Physical-Mode Quantization of the Free Electromagnetic Field

Physical-mode quantization begins after the Maxwell constraints have been solved and their gauge directions quotiented. In the source-free theory, Coulomb or radiation gauge leaves one transverse canonical pair; quantizing that pair and selecting the standard positive-frequency Minkowski vacuum gives exactly two positive-norm photon oscillators at every nonzero momentum. The tradeoff is visible from the outset: the transverse projector is spatially nonlocal, and Lorentz covariance is no longer manifest.

Required background. Maxwell Constraints as a Worked Application supplies the reduced transverse canonical pair, projector bracket, and boundary and zero-mode assumptions. Massive and Massless Spin-One Polarizations supplies the two transverse polarization vectors and their completeness relation. Canonical Quantization: Algebra, Representation, and State supplies the CCR construction and the distinction among algebra, representation, and state.

The transverse Maxwell phase space

Work with source-free Maxwell theory on a fixed $\mathbb R^3$ time slice. Fields and admitted gauge parameters are taken to fall off rapidly enough that surface terms vanish and $\nabla^2$ has no retained harmonic or zero mode. All polarization formulas below are for $\mathbf k\neq0$. Spatial indices on the reduced variables are moved with $\delta_{ij}$.

Coulomb gauge imposes $\partial_iA_i=0$, and preserving it in time gives $\partial_i\dot A_i=\partial_t(\partial_iA_i)=0$. Since $\pi^i=\dot A_i-\partial_iA_0$, Gauss law then gives

$$0=\partial_i\pi^i=-\nabla^2A_0.$$

The falloff condition therefore fixes $A_0=0$. In this bounded free setting, Coulomb gauge has become radiation gauge,

$$A_0=0, \qquad \partial_iA_i=0.$$

A residual transformation $A_\mu\mapsto A_\mu+\partial_\mu\alpha$ must obey $\nabla^2\alpha=0$ to preserve Coulomb gauge and $\dot\alpha=0$ to preserve $A_0=0$. No nontrivial decaying solution remains; a spacetime constant, if admitted, acts trivially on $A_\mu$. Boundary, harmonic, and global modes can invalidate this conclusion and are not silently discarded in other settings. The Coulomb-gauge elimination of $A_0$ and the role of its boundary condition are developed in Srednicki 2006, § 55, printed pp. 335–336, author manuscript.

The resulting classical variables are the divergence-free pair

$$(A_i^T,\pi_T^i), \qquad \partial_iA_i^T=\partial_i\pi_T^i=0,$$

with reduced Hamiltonian and bracket

$$H_{\mathrm{red}} =\frac12\int\mathrm d^3\mathbf x\, \left(\boldsymbol\pi_T^2+\mathbf B^2\right),$$

where $\mathbf B^2\equiv\frac12F_{ij}F_{ij}$ on the spatial slice.

$$\begin{aligned} \{A_i^T(\mathbf x),\pi_T^j(\mathbf y)\} &=\mathsf P_i{}^j \delta^{(3)}(\mathbf x-\mathbf y), \\ \mathsf P_i{}^j &=\delta_i{}^j -\frac{\partial_i\partial^j}{\nabla^2}. \end{aligned}$$

Thus the constraints and quotient have already removed $A_0$, its conjugate, and the longitudinal spatial pair. Quantization need not introduce them again. Weinberg 1995, § 8.3, pp. 347–350 derives the Coulomb-gauge transverse commutator and Hamiltonian from the corresponding constrained canonical system.

Quantizing the transverse canonical pair

Promote the reduced bracket to the equal-time field algebra

$$\begin{aligned} [\widehat A_i^T(t,\mathbf x), \widehat\pi_T^j(t,\mathbf y)] &=i\mathsf P_i{}^j \delta^{(3)}(\mathbf x-\mathbf y), \\ [\widehat A_i^T,\widehat A_j^T] &=[\widehat\pi_T^i,\widehat\pi_T^j]=0. \end{aligned}$$

These are distributional equations: the fields must be spatially smeared on a common invariant domain. The first kernel is not an ordinary pointwise delta. Its momentum representation is

$$\mathsf P_i{}^j\delta^{(3)}(\mathbf r) =\int\frac{\mathrm d^3\mathbf k}{(2\pi)^3} \left( \delta_i{}^j-\frac{k_i k^j}{\mathbf k^2} \right)e^{i\mathbf k\cdot\mathbf r}.$$

Set $k^0=\omega_{\mathbf k}=|\mathbf k|$ and use the site-wide on-shell normalization package

$$\int_{\mathbf k} \equiv \int\frac{\mathrm d^3\mathbf k} {(2\pi)^3\,2\omega_{\mathbf k}}, \qquad k\cdot x=\omega_{\mathbf k}t-\mathbf k\cdot\mathbf x.$$

For each nonzero $\mathbf k$, choose two spatial polarization vectors $e_i^{(\lambda)}(\mathbf k)$ satisfying

$$\begin{aligned} k^i e_i^{(\lambda)}&=0, &e_i^{(\lambda)*}e_i^{(\lambda')}&=\delta_{\lambda\lambda'}, \\ \sum_{\lambda=1}^2 e_i^{(\lambda)}e^{(\lambda)*j} &=P_i{}^j(\mathbf k), &P_i{}^j(\mathbf k)&= \delta_i{}^j-\frac{k_i k^j}{\mathbf k^2}. \end{aligned}$$

No continuous real linear-polarization frame, and no continuous frame for either helicity line separately, can be chosen globally over all momentum directions. Such bases may be chosen patchwise or merely measurably for the mode formulas, while $P_i{}^j(\mathbf k)$ is global and independent of that choice.

The Hermitian transverse field and its momentum are

$$\begin{aligned} \widehat A_i^T(x) &=\sum_{\lambda=1}^2\int_{\mathbf k} \left[ \begin{aligned} &e_i^{(\lambda)}(\mathbf k) a_\lambda(\mathbf k)e^{-ik\cdot x} \\ &+e_i^{(\lambda)*}(\mathbf k) a_\lambda^\dagger(\mathbf k)e^{ik\cdot x} \end{aligned} \right], \\ \widehat\pi_T^i(x) &=-i\sum_{\lambda=1}^2\int_{\mathbf k} \omega_{\mathbf k} \left[ \begin{aligned} &e^{(\lambda)i}(\mathbf k) a_\lambda(\mathbf k)e^{-ik\cdot x} \\ &-e^{(\lambda)*i}(\mathbf k) a_\lambda^\dagger(\mathbf k)e^{ik\cdot x} \end{aligned} \right]. \end{aligned}$$

Hermiticity exchanges the two terms; it does not impose $a_\lambda(-\mathbf k)=a_\lambda^\dagger(\mathbf k)$. The independent annihilation modes at opposite momenta remain distinct.

The oscillator algebra matched to this expansion is

$$\begin{aligned} [a_\lambda(\mathbf k), a_{\lambda'}^\dagger(\mathbf q)] &=(2\pi)^3\,2\omega_{\mathbf k}\, \delta_{\lambda\lambda'} \delta^{(3)}(\mathbf k-\mathbf q), \\ [a_\lambda,a_{\lambda'}] &=[a_\lambda^\dagger,a_{\lambda'}^\dagger]=0. \end{aligned}$$

This normalization can be checked without analogy to a scalar field. At equal times, the two mixed terms in $[\widehat A_i^T,\widehat\pi_T^j]$ give

$$\begin{aligned} &\frac{i}{2} \int\frac{\mathrm d^3\mathbf k}{(2\pi)^3} P_i{}^j(\mathbf k) \left[ e^{i\mathbf k\cdot\mathbf r} +e^{-i\mathbf k\cdot\mathbf r} \right] \\ &\qquad= i\mathsf P_i{}^j\delta^{(3)}(\mathbf r), \qquad \mathbf r=\mathbf x-\mathbf y, \end{aligned}$$

because $P_i{}^j(-\mathbf k)=P_i{}^j(\mathbf k)$. Each term supplies half the transverse delta, fixing simultaneously the measure, field coefficient, oscillator commutator, and conjugations. Srednicki 2006, § 55, printed pp. 336–338, author manuscript gives this mode expansion, completeness relation, transverse field commutator, and oscillator algebra in the same normalization.

The two-polarization Fock space

The CCR algebra does not choose a representation or a state. For a two-component square-integrable wave packet, define

$$\begin{aligned} a(f) &=\sum_{\lambda=1}^2\int_{\mathbf k} f_\lambda(\mathbf k)^*a_\lambda(\mathbf k), \\ a^\dagger(f) &=\sum_{\lambda=1}^2\int_{\mathbf k} f_\lambda(\mathbf k)a_\lambda^\dagger(\mathbf k), \\ \mathcal H_1 &=L^2\!\left( \frac{\mathrm d^3\mathbf k} {(2\pi)^3\,2\omega_{\mathbf k}} \right)\otimes\mathbb C^2, \end{aligned}$$

and make the additional standard Minkowski choice: positive-frequency modes are the annihilation modes, and the normalized vacuum obeys $a(f)|0\rangle=0$ for every $f\in\mathcal H_1$.

The normalizable one-photon state $|f\rangle=a^\dagger(f)|0\rangle$ has

$$\langle f|g\rangle =\sum_{\lambda=1}^2\int_{\mathbf k} f_\lambda(\mathbf k)^*g_\lambda(\mathbf k).$$

The physical Hilbert space selected here is the symmetric Fock space $\mathcal F_s(\mathcal H_1)$. Its sector construction and operator-domain qualifications are developed on Fock Space, Vacuum, and Particle Number. Sharp-momentum kets are generalized states,

$$\begin{aligned} |\mathbf k,\lambda\rangle &=a_\lambda^\dagger(\mathbf k)|0\rangle, \\ \langle\mathbf q,\lambda'|\mathbf k,\lambda\rangle &=(2\pi)^3\,2\omega_{\mathbf k}\, \delta_{\lambda'\lambda} \delta^{(3)}(\mathbf q-\mathbf k). \end{aligned}$$

A unitary change of the two-dimensional polarization basis preserves the algebra and norm. Within any momentum-space patch, an oriented real transverse frame may be recombined into circular polarizations with helicities $h=+1$ and $h=-1$. There is no timelike or longitudinal photon oscillator in this reduced description. The positive Hilbert norm above must not be confused with the negative Lorentz norm of a spacelike four-polarization. Weinberg 1995, § 8.4, pp. 351–353 constructs the two-polarization free field and its oscillator algebra from the transverse canonical system.

Hamiltonian and checks

Products of continuum fields at the same point require regulation. Put the theory in a periodic box of volume $V=L^3$, omit the zero mode, and retain a finite inversion-symmetric momentum set $K_\Lambda$. Define ordinary oscillators by

$$a_\lambda(\mathbf k) =\sqrt{2\omega_{\mathbf k}V}\,b_{\lambda\mathbf k}, \qquad [b_{\lambda\mathbf k},b_{\lambda'\mathbf q}^\dagger] =\delta_{\lambda\lambda'}\delta_{\mathbf k\mathbf q}.$$

Substitution into the regulated reduced Hamiltonian gives

$$\begin{aligned} H_\Lambda &=\sum_{\mathbf k\in K_\Lambda} \sum_{\lambda=1}^2 \omega_{\mathbf k} \left( b_{\lambda\mathbf k}^\dagger b_{\lambda\mathbf k} +\frac12 \right), \\ E_{0,\Lambda} &=\frac12 \sum_{\mathbf k\in K_\Lambda} \sum_{\lambda=1}^2 \omega_{\mathbf k}. \end{aligned}$$

The $bb$ and $b^\dagger b^\dagger$ terms cancel because $\omega_{\mathbf k}^2=\mathbf k^2$; the transverse electric and magnetic terms supply equal on-shell oscillator energies. There are two copies of the usual zero-point contribution, one for each physical polarization.

Normal ordering relative to the chosen vacuum removes this finite regulated constant. In the continuum, it is cleaner to define the excitation generators directly on their natural finite-particle domains:

$$\begin{aligned} H_{\mathrm{exc}} &=\sum_{\lambda=1}^2\int_{\mathbf k} \omega_{\mathbf k} a_\lambda^\dagger(\mathbf k)a_\lambda(\mathbf k), \\ \mathbf P_{\mathrm{exc}} &=\sum_{\lambda=1}^2\int_{\mathbf k} \mathbf k\, a_\lambda^\dagger(\mathbf k)a_\lambda(\mathbf k), \\ N &=\sum_{\lambda=1}^2\int_{\mathbf k} a_\lambda^\dagger(\mathbf k)a_\lambda(\mathbf k). \end{aligned}$$

These continuum expressions are distributional shorthand for second quantization on wave packets. Here $N$ counts free-photon occupation in the selected representation; it is not a gauge charge. The generators obey

$$\begin{aligned} [H_{\mathrm{exc}},a_\lambda^\dagger(\mathbf k)] &=\omega_{\mathbf k}a_\lambda^\dagger(\mathbf k), \\ [\mathbf P_{\mathrm{exc}},a_\lambda^\dagger(\mathbf k)] &=\mathbf k\,a_\lambda^\dagger(\mathbf k), \end{aligned}$$

so every finite-particle excitation has nonnegative energy. This is a statement about the selected physical Fock representation. It is not obtained by subtracting a formal infinity, and normal ordering is not a replacement for vacuum-energy or composite-operator renormalization; those distinctions are developed on Normal Ordering and Vacuum Terms.

Two further checks close the construction. First, the mode expansion obeys

$$\dot{\widehat A}_i^T=\widehat\pi_{Ti}, \qquad \dot{\widehat\pi}_{Ti}=\nabla^2\widehat A_i^T,$$

and hence the source-free wave equation. Second, $\operatorname{tr}P(\mathbf k)=2$: every nonzero momentum supports exactly two oscillator families. The oscillator Hamiltonian and its two-polarization zero-point term are given explicitly in Srednicki 2006, § 55, printed pp. 337–338, author manuscript.

What the reduced description hides

Physical-mode quantization makes positivity and the two-helicity content transparent because it never introduces unphysical oscillators. It also hides three structures that a covariant treatment must recover.

First, $\mathsf P_i{}^j$ contains $\nabla^{-2}$. The representative $A_i^T=\mathsf P_i{}^jA_j$ is therefore spatially nonlocal and need not have a vanishing commutator at spacelike separation. By contrast, the free field-strength commutator is obtained by differentiating the massless Pauli–Jordan distribution, so $[F_{\mu\nu}(x),F_{\rho\sigma}(y)]=0$ at spacelike separation. Locality of the gauge-invariant field strength does not turn the projected potential into a local four-vector. Spacelike Compatibility and Local Observables develops the field-versus-observable locality qualification.

Second, the time slice, the condition $A_0=0$, and the spatial transverse projector obscure Lorentz covariance. A boost of a radiation-gauge representative generally requires a compensating gauge transformation before it satisfies the same gauge conditions again. The physical helicities and their Fock norms are unchanged, but covariance is not manifest in the chosen coordinates.

Third, the reduction used the source-free equation and strong falloff. Charged matter makes $A_0$ solve a nontrivial Poisson equation and produces an instantaneous Coulomb term. Boundaries, nontrivial topology, and retained zero or harmonic modes can add residual transformations or physical sectors that the plane-wave Fock space does not contain.

The next chapter step is Covariant Free-Photon Quantization and Propagator, which introduces the four-component auxiliary description and compares its physical quotient with the two transverse modes found here. Charged matter and non-Abelian dynamics belong to Dynamical Gauge Fields and Matter. General gauge fixing belongs to Gauge Fixing, BRST, and BV, with the cohomological construction developed on The BRST Differential and Gauge-Fixed Complex.

Common pitfalls

Imposing $a_\lambda(-\mathbf k)=a_\lambda^\dagger(\mathbf k)$. The Hermitian field pairs each positive-frequency mode with its adjoint negative-frequency mode. Annihilators at $\mathbf k$ and $-\mathbf k$ are independent.

Claiming that the CCR select the photon vacuum. The transverse CCR define an algebra. The positive-frequency split, Fock representation, and Minkowski vacuum are additional choices made here.

Calling Coulomb gauge complete without stating boundary conditions. The residual equation is $\nabla^2\alpha=0$, supplemented here by $\dot\alpha=0$. Falloff removes its nontrivial solutions on $\mathbb R^3$; other boundaries or topologies need a separate analysis.

Reading a spacelike polarization norm as a negative state norm. The Lorentz norm belongs to a four-vector representative. The one-photon Hilbert norm is the positive $L^2$ norm displayed above.

Check your understanding

1. Starting from the mode expansion, recover the coefficient of the transverse delta in $[\widehat A_i^T(t,\mathbf x),\widehat\pi_T^j(t,\mathbf y)]$.

   Solution

   The annihilator in $\widehat A_i^T$ commutes only with the creator in $\widehat\pi_T^j$, producing one factor $iP_i{}^j(\mathbf k)/2$. The creator–annihilator cross-term produces the same factor with $\mathbf k\mapsto-\mathbf k$. Since the projector is even, the two half-projectors add to $i\mathsf P_i{}^j\delta^{(3)}(\mathbf x-\mathbf y)$.

2. Determine the residual gauge transformations that preserve both radiation gauge conditions under the assumptions of this page.

   Solution

   From $\partial_i(A_i+\partial_i\alpha)=0$ one obtains $\nabla^2\alpha=0$. From $A_0+\dot\alpha=0$ one obtains $\dot\alpha=0$. A time-independent harmonic function that decays at spatial infinity is zero; an admitted spacetime constant has vanishing derivative and acts trivially.

3. Show that a regulated one-photon state has positive excitation energy.

   Solution

   In the box, $H_{\mathrm{exc},\Lambda}=\sum_{\mathbf k,\lambda} \omega_{\mathbf k}b_{\lambda\mathbf k}^\dagger b_{\lambda\mathbf k}$. Therefore $[H_{\mathrm{exc},\Lambda},b_{\lambda\mathbf k}^\dagger] =\omega_{\mathbf k}b_{\lambda\mathbf k}^\dagger$. Acting on the vacuum gives energy $\omega_{\mathbf k}=|\mathbf k|>0$; the two values of $\lambda$ have the same positive energy.

References

• Srednicki, Mark. Quantum Field Theory. Author-hosted manuscript, University of California, Santa Barbara, 2006. Author’s manuscript page.
• Weinberg, Steven. The Quantum Theory of Fields. Volume I: Foundations. Cambridge: Cambridge University Press, 1995. First edition; 2005 paperback, 2012 printing consulted. DOI.