University Quantum Mechanics II · Spin and Charge in Electromagnetic Fields · 3.6
Minimal coupling: H = (p − qA)²/2m + qφ
Every orbital magnetic effect in the rest of this course enters through one substitution, p → p − qA. Make it carefully: learn the order in which p and A must be written, which momentum is physical, why the Lorentz force comes back out exactly, and which commutator the Landau levels are built on.
Build the model
Connect the measurement to the mechanism.
Quantum Mechanics I never said how a charge feels a magnetic field, and nothing inside the Schrödinger equation can supply it. Minimal coupling is the postulate that fills the gap: carry over the classical Hamiltonian H = (p − qA)²/2m + qφ, keep p = −iħ∇ as the canonical momentum with [xᵢ, pⱼ] = iħδᵢⱼ, and let the potentials rather than the fields enter. Its credentials are then checked, not assumed.
The Heisenberg equations make (p − qA)/m the velocity and return the Lorentz force as a symmetrised, Hermitian operator, so in uniform fields the centre of any packet runs the classical orbit exactly, turning at ωc = |q|B/m with no ħ in it. The price is that one operator can no longer do two jobs: p still generates translations but shifts with the gauge, while the measurable mv = p − qA has components that fail to commute, [vₓ, vy] = iħqBz/m², which puts a floor of ħωc/2 under the transverse kinetic energy. The model has three edges.
It is non-relativistic, missing corrections of relative order (v/c)², α² for an atomic electron, that the Dirac equation supplies later. Its fields are classical, so no photon is ever exchanged. And it couples charge, not spin: the spin moment is added by hand, the Pauli form of the kinetic energy fixes g = 2, and QED raises that by α/π.
- Simple definition
- Minimal coupling places a charge q in electromagnetic fields by replacing the canonical momentum p = −iħ∇ with p − qA in the kinetic energy and adding the potential energy qφ, which makes (p − qA)/m, not p/m, the velocity.
- Example
- An electron in B = 1.00 T along z, in the gauge A = (0, Bx, 0), with a real wavefunction centred at x = 1.00 μm has ⟨p⟩ = 0, yet ⟨vy⟩ = (⟨py⟩ − q⟨Ay⟩)/mₑ = eBx/mₑ = 1.76 × 10⁵ m s⁻¹.
The one postulate: kinetic energy built from p − qA instead of p. Everything below follows from it and the unchanged [xᵢ, pⱼ] = iħδᵢⱼ.
p = −iħ∇ in kg m s⁻¹, A in T m, φ in V, q in C, m in kg. Since 1 T = 1 kg C⁻¹ s⁻¹, qA is a momentum.
Act on ψ and use the product rule. Drop the extra term and the expanded Hamiltonian stops being Hermitian whenever ∇⋅A ≠ 0.
∇⋅A in T, each side in kg m² s⁻¹ T. The two orders agree only in a Coulomb gauge, ∇⋅A = 0.
Read it term by term: p²/2m is the free particle, the A⋅p term is linear in the field, q²A²/2m is quadratic and never negative, and the ∇⋅A term is pure ordering, carried only when the gauge is not Coulomb.
Every term in J. The third vanishes in a Coulomb gauge, and the symmetric and Landau gauges of a uniform B are both Coulomb.
A phase gradient in ψ is not a velocity. m⟨v⟩ = ⟨p⟩ − q⟨A⟩ is what a detector measures, whatever gauge ψ was written in.
v in m s⁻¹. mv = p − qA is the mechanical momentum and is gauge-invariant; the canonical p shifts with A.
Derived from the Heisenberg equation. The half-sum is Hermitian; in a uniform B it is exactly v × B, and its mean is the classical force.
E = −∇φ − ∂A/∂t in V m⁻¹, B = ∇ × A in T. The half-sum equals v × B + (iħ/2m)∇ × B.
Robertson's bound gives ΔvₓΔvy ≥ ħ|q|B/2m², so the mean transverse kinetic energy is at least ħωc/2: the seed of the Landau levels.
Bz in T, each side in m² s⁻². The canonical pₓ and py commute; the mechanical components do not.
A postulate borrowed from classical mechanics
Nothing in Quantum Mechanics I says how a particle feels a magnetic field, so the coupling is postulated, and the postulate is borrowed. The classical Lagrangian L = ½mv² − qφ + qv⋅A reproduces the Lorentz force; its canonical momentum is ∂L/∂v = mv + qA, and the Legendre transform gives H = (p − qA)²/2m + qφ. Quantisation keeps the canonical momentum: p = −iħ∇ and [xᵢ, pⱼ] = iħδᵢⱼ are exactly as before, and all of the change goes into H. The rule is called minimal because the field enters only through p → p − qA, never through a term built from B itself, so a spin moment, which gauge invariance would also allow, must be added separately. Check the dimensions first: [qA] = C × T m, and 1 T = 1 kg C⁻¹ s⁻¹, so qA is a momentum in kg m s⁻¹ and p − qA is a legal difference. Two tests decide whether the postulate is right, carried out here and in the next topic: the Heisenberg equations must return the Lorentz force, and nothing measurable may depend on the gauge chosen for A and φ.
Expand the square in the right order
Multiplying out (p − qA)⋅(p − qA) produces both p⋅A and A⋅p, and they are different operators. Act on ψ: p⋅(Aψ) = −iħ∇⋅(Aψ) = −iħ(∇⋅A)ψ + A⋅(−iħ∇ψ), so p⋅A = A⋅p − iħ(∇⋅A), and (p − qA)² = p² − 2qA⋅p + iħq(∇⋅A) + q²A². The iħ term is what keeps H Hermitian: (A⋅p)† = p⋅A, so −2qA⋅p alone is Hermitian only when ∇⋅A = 0, and iħq(∇⋅A) repairs exactly the difference. It vanishes as ħ → 0, as a pure ordering effect must. In a Coulomb gauge, ∇⋅A = 0, the orders agree and H = p²/2m − (q/m)A⋅p + q²A²/2m + qφ. Both standard gauges of a uniform field, the symmetric A = (B/2)(−y, x, 0) and the Landau A = (0, Bx, 0), are Coulomb gauges, which is why the term is rarely seen and easily forgotten when a problem hands over a gauge that is not. For a weak uniform field the linear term becomes the orbital Zeeman coupling and the quadratic one the diamagnetic term; their sizes in an atom are the business of a later topic.
The velocity is (p − qA)/m, and p is no longer mv
Apply the Heisenberg equation, dxᵢ/dt = (i/ħ)[H, xᵢ]. Only the kinetic term fails to commute with xᵢ, and because A commutes with position, [(pⱼ − qAⱼ)², xᵢ] = −2iħ(pᵢ − qAᵢ)δᵢⱼ. So v = (p − qA)/m, derived. The two roles one operator played in Quantum Mechanics I are now split. The canonical p = −iħ∇ still generates translations and still obeys [xᵢ, pⱼ] = iħδᵢⱼ, but it equals mv + qA. The mechanical momentum mv = p − qA is what a time-of-flight detector or a deflection measures; p is what the phase gradient of ψ reports. Because A can be shifted by any gradient ∇χ without changing B, ⟨p⟩ shifts by q⟨∇χ⟩ with it, and the phase of ψ must move to keep m⟨v⟩ fixed, which is the subject of the next topic. So a value of ⟨p⟩ means nothing until the gauge is named. Note that [xᵢ, πⱼ] = iħδᵢⱼ still holds for π = p − qA, since A commutes with x: the two momenta are told apart only by commutators among their own components, and by how they respond to a change of gauge.
The Lorentz force comes back, symmetrised
Differentiate once more, with π = p − qA. The pieces are [πᵢ, πⱼ] = iħqεᵢⱼₖBₖ, [φ, πᵢ] = iħ∂ᵢφ and the explicit time dependence of A, and m dv/dt = (i/ħ)[H, π] − q∂A/∂t assembles them into m dv/dt = q[E + (v × B − B × v)/2], with E = −∇φ − ∂A/∂t and B = ∇ × A. Potentials went in; only fields come out. This is derived, and it is the test the postulate had to pass: as ħ → 0 the operators commute and it is the classical Lorentz force. Why two orderings? For operators (v × B)† = −B × v, and vⱼ does not commute with Bₖ(r): [vⱼ, Bₖ] = −(iħ/m)∂ⱼBₖ. So v × B alone is not Hermitian, and the half-sum is. The two differ by (iħ/2m)∇ × B, which vanishes wherever no current flows and E is steady, as in the gap of a magnet, so there the ordering is invisible. Dimensions agree: ħ∇ × B/m is J s × T m⁻¹ ÷ kg = m s⁻¹ T, the unit of v × B.
Uniform fields: the packet's centre runs the classical orbit
In uniform E and B the fields are numbers, not operators, so the half-sum is exactly v × B and the force is linear in v. Expectation values then give d⟨r⟩/dt = ⟨v⟩ and m d⟨v⟩/dt = q(E + ⟨v⟩ × B): Newton's equations for the centre, exact for a packet of any width or shape, with no Ehrenfest proviso to meet. So ⟨v⟩ turns about B at ωc = |q|B/m, with no ħ, clockwise seen from the tip of B for a positive charge, and with E ⊥ B the centre drifts at E × B/B². The measured check: fc = eB/2πm is 27.99 GHz T⁻¹ for the electron and 15.25 MHz T⁻¹ for the proton, and their ratio, 1836.15, is the measured mₚ/mₑ. Penning traps rely on this exactness, since their quadrupole E is linear in r, and frequency ratios taken in one trap field underlie the most precise masses of both particles: the proton's cyclotron frequency against a carbon ion's, and the electron's spin precession against the cyclotron frequency of the ion that binds it. In a non-uniform field ⟨B(r)⟩ ≠ B(⟨r⟩), and the orbit is classical only while the packet is narrow compared with the distance over which B changes. Relativistic corrections, of relative order (v/c)², α² for an atomic electron, are quoted later from the Dirac equation.
Velocities that do not commute, and what minimal leaves out
Canonical components commute, [pₓ, py] = 0; mechanical ones do not: [πₓ, πy] = −q[pₓ, Ay] − q[Aₓ, py] = iħq(∂ₓAy − ∂yAₓ) = iħqBz, so [vₓ, vy] = iħqBz/m², gauge-invariant as it must be. Robertson's inequality gives ΔvₓΔvy ≥ ħ|q|B/2m², 1.02 × 10⁷ m² s⁻² for an electron in 1.00 T, and since ⟨vₓ²⟩ + ⟨vy²⟩ ≥ 2ΔvₓΔvy, the mean transverse kinetic energy is at least ħωc/2, which is 57.9 μeV at 1.00 T. The bound is derived here; that the lowest Landau level reaches it is shown in the topic that closes this section. The same commutator, written π × π = iħqB, settles spin. Minimal coupling gives spin no coupling at all, but writing the kinetic energy as (σ⋅π)²/2m, identical when A = 0, gives π²/2m − (qħ/2m)σ⋅B: a moment with g = 2. That form is the Dirac equation's non-relativistic limit, quoted. The measured gₑ = 2.00231930 exceeds 2 by 0.0023193, and QED's leading term α/π = 0.0023228 accounts for it to 0.15%, the rest arriving at order α²; both are quoted.
Change one variable at a time
Make the relationship visible.
Set λ = 1 and step ωc t: the tip of p slides along a flat line, since A = (0, Bx, 0) leaves H free of y and py is conserved; λ = 0 conserves pₓ, λ = 0.5 only the canonical Lz. Then move λ alone: mv never changes while p swings, the mark of a gauge-dependent quantity.
CANONICAL pₓ-0.50 mv
CANONICAL py-0.18 mv
|qA| AT THE PARTICLE0.68 mv
CANONICAL Lz ABOUT O-0.25 mvR
Live interpretationCANONICAL pₓ: −0.50 mv. CANONICAL py: −0.18 mv. |qA| AT THE PARTICLE: 0.68 mv. CANONICAL Lz ABOUT O: −0.25 mvR
Catch the common trap
Explain before calculating.
A wave packet of charge q moves in uniform, static fields E and B under H = (p − qA)²/2m + qφ. Which statement about the Heisenberg equations for its centre is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA proton's wave packet is on average at rest, ⟨v⟩ = 0, centred at (x, y) = (3.00 μm, −2.00 μm) in a uniform field of 0.200 T along z. Find ⟨p⟩ in three gauges for this field: symmetric A = (B/2)(−y, x, 0), and the Landau gauges A = (0, Bx, 0) and A = (−By, 0, 0). What do the answers say about the proton's motion?
- Each gauge gives the same field: Bz = ∂ₓAy − ∂yAₓ is B/2 + B/2, B − 0 and 0 + B. They differ by gradients, for instance (0, Bx, 0) − (B/2)(−y, x, 0) = ∇(Bxy/2).
- Canonical and mechanical momenta differ by qA: ⟨p⟩ = m⟨v⟩ + q⟨A(r)⟩. Each A is linear in r, so ⟨A(r)⟩ = A(⟨r⟩) exactly, and with ⟨v⟩ = 0, ⟨p⟩ = qA(3.00 μm, −2.00 μm). Here qB = 1.602 × 10⁻¹⁹ C × 0.200 T = 3.20 × 10⁻²⁰ kg s⁻¹.
- Symmetric gauge: ⟨p⟩ = (qB/2)(−y, x) = 1.60 × 10⁻²⁰ kg s⁻¹ × (2.00 μm, 3.00 μm) = (3.20 × 10⁻²⁶, 4.81 × 10⁻²⁶) kg m s⁻¹.
- Landau A = (0, Bx, 0): ⟨p⟩ = (0, qBx) = (0, 9.61 × 10⁻²⁶) kg m s⁻¹. Landau A = (−By, 0, 0): ⟨p⟩ = (−qBy, 0) = (6.41 × 10⁻²⁶, 0) kg m s⁻¹. Units: C × T × m = kg m s⁻¹, a momentum, as it must be.
- Read as mv, 9.61 × 10⁻²⁶ kg m s⁻¹ would mean 57.5 m s⁻¹ for a proton, yet m⟨v⟩ = 0 in every gauge. The canonical momentum records where the proton sits relative to the gauge, not how it moves.
Answer⟨p⟩ = (3.20, 4.81) × 10⁻²⁶ kg m s⁻¹ in the symmetric gauge, (0, 9.61 × 10⁻²⁶) and (6.41 × 10⁻²⁶, 0) kg m s⁻¹ in the Landau gauges, while m⟨v⟩ = 0 in all three.
MediumAn electron is released with ⟨v⟩ = 0 and ⟨x⟩ = 0 at t = 0 in a uniform field E = 800 V m⁻¹ along +x, with no magnetic field. Describe the field once by φ = −Ex, A = 0, and once by φ = 0, A = −Et x̂. Use the Heisenberg equations to find, at t = 1.50 ns, ⟨vₓ⟩, ⟨pₓ⟩ and the change in ⟨H⟩ in each description.
- Both describe one field: E = −∇φ − ∂A/∂t gives +E x̂ either way, and ∇ × A = 0. With q = −e, the first gives H₁ = p²/2m + eEx; in the second qAₓ = eEt, so H₂ = (pₓ − eEt)²/2m + (py² + pz²)/2m.
- Description 1: dpₓ/dt = (i/ħ)[H₁, pₓ] = −eE, and v = p/m because A = 0. So ⟨pₓ⟩ = −eEt and ⟨vₓ⟩ = −eEt/m.
- Description 2: H₂ contains no x, so [H₂, pₓ] = 0 and ⟨pₓ⟩ keeps its value at t = 0, when A = 0 and ⟨pₓ⟩ = m⟨vₓ⟩ = 0. Hence ⟨pₓ⟩ = 0 for all t, while ⟨vₓ⟩ = (⟨pₓ⟩ − eEt)/m = −eEt/m, the same as before.
- Numbers: eEt = 1.602 × 10⁻¹⁹ C × 800 V m⁻¹ × 1.50 × 10⁻⁹ s = 1.92 × 10⁻²⁵ kg m s⁻¹, so ⟨vₓ⟩ = −1.92 × 10⁻²⁵ ÷ 9.109 × 10⁻³¹ = −2.11 × 10⁵ m s⁻¹. ⟨pₓ⟩ is −1.92 × 10⁻²⁵ kg m s⁻¹ in description 1 and 0 in description 2.
- Energy: dπ/dt = qE is a number, so the spread of v never changes and the kinetic energy rises by exactly (eEt)²/2m = 2.03 × 10⁻²⁰ J = 0.127 eV. In description 1 the term eE⟨x⟩, with ⟨x⟩ = −eEt²/2m = −0.158 mm, falls by the same 0.127 eV, so ⟨H₁⟩ is conserved: H₁ has no explicit t. In description 2, d⟨H₂⟩/dt = ⟨∂H₂/∂t⟩ ≠ 0 and ⟨H₂⟩ rises by 0.127 eV.
Answer⟨vₓ⟩ = −2.11 × 10⁵ m s⁻¹ in both. ⟨pₓ⟩ = −1.92 × 10⁻²⁵ kg m s⁻¹ with the scalar potential but stays 0 with the vector potential, and ⟨H⟩ is unchanged in the first description but rises by 0.127 eV in the second.
HardA problem supplies A = (B/2)(−y, x, 0) + κ(x, y, 0) for an electron, with B = 0.20 T and κ = 0.40 mT. Confirm the magnetic field, expand (p − qA)² correctly, then suppose a student drops the ordering term and uses Hₛ = p²/2m − (q/m)A⋅p + q²A²/2m. Show that the total probability N = ⟨ψ|ψ⟩ then changes exponentially, and find how long it takes to grow by 1.0 %.
- κ(x, y, 0) = ∇χ with χ = κ(x² + y²)/2, and a gradient has no curl, so ∇ × A = B ẑ with B = 0.20 T, as for the symmetric gauge alone. But ∇⋅A = 0 + 2κ = 0.80 mT, so this is not a Coulomb gauge.
- With p⋅A = A⋅p − iħ(∇⋅A): (p − qA)² = p² − 2qA⋅p + iħq(∇⋅A) + q²A² = p² − 2qA⋅p + 2iħqκ + q²A². Divided by 2m, the correct H contains the constant iħqκ/m, so Hₛ = H − iħqκ/m.
- The norm obeys dN/dt = (1/iħ)⟨ψ|(Hₛ − Hₛ†)|ψ⟩. H is Hermitian, so Hₛ − Hₛ† = −2iħqκ/m, and dN/dt = −(2qκ/m)N, giving N(t) = N(0)e(−2qκt/m).
- For the electron q = −e, so N grows at Γ = 2eκ/mₑ = 2 × 1.602 × 10⁻¹⁹ C × 4.0 × 10⁻⁴ T ÷ 9.109 × 10⁻³¹ kg = 1.41 × 10⁸ s⁻¹. Units: C T kg⁻¹ = s⁻¹, since 1 T = 1 kg C⁻¹ s⁻¹.
- N reaches 1.010 N(0) after t = ln(1.010)/Γ = 0.00995 ÷ 1.41 × 10⁸ s⁻¹ = 7.1 × 10⁻¹¹ s. The rate depends on κ, a gauge choice that alters no field, and not on B at all: a prediction set by an unobservable choice condemns Hₛ. As κ → 0 the gauge is Coulomb and the growth vanishes, and the correct H conserves N in every gauge.
AnswerThe field is still 0.20 T along z, but ∇⋅A = 0.80 mT. The dropped term is iħqκ/m; without it N grows as e(2eκt/mₑ) at 1.41 × 10⁸ s⁻¹, reaching 1.010 N(0) after about 71 ps.