University Quantum Field Theory I · Covariant Quantisation of the Electromagnetic Field · 8.3

The Gupta-Bleuler condition and the physical subspace

Feynman gauge left the photon with four polarisations, one of them of negative norm. This topic decides which states may describe the world, and shows that whatever they hold beyond two transverse photons is invisible to every gauge-invariant quantity of the free field, and stays so while charge is conserved; the all-orders proof waits for Unit 12.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Topic 8.02 quantised all four components of A^μ in Feynman gauge and paid for covariance with timelike photons of norm −1. Something must now pick out the physical states, and the obvious candidate, the Lorenz condition ∂⋅A = 0 that the gauge-fixing term imitates, fails: π⁰ = −∂⋅A, so the canonical commutator gives [A₀, ∂⋅A] = −iδ³, and even as a condition on states it rejects the vacuum. Gupta and Bleuler keep half of it, ∂⋅A⁽⁺⁾|ψ⟩ = 0, which still gives ⟨ψ|∂⋅A|ψ⟩ = 0. For a momentum along z it reads [a(k, 0) − a(k, 3)]|ψ⟩ = 0, derived.

The combination b = [a(0) − a(3)]/√2 commutes with its own adjoint, so the condition admits timelike and longitudinal quanta only as b† quanta, of zero norm, and these add nothing to ⟨H⟩, ⟨P⟩ or any expectation built from F_μν, while ⟨A_μ⟩ moves by a gauge term ∂_μΛ, all shown. States differing by them are physically identical, and the classes they form are UAQM-11.04's transverse photons: two per momentum, positive norm. A conserved current keeps □(∂⋅A) = 0, so the condition holds at all times, shown. The model stops in two places.

It needs ∂⋅j = 0 exactly: a source that fails to conserve charge pushes states out of the physical subspace. And that interactions never leak probability into null states, to all orders, is proved only in Unit 12.

Simple definition: The Gupta-Bleuler condition admits a state of the covariantly quantised photon field only if the positive-frequency part of ∂⋅A annihilates it, which allows timelike and longitudinal photons only in the zero-norm combination a(0)† − a(3)†.
Example: For k along z with |k| = 1.00 eV, [a(2)† + 2(a(0)† − a(3)†)]|0⟩ is physical: a(0) − a(3) sends the pair to 2(−1 + 1)|0⟩ = 0, and its norm 1 + 4(−1 + 1) = 1 and energy 1.00 eV are those of a(2)†|0⟩ alone, while a(3)†|0⟩, of norm +1, fails.

The Lorenz condition fails as an operator[A₀(t, x), ∂⋅A(t, y)] = −i δ³(x − y)

Derived from [A₀, π⁰] = iδ³. An operator with a non-zero commutator is not zero, and ∂⋅A = 0 would put back π⁰ = 0, topic 8.01's defect.

Feynman gauge, π⁰ = −∂⋅A; ħ = c = 1, A₀ in GeV, ∂⋅A in GeV², δ³ in GeV³, so both sides carry GeV³

The Gupta-Bleuler condition for one momentum[a(k, 0) − a(k, 3)]|ψ⟩ = 0, with k^μ = |k|(1, 0, 0, 1)

a(k, λ) in GeV⁻¹ removes a photon of polarisation ε(λ); ε(0) = (1, 0, 0, 0), ε(3) = (0, 0, 0, 1), so a(k, 3) = −a₃(k) of 8.02

The null combination[b, b†] = 0, with b = [a(0) − a(3)]/√2

Derived: (−1 + 1)/2 = 0. As b commutes with b†, b|ψ⟩ = 0 survives any number of b† quanta, each adding zero norm, while a lone a(0)† or a(3)† carries c† and fails.

one box mode, [a(0), a(0)†] = −1 and [a(3), a(3)†] = +1; b, b† dimensionless; the partner c = [a(0) + a(3)]/√2 has [b, c†] = −1

One mode's energy in light-cone formHₖ = |k|[a(1)†a(1) + a(2)†a(2) − b†c − c†b]

Derived from 8.02's Hamiltonian. On a physical state b kills the ket and b† the bra, so ⟨H⟩ is the transverse photons' energy and nothing else.

Hₖ and |k| in GeV; a(1), a(2) transverse; the last two terms are |k|[a(3)†a(3) − a(0)†a(0)] rewritten

Zero-norm quanta move A by a gauge termδ⟨A^μ(x)⟩ = ∂^μΛ(x), with □Λ = 0

Shown: the shift lies along k^μ, so it cancels in ⟨F_μν⟩ and keeps ∂⋅⟨A⟩ = 0. The potentials differ; every field strength agrees.

Λ real, mass dimension 0; for a coherent admixture e^{(w(a(0)† − a(3)†))}, Λ = (2Nₖ/|k|) Im[w e^(−ik⋅x)], Nₖ = (2|k|V)^(−1/2)

Why the condition lasts□(∂⋅A) = ∂_μ j^μ, which vanishes for a conserved current

Derived by taking ∂_ν of the field equation. ∂⋅A stays a free massless field, so its positive-frequency part, and the physical subspace with it, never changes.

□ = ∂ₜ² − ∇² in GeV², ∂⋅A in GeV², j^μ in GeV³; from ∂_μF^μν + ∂^ν(∂⋅A) = j^ν, Feynman gauge

The Lorenz condition cannot be an operator equation

Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. Topic 8.02, quoted, added −½(∂⋅A)² to −¼F² and quantised all four components at ξ = 1: the field equation is □A^μ = 0, and per box mode [a(λ), a(λ′)†] = −g^(λλ′), so a timelike photon has norm −1. Since □A^μ = 0 is the Lorenz-gauge wave equation, it is tempting to finish by imposing ∂⋅A = 0. It cannot be done. The momentum of A₀ is now π⁰ = −∂⋅A, so [A₀(t, x), π⁰(t, y)] = iδ³(x − y) becomes [A₀(t, x), ∂⋅A(t, y)] = −iδ³(x − y), derived. An operator with a non-zero commutator is not zero, and setting it to zero would restore π⁰ = 0, the defect of 8.01. As a condition on states, ∂⋅A|ψ⟩ = 0, it fails differently: ∂⋅A|0⟩ = ∂⋅A⁽⁻⁾|0⟩ creates a(0)† − a(3)† quanta, a vector of zero norm but not the zero vector, since a creation operator annihilates no non-zero state. Not even the vacuum would qualify. Dimensions check: A₀ carries GeV and ∂⋅A GeV², so the left side carries GeV³, as δ³ does.

Keep only the annihilating half

Split ∂⋅A = ∂⋅A⁽⁺⁾ + ∂⋅A⁽⁻⁾ into annihilation and creation parts and demand only ∂⋅A⁽⁺⁾|ψ⟩ = 0. As ⟨ψ|∂⋅A⁽⁻⁾ is the adjoint of ∂⋅A⁽⁺⁾|ψ⟩, every allowed state has ⟨ψ|∂⋅A|ψ⟩ = 0: the Lorenz condition survives for means, which is all a classical field needs, and the vacuum passes. With 8.02's expansion A_μ = ∫ d³k/((2π)³2|k|) [a_μ(k)e^(−ik⋅x) + h.c.], the derivative brings down −ik^μ, and distinct k give independent exponentials, so the condition holds momentum by momentum: k^μa_μ(k)|ψ⟩ = 0. For k^μ = |k|(1, 0, 0, 1) that is |k|(a₀ + a₃)|ψ⟩ = 0 in 8.02's lower-index operators. Now name each operator by its polarisation vector, ε(0) = (1, 0, 0, 0) and ε(3) = (0, 0, 0, 1), written with upper indices. Then a(k, 0) = a₀, while a(k, 3), the coefficient in A³ = −A₃, is −a₃. The condition becomes [a(k, 0) − a(k, 3)]|ψ⟩ = 0, derived; the two transverse operators do not appear at all.

The light-cone pair decides who passes

For one mode, [a(0), a(0)†] = −1 and [a(3), a(3)†] = +1. Rotate to b = [a(0) − a(3)]/√2 and c = [a(0) + a(3)]/√2: then [b, b†] = (−1 + 1)/2 = 0, [c, c†] = 0 and [b, c†] = (−1 − 1)/2 = −1, derived. The condition is b|ψ⟩ = 0. Because b commutes with b†, b(b†)ⁿ|φ⟩ = (b†)ⁿb|φ⟩ = 0 for any state |φ⟩ of transverse photons alone, so the physical states are |ψ⟩ = Σₙ(b†)ⁿ|φₙ⟩, with any number of b quanta. A c quantum fails, b c†|φ⟩ = −|φ⟩, and so does a lone timelike or longitudinal photon, because a(0)† = (b† + c†)/√2 and a(3)† = (c† − b†)/√2 both contain c†. So a(3)†|0⟩, norm +1, is excluded as firmly as a(0)†|0⟩, norm −1. For norms, move each b to the right through the b†'s: ⟨ψ|ψ⟩ = ⟨φ₀|φ₀⟩ ≥ 0, set by the part with no b quanta. The physical subspace has no negative norms, derived; what it still has are non-zero states of zero norm, such as b†|0⟩.

What the zero-norm quanta cannot change

One mode's energy is Hₖ = |k|[Σᵢ a(i)†a(i) − a(0)†a(0)], quoted from 8.02, and in the light-cone pair its timelike and longitudinal terms become −|k|(b†c + c†b), derived. On a physical state b annihilates the ket and b† the bra, so ⟨H⟩ is the transverse photons' energy alone; the momentum Pₖ = k̂Hₖ of the mode inherits the same result. For the field strength, the 0–3 part of a box mode's A^μ⁽⁺⁾ is Nₖ[(1, 0, 0, −1)b + (k^μ/|k|)c]e^(−ik⋅x)/√2 with Nₖ = (2|k|V)^(−1/2), and in F_μν the c term carries k_μk_ν − k_νk_μ = 0. So F_μν holds only b, b† and transverse operators, and every expectation built from F equals its value in |φ₀⟩, shown. The potential does move. The coherent admixture e^{(w(a(0)† − a(3)†))} gives a(0) and a(3) the same eigenvalue −w, so ⟨A^μ⟩ gains −2Nₖ Re[w e^(−ik⋅x)]k^μ/|k|, which is ∂^μΛ with Λ = (2Nₖ/|k|) Im[w e^(−ik⋅x)] and □Λ = −k²Λ = 0: a residual gauge transformation, invisible in E and B.

Identify the null states: two photons remain

Within the physical subspace the inner product is positive semi-definite, so the Cauchy-Schwarz inequality holds there, and a null physical state |χ⟩ obeys |⟨ψ|χ⟩|² ≤ ⟨ψ|ψ⟩⟨χ|χ⟩ = 0 for every physical |ψ⟩. Adding it changes no norm, no transition amplitude to a physical state and no expectation built from F. So declare |ψ⟩ and |ψ⟩ + |χ⟩ the same state. Each class holds exactly one member with no b quanta, built from a(1)† and a(2)† alone, so the classes form UAQM-11.04's Fock space of transverse photons, quoted: positive norm, two helicities per momentum. The transverse space is reached as a quotient, not as a subspace; Coulomb gauge simply chose that representative from the start. The count of two meets measurement: with two states per mode Planck's law gives σ = 5.670 × 10⁻⁸ W m⁻² K⁻⁴, which radiometry confirms (quoted in 8.02), and the cosmic background at its measured 2.7255 K holds 410.7 photons per cm³. Four positive-norm polarisations would double both.

A conserved current keeps the condition true

A condition imposed on free fields must survive a source. With −j⋅A added, the Feynman-gauge field equation is ∂_μF^μν + ∂^ν(∂⋅A) = j^ν, which is □A^ν = j^ν. Take ∂_ν of it: ∂_ν∂_μF^μν = 0 by antisymmetry, so □(∂⋅A) = ∂_μj^μ, derived, and a conserved current leaves ∂⋅A a free massless field even while A^μ itself is driven. Its positive-frequency part is then the same operator at every time, and a state that obeys the condition once obeys it always. Dimensions: both sides carry GeV⁴. In modes, a c-number source leaves a(λ) with the eigenvalue i g^(λλ) ε(λ)⋅J, no sum on λ, where J^μ is the current's Fourier amplitude at k: a(0) gets iJ⁰ and a(3) gets iJ³, so a(0) − a(3) acquires i(J⁰ − J³) = i k⋅J/|k|, which conservation sets to zero on the light cone. A source with ∂⋅j ≠ 0 would create c quanta, and the Hard example shows ⟨H⟩ then falling below the vacuum's. The free limit j → 0 is safe trivially. The Dirac current enters in Unit 9, and the all-orders argument in Unit 12.

Change one variable at a time

Make the relationship visible.

Interactive model

Mixing angle θ of a(0)† and a(3)†0 °

Transverse amplitude t0.5

Start at θ = 0°, a lone timelike photon: with t = 0.5 the norm is −0.75. Slide θ to −45°: the dashed curve reaches zero, the pair adds no norm and the norm is t² alone. At +45° the norm is t² again, but r² = 2: zero norm is not the test.

NORM ⟨ψ|ψ⟩ = ⟨H⟩/|k|-0.75

0–3 PAIR SHARE −cos 2θ-1.00

TRANSVERSE SHARE t²0.25

GUPTA-BLEULER RESIDUAL r1.00

Live interpretationNORM ⟨ψ|ψ⟩ = ⟨H⟩/|k|: −0.75. 0–3 PAIR SHARE −cos 2θ: −1.00. TRANSVERSE SHARE t²: 0.25. GUPTA-BLEULER RESIDUAL r: 1.00

Catch the common trap

Explain before calculating.

Misconception check

The usual misreading is that the Gupta-Bleuler condition deletes the timelike and longitudinal photons, leaving transverse ones only. It does not. Timelike and longitudinal quanta stay in physical states, provided they come as a(0)† − a(3)†: [1 + a(0)† − a(3)†]a(1)†|0⟩ is physical, and its norm 1 and energy |k| are those of a(1)†|0⟩. Nor is the longitudinal photon admitted for its positive norm: a(3)†|0⟩ has norm +1 and fails, because a(3)† contains the partner c†, which the condition rejects. The mirror-image slip takes zero norm as the test: (a(0)† + a(3)†)|0⟩ has norm −1 + 1 = 0, yet a(0) − a(3) sends it to −2|0⟩. The transverse Fock space appears only after states that differ by null vectors are identified: it is a quotient of the physical subspace, not a subspace of the four-photon space.

For a momentum along z, |ψ⟩ = [0.8 a(1)† + 0.6 a(0)† + β a(3)†]|0⟩, where a(λ) removes the photon of polarisation vector ε(λ), with [a(0), a(0)†] = −1 and [a(3), a(3)†] = +1. Which value of β makes |ψ⟩ obey the Gupta-Bleuler condition, and what is ⟨ψ|ψ⟩?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA mode has momentum along z with |k| = 2.48 eV. Test a(3)†|0⟩, [a(0)† + a(3)†]|0⟩, [a(0)† − a(3)†]|0⟩ and [a(1)† + 0.5(a(0)† − a(3)†)]|0⟩ against the Gupta-Bleuler condition, give each one's norm and ⟨ψ|H|ψ⟩, and find the wavelength of the photon the last one describes.

Per mode [a(0), a(0)†] = −1 and [a(3), a(3)†] = +1, and the condition is [a(0) − a(3)]|ψ⟩ = 0. Every one-quantum state of the mode obeys H|ψ⟩ = |k||ψ⟩, so ⟨ψ|H|ψ⟩ = |k|⟨ψ|ψ⟩.
a(3)†|0⟩: [a(0) − a(3)]a(3)†|0⟩ = −|0⟩, so it fails, with norm +1 and ⟨ψ|H|ψ⟩ = 2.48 eV. Positive norm does not make it physical.
[a(0)† + a(3)†]|0⟩: the condition gives (−1 − 1)|0⟩ = −2|0⟩, so it fails, although its norm is −1 + 1 = 0 and ⟨ψ|H|ψ⟩ = 0.
[a(0)† − a(3)†]|0⟩: the condition gives (−1 + 1)|0⟩ = 0, so it passes, with norm 0 and ⟨ψ|H|ψ⟩ = 0: a physical state that is invisible.
[a(1)† + 0.5(a(0)† − a(3)†)]|0⟩: a(1)† commutes with a(0) − a(3), so it passes; the cross terms vanish and the pair adds 0.25(−1 + 1) = 0, so the norm is 1 and ⟨ψ|H|ψ⟩ = 2.48 eV, those of a(1)†|0⟩.
Restore units: λ = 2πħc/|k| = 1239.8 eV nm ÷ 2.48 eV = 500 nm, a green photon. ħc divided by an energy is a length, as it must be.

AnswerOnly [a(0)† − a(3)†]|0⟩ (norm 0, ⟨H⟩ = 0) and the transverse mixture (norm 1, ⟨H⟩ = 2.48 eV, λ = 500 nm) are physical; a(3)†|0⟩ (norm +1) and [a(0)† + a(3)†]|0⟩ (norm 0) fail.

MediumOne box mode along z with |k| = 1.50 eV is in |ψ⟩ = N exp[u a(1)† + w(a(0)† − a(3)†)]|0⟩ with u = 2.00 and w = 1.50. Show that |ψ⟩ is physical, find N, ⟨H⟩ and the expectation values ⟨a(0)†a(0)⟩ and ⟨a(3)†a(3)⟩, and show that the w part of ⟨A^μ⟩ is a gauge term that leaves ⟨E_z⟩ = 0.

Commute each annihilator through the exponent: a(1)|ψ⟩ = u|ψ⟩, a(0)|ψ⟩ = w[a(0), a(0)†]|ψ⟩ = −w|ψ⟩, and a(3)|ψ⟩ = −w[a(3), a(3)†]|ψ⟩ = −w|ψ⟩. So [a(0) − a(3)]|ψ⟩ = (−w + w)|ψ⟩ = 0: physical.
Norm: ⟨0|e^{(u a(1) + w(a(0) − a(3)))}e^{(u a(1)† + w(a(0)† − a(3)†))}|0⟩ = exp[u² + w²(−1 + 1)] = exp(4.00). So N = e^(−2.00) = 0.135, whatever w is.
Expectations: ⟨a(1)†a(1)⟩ = u² = 4.00, the mean transverse photon number, and ⟨a(0)†a(0)⟩ = ⟨a(3)†a(3)⟩ = w² = 2.25, which are not photon counts, since the timelike number operator is −a(0)†a(0). With H = |k|[a(1)†a(1) + a(2)†a(2) + a(3)†a(3) − a(0)†a(0)], ⟨H⟩ = 1.50 eV × (4.00 + 2.25 − 2.25) = 6.00 eV, the transverse photons' energy alone.
Potential: ⟨A^μ⟩ = 2Nₖ Re[u e^(−ik⋅x)]ε^μ(1) − 2Nₖ Re[w e^(−ik⋅x)][ε^μ(0) + ε^μ(3)], with Nₖ = (2|k|V)^(−1/2). As ε(0) + ε(3) = k/|k|, the second term is ∂^μΛ with Λ = −(3.00Nₖ/|k|) sin(k⋅x): ∂^μ sin(k⋅x) = k^μ cos(k⋅x). Λ is dimensionless, and □Λ = −k²Λ = 0.
So ⟨A⁰⟩ = ⟨A³⟩ = −3.00Nₖ cos[|k|(t − z)], at 3.00/4.00 = 0.75 of ⟨A¹⟩'s amplitude 4.00Nₖ. Yet ⟨E_z⟩ = −∂_z⟨A⁰⟩ − ∂ₜ⟨A³⟩ = 0, since both are one function of t − z, and ⟨Eₓ⟩, ⟨B_y⟩ come from u alone.

AnswerPhysical; N = e^(−2.00) = 0.135; ⟨H⟩ = 6.00 eV, as ⟨a(0)†a(0)⟩ = ⟨a(3)†a(3)⟩ = 2.25 cancel; ⟨A⁰⟩ = ⟨A³⟩ oscillate at 0.75 of the transverse amplitude as the gauge term ∂^μΛ, and ⟨E_z⟩ = 0.

HardA c-number current, coupled through −j⋅A, leaves one mode along z with |k| = 2.00 eV in |out⟩ = N exp[Σ_λ ζ_λ a(λ)†]|0⟩, ζ_λ = −i ε(λ)⋅J, quoted from the Dyson series for a classical source, with J^μ the current's dimensionless Fourier amplitude. Take J = (0.30, 0.40, 0, 0.30) for source 1 and J = (0.50, 0.40, 0, 0.10) for source 2. Test each |out⟩ and find ⟨H⟩.

With ε(0)⋅J = J⁰ and ε(i)⋅J = −Jⁱ: ζ₀ = −iJ⁰, ζ₁ = iJ¹, ζ₂ = 0 and ζ₃ = iJ³. Commuting through the exponent, a(λ)|out⟩ = −g^(λλ)ζ_λ|out⟩, so a(0)|out⟩ = iJ⁰|out⟩ and a(3)|out⟩ = iJ³|out⟩.
Hence [a(0) − a(3)]|out⟩ = i(J⁰ − J³)|out⟩ = (i k⋅J/|k|)|out⟩. Source 1: J⁰ − J³ = 0, physical. Source 2: 0.40i|out⟩ ≠ 0, so the condition fails; k⋅J = 2.00 eV × 0.40 = 0.80 eV, the on-shell Fourier component of ∂⋅j.
⟨a(λ)†a(λ)⟩ = |ζ_λ|², so ⟨H⟩ = |k|(|J¹|² + |J²|² + |J³|² − |J⁰|²) = −|k|J⋅J for real J. Source 1: 2.00 eV × (0.16 + 0.09 − 0.09) = 0.320 eV. Source 2: 2.00 eV × (0.16 + 0.01 − 0.25) = −0.160 eV.
Source 2's state lies 0.160 eV below the vacuum, impossible for anything physical: its 0–3 part is [(ζ₀ − ζ₃)b† + (ζ₀ + ζ₃)c†]/√2 with ζ₀ + ζ₃ = −0.40i, so the source has made c quanta.
Source 1's 0–3 part is −0.30i[a(0)† − a(3)†], pure b quanta. Drop it to reach the class representative N′exp(0.40i a(1)†)|0⟩: 0.16 transverse photons on average, carrying 0.16 × 2.00 eV = 0.320 eV, the same answer, now with no unphysical photon in sight.
This is □(∂⋅A) = ∂⋅j mode by mode: J⁰ = J³ is charge conservation on the light cone. Dimensions: J is dimensionless and |k| an energy, so ⟨H⟩ is in eV. As J → 0, ⟨H⟩ → 0 and |out⟩ → |0⟩, the free limit.

AnswerSource 1 is physical with ⟨H⟩ = 0.320 eV, carried by 0.16 transverse photons; source 2 gives [a(0) − a(3)]|out⟩ = 0.40i|out⟩ and ⟨H⟩ = −0.160 eV, because it fails to conserve charge at this k.