advanced33 min studyLesson 658 in Physics

Poincaré generators in field theory

Connect Noether stress energy to quantum translations, rotations, and boosts, with explicit generator signs and domain assumptions.

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What you’ll understand

  • Identify total four-momentum and the role of normal-ordered stress energy.
  • Define rotation and boost generators with a declared sign convention.
  • Check a many-particle four-momentum and a conserved energy-center boost charge.

A relativistic field theory needs more than a positive Hamiltonian: its generators must fit into a consistent representation of spacetime symmetry. Time translations, spatial translations, rotations, and boosts act on the same state space and satisfy linked commutation relations. Noether currents identify candidate generators, while quantization requires suitable operator definitions and domains. Free fields provide explicit examples in which the total four-momentum is the sum of occupied one-particle four-momenta and the vacuum can be chosen invariant.

Stress energy generates translations

For a real scalar, the translation current gives a symmetric stress tensor built from derivatives and the Lagrangian. Its zero-zero component is energy density and its zero-spatial components are momentum density with raised spacetime indices. Spatially integrating them gives the total four-momentum when surface flux vanishes. The metric matters: raising a spatial derivative changes its sign, so a three-vector gradient cannot be substituted blindly for every indexed derivative.

In the free quantum theory, local stress components are composite distributions defined through a vacuum subtraction or equivalent Wick prescription. Their spatial integrals act on a suitable dense domain, with limits and boundary terms controlled. The integrated translation generators have commuting components in the ordinary Poincaré representation. Their commutators with a scalar field give spacetime derivatives. This relation specifies the active translation convention and checks that the Hamiltonian agrees with the previously derived Heisenberg evolution.

Tμν=∂μϕ ∂νϕ−ημνL,Pμ=∫d3x :T0μ:,[Pμ,ϕ(x)]=−i∂μϕ(x)T^{\mu\nu}=\partial^\mu\phi\,\partial^\nu\phi-\eta^{\mu\nu}\mathcal L,\qquad P^\mu=\int d^3x\,:T^{0\mu}:,\qquad [P^\mu,\phi(x)]=-i\partial^\mu\phi(x)
We use ℏ = c = 1, η = diag(1,−1,−1,−1), and U(a) = exp(iaμPμ), so U(a)φ(x)U(a)⁻¹ = φ(x+a). T has energy dimension four; P dimension one. The tensor formula is classical or an appropriately defined free Wick product.↔ Scroll wide equations horizontally.

Rotations and boosts involve different moments

Rotations use the spatial moment of momentum density. A scalar has no independent intrinsic spin term in its canonical Lorentz current; spinor and vector fields require the corresponding spin current or an improved stress tensor. Improvements can rearrange local densities without changing total charges when their surface contributions vanish. Splitting angular momentum into orbital and spin pieces can depend on gauge and convention, while the total physical rotation generator is the relevant conserved quantity.

Define the boost generator as the spatial moment of energy density minus time times total momentum. This chooses the sign opposite to a frequently used upper-index zero-spatial Lorentz generator. Its conservation states that the center of energy moves with velocity total momentum divided by total energy for an isolated state. The center of energy is a weighted moment of a field expectation, not a new claim that every relativistic field has a simple covariant single-particle position operator.

Ji=ϵijk∫d3x xj:T0k:,Ki=∫d3x xi:T00:−tPiJ_i=\epsilon_{ijk}\int d^3x\,x_j:T^{0k}:,\qquad K_i=\int d^3x\,x_i:T^{00}:-tP_i
Here spatial subscripts on J,K,P denote Euclidean three-vector components, with ε123 = +1. J and K are dimensionless in natural units. The sign of K is explicitly fixed by this expression. Finite energy moments and vanishing boundary flux are assumed.↔ Scroll wide equations horizontally.

The algebra constrains the full dynamics

Rotations make momentum transform as a vector. Boosts mix energy and momentum, and two boosts along different directions differ by a rotation. Their commutation relations are therefore not independent optional checks. A cutoff that singles out a frame may obscure or break these relations before a symmetry-restoring continuum prescription is applied. Agreement of one mode's dispersion with the mass shell does not alone establish the full generator algebra.

For free particles, total four-momentum adds across occupied modes, including antiparticles with positive energy. The invariant mass of a multiparticle state depends on relative motion and can exceed the sum of rest masses. In an interacting theory, boosts generally contain interaction contributions as well as the Hamiltonian, while total asymptotic momenta still label scattering states under suitable assumptions. Writing classical Noether integrals is consequently a formal starting point; constructing self-adjoint quantum generators and a unitary representation is an additional mathematical task.

[Ji,Pj]=iϵijkPk,[Ki,H]=iPi,[Ki,Pj]=iδijH,[Ki,Kj]=−iϵijkJk[J_i,P_j]=i\epsilon_{ijk}P_k,\quad [K_i,H]=iP_i,\quad [K_i,P_j]=i\delta_{ij}H,\quad [K_i,K_j]=-i\epsilon_{ijk}J_k
These signs use K = ∫xT⁰⁰−tP and the stated active translation convention. H = P⁰. Relations are operator identities on a common invariant domain in a valid representation, not unrestricted multiplication rules for arbitrary singular local products.↔ Scroll wide equations horizontally.

Add on-shell momenta before taking an invariant

Two quanta of a free scalar of mass 4 occupy normalized distinct box modes with momenta (3,0,0) and (−3,0,0). Find total energy, momentum, and invariant mass in natural units.

  1. Each quantum has positive energy 32+42=5\sqrt{3^2+4^2}=5. The opposite spatial directions do not reverse either energy.

  2. The total four-momentum is (10,0,0,0)(10,0,0,0) because energies add while the spatial momenta cancel.

  3. The invariant mass is H2−P2=10\sqrt{H^2-\mathbf P^2}=10. It exceeds the rest-mass sum 8 by the pair’s kinetic energy in its center-of-momentum frame.

The result

The state has total energy 10, spatial momentum zero, and invariant mass 10, with the box-mode normalization avoiding squared continuum deltas.

Explore the model

Interactive lab

Interactive lab · spacetime

A boost preserves the event interval

Reset and set BOTH Relative velocity β to 0 and Event time ct to 1; the fixed event has x = 0.6 and interval 0.64. Then Reset and set BOTH β to 0.6 and ct to 1; predict ct′ = 0.8, x′ = 0, and the same interval. The instrument transforms that one event in units c = 1. It does not construct quantum Poincaré generators, spin transformations, or an interacting boost operator.

A boost preserves the event intervalA fixed x = 0.6 event is Lorentz transformed with c = 1. Playback highlights the relation between coordinate descriptions; it does not move the event. Current readings: Lorentz factor γ: 1.12; Event x′: 0.168; Event ct′: 0.817. All parameters are adjustable with the labeled controls below.−4.69−2.3502.354.69−2.2−1.101.12.2x or x′ · c = 1ct or ct′EE′ct² − x² = 0.64
Original eventTransformed eventLight cone
Lorentz factor γ
1.12
Event x′
0.168
Event ct′
0.817
t = 0 s

A fixed x = 0.6 event is Lorentz transformed with c = 1. Playback highlights the relation between coordinate descriptions; it does not move the event. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forLorentz geometry provides a consistency check, while a quantum field representation must also realize the generator algebra.

Check your understanding

With Kᵢ = ∫xᵢT⁰⁰ d³x − tPᵢ, which commutator expresses the mixing of momentum with energy?

Try it yourself

An isolated state has constant mean energy E = 5 and x momentum P = 3. Its energy center is initially xE = 1. Using K = E xE − tP, find xE at t = 2 and verify K remains fixed. Use consistent natural length and time units.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Conservation of K gives energy-center velocity P/E when energy is constant.

Compare with the solution

Initially K = 5. The energy-center velocity is P/E = 3/5, so at t = 2 its position is xE=1+(3/5)(2)=2.2x_E=1+(3/5)(2)=2.2. Substitution gives K=5(2.2)−2(3)=5K=5(2.2)-2(3)=5, unchanged. This uses a finite energy moment and a closed system; outgoing boundary flux would invalidate the isolated-state conservation argument.

Take it with you

  • Stress energy identifies total translations only after its quantum composite products are defined.
  • Boost signs and spatial-index conventions must accompany the displayed algebra.
  • Multiparticle invariants and the conserved energy center provide concrete checks of relativistic generators.

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