = +½, as if a strong field had held it there and been switched off instantly. Find the probability P(t) of still finding that state, its minimum and when it first occurs, and how ⟨J₂z⟩ and ⟨J₁z⟩ move.
- The state has M = 1, so it lives in the block with |j = 2, M = 1⟩ = (√3/2)|½, ½⟩ + ½|3/2, −½⟩ and |j = 1, M = 1⟩ = −½|½, ½⟩ + (√3/2)|3/2, −½⟩. The matrix is orthogonal, so invert by transposing: |½, ½⟩ = (√3/2)|j = 2⟩ − ½|j = 1⟩.
- The coupled states are stationary, with E₂ = 3a/4 and E₁ = −5a/4, so |ψ(t)⟩ = (√3/2)e(−iE₂t/ħ)|j = 2⟩ − ½e(−iE₁t/ħ)|j = 1⟩.
- Project back: ⟨½, ½|ψ(t)⟩ = ¾e(−iE₂t/ħ) + ¼e(−iE₁t/ħ). Squaring, P(t) = (¾)² + (¼)² + 2(¾)(¼)cos(ΔE t/ħ) = 5/8 + (3/8)cos(2πft), with ΔE = E₂ − E₁ = 2a and f = 2a/h = 6.834 GHz.
- The minimum is 5/8 − 3/8 = 0.25, first when 2πft = π: t = 1/(2f) = 1/(1.3668 × 10¹⁰ Hz) = 7.32 × 10⁻¹¹ s = 73.2 ps. The missing 0.75 is in |3/2, −½⟩, the only other state with M = 1.
- ⟨J₂z⟩ = (ħ/2)P − (ħ/2)(1 − P) = (P − ½)ħ swings from +ħ/2 to −ħ/4 and back. H commutes with Jz, so ⟨Jz⟩ = ħ throughout and ⟨J₁z⟩ = ħ − ⟨J₂z⟩ swings from ħ/2 to 5ħ/4: angular momentum flows between the parts, never out of the pair.
- Checks: P(0) = 5/8 + 3/8 = 1; ΔE t/ħ is dimensionless; and the stretched state |3/2, ½⟩, an eigenstate of H, would give P = 1 at all t.
AnswerP(t) = 5/8 + (3/8)cos(2πft) with f = 6.834 GHz; the minimum 0.25 first comes at 73.2 ps. ⟨J₂z⟩ swings between +ħ/2 and −ħ/4 while ⟨J₁z⟩ swings between ħ/2 and 5ħ/4, their sum fixed at ħ.