University Quantum Field Theory I · Natural Units, Relativistic States and the Feynman Propagator · 1.3

Invariant normalisation and the Poincaré group on Fock space

Every rate in this course will be a Lorentz scalar divided by a few factors of 2E, and this is where they are earned. We take Advanced QM's box modes to the continuum, normalise states so every frame agrees on them, and build the operators that boost and translate them, so you can see which parts of a calculation may depend on the frame.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Advanced QM quantised the scalar field in a box, and a box is exactly what a boost changes: it contracts, and its allowed momenta move, so the Kronecker normalisation ⟨k|k′⟩ = δₖₖ′ means something different in every frame. Letting the box grow removes the box but exposes a subtler problem: the continuum δ³(p − q) is not invariant either, because a boost stretches every interval of momentum along its direction by E′/E, while d³p/(2Eₚ) is invariant, being d⁴p δ(p² − m²) θ(p⁰) in disguise. So the operators are rescaled to a(p) = √(2EₚV) aₖ, states are normalised to ⟨p|q⟩ = (2π)³ 2Eₚ δ³(p − q), and every frame computes the same norms.

With that settled, Noether's theorem supplies ten conserved charges, P^μ and M^μν, which become bilinears in a†a and exponentiate to a unitary U(Λ, a) carrying |p⟩ to e^(i(Λp)⋅a)|Λp⟩ with no Jacobian factor: unitarity is the invariance of the normalisation, nothing more. The construction stops in two places. It is for spin zero, and spin needs the little group of the next topic.

And it is for free fields: once an interaction enters, P and J keep their free form while H and the boosts must carry it, because [Kᵢ, Pⱼ] = −iδᵢⱼ H holds only with the full H.

Simple definition: Invariant normalisation is the choice ⟨p|q⟩ = (2π)³ 2Eₚ δ³(p − q) for one-particle states, the one normalisation every inertial frame agrees on, so that a Lorentz transformation acts on states as a unitary operator that simply relabels p as Λp.
Example: A π⁺ at rest in a co-moving box of volume V has ⟨p|p⟩ = 2mV. Seen from a frame where it moves at p = 300 MeV, E = 330.9 MeV = 2.371 times m while the box contracts to V/2.371, so 2E × V/2.371 is again 2mV: one norm in both frames.

Continuum ladder operators[a(p), a†(q)] = (2π)³ 2Eₚ δ³(p − q)

Derived from UAQM-09.02's [aₖ, aₖ′†] = δₖₖ′ as V → ∞, with Vδₖₖ′ → (2π)³δ³(p − q). The √(2Eₚ) is chosen so that a†(p) makes a state every frame normalises alike.

a(p) = √(2EₚV) aₖ has mass dimension −1 (GeV⁻¹); Eₚ = √(p² + m²) in GeV; δ³ in GeV⁻³

Lorentz-invariant measured³p/((2π)³ 2Eₚ) = d⁴p/(2π)³ δ(p² − m²) θ(p⁰)

Derived: d⁴p, p² and the sign of p⁰ are each unchanged by an orthochronous Λ, so the left side is too. Its 1/Eₚ cancels the E′/E by which a boost stretches d³p.

d³p in GeV³ and Eₚ in GeV, so the measure has mass dimension 2; θ(p⁰) keeps the positive-energy sheet

Invariant one-particle normalisation⟨p|q⟩ = (2π)³ 2Eₚ δ³(p − q)

Invariant because 2Eₚδ³(p − q) is the inverse of the invariant measure. With δ³(p − q) alone a boost would rescale every norm by E/E′.

|p⟩ = a†(p)|0⟩, mass dimension −1 (GeV⁻¹); in a box of volume V, ⟨p|p⟩ = 2EₚV

Poincaré transformation of a one-particle stateU(Λ, a)|p⟩ = e^(i(Λp)⋅a) |Λp⟩

Derived from U φ(x) U⁻¹ = φ(Λx + a) by changing variables inside the invariant measure: the boost relabels p, the translation adds a phase, and no norm changes.

U(Λ, a) = e^(iP⋅a) U(Λ); a in GeV⁻¹, with 1 GeV⁻¹ = 0.1973 fm, so (Λp)⋅a is a phase in radians

Boost generator as a Noether chargeK = tP − ∫d³x x T⁰⁰

Derived with UAQM-09.01's theorem. It depends on t explicitly, so it is conserved although [K, H] ≠ 0, and its conservation says the centre of energy moves uniformly at P/H.

Kⁱ = M⁽⁰ⁱ⁾, dimensionless; t and x in GeV⁻¹, P in GeV, energy density T⁰⁰ in GeV⁴

Translations joined to the Lorentz algebra[M^μν, P^ρ] = i(g^νρ P^μ − g^μρ P^ν)

Derived on one-particle wavefunctions and added to UAQM-05.01's quoted Lorentz algebra. Because it holds with the full H, it forces the boosts to carry any interaction.

M^μν dimensionless, P^ρ in GeV; in components [Kᵢ, H] = −iPᵢ, [Kᵢ, Pⱼ] = −iδᵢⱼ H, [Jᵢ, Pⱼ] = iεᵢⱼₖ Pₖ

From box modes to continuum operators

Natural units ħ = c = 1 and the metric (+, −, −, −) throughout, with p⋅x = Eₚ t − p⃗⋅x⃗. Quoted from UAQM-09.02: in a periodic box of volume V = L³, φ(x) = Σₖ (2EₖV)^(−1/2)[aₖ e^(−ik⋅x) + aₖ† e^(ik⋅x)] with [aₖ, aₖ′†] = δₖₖ′. The modes sit 2π/L apart, 62.0 MeV for a 20 fm box, so as L grows Σₖ becomes V∫d³p/(2π)³ and δₖₖ′ becomes (2π)³δ³(p − q)/V, the replacement that keeps Σₖ δₖₖ′ = 1. Rescale a(p) = √(2EₚV) aₖ. The field becomes φ(x) = ∫d³p/((2π)³2Eₚ)[a(p)e^(−ip⋅x) + a†(p)e^(ip⋅x)] and the algebra [a(p), a†(q)] = (2π)³2Eₚδ³(p − q), both derived. Dimensions check: φ has mass dimension 1 and the measure 2, so a(p) has −1, and the commutator's −2 is matched by 2Eₚδ³, of 1 − 3. The √V in the rescaling is forced by the limit; the √(2Eₚ) is a choice, and the next two sections show why it is the right one.

Why d³p/2Eₚ, and not d³p

d³p alone is not invariant. Boost along z with velocity β: p′_z = γ(p_z + βEₚ) with the transverse components fixed, and since ∂Eₚ/∂p_z = p_z/Eₚ, dp′_z/dp_z = γ(1 + βp_z/Eₚ) = E′/E. Every cell of momentum space stretches by exactly the factor its energy grows, so d³p/Eₚ does not change. The covariant proof takes one line: δ(p² − m²) = [δ(p⁰ − Eₚ) + δ(p⁰ + Eₚ)]/(2Eₚ), and θ(p⁰) keeps the first term, so integrating d⁴p δ(p² − m²) θ(p⁰) over p⁰ leaves d³p/(2Eₚ), derived. Each factor is invariant: det Λ = 1, p² is a scalar, and no orthochronous Λ flips the sign of p⁰ for timelike p. A proton boosted from rest to β = 0.6 shows it: d³p grows by γ = 1.25, and E grows from 938.27 to 1172.8 MeV, by the same 1.25. Limits: as m → 0 the hyperboloid closes onto the light cone and d³p/(2|p|) is still invariant; as p → 0, Eₚ → m, a constant, which is why non-relativistic quantum mechanics never needed it.

States every frame normalises alike

Define |p⟩ = a†(p)|0⟩. The commutator gives ⟨p|q⟩ = (2π)³2Eₚδ³(p − q), derived, and it is invariant because 2Eₚδ³(p − q) undoes the invariant measure: ∫d³p/(2Eₚ) × 2Eₚδ³(p − q) f(p) = f(q) in every frame. Directly, δ(p′_z − q′_z) = (E/E′)δ(p_z − q_z), so E′δ³(p′ − q′) = Eδ³(p − q). Two consequences, shown: the one-particle identity is ∫d³p/((2π)³2Eₚ) |p⟩⟨p|, and ⟨0|φ(x)|p⟩ = e^(−ip⋅x), a wavefunction with no frame-dependent prefactor. In a box moving with the particle ⟨p|p⟩ = 2EₚV, and a boost from its rest frame raises E by γ while contracting V by 1/γ, so the product never changes. The measured trace is time dilation. Unit 5 divides a Lorentz-invariant integral by the decaying state's 2E, quoted ahead, so a rate falls as m/E: muons circulating in CERN's storage ring with momentum 3.094 GeV have γ = 29.30, a predicted lifetime of 64.4 μs, and 64.4 μs was measured in 1977.

Ten Noether charges, written in modes

UAQM-09.01's theorem, quoted, turns each continuous symmetry into a conserved current. Translations give T^μν and P^μ = ∫d³x T^(0μ). Lorentz transformations, under which a scalar changes only through its argument, give M^μν = ∫d³x (x^μT^(0ν) − x^νT^(0μ)) with no spin term, a two-line application of the theorem: rotations Jⁱ = ½ε^ijk M^jk and boosts Kⁱ = M⁽⁰ⁱ⁾, so K = tP − ∫d³x x T⁰⁰. Substituting the modes and normal ordering gives P^μ = ∫d³p/((2π)³2Eₚ) p^μ a†(p)a(p), derived; the same steps give J = −i∫d³p/((2π)³2Eₚ) a†(p)(p × ∇ₚ)a(p) and K = −i∫d³p/((2π)³2Eₚ) Eₚ a†(p)∇ₚ a(p), stated, the explicit t in K cancelling against the phases e^{(∓iEₚₜ)} of the modes. A check on one particle: at t = 0, −∫x T⁰⁰ on a narrow packet is −E⟨x⟩, and in momentum space x = i∇ₚ, so K acts as −iEₚ∇ₚ. The measure earns its keep here: ∫d³p/(2Eₚ) ψ*(−iEₚ∇ₚχ) = −(i/2)∫d³p ψ*∇ₚχ, Hermitian after one integration by parts. With d³p as the measure, K would need symmetrising.

The unitary U(Λ, a) and the Poincaré algebra

Exponentiate: U(1, a) = e^(iP⋅a) and, for Λ = 1 + ω, U(Λ) = e^{(−(i/2)ω_μν M^μν)}, so a boost of rapidity η along z is e^(−iηK_z). Hermitian generators make U(Λ, a) unitary, and U φ(x) U⁻¹ = φ(Λx + a), checked infinitesimally from [P^μ, φ] = −i∂^μφ and [K_z, φ] = i(z∂ₜ + t∂_z)φ. On modes, p⋅(Λx) = (Λ⁻¹p)⋅x, and relabelling p → Λp inside the invariant measure gives U a(p) U⁻¹ = e^{(−i(Λp)⋅a)} a(Λp), derived. With U|0⟩ = |0⟩ this is U(Λ, a)|p⟩ = e^(i(Λp)⋅a)|Λp⟩, and ⟨Λp|Λq⟩ = ⟨p|q⟩ by the previous section. The Lorentz algebra is UAQM-05.01's, quoted: [Jᵢ, Jⱼ] = iεᵢⱼₖ Jₖ, [Jᵢ, Kⱼ] = iεᵢⱼₖ Kₖ, [Kₓ, K_y] = −iJ_z. Letting the one-particle forms act on a wavefunction adds, derived, [P^μ, P^ν] = 0, [Jᵢ, Pⱼ] = iεᵢⱼₖ Pₖ, [Jᵢ, H] = 0, [Kᵢ, H] = −iPᵢ and [Kᵢ, Pⱼ] = −iδᵢⱼ H. So a boost and a translation do not commute: U(Λ) e^(iP⋅a) U(Λ)⁻¹ = e^(iP⋅Λa).

What changes when the fields interact

Everything so far is free. Add a non-derivative interaction and Noether's theorem still supplies ten conserved charges, but they split, stated here for Unit 2. P = −∫d³x π∇φ and J are built from T⁽⁰ⁱ⁾ = π∂^iφ, which has no g⁽⁰ⁱ⁾ℒ term and never sees the interaction, so they keep their free mode forms. H = ∫d³x T⁰⁰ does see it, and so does K, which at t = 0 becomes K₀ − ∫d³x x ℋᵢₙₜ. Geometry explains the split: space translations and rotations map the surface t = 0 to itself and need no equation of motion, while time translations and boosts move or tilt it. The algebra demands it too: [Kᵢ, Pⱼ] = −iδᵢⱼ H must hold with the full H, and P is free, so K cannot be. That is why topic 2.04 must prove, using microcausality, that the scattering operator commutes with the free U(Λ), rather than assume it. The construction also assumed spin zero; with spin, U(Λ) acting on |p, σ⟩ mixes σ by a Wigner rotation, the next topic's business.

Change one variable at a time

Make the relationship visible.

Interactive model

Packet rapidity y₀0.3

Boost rapidity η0.8

Mass m0.5 GeV

Set y₀ = 0 and drag η to 1.0: the lower bar is cosh 1 = 1.543 times the upper, exactly the energy ratio, so Δp_z/E is unchanged. Then shrink m: the hyperbola closes onto the light cone, yet every ratio stays put, because only rapidities enter them.

ENERGY RATIO E′/E1.596

STRETCH Δp′_z/Δp_z1.596

Δp_z/E AFTER THE BOOST0.505

δ³-NORM FACTOR E/E′0.627

Live interpretationENERGY RATIO E′/E: 1.596. STRETCH Δp′_z/Δp_z: 1.596. Δp_z/E AFTER THE BOOST: 0.505. δ³-NORM FACTOR E/E′: 0.627

Catch the common trap

Explain before calculating.

Misconception check

“A symmetry generator commutes with the Hamiltonian, so [Kᵢ, H] = −iPᵢ means boosts are not a symmetry of the quantum theory.” The premise holds only for generators with no explicit time dependence. K = tP − ∫d³x x T⁰⁰ carries an explicit t, and its conservation reads dK/dt = ∂K/∂t + i[H, K] = P − P = 0: the commutator is not zero, it is exactly what cancels the explicit term. Nor could it be zero. Energy is the time component of a four-vector, so boosting a state must change its energy, and a K commuting with H would leave every energy eigenvalue where it was. Lorentz invariance means that the ten generators close on the Poincaré algebra with the full H, not that each commutes with H, and it is the same commutator that forces K to carry the interaction once there is one.

A colleague keeps quantum mechanics' normalisation ⟨p|q⟩ = δ³(p − q) and defines a boost by U(Λ)|p⟩ = |Λp⟩. A π⁺ at rest is boosted until its energy is twice its mass. What goes wrong, and what is the fix?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA K⁺ (m = 493.68 MeV) moves along z with p_z = 493.68 MeV. A spectrometer accepts a momentum box Δpₓ = Δp_y = 2.0 MeV, Δp_z = 4.0 MeV around it. Find Δ³p and Δ³p/(2E) in the laboratory and in the kaon's rest frame, and say which a Lorentz-invariant calculation may use.

Laboratory: E = √(p² + m²) = √2 × 493.68 = 698.17 MeV. Δ³p = 2.0 × 2.0 × 4.0 = 16.0 MeV³, so Δ³p/(2E) = 16.0/1396.34 = 0.011459 MeV².
Boost to the rest frame: β = p/E = 0.70711 and γ = E/m = 1.41421. The transverse components are untouched, and p′_z = γ(p_z − βEₚ) gives dp′_z/dp_z = γ(1 − βp_z/Eₚ) = γ(1 − β²) = 1/γ = 0.70711 at the centre of the box, which is E′/E = 493.68/698.17.
Rest frame: Δp′_z = 4.0 × 0.70711 = 2.8284 MeV, Δ³p′ = 11.314 MeV³, E′ = 493.68 MeV, and Δ³p′/(2E′) = 11.314/987.36 = 0.011459 MeV², the laboratory value.
Checks: the measure has mass dimension 2, as MeV² says. Δ³p fell by 1/√2 and so did E, so only their ratio is a scalar; an integral over d³p without the 1/(2Eₚ) gives a frame-dependent answer.

AnswerΔ³p = 16.0 MeV³ in the laboratory and 11.3 MeV³ at rest; Δ³p/(2E) = 0.0115 MeV² in both, so only the latter belongs in an invariant calculation.

MediumA π⁰ (m = 134.98 MeV) at rest is in the state |p⟩. Apply U(Λ, a) = e^(iP⋅a)U(Λ), with Λ a boost of rapidity η = 0.50 along z and a = (3.00 fm, 0, 0, 2.00 fm) in (ct, x, y, z). Find the final state and its phase, check that the phase is a Lorentz scalar, and compare the phase when the translation acts before the boost.

U(Λ)|p⟩ = |Λp⟩ with Λp = (m cosh η, 0, 0, m sinh η) = (152.21, 0, 0, 70.34) MeV, using cosh 0.5 = 1.12763 and sinh 0.5 = 0.52110. No √(E′/E) appears, because ⟨p|q⟩ carries 2Eₚ.
The translation multiplies by e^(i(Λp)⋅a): (Λp)⋅a = 152.21 × 3.00 − 70.34 × 2.00 = 315.95 MeV fm, and dividing by ħc = 197.327 MeV fm gives 1.6011 rad.
Scalar check: (Λp)⋅a = p⋅(Λ⁻¹a). The time component of Λ⁻¹a is 3.00 cosh η − 2.00 sinh η = 2.3407 fm, and p⋅(Λ⁻¹a) = 134.98 × 2.3407 = 315.95 MeV fm, the same number.
Translation first: U(Λ)e^(iP⋅a)|p⟩ = e^(ip⋅a)|Λp⟩, with p⋅a = 134.98 × 3.00 = 404.94 MeV fm, or 2.0521 rad. The final momentum agrees but the phase differs by 0.4510 rad, because U(Λ)e^(iP⋅a)U(Λ)⁻¹ = e^(iP⋅Λa): boosts and translations do not commute, [K_z, P_z] = −iH.

AnswerEither order gives |Λp⟩ with E′ = 152.2 MeV and p′_z = 70.3 MeV; boost then translate gives the phase 1.601 rad, translate then boost 2.052 rad.

HardAt t = 0 a free proton (938.27 MeV) sits at rest at the origin and a free π⁺ (139.57 MeV) with momentum 300.0 MeV along x is centred at x = 2.00 fm, each a packet narrow in momentum. Evaluate Kₓ = tPₓ − ∫d³x x T⁰⁰ at t = 0, use its conservation to predict the centre of energy at t = 10.0 fm/c, check the prediction against the packets' own motion, and compare the mass-weighted centre.

Energies: E_π = √(300.0² + 139.57²) = 330.88 MeV, so H = 938.27 + 330.88 = 1269.15 MeV and Pₓ = 300.0 MeV. K is bilinear in a†a, so on two free quanta it is the sum of one-body charges, and for narrow packets ∫x T⁰⁰ becomes Σ Eᵢ⟨xᵢ⟩.
At t = 0: Kₓ = −(938.27 × 0 + 330.88 × 2.00) MeV fm = −661.76 MeV fm. Dividing by ħc = 197.327 MeV fm gives Kₓ = −3.354, dimensionless, as a generator multiplying a rapidity must be.
Conservation: Kₓ = tPₓ − H X_E, with X_E = ∫x T⁰⁰/H the centre of energy, so X_E(t) = (tPₓ − Kₓ)/H. At t = 10.0 fm/c, X_E = (3000.0 + 661.76)/1269.15 = 2.885 fm, moving at Pₓ/H = 0.2364.
Direct check: the pion moves at p/E = 0.9067 and reaches 2.00 + 9.067 = 11.067 fm, so X_E = 330.88 × 11.067/1269.15 = 2.885 fm, the prediction, and Kₓ is unchanged to the four figures kept.
Mass-weighted centre: Σmᵢ xᵢ/Σmᵢ moves at 139.57 × 0.9067/1077.84 = 0.1174, neither Pₓ/H nor Pₓ/M = 0.2783. The point that moves uniformly is the centre of energy, the one the boost charge fixes. With an interaction, ℋᵢₙₜ would add its own term to ∫x T⁰⁰.

AnswerKₓ = −662 MeV fm, or −3.35 in natural units; X_E = 2.89 fm at t = 10.0 fm/c, moving at 0.236c, while the mass-weighted centre moves at only 0.117c.