University Quantum Mechanics II · Spin and Charge in Electromagnetic Fields · 3.2
The spin magnetic moment and the Hamiltonian H = −μ⋅B
Every spin-in-a-field problem in the rest of this course starts from one line, H = −μ⋅B. This topic earns that line: where the classical ratio q/2m comes from, why spin roughly doubles it, which numbers you take on trust from Dirac, QED and experiment, and why a uniform field twists a spin without ever pushing it.
Build the model
Connect the measurement to the mechanism.
A charge going round a loop carries a magnetic moment fixed by its angular momentum: μ = IA = (q/2m)L whatever the radius or speed, which for the electron's orbit is −(μB/ħ)L, one Bohr magneton μB = eħ/2mₑ per unit ħ. Spin has no loop, so that argument cannot be carried over, but rotation symmetry does most of the work: on the two-dimensional spin space the only operator that turns like a vector is S itself, up to a constant, so every spin moment is μ = γS with one constant γ. For the electron γ = −gₑ e/2mₑ with gₑ = 2.00231930, twice the loop value and a little more.
The 2 is quoted from the Dirac equation and the 0.00232 from quantum electrodynamics, whose leading term α/π is first order in the fine-structure constant; neither is derived here, and the proton's gₚ = 5.586 is simply measured. With γ in hand, H = −μ⋅B = −γS⋅B is the complete Hamiltonian of a spin whose position is ignored: a 2 × 2 matrix with levels ±½ħ|γ|B, 28.02 GHz apart per tesla for the electron and 42.58 MHz for the proton. Because a uniform B leaves it independent of position, it can exert a torque, μ × B, but no force; a force, −∇H, needs B to vary in space.
The model stops being the whole story once anything else couples to the spin: a field gradient ties it to position, spin–orbit coupling to the orbit, the hyperfine interaction to the nucleus.
- Simple definition
- The spin magnetic moment is the vector operator μ = γS through which a particle's spin couples to a magnetic field, H = −μ⋅B; for the electron γ = −gₑ e/2mₑ with gₑ = 2.0023, so μ points opposite to S.
- Example
- An electron spin in B = 1.00 T along z has H = (gₑ μB B/2)σz, with levels ±57.95 μeV: spin along B is the upper level, and the 115.9 μeV gap is a photon of 28.02 GHz.
Derived from μ = IA: radius and speed cancel, so μ/L depends on q/m alone. This is g = 1, the value every later g-factor is measured against.
μ in A m² (= J T⁻¹), q in C, m in kg, L in J s. Holds for any body whose charge and mass share one distribution.
The moment of one unit ħ of orbital angular momentum: μL = −(μB/ħ)L, so μz = −mₗ μB for an electron.
e in C, ħ in J s, mₑ in kg. As energy per field, 5.788 × 10⁻⁵ eV T⁻¹, or 57.88 μeV T⁻¹.
Rotation symmetry forces μ ∝ S, leaving one constant to measure. The minus sign, from the negative charge, puts μ opposite to S.
S in J s, μ in J T⁻¹. The 2 is quoted from the Dirac equation, the 0.00231930 from QED; neither is derived here.
Positive, so the proton's moment lies along its spin, and 658 times weaker than the electron's: proton resonance runs at MHz where electron resonance runs at GHz.
μN = eħ/2mₚ = 5.051 × 10⁻²⁷ J T⁻¹, which is μB/1836. gₚ is measured; no model at this level gives it.
The complete Hamiltonian of a spin with its position ignored, solved by the two-state methods of Quantum Mechanics I. Spin along B is the upper level.
A 2 × 2 matrix in J. Levels ±½gₑ μB|B|, a gap of 115.9 μeV at 1 T, which is 28.02 GHz.
From the Heisenberg equation for p: a Hamiltonian with no position dependence conserves momentum, so a uniform field can only twist the atom's moment, never move the atom.
Force in N, on a neutral atom, where −μ⋅B is the only term in H that depends on position. Zero in a uniform field. On a magnet's symmetry axis Fz = μz ∂Bz/∂z, and J T⁻¹ × T m⁻¹ = N.
A circulating charge fixes the ratio μ/L
Start where the moment is not in doubt. A charge q on a circle of radius r at speed v is a current qv/2πr enclosing πr², so μ = IA = qvr/2, while L = mvr. The ratio μ/L = q/2m has lost r and v, and summing over any body whose charge and mass share one distribution keeps it: μ = (q/2m)L, derived. Dimensions: C kg⁻¹ × J s = A m², and since 1 T = 1 N A⁻¹ m⁻¹, A m² = J T⁻¹. For the electron's orbit q = −e, so μ = −(μB/ħ)L and μz = −mₗ μB, antiparallel to L. As a check put the electron on the Bohr orbit, r = a₀ at v = αc = 2.19 × 10⁶ m s⁻¹: μ = e(αc)a₀/2 = 9.274 × 10⁻²⁴ J T⁻¹, exactly one μB, so an atomic moment is first order in α in units of e c a₀. The measured check is the normal Zeeman effect: cadmium's red 643.8 nm line, from states with no net spin, splits by μB B, 14.0 GHz per tesla, as g = 1 requires. The orbital moment is taken from the loop here; minimal coupling derives it quantum mechanically in the orbital Zeeman topic.
Rotations fix the direction of a spin moment, not its size
Spin has no loop, so μ = (q/2m)L cannot simply be carried over. The form of a spin moment is forced all the same. On the spin-½ space every Hermitian operator is aI + b⋅σ, so each component is μⱼ = aⱼI + Σₖ bⱼₖσₖ. Rotate the spin state and ⟨μ⟩ must turn with ⟨S⟩ by the same rotation, since empty space offers no other direction. The constant part aⱼ is the same in every state, so it would have to equal its own rotated image under every rotation, forcing a = 0; and the 3 × 3 matrix bⱼₖ must commute with every rotation, which only a multiple of the identity does. So μ = γS, derived, with one constant γ carrying everything specific to the particle. Write γ = gq/2m, so that g compares the moment with the loop value and g = 1 is the classical guess. For the electron the measured value is gₑ = 2.00231930, and μ = −gₑ(μB/ħ)S: the moment points against the spin and is twice as large per unit angular momentum as an orbit's. That factor of 2 is why the anomalous Zeeman effect resisted every model built from orbits alone.
The 2 and the 0.0023 are quoted, from Dirac and QED
Nothing in non-relativistic quantum mechanics predicts gₑ; it is an input. The Dirac equation, relativistic quantum mechanics, gives exactly g = 2 for a point particle of spin ½. Writing the kinetic energy of a spin-½ charge as (σ⋅π)²/2m, with π = p − qA its kinetic momentum, produces the same 2, but that rewriting is a choice only the Dirac equation justifies, so here the 2 is quoted. The rest is quantum electrodynamics. Its expansion of the anomaly aₑ = (gₑ − 2)/2 begins with Schwinger's α/2π = 1.16141 × 10⁻³, first order in α; the measured value is aₑ = 1.15965 × 10⁻³. The leading term is 0.15% high, and the second-order term, −0.328(α/π)², lowers it by 1.77 × 10⁻⁶ to 1.15964 × 10⁻³. Carried to fifth order with α from atom interferometry, theory meets the Penning-trap measurement to about a part in 10⁹, among the most precise agreements in physics. In this course quote gₑ = 2.0023, and use g = 2 only where 0.1% does not matter.
Protons and neutrons: g from experiment
For a nucleon the natural unit is the nuclear magneton μN = eħ/2mₚ = 5.051 × 10⁻²⁷ J T⁻¹, smaller than μB by mₚ/mₑ = 1836, and μ = g(μN/ħ)S. A point Dirac proton would have g = 2. The measured gₚ is 5.586, a moment of 2.793 μN, and the neutron, with no charge at all, has gₙ = −3.826. Both are quoted from experiment: they come from charged constituents inside the nucleon, and no model at this level supplies them. The signs matter: the proton's moment lies along its spin, the electron's and the neutron's against it. The sizes matter more. The moment ratio (gₑ μB)/(gₚ μN) is 658.2, so in 1 T an electron spin's levels are 28.02 GHz apart and a proton's 42.58 MHz. That second number is measured directly, since proton resonance in water is how laboratory and MRI magnets are calibrated, and the ratio 658.21 is itself a measured constant. Nuclear moments are thus about a thousand times weaker than electronic ones, which is why hyperfine structure, a coupling to the nuclear moment, is so much smaller than fine structure.
H = −μ⋅B is the whole Hamiltonian of a lone spin
If the spin couples to nothing but a uniform field, its state lives in ℂ² and H = −μ⋅B = −γS⋅B is a 2 × 2 matrix. That H is postulated, the energy of a classical dipole promoted to an operator, and the measured resonances below are its test. For the electron H = (gₑ μB/2)σ⋅B: with B along z it is diagonal with levels ±½gₑ μB B, and the upper one has the spin along B, because the moment points the other way. A spin prepared along or against B is stationary; any other state is a superposition, evolved by U(t) = exp(−iHt/ħ) exactly as in Quantum Mechanics I. At 1 T the gap is 115.9 μeV, or 28.02 GHz, and an X-band spectrometer at 9.5 GHz finds free radicals, whose g lies within about 0.2% of gₑ, resonating near the predicted 0.339 T. Limits: at B = 0 the levels merge, since a free spin has no preferred axis, and g = 1 would halve the gap. Why may position be ignored? In a uniform field the full H is Horb ⊗ I + I ⊗ Hₛₚᵢₙ, so the spin evolves on its own. That fails once something couples the spin to anything else: a field gradient to position, spin–orbit coupling to the orbit, the hyperfine interaction to the nucleus.
A uniform field twists the moment but cannot push it
For a neutral carrier such as a silver atom, whose moment is that of one unpaired electron spin, −μ⋅B is the only term in H that depends on position, and the Heisenberg equation of the previous topic gives dp/dt = (i/ħ)[H, p] = ∇(μ⋅B). The spin operators commute with p, so only the position dependence of B contributes, and in a uniform field the force is zero for every spin state, derived. What a uniform field does exert is the classical torque τ = μ × B, perpendicular to B; and since [H, S⋅B] = 0, the component of S along B never changes. The torque therefore cannot align the spin, only swing it around B, the precession the next topic derives. A force needs a gradient: on a magnet's symmetry axis, where Bₓ and By vanish at every z, Fz = μz ∂Bz/∂z, the Stern–Gerlach force of Quantum Mechanics I. For one electron's spin moment, ½gₑ μB = 9.285 × 10⁻²⁴ J T⁻¹, a gradient of 800 T m⁻¹ gives 7.43 × 10⁻²¹ N, which accelerates a silver atom at 4.15 × 10⁴ m s⁻². A real magnet cannot supply ∂Bz/∂z alone, because ∇⋅B = 0 forces a transverse gradient as well; why a strong holding field makes that harmless belongs to the Stern–Gerlach topic.
Change one variable at a time
Make the relationship visible.
With the gradient at 0, sweep θ: ⟨H⟩ and the torque change but the force bar never appears, because a uniform field cannot push. Then add a gradient: the force switches on, reverses at θ = 90° with ⟨μz⟩, and peaks at 0° and 180°, just where the dashed torque curve falls to zero.
ENERGY ⟨H⟩29.0 μeV
LEVEL GAP ΔE/h28.02 GHz
TORQUE |⟨μ⟩ × B|8.04 × 10⁻²⁴ N m
MEAN FORCE ⟨Fz⟩0.00 × 10⁻²¹ N
Live interpretationENERGY ⟨H⟩: 29.0 μeV. LEVEL GAP ΔE/h: 28.02 GHz. TORQUE |⟨μ⟩ × B|: 8.04 × 10⁻²⁴ N m. MEAN FORCE ⟨Fz⟩: 0.00 × 10⁻²¹ N
Catch the common trap
Explain before calculating.
An electron is held in a uniform field B = 0.50 T along +z and prepared in |↑⟩, the state with Sz = +ħ/2. Treating it as a spin alone, what is its energy, and what does the field do to it?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron spin, its position ignored, sits in a uniform field of 0.250 T pointing along −z. Write H as a 2 × 2 matrix in the Sz basis, find both energies in μeV, say which Sz state is the ground state, and find the frequency of the photon that flips it.
- μ = −gₑ(μB/ħ)S, so H = −μ⋅B = (gₑ μB/ħ)S⋅B. With B = −Bẑ and Sz = (ħ/2)σz, H = −(gₑ μB B/2)σz: diagonal in the Sz basis, with entry −gₑ μB B/2 for |↑⟩ and +gₑ μB B/2 for |↓⟩.
- gₑ μB B/2 = 2.00232 × 5.7884 × 10⁻⁵ eV T⁻¹ × 0.250 T ÷ 2 = 1.4488 × 10⁻⁵ eV = 14.49 μeV. Dimensions: eV T⁻¹ × T = eV.
- So E = −14.49 μeV for |↑⟩ and +14.49 μeV for |↓⟩. The ground state is |↑⟩: its spin points along +z, against the field, which puts its moment along B. Reverse B and the labels swap, as they must.
- The gap is 28.98 μeV, so f = ΔE/h = 2.8975 × 10⁻⁵ eV ÷ 4.1357 × 10⁻¹⁵ eV s = 7.006 × 10⁹ Hz. Check: 28.02 GHz T⁻¹ × 0.250 T = 7.01 GHz.
AnswerH = −(14.49 μeV)σz. E = −14.49 μeV for |↑⟩, the ground state, and +14.49 μeV for |↓⟩; the flip photon has f = 7.01 GHz.
MediumA sample at 310 K sits in a 3.00 T field. Treating each spin as governed by H = −μ⋅B, find the fractional excess P = (Nlow − Nhigh)/N in the lower level for proton spins and for unpaired electron spins, and say which way the net spin and the net moment point for each.
- Gaps: for the proton ΔEₚ = gₚ μN B = 5.5857 × 3.1525 × 10⁻⁸ eV T⁻¹ × 3.00 T = 5.283 × 10⁻⁷ eV (127.7 MHz); for the electron ΔEₑ = gₑ μB B = 2.00232 × 5.7884 × 10⁻⁵ eV T⁻¹ × 3.00 T = 3.477 × 10⁻⁴ eV (84.07 GHz).
- Boltzmann populations give Nlow/Nhigh = e(ΔE/kT), so P = tanh(ΔE/2kT). Here kT = 8.617 × 10⁻⁵ eV K⁻¹ × 310 K = 2.671 × 10⁻² eV, and 2kT = 5.343 × 10⁻² eV.
- Protons: ΔE/2kT = 5.283 × 10⁻⁷ ÷ 5.343 × 10⁻² = 9.89 × 10⁻⁶, small enough that tanh x ≈ x, so Pₚ = 9.9 × 10⁻⁶: about ten spare spins per million, which is all an MRI scanner has to work with.
- Electrons: ΔE/2kT = 3.477 × 10⁻⁴ ÷ 5.343 × 10⁻² = 6.51 × 10⁻³, so Pₑ = 6.5 × 10⁻³, larger by 658, the ratio of the moments.
- Directions: E = −μ⋅B puts the moment along B in the lower level for both. The proton's moment lies along its spin, so its net spin points along B; the electron's lies against its spin, so its net spin points against B. The net moment is along B in both cases, which is paramagnetism.
AnswerPₚ = 9.9 × 10⁻⁶ and Pₑ = 6.5 × 10⁻³, a ratio of 658. Both net moments point along B; the net proton spin points along B and the net electron spin against it.
HardA single electron is held in a Penning trap with B = 4.20 T. Ignoring the trap's electric field and relativistic shifts, find the cyclotron frequency fc = eB/2πmₑ and the spin-flip frequency fₛ from H = −μ⋅B, show that their difference is fₐ = [(gₑ − 2)/2]fc, and compare its measured value with QED's leading prediction.
- The spin gap is gₑ μB B, and μB/h = eħ/(2mₑ × 2πħ) = e/4πmₑ, so fₛ = gₑ eB/4πmₑ = (gₑ/2)fc. With g = 2 exactly the spin would flip at precisely the cyclotron frequency; the two clocks agree only as far as gₑ = 2.
- fc = 1.602177 × 10⁻¹⁹ C × 4.20 T ÷ (2π × 9.109384 × 10⁻³¹ kg) = 1.175685 × 10¹¹ Hz = 117.5685 GHz. Units: C T kg⁻¹ = s⁻¹, since 1 T = 1 kg C⁻¹ s⁻¹.
- fₛ = (gₑ/2)fc = 1.00115965 × 117.5685 GHz = 117.7048 GHz, and fₐ = fₛ − fc = aₑ fc with aₑ = (gₑ − 2)/2 = 1.15965 × 10⁻³, so fₐ = 136.34 MHz. As gₑ → 2, fₐ → 0, the limiting case.
- QED's leading term, quoted, is aₑ = α/2π = 1.16141 × 10⁻³, first order in α. It predicts fₐ = 136.55 MHz, 0.21 MHz or 0.15% high. The second-order term, −0.328(α/π)² = −1.77 × 10⁻⁶, brings the prediction to 136.337 MHz, within 0.002 MHz of the measurement.
- Measuring fₐ rather than fₛ is the point: fₐ/fc gives aₑ directly, and since gₑ = 2(1 + aₑ), fixing fₐ to 1 part in 10⁶ fixes gₑ to about 1.2 parts in 10⁹.
Answerfc = 117.57 GHz, fₛ = 117.70 GHz and fₐ = 136.34 MHz. The α/2π term alone predicts 136.55 MHz, 0.15% high; adding the α² term gives 136.34 MHz.