University Quantum Mechanics II · Spin and Charge in Electromagnetic Fields · 3.7
Gauge transformations and the phase of the wavefunction
Minimal coupling put A and φ into the Hamiltonian, and neither is unique. This lesson works out what ψ must do when they are changed, sorts what you calculate into gauge-dependent bookkeeping and gauge-invariant prediction, and names the one assumption, a single-valued Λ, that Unit 14 breaks.
Build the model
Connect the measurement to the mechanism.
Minimal coupling wrote the Hamiltonian in terms of A and φ, and those are not unique: any smooth, single-valued Λ(r, t) turns them into A + ∇Λ and φ − ∂Λ/∂t without touching E or B. A classical particle, pushed only by E and B, never notices. A wavefunction has to, because its Hamiltonian contains the potentials themselves. The resolution derived here is that ψ changes as well, by the local phase factor e(iqΛ/ħ), so that potentials and wavefunction transform together as one symmetry.
The phase drops out of |ψ|², but not out of everything: it shifts the canonical momentum −iħ∇ by q∇Λ. The wavelength of ψ, its phase at a point and ⟨p⟩ are therefore conventions, while the mechanical momentum p − qA, the velocity, and the probability current, once it carries its −(q/m)A|ψ|² term, come out the same in every gauge. A time-dependent Λ also moves the zero of energy, which is why only energy differences are measured. The law is exact: no expansion in α or v/c enters, and the Dirac equation, quoted from relativistic quantum mechanics, keeps it unchanged.
What it needs is a single-valued Λ. In a field-free region that encloses magnetic flux, the Λ that would remove A must wind by the flux on every circuit, and unless that flux is a whole multiple of h/|q| the potential cannot be transformed away, even where no force acts.
- Simple definition
- A gauge transformation replaces A by A + ∇Λ and φ by φ − ∂Λ/∂t for a smooth, single-valued Λ(r, t), and multiplies ψ by e(iqΛ/ħ); the new potentials and wavefunction describe exactly the same physics as the old.
- Example
- For an electron, Λ = A₀x with A₀ = 1.0 × 10⁻⁵ T m multiplies ψ by e(−ieA₀x/ħ), a phase that turns by eA₀/ħ = 1.52 × 10¹⁰ rad per metre, 15.2 rad per nanometre, while |ψ|², the velocity and the kinetic energy stay exactly as they were.
Derived: B′ = B because ∇ × ∇Λ = 0, and E′ = E because the mixed partial derivatives of Λ commute. The Lorentz force never sees Λ; the Hamiltonian does.
Λ(r, t) in Wb = V s = T m²; A in T m and φ in V; Λ smooth and single-valued
Derived: (p − qA′)ψ′ = e(iqΛ/ħ)(p − qA)ψ, and the same holds for iħ∂/∂t − qφ′, so ψ′ solves the Schrödinger equation with the new potentials.
Λ = Λ(r, t) in Wb, q in C, ħ in J s, so qΛ/ħ is a pure number; for an electron q = −e, and h/e = 4.14 × 10⁻¹⁵ Wb turns it by 2π
Derived. Λ = −φ₀t raises every level by qφ₀, Quantum Mechanics I's statement that the zero of potential energy only changes how fast the overall phase turns.
H in J; q ∂Λ/∂t in C × V = J. For a static Λ the spectra of H′ and H are identical
Derived. ⟨A⟩ shifts by the same ⟨∇Λ⟩, so the mechanical momentum ⟨p − qA⟩ = m⟨v⟩ is invariant: it, not the ħk read off the ripples of ψ, gives the velocity.
p = −iħ∇ in kg m s⁻¹; q∇Λ in C × T m = kg m s⁻¹
Derived from continuity with p − qA in place of p. The first term alone gains (q/m)∇Λ|ψ|² in a gauge change; the A term removes it, so only the sum is observable.
j in m⁻² s⁻¹ for |ψ|² in m⁻³, m in kg; with ψ = |ψ|e(iθ), j = |ψ|²(ħ∇θ − qA)/m
Derived. Around enclosed flux Φ the integral gains Φ per circuit and the phase qΦ/ħ, so unless Φ = nh/|q| the potential cannot be gauged away. Unit 14 takes up the consequence.
Λ in Wb; needs a simply connected region with B = 0, where the integral is path-independent; ψ₀ solves the problem with A = 0
Two sets of potentials, one set of fields
Classically the potentials are scaffolding. For any smooth, single-valued Λ(r, t), replacing A by A + ∇Λ and φ by φ − ∂Λ/∂t leaves the fields alone: B′ = ∇ × A + ∇ × ∇Λ = B because the curl of a gradient vanishes, and E′ = −∇φ + ∇(∂Λ/∂t) − ∂A/∂t − ∂(∇Λ)/∂t = E because the mixed partial derivatives of Λ commute. Λ is measured in webers, V s = T m², so that ∇Λ is a vector potential in T m and ∂Λ/∂t a potential in V. Three gauge functions recur in this unit. Λ = −φ₀t adds the constant φ₀ to φ; Λ = A₀⋅r adds the constant vector A₀ to A; and Λ = Bxy/2 turns the symmetric-gauge potential (B/2)(−y, x, 0) of a uniform field along z into the Landau-gauge potential (0, Bx, 0). The Lorentz force contains only E and B, so a classical trajectory never notices the choice. The minimal-coupling Hamiltonian postulated in the previous lesson contains A and φ themselves, which is why the question of what ψ does is not idle.
Passing p − qA through a phase
Try ψ′ = e(iqΛ/ħ)ψ and push each operator of the Schrödinger equation through the exponential. The gradient acts on the phase as well as on ψ: −iħ∇(e(iqΛ/ħ)ψ) = e(iqΛ/ħ)(−iħ∇ + q∇Λ)ψ. Subtracting qA′ = qA + q∇Λ cancels the extra term exactly, so (p − qA′)ψ′ = e(iqΛ/ħ)(p − qA)ψ: the combination p − qA transforms like ψ itself, and applying it twice carries the kinetic term through. The time side works the same way: iħ∂ψ′/∂t = e(iqΛ/ħ)(iħ∂/∂t − q∂Λ/∂t)ψ, and subtracting qφ′ = qφ − q∂Λ/∂t removes the extra piece. Every term of the new equation is the old one times the same phase, so ψ′ solves the problem with the primed potentials whenever ψ solves it with the old ones. The sign is fixed too: with A′ = A + ∇Λ the phase is +qΛ/ħ, which for an electron is −eΛ/ħ. It is dimensionless, coulombs times volt seconds over joule seconds, and it vanishes as q → 0, so a neutron's ψ is untouched. The spin term −μ⋅B of this unit contains B alone, so both components of a spinor take the same phase. No approximation entered, so the law holds at every order in α.
A time-dependent Λ moves the zero of energy
When Λ depends on time the Hamiltonian changes by more than a unitary rotation: with U = e(iqΛ/ħ), the same calculation gives H′ = UHU† − q∂Λ/∂t. For a static Λ the spectra of H′ and H coincide. The simplest time-dependent case is a fact from Quantum Mechanics I: Λ = −φ₀t raises φ by φ₀ everywhere and multiplies ψ by e(−iqφ₀t/ħ), so every level rises by qφ₀ and only differences can be measured. A uniform field E₀ along x shows the full effect. Describe it by φ = −E₀x and A = 0; the gauge function Λ = −E₀xt turns this into φ′ = 0 and A′ = −E₀t x̂. The second Hamiltonian, (p + qE₀t x̂)²/(2m), does not contain x, so the Heisenberg equation gives dp/dt = 0: the canonical momentum is conserved in one gauge and grows at qE₀ in the other. The mechanical momentum p − qA grows at qE₀ in both. For an electron in E₀ = 1.0 × 10³ V m⁻¹ it changes by 1.60 × 10⁻²⁵ kg m s⁻¹ in 1.0 ns, a speed change of 1.76 × 10⁵ m s⁻¹, whichever gauge keeps the books.
Invariant predictions, gauge-dependent bookkeeping
Sort every quantity by whether it survives the phase. |ψ′|² = |ψ|² because |e(iqΛ/ħ)| = 1, so every position probability is invariant. The canonical momentum is not: U†pU = p + q∇Λ, so ⟨p⟩′ = ⟨ψ′|p|ψ′⟩ = ⟨p⟩ + q⟨∇Λ⟩, whereas the mechanical momentum p − qA keeps its expectation value because A shifts by exactly the same ∇Λ. The velocity is ⟨p − qA⟩/m, not ⟨p⟩/m, and the difference need not be small. A 100 eV electron has mechanical momentum 5.40 × 10⁻²⁴ kg m s⁻¹, while an electron 1.0 cm from the axis of a 1.0 T field in the symmetric gauge has |qA| = eBr/2 = 8.0 × 10⁻²² kg m s⁻¹, about 150 times larger. Move the axis about which the symmetric gauge is written, itself a gauge change by a constant vector, and that number changes while nothing physical does. So the local wavelength of ψ, its phase at a point and the value of ⟨p⟩ are bookkeeping. Operators built from r, p − qA, E and B, such as position, velocity, kinetic energy and orbital magnetic moment, carry the predictions, and their expectation values agree in every gauge.
The current needs its −(q/m)A|ψ|² term
Repeating the continuity argument of Quantum Mechanics I with p − qA in place of p gives the probability current j = Re[ψ*(p − qA)ψ]/m = (ħ/m) Im(ψ*∇ψ) − (q/m)A|ψ|², in m⁻² s⁻¹ for a three-dimensional ψ. The second term is not optional. Under a gauge change Im(ψ′*∇ψ′) = Im(ψ*∇ψ) + (q/ħ)|ψ|²∇Λ, so the first term alone gains (q/m)|ψ|²∇Λ, and the A term, now built from A + ∇Λ, subtracts exactly that. With ψ = |ψ|e(iθ) the structure is plain: j = |ψ|²(ħ∇θ − qA)/m, a density times a velocity, in which the phase gradient and the vector potential appear only in the invariant combination. For the 100 eV beam of the easy example below, described with a constant A = 1.0 × 10⁻⁴ T m along the beam, the canonical part alone flows backwards at −1.17 × 10⁷ m s⁻¹ times |ψ|², the A term adds +1.76 × 10⁷ m s⁻¹, and the sum is the physical 5.93 × 10⁶ m s⁻¹. Only the corrected current obeys ∂|ψ|²/∂t + ∇⋅j = 0 when A is present, and only it can be compared with a count of particles crossing a surface.
Where Λ must be single-valued
Every step assumed a smooth, single-valued Λ, so that ψ′ is single-valued whenever ψ is. Within that assumption a simply connected region with B = 0 carries no information in A: the line integral of A from a fixed point r₀ to r is then independent of path, it defines Λ with A = ∇Λ, and ψ = e(iqΛ/ħ)ψ₀, where ψ₀ is the solution with no potential at all. Outside a long solenoid of flux Φ the field also vanishes, but A = Φ/(2πr) along the azimuthal direction, the gradient of Φ times the angle around the solenoid over 2π, and an angle is not single-valued. Stokes' theorem makes the obstruction exact: ∮A⋅dl = Φ around any loop enclosing the solenoid, while every single-valued gradient has zero circulation. The would-be phase jumps by qΦ/ħ per circuit, harmless only when Φ is a whole multiple of h/|q|, 4.14 × 10⁻¹⁵ Wb for an electron. Superconducting rings, whose carriers have charge −2e, trap flux in steps of h/2e = 2.07 × 10⁻¹⁵ Wb, as measured in 1961. For any other flux A cannot be removed where no force acts, and Unit 14 shows that interference detects it.
Change one variable at a time
Make the relationship visible.
Raise the gauge amplitude from 0: the ripples of Re ψ′ bunch and stretch and the dashed canonical current swells, but ±|ψ| and the solid current never move. Then set k₀ = 1, κ = 2 and the amplitude to 1.5: the canonical part runs backwards on the flanks while the electron still moves forward.
CANONICAL ⟨p⟩/ħ2.12 nm⁻¹
MECHANICAL ⟨p − qA⟩/ħ2.00 nm⁻¹
ELECTRON SPEED ⟨v⟩232 km s⁻¹
PEAK |A′|, ELECTRON0.987 μT m
Live interpretationCANONICAL ⟨p⟩/ħ: 2.12 nm⁻¹. MECHANICAL ⟨p − qA⟩/ħ: 2.00 nm⁻¹. ELECTRON SPEED ⟨v⟩: 232 km s⁻¹. PEAK |A′|, ELECTRON: 0.987 μT m
Catch the common trap
Explain before calculating.
A field-free region is described with A = 0 and φ = 0, and an electron there has wavefunction ψ. The region is re-described with the gauge function Λ = A₀x, where A₀ is a positive constant. Which statement about the new description is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA 100 eV electron beam travels along +x through a field-free region described with A = 0 and φ = 0, so ψ = e(ik₀x). Re-describe the region with the gauge function Λ = A₀x, A₀ = 1.0 × 10⁻⁴ T m. Find the new potentials, the new wavefunction and its wavenumber, and show that the mechanical momentum and kinetic energy are unchanged.
- Original wavenumber: p₀ = √(2mE) = √(2 × 9.109 × 10⁻³¹ kg × 1.602 × 10⁻¹⁷ J) = 5.40 × 10⁻²⁴ kg m s⁻¹, so k₀ = p₀/ħ = 5.12 × 10¹⁰ m⁻¹ and v = p₀/m = 5.93 × 10⁶ m s⁻¹.
- New potentials: A′ = ∇Λ = A₀ x̂ = 1.0 × 10⁻⁴ T m along x, and φ′ = −∂Λ/∂t = 0. The curl of a constant vector is zero, so the region is still field-free.
- New wavefunction, with q = −e: ψ′ = e(−ieA₀x/ħ) e(ik₀x) = e(ik′x), where eA₀/ħ = 1.602 × 10⁻¹⁹ × 1.0 × 10⁻⁴/1.055 × 10⁻³⁴ = 1.52 × 10¹¹ m⁻¹, so k′ = 5.12 × 10¹⁰ − 1.52 × 10¹¹ = −1.01 × 10¹¹ m⁻¹. The canonical momentum ħk′ = −1.06 × 10⁻²³ kg m s⁻¹ now points backwards.
- Mechanical momentum: ħk′ − qA′ = ħk′ + eA₀ = −1.06 × 10⁻²³ + 1.60 × 10⁻²³ = 5.40 × 10⁻²⁴ kg m s⁻¹, exactly ħk₀. The kinetic energy (p − qA′)²/(2m) is therefore still 100 eV, and |ψ′|² = |ψ|² = 1.
- Check by limits: as A₀ → 0 the extra phase vanishes, and for a neutral particle it would vanish for any A₀. Dimensionally eA₀/ħ is C × T m/(J s) = m⁻¹, a wavenumber as it must be.
AnswerA′ = 1.0 × 10⁻⁴ T m along x with B = 0; ψ′ = e(ik′x) with k′ = −1.01 × 10¹¹ m⁻¹, so the canonical momentum reverses, while p − qA′ = 5.40 × 10⁻²⁴ kg m s⁻¹, v = 5.93 × 10⁶ m s⁻¹ and 100 eV are unchanged.
MediumA uniform field B = 2.0 T along z can be described by the symmetric-gauge potential A = (B/2)(−y, x, 0) or by the Landau-gauge potential A′ = (0, Bx, 0). Find the gauge function Λ that takes the first to the second, the phase it gives an electron's wavefunction at x = y = 40 nm, and the change in ⟨p⟩ for an electron packet centred at x = 0.50 μm, y = 0.
- Same field: for A the z-component of the curl is ∂(Bx/2)/∂x − ∂(−By/2)/∂y = B/2 + B/2 = B, and for A′ it is ∂(Bx)/∂x − 0 = B. The difference of the two potentials must therefore be a gradient.
- A′ − A = (By/2, Bx/2, 0), which is ∇(Bxy/2). So Λ = Bxy/2, up to an irrelevant constant, in T m² = Wb as a gauge function must be.
- Phase, with q = −e: qΛ/ħ = −eBxy/(2ħ). With e/ħ = 1.519 × 10¹⁵ Wb⁻¹, B = 2.0 T and xy = (40 × 10⁻⁹ m)² = 1.6 × 10⁻¹⁵ m², it is −1.519 × 10¹⁵ × 2.0 × 1.6 × 10⁻¹⁵/2 = −2.43 rad. In terms of the magnetic length ℓ = √(ħ/eB) = 18.1 nm it is −xy/(2ℓ²), of order one across a patch a magnetic length wide.
- Canonical momentum: ⟨p⟩′ = ⟨p⟩ + q⟨∇Λ⟩ = ⟨p⟩ − (eB/2)(⟨y⟩, ⟨x⟩, 0). Because ∇Λ is linear in position, its average is its value at the packet's centre, whatever the packet's shape. With ⟨y⟩ = 0 the x-component is unchanged, and the y-component changes by −eB⟨x⟩/2 = −1.602 × 10⁻¹⁹ × 2.0 × 0.50 × 10⁻⁶/2 = −8.0 × 10⁻²⁶ kg m s⁻¹.
- Read as mv, that shift would be a velocity of 8.0 × 10⁻²⁶/9.11 × 10⁻³¹ = 8.8 × 10⁴ m s⁻¹ along −y, produced by relabelling alone. The mechanical momentum ⟨p + eA⟩ changes by −eB⟨x⟩/2 + e(B⟨x⟩ − B⟨x⟩/2) = 0 along y, and |ψ|² is untouched.
AnswerΛ = Bxy/2; the electron's phase at (40 nm, 40 nm) changes by −2.43 rad; ⟨p⟩ changes by −8.0 × 10⁻²⁶ kg m s⁻¹ along y and not at all along x, while the mechanical momentum and |ψ|² are unchanged.
HardDeep inside a thick superconducting ring the paired electrons share one wavefunction ψ = √ρ e(iθ) with carrier charge q = −2e, a model quoted from the theory of superconductivity, and no current flows in the bulk. From the gauge-invariant current and the single-valuedness of ψ, find the fluxes Φ a loop deep in the metal can enclose, evaluate the step against experiment, and find the field that threads one step through a hole of radius 5.0 μm.
- In polar form the gauge-invariant current is j = ρ(ħ∇θ − qA)/m*, with m* the carrier mass, which drops out. In the bulk j = 0, so ħ∇θ = qA at every point of a loop C lying there. Neither term is gauge-invariant on its own; their difference is, which is why the condition means something.
- Integrate around C and use Stokes' theorem: ħ∮∇θ⋅dl = q∮A⋅dl = qΦ, where Φ is the flux through C.
- ψ is single-valued, so θ can change only by a whole multiple of 2π around C: ∮∇θ⋅dl = 2πn. Hence qΦ = 2πnħ = nh, and |Φ| = |n|h/|q| for integer n. This is the winding gauge function made concrete: it multiplies ψ by e(iqΦ/ħ) per circuit, which is 1 only for these fluxes. A loop that can shrink to a point inside the metal has n = 0, so solid metal holds Φ = 0, the Meissner limit; only the hole allows n ≠ 0.
- With |q| = 2e: h/2e = 6.626 × 10⁻³⁴ J s/(2 × 1.602 × 10⁻¹⁹ C) = 2.07 × 10⁻¹⁵ Wb, and J s C⁻¹ = V s = Wb confirms the unit. London had predicted h/e = 4.14 × 10⁻¹⁵ Wb for single-electron carriers; the 1961 measurements of Deaver and Fairbank and of Doll and Näbauer found steps near h/2e, evidence that the carriers are pairs.
- One step through the hole: B = Φ/πr² = 2.07 × 10⁻¹⁵ Wb/[π(5.0 × 10⁻⁶ m)²] = 2.07 × 10⁻¹⁵/7.85 × 10⁻¹¹ = 2.6 × 10⁻⁵ T, comparable to the Earth's field, so a single step is easily swamped unless the ring is small and well shielded.
AnswerΦ = nh/2e, steps of 2.07 × 10⁻¹⁵ Wb, as measured in 1961; single-electron carriers would give h/e = 4.14 × 10⁻¹⁵ Wb. One step through a 5.0 μm hole needs B ≈ 2.6 × 10⁻⁵ T.