University Quantum Mechanics II · Spin and Charge in Electromagnetic Fields · 3.8
Orbital Zeeman and diamagnetic terms in a weak field
Here minimal coupling stops being a postulate and starts predicting spectra. Expand one square about the nucleus and two terms fall out: a linear one you know as a magnetic moment, and a quadratic one you were probably told to drop. You will size both, find the atoms where the second is the whole answer, and hand Unit 9 its operator.
Build the model
Connect the measurement to the mechanism.
Take the minimal-coupling Hamiltonian of topic 03.06, (p − qA)²/2m + V(r), and describe a uniform field by the symmetric gauge A = ½B × r centred on the nucleus. Because ∇⋅A = 0 the square expands cleanly into p²/2m, a term linear in B, −(q/2m)B⋅L, and a term quadratic in B, (q²B²/8m)(x² + y²). The two behave nothing alike. The linear term is an orbital magnetic moment with gL = 1; it commutes with any central H₀, so it moves each level by exactly mₗμB B and mixes nothing.
The quadratic term needs no moment: it is never negative, so it raises every level, by an amount set by the area of the orbit across B. Measured in Zeeman steps it is the flux through that area in units of 2h/e, about 2 × 10⁻⁶ for ground-state hydrogen in 1 T. It therefore hides in most atoms but rules two places: closed shells, where L = S = 0 remove the linear terms and it is the entire response, the diamagnetism of every noble gas; and Rydberg orbits, whose area grows as n⁴ until near n = 30 it matches a Zeeman step in 1 T. The model stops where its premise does.
Once the quadratic term nears the gap to the next shell, neither the Coulomb field nor B is a small correction to the other, and the spectrum belongs to neither the atom nor the Landau levels of topic 03.09. Spin is not in the square at all; it enters with a gₑ this lesson quotes.
- Simple definition
- In a weak uniform field B along z, the minimal-coupling kinetic energy of a bound electron gains an orbital Zeeman term (μB/ħ)B Lz, linear in B, and a diamagnetic term (e²B²/8mₑ)(x² + y²), quadratic in B and never negative.
- Example
- In 1.00 T the circular 3d state of hydrogen, mₗ = +2, rises by 2μB B = 1.16 × 10⁻⁴ eV through the orbital Zeeman term and by (e²B²/8mₑ) × 108a₀² = 6.65 × 10⁻⁹ eV through the diamagnetic term, 1.1 × 10⁻⁴ of one Zeeman step.
∇⋅A = 0 makes p⋅A = A⋅p, so the two cross terms merge into ½B⋅L; centring r on the nucleus keeps H₀ rotation-invariant about the same point.
A in T m, B in T, r in m measured from the nucleus; ∇ × A = B and ∇⋅A = 0
It commutes with any central H₀, so level n l mₗ moves by exactly mₗμB B, 14.0 GHz per tesla per unit of mₗ, and no states mix.
μB = eħ/2mₑ = 9.274 × 10⁻²⁴ J T⁻¹ = 5.788 × 10⁻⁵ eV T⁻¹; L in J s
Never negative, so it raises every level. To first order the shift is its expectation value: 1.23 × 10⁻¹⁰ eV for 1s hydrogen in 1 T, where ⟨x² + y²⟩ = 2a₀².
x² + y² in m², squared distance from the axis along B through the nucleus; e²a₀²/8mₑ = 6.16 × 10⁻¹¹ eV T⁻²
1s hydrogen in 1 T holds 4 × 10⁻⁶ flux quanta. A circular orbit, ⟨x² + y²⟩ = n³(n + 1)a₀², holds two in 1 T, and so matches one Zeeman step, near n = 31.
dimensionless; Φ = πB⟨x² + y²⟩ in Wb is the flux through the orbit, and h/e = 4.14 × 10⁻¹⁵ Wb
It is μ₀NA/B times the induced moment −∂ΔE/∂B of an L = S = 0 atom, whose diamagnetic shift ΔE grows as B². Per atom it equals −(2π/3)α²a₀Σ⟨rᵢ²⟩, an effect of order α² times an atomic volume.
χₘ in m³ mol⁻¹; the sum runs over the atom's electrons, ⟨rᵢ²⟩ in m²; −9.95 × 10⁻¹² m³ mol⁻¹ per a₀² of sum
For the electron the last term is (2μB/ħ)S⋅B, so the weak-field operator becomes (μB/ħ)(L + 2S)⋅B with g = 2 exactly; measured gₑ = 2.00232.
σ the Pauli matrices; the identity is derived from [πₓ, πy] = iqħBz with π = p − qA
Expand the square in the symmetric gauge
Start from H = (p − qA)²/2m + V(r), postulated in topic 03.06. Multiplying out gives p² − q(p⋅A + A⋅p) + q²A², and the order matters: acting on ψ, p⋅(Aψ) = A⋅pψ − iħ(∇⋅A)ψ. The symmetric gauge A = ½B × r has ∇⋅A = 0, so the two cross terms are equal and sum to 2A⋅p. The triple product then becomes angular momentum, A⋅p = ½(B × r)⋅p = ½B⋅(r × p) = ½B⋅L, legitimate because B is constant, while A² = ¼|B × r|² = ¼B²(x² + y²) for B along z. So H = p²/2m + V − (q/2m)B⋅L + (q²B²/8m)(x² + y²), derived with no approximation; weak field only means treating the last two terms as small. One choice hides in it, the origin. Moving it by d shifts A by the constant −½B × d, the gradient of −½(B × d)⋅r, which is a gauge change (topic 03.07): the spectrum is unchanged, but the tidy split into B⋅L and x² + y² now refers to the wrong point. Centre r on the nucleus, where V is spherical, and Lz commutes with H₀.
The linear term shifts levels and mixes nothing
For the electron q = −e, so the linear term is (e/2mₑ)B⋅L = (μB/ħ)B⋅L, with μB = eħ/2mₑ = 5.788 × 10⁻⁵ eV T⁻¹. It is −μL⋅B for an orbital moment μL = −(e/2mₑ)L, the moment of a classical current loop, so the orbital g-factor of 1 is derived, not quoted. With B along z the term is (μB B/ħ)Lz, and Lz commutes with any central H₀, so each |n l mₗ⟩ stays an eigenstate and moves by exactly mₗμB B. Dimensions: eB/mₑ is in s⁻¹, and ħ times it is an energy. Limit: the level spacing corresponds to the angular frequency eB/2mₑ, half the cyclotron frequency, which is Larmor's classical theorem that a weak field adds a uniform rotation to the orbit. The measured test needs a line between two spin-zero levels, so that no spin term competes: the red cadmium line at 643.8 nm splits into the normal Zeeman triplet, components 0.0194 nm apart per tesla, exactly the λ²μB B/hc this term predicts.
The quadratic term needs no moment and always pushes up
(e²B²/8mₑ)(x² + y²) is a positive operator, so every level rises. To first order, the rule time-independent perturbation theory will justify later, the shift is (e²B²/8mₑ)⟨x² + y²⟩. Only motion across B counts: a spherical state has ⟨x² + y²⟩ = (2/3)⟨r²⟩, and integrating r² against the 1s density gives ⟨r²⟩ = 3a₀², so ⟨x² + y²⟩ = 2a₀² and the shift is e²B²a₀²/4mₑ = 1.23 × 10⁻¹⁰ eV at 1 T. Dimensions: (C T m)²/kg = (kg s⁻¹ m)²/kg = J. That is 2.13 × 10⁻⁶ of μB B, and also exactly μB B/Eₕ with Eₕ = 27.2 eV, so for 1s the energies run Eₕ : μB B : ΔEdia = 1 : x : x², with x = B/(4.70 × 10⁵ T). The ratio has a picture. eB⟨x² + y²⟩/4ħ equals Φ/(2h/e), where Φ = πB⟨x² + y²⟩ is the flux through the orbit's mean cross-section, so ground-state hydrogen in 1 T encloses 4.3 × 10⁻⁶ flux quanta h/e.
Rydberg orbits: the n⁴ that makes it matter
Big orbits catch more flux. Take a circular state, l = mₗ = n − 1, whose radial function r(n−1)e(−r/na₀) has no nodes: two factorial integrals give ⟨r²⟩ = (na₀/2)²(2n + 2)(2n + 1), and |Yₗₗ|² ∝ sin(2l)θ gives ⟨sin²θ⟩ = (2l + 2)/(2l + 3). Together ⟨x² + y²⟩ = n³(n + 1)a₀², derived, and 2a₀² at n = 1 as it must be. One Zeeman step is overtaken when 1.06 × 10⁻⁶ n³(n + 1)(B/1 T) passes 1, at n = 31 in 1 T. The linear term stays harmless however large, since it moves a whole mₗ manifold rigidly. The quadratic term does not commute with H₀: x² + y² links l to l ± 2 and n to other n, and in hydrogen, states of one n and mₗ whose l differ by 2 are degenerate, so beyond circular states the expectation value must give way to diagonalising within the shell. Near the gap Eₕ/n³ to the next shell first order fails outright. Garton and Tomkins saw that regime in 1969: barium in about 2.5 T absorbed in resonances near the ionisation limit spaced by close to 1.5ħωc.
Closed shells: the quadratic term is the whole response
A closed-shell atom such as helium has L = 0 and S = 0, so Lz|ψ⟩ = Sz|ψ⟩ = 0: the linear terms vanish on the state itself, not merely on average, and so they give no second-order shift either. What survives is Σᵢ(e²B²/8mₑ)(xᵢ² + yᵢ²), which spherical symmetry turns into ΔE = (e²B²/12mₑ)Σᵢ⟨rᵢ²⟩. Because ΔE grows as B², the atom carries an induced moment μ = −∂ΔE/∂B = −(e²B/6mₑ)Σ⟨rᵢ²⟩, pointing against B: Lenz's law, and the sign of diamagnetism. Multiplying by μ₀NA/B gives the molar susceptibility χₘ = −μ₀NA e²Σ⟨rᵢ²⟩/(6mₑ), −9.95 × 10⁻¹² m³ mol⁻¹ for every a₀² in the sum, all derived. Its order in α appears once μ₀e²/(4πmₑ) is recognised as the classical electron radius rₑ = α²a₀: per atom χ is −(2π/3)α²a₀Σ⟨rᵢ²⟩, α² times an atomic volume. That is why the volume susceptibilities of diamagnetic solids sit near −10⁻⁵, and why the outermost electrons, with the largest ⟨r²⟩, dominate.
Adding spin: where the 2 in L + 2S comes from
Spin is not in (p − qA)²/2m. Topic 03.02 added (gₑμB/ħ)S⋅B by hand with gₑ measured, and rounding gₑ to 2 gives the operator Unit 9 diagonalises, (μB/ħ)(L + 2S)⋅B, plus the diamagnetic term. The 2 can be produced instead of inserted. For any vectors (σ⋅a)(σ⋅b) = a⋅b + iσ⋅(a × b), and with π = p − qA the components fail to commute, [πₓ, πy] = iqħBz, so π × π = iqħB and (σ⋅π)²/2m = π²/2m − (qħ/2m)σ⋅B, which for the electron is π²/2m + (2μB/ħ)S⋅B. The algebra is derived; the starting form is quoted from the Dirac equation's non-relativistic limit, since at A = 0 it equals p²/2m, so the Schrödinger equation alone cannot choose between the two forms. Measured gₑ = 2.0023193, and quantum electrodynamics, quoted, gives gₑ/2 − 1 = α/2π + … ≈ 0.00116, first order in α: 32.5 MHz per tesla on a 28.0 GHz T⁻¹ spin resonance. In n = 2 hydrogen μB B meets the measured 10.97 GHz fine structure near 0.78 T, which is why Unit 9 needs two regimes.
Change one variable at a time
Make the relationship visible.
Start at n = 1 and B = 1 T: the slope-2 line sits 5.7 decades under μB B. Raise n and it climbs about four decades per tenfold n, meeting μB B at 1 T between n = 30 and 31. Then push B up: soon after, it crosses the dashed shell gap, and neither term is a perturbation any more.
FIELD B1.00 T
LOG₁₀(DIAMAGNETIC ÷ μB B)-5.67
EQUAL TO μB B AT470103.50 T
REACHES SHELL GAP AT287880.18 T
Live interpretationFIELD B: 1.00 T. LOG₁₀(DIAMAGNETIC ÷ μB B): −5.67. EQUAL TO μB B AT: 470103.50 T. REACHES SHELL GAP AT: 287880.18 T
Catch the common trap
Explain before calculating.
A hydrogen atom in its 1s ground state, where ⟨r²⟩ = 3a₀², sits in a weak uniform field B along z. Ignoring spin, what does first-order perturbation theory, with each term of H acting once, give for the change in its energy?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA hydrogen atom in its 2p level sits in a uniform 1.50 T field along z. Ignoring spin and fine structure, find the orbital Zeeman and diamagnetic shifts of mₗ = +1, 0 and −1, and say, as a frequency, how far the diamagnetic term lifts the midpoint of the mₗ = ±1 pair above the mₗ = 0 level.
- Linear term: mₗμB B = mₗ × 5.788 × 10⁻⁵ eV T⁻¹ × 1.50 T = mₗ × 8.683 × 10⁻⁵ eV, or mₗ × 21.0 GHz after dividing by h. It is exact, because Lz commutes with the Coulomb Hamiltonian.
- Radial part: the 2p function r e(−r/2a₀) is nodeless, so ⟨r²⟩ = ∫r⁶e(−r/a₀)dr ÷ ∫r⁴e(−r/a₀)dr = (6!/4!)a₀² = 30a₀².
- Angular part: |Y₁₀|² ∝ cos²θ gives ⟨cos²θ⟩ = 3/5, so ⟨sin²θ⟩ = 2/5; |Y₁,±₁|² ∝ sin²θ gives ⟨sin²θ⟩ = 4/5. Hence ⟨x² + y²⟩ = 12a₀² for mₗ = 0 and 24a₀² for mₗ = ±1, averaging over mₗ to (2/3) × 30a₀² = 20a₀², as the isotropy of Σ|Y₁ₘ|² requires.
- Diamagnetic shifts: (e²B²/8mₑ)a₀² = 6.156 × 10⁻¹¹ eV T⁻² × (1.50 T)² = 1.385 × 10⁻¹⁰ eV, so mₗ = ±1 rise by 24 × 1.385 × 10⁻¹⁰ eV = 3.32 × 10⁻⁹ eV (804 kHz) and mₗ = 0 by 1.66 × 10⁻⁹ eV (402 kHz).
- First order is safe here: 2s shares mₗ = 0 with 2p₀ and the same energy, but it has opposite parity and x² + y² is even, so the two are not coupled.
- The ±1 pair keeps its 42.0 GHz separation, but its midpoint now sits 804 − 402 = 402 kHz above mₗ = 0, an asymmetry 1.9 × 10⁻⁵ of the 21.0 GHz Zeeman step.
AnswerLinear: mₗ × 8.68 × 10⁻⁵ eV (mₗ × 21.0 GHz). Diamagnetic: +3.32 × 10⁻⁹ eV (804 kHz) for mₗ = ±1 and +1.66 × 10⁻⁹ eV (402 kHz) for mₗ = 0, so the ±1 midpoint sits 402 kHz above mₗ = 0.
MediumHydrogen is prepared in the circular Rydberg state n = 20, l = mₗ = 19. Using its nodeless radial function r¹⁹e(−r/20a₀) and |Y₁₉,₁₉|² ∝ sin³⁸θ, find ⟨x² + y²⟩, the field B along z at which its diamagnetic shift equals one Zeeman step μB B, and whether first order still holds there.
- Radial: with ∫rᵏe(−r/L)dr = k! L(k+1) and L = 10a₀, ⟨r²⟩ = ∫r⁴²e(−r/10a₀)dr ÷ ∫r⁴⁰e(−r/10a₀)dr = 42 × 41 × (10a₀)² = 172 200a₀².
- Angular: ⟨sin²θ⟩ = ∫sin⁴¹θ dθ ÷ ∫sin³⁹θ dθ = 40/41 by the Wallis ratio, so ⟨x² + y²⟩ = 172 200a₀² × 40/41 = 168 000a₀², which is n³(n + 1)a₀² as the general circular-state result says.
- Set (e²B²/8mₑ)⟨x² + y²⟩ = μB B = eħB/2mₑ. One power of B cancels, leaving B = 4ħ/(e⟨x² + y²⟩) = 4 × 1.0546 × 10⁻³⁴ J s ÷ (1.602 × 10⁻¹⁹ C × 168 000 × 2.800 × 10⁻²¹ m²) = 5.60 T.
- Flux check: πB⟨x² + y²⟩ = π × 4ħ/e = 2h/e = 8.27 × 10⁻¹⁵ Wb, two flux quanta, as Φ/(2h/e) = 1 requires; with the rounded field, π × 5.60 T × 4.704 × 10⁻¹⁶ m² = 8.28 × 10⁻¹⁵ Wb, the same to rounding.
- Both shifts are then μB B = 5.788 × 10⁻⁵ eV T⁻¹ × 5.60 T = 3.24 × 10⁻⁴ eV. This is the lowest state with mₗ = 19, so the diamagnetic term can mix it only with n ≥ 21, and the nearest gap is 13.606 eV × (1/20² − 1/21²) = 3.16 × 10⁻³ eV. The shift is 10% of it, so first order is still fair.
- The linear shift 19μB B = 6.16 × 10⁻³ eV is twice that gap, yet harmless: every mₗ = 19 state receives it, so it mixes nothing.
Answer⟨x² + y²⟩ = 1.68 × 10⁵ a₀². The diamagnetic shift equals μB B = 3.2 × 10⁻⁴ eV at B = 5.6 T, when the orbit holds two flux quanta; that is 10% of the 3.2 × 10⁻³ eV gap to n = 21, so first order still holds.
HardHelium's ground state has L = 0 and S = 0, and tables built from accurate variational wavefunctions give ⟨r²⟩ = 1.1935a₀² for each electron. Find its energy shift and induced moment in 10.0 T, derive its molar susceptibility and compare it with the measured −1.88 × 10⁻⁶ cm³ mol⁻¹ (CGS), then repeat with a screened hydrogenic 1s² of Zeff = 27/16 and say what the comparison tests.
- Lz|ψ⟩ = Sz|ψ⟩ = 0, so both linear terms vanish on the state. What is left is Σᵢ(e²B²/8mₑ)(xᵢ² + yᵢ²), and for a spherical state ⟨xᵢ² + yᵢ²⟩ = (2/3)⟨rᵢ²⟩, so ΔE = (e²B²/12mₑ)Σ⟨rᵢ²⟩ with Σ⟨rᵢ²⟩ = 2.387a₀².
- Numbers: e²a₀²/(12mₑ) = (2/3) × 6.156 × 10⁻¹¹ eV T⁻² = 4.104 × 10⁻¹¹ eV T⁻², so at 10.0 T, ΔE = 4.104 × 10⁻¹¹ eV T⁻² × 100 T² × 2.387 = 9.80 × 10⁻⁹ eV, positive as a diamagnetic shift must be.
- Induced moment: ΔE ∝ B², so μ = −∂ΔE/∂B = −2ΔE/B = −1.96 × 10⁻⁹ eV T⁻¹ = −3.14 × 10⁻²⁸ J T⁻¹, or −3.4 × 10⁻⁵ μB, pointing against the field.
- Susceptibility: χₘ = μ₀NAμ/B = −μ₀NA e²Σ⟨rᵢ²⟩/(6mₑ) = −9.953 × 10⁻¹² m³ mol⁻¹ × 2.387 = −2.38 × 10⁻¹¹ m³ mol⁻¹. The measured value converts as 4π × 10⁻⁶ × (−1.88 × 10⁻⁶) = −2.36 × 10⁻¹¹ m³ mol⁻¹: agreement within 1%.
- Order in α: μ₀e²/(4πmₑ) = rₑ = α²a₀ = 2.82 × 10⁻¹⁵ m, so per atom χ = −(2π/3)α²a₀Σ⟨rᵢ²⟩ = −3.94 × 10⁻³⁵ m³, and NA times that returns the molar value.
- Screened estimate, quoted from the variational method later in the course: a hydrogenic 1s with Zeff = 27/16 has ⟨r²⟩ = 3a₀²/Zeff² = 1.0535a₀², giving χₘ = −2.10 × 10⁻¹¹ m³ mol⁻¹, 12% low, though the same function misses the energy by only 2%. r² weights the outer tail the energy barely samples, so χ tests the wavefunction where the energy cannot.
AnswerAt 10.0 T, ΔE = 9.8 × 10⁻⁹ eV and μ = −3.1 × 10⁻²⁸ J T⁻¹, against B. χₘ = −2.38 × 10⁻¹¹ m³ mol⁻¹ against −2.36 × 10⁻¹¹ measured; Zeff = 27/16 gives −2.10 × 10⁻¹¹, 12% low.