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advanced33 min studyLesson 657 in Physics

Free photon polarizations and gauge redundancy

Reduce free Maxwell theory to transverse oscillators and keep canonical brackets consistent with Gauss constraints and gauge-dependent potentials.

Before you begin
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The question for today

What you’ll understand

  • Count physical photon polarizations after constraints and gauge redundancy.
  • Use the transverse polarization projector and positive photon energies.
  • Distinguish a nonlocal gauge-fixed potential bracket from locality of electromagnetic observables.

A four-component electromagnetic potential does not describe four physical photon oscillators. Maxwell theory has a constraint and a gauge redundancy, and quantization must respect both. In the free source-free sector, radiation gauge exposes two transverse polarizations with positive energy. It also makes the canonical potential algebra spatially nonlocal through a transverse projector. This feature reflects the constrained choice of coordinates; it does not permit gauge-invariant electric or magnetic measurements to signal outside the light cone.

Gauss law removes independent longitudinal data

Use natural units and the plus, minus, minus, minus metric, with the usual three-vector electric field equal to minus the time derivative of the vector potential minus the gradient of scalar potential. The Maxwell Lagrangian has no time derivative of scalar potential, so its conjugate momentum vanishes. Varying that potential imposes Gauss law. It acts as a constraint rather than as another independent wave equation with freely chosen initial velocity.

Choose vanishing divergence of the spatial vector potential. In a source-free region with suitable decay or boundary conditions and no residual zero-mode complication, the scalar potential can then be set to zero. Two transverse spatial components remain. With charged matter present, the scalar potential instead solves a sourced constraint and cannot generally be set to zero in the same gauge. Global boundary modes, finite-volume zero modes, and nontrivial topology also require separate treatment rather than automatic use of the nonzero-momentum projector.

L=−14FμνFμν=12(E2−B2),π0=0,∇⋅E=0,∇⋅A=0\mathcal L=-\frac14F_{\mu\nu}F^{\mu\nu}=\frac12(\mathbf E^2-\mathbf B^2),\quad \pi^0=0,\quad \nabla\cdot\mathbf E=0,\quad \nabla\cdot\mathbf A=0
Here E = −∂tA−∇A⁰ and B = ∇×A are conventional physical three-vectors. The displayed Gauss law is source-free. A has energy dimension one, F dimension two. Radiation gauge with A⁰ = 0 additionally needs the stated boundary and residual-gauge conditions.↔ Scroll wide equations horizontally.

Polarization completeness is transverse

For each nonzero momentum, choose two orthonormal polarization vectors perpendicular to its direction. Their outer products sum to the projector onto the transverse plane, not to the full three-dimensional identity. A different transverse basis redistributes amplitudes among the two labels without adding a third propagating oscillator. Circular polarizations are complex combinations of the same two degrees of freedom; their use requires a complex conjugate in the completeness sum.

The free vector potential expands with these polarizations and the same inverse square root of twice positive frequency used for a massless oscillator. Bosonic ladders have a delta in polarization as well as momentum. After subtracting the reference vacuum energy, the Hamiltonian counts photons with energy equal to momentum magnitude. Helicity labels plus and minus one describe the physical massless particle states. A longitudinal potential component can encode a constraint or gauge change, but it is not a third free photon polarization.

∑λ=12ϵi(λ)(k)ϵj(λ)∗(k)=δij−kikj∣k∣2,:H:=∑λ=12∫d3k(2π)3∣k∣akλ†akλ\sum_{\lambda=1}^2\epsilon_i^{(\lambda)}(\mathbf k)\epsilon_j^{(\lambda)*}(\mathbf k)=\delta_{ij}-\frac{k_ik_j}{|\mathbf k|^2},\qquad :H:=\sum_{\lambda=1}^2\int\frac{d^3k}{(2\pi)^3}|\mathbf k|a_{\mathbf k\lambda}^{\dagger}a_{\mathbf k\lambda}
The polarizations have Euclidean unit norm and k·ε = 0, with k ≠ 0. [aₖλ,a†qσ] = (2π)³δλσδ³(k−q). H has energy units. These are free source-free physical modes, with a specified treatment of the zero-momentum sector.↔ Scroll wide equations horizontally.

Constrained brackets preserve the chosen coordinates

Applying an unconstrained scalar-style identity bracket to every spatial component conflicts with the imposed divergenceless conditions. The correct reduced canonical bracket contains the transverse delta kernel. Taking a spatial divergence then gives zero, as the constraint requires. With the conventional physical spatial vector potential used here, its reduced momentum is its time derivative, which equals minus the transverse electric field. This sign must be kept separate from conventions that use lowered spatial components of the four-potential.

The inverse Laplacian implicit in the transverse projector is spatially nonlocal. Consequently a gauge-fixed potential is not itself a local gauge-invariant observable. Field strengths obey the appropriate causal commutation relations after the full constrained construction. Covariant quantization offers an alternative using extra formal polarizations and a physical-state condition or gauge-cohomology construction. Those auxiliary states are not four positive-norm physical photons, and gauge fixing should not be confused with changing electromagnetic fields by an observable operation.

[AiT(t,x),ΠjT(t,y)]=i∫d3k(2π)3(δij−kikj∣k∣2)eik⋅(x−y),ΠT=A˙T=−ET[A_i^{T}(t,\mathbf x),\Pi_j^{T}(t,\mathbf y)]=i\int\frac{d^3k}{(2\pi)^3}\left(\delta_{ij}-\frac{k_ik_j}{|\mathbf k|^2}\right)e^{i\mathbf k\cdot(\mathbf x-\mathbf y)},\quad \Pi^{T}=\dot{\mathbf A}^{T}=-\mathbf E^{T}
Indices i,j here label ordinary Euclidean spatial vector components. The reduced bracket is a distribution with an appropriate inverse-Laplacian boundary prescription. It is not the unconstrained δijδ³ bracket and is not a claim of nonlocal signaling by Fμν.↔ Scroll wide equations horizontally.

Work it through

Project onto the physical plane

For a nonzero photon momentum along +z, apply the polarization completeness projector to a vector v = (2,−1,3). Find the projected vector and its squared Euclidean length, and check which component is removed.

  1. The unit momentum direction is (0,0,1), so PT=1−k^k^T=diag⁡(1,1,0)P_T=\mathbf1-\hat{\mathbf k}\hat{\mathbf k}^{T}=\operatorname{diag}(1,1,0).

  2. Multiplication gives PTv=(2,−1,0)P_Tv=(2,-1,0). The longitudinal z component has been removed.

  3. The transverse squared length is 22+(−1)2=52^2+(-1)^2=5. The original squared length was 14, with the removed longitudinal contribution equal to 9.

The result

The projector gives (2,−1,0), of squared length 5. It retains two physical transverse components for this momentum.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

A component depends on the selected direction

Reset and set BOTH Angle to 0 degrees and Magnitude to 2; predict x projection 2. Then Reset and set BOTH Angle to 90 degrees and Magnitude to 2; predict x projection 0. This planar vector illustrates component projection only. It does not impose three-dimensional transversality, quantize photons, solve Gauss law, or model a gauge transformation.

A component depends on the selected directionPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. 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 forA projection removes a specified component; identifying the physical projection requires the field constraints.

Pause & reason

Check your understanding

In free Maxwell theory at nonzero momentum, what does the physical polarization sum equal?

Your turn

Try it yourself

In a regulated nonzero-momentum mode along +z with |k| = 4, place two photons in one transverse polarization and one in the other. Find normal-ordered energy and z momentum. What is the invariant mass of this exactly collinear three-photon state?

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

Give me a hint

Each photon contributes the same positive four-momentum independently of polarization.

Compare with the solution

The total photon count is three, so H=3(4)=12H=3(4)=12 and Pz = 12, with transverse momentum zero. The total invariant is H2−P2=144−144=0H^2-\mathbf P^2=144-144=0, so the invariant mass is zero. This statement depends on exact collinearity; two noncollinear positive-energy photon momenta can have a timelike total even though each individual photon is massless.

Keep the idea

Take it with you

  • Gauss law and gauge redundancy prevent four independent physical photon oscillators.
  • Two transverse polarizations produce a projector and a positive free Hamiltonian.
  • Gauge-fixed potential brackets can be nonlocal while gauge-invariant electromagnetic observables remain causal.

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