Covariant Free-Photon Quantization and Propagator

Covariant quantization keeps the local four-component potential and yields a manifestly covariant photon propagator, but its four oscillator families cannot all carry positive norm. In the free Abelian theory, the Gupta–Bleuler construction resolves the tension in three steps: build an indefinite auxiliary space, impose the positive-frequency Lorenz condition to obtain a positive-semidefinite subspace, and quotient its null radical. The completed quotient is a positive Hilbert space with the same two photon helicities as transverse quantization.

Required background. Massive and Massless Spin-One Polarizations supplies the massless polarization quotient and its two helicities. Physical-Mode Quantization of the Free Electromagnetic Field supplies the positive two-polarization Fock space against which the covariant construction will be compared.

Helpful background. Spacelike Compatibility and Local Observables separates locality of a gauge-fixed potential from locality of gauge-invariant observables.

The covariant gauge-fixed Maxwell field

Work with the free Abelian field on four-dimensional Minkowski space in the standard Minkowski vacuum. Fields are smeared with rapidly decreasing test functions, spatial boundary terms vanish, and $\mathbf k=0$, harmonic, boundary, and global sectors are excluded. External conserved currents will be used only as a diagnostic of the free propagator, not as an interacting matter theory.

The linear covariant family is

$$\mathcal L_\xi =-\frac14F_{\mu\nu}F^{\mu\nu} -\frac1{2\xi}(\partial_\mu A^\mu)^2, \qquad \xi\neq0.$$

Integrating by parts under the stated falloff gives

$$\mathcal L_\xi \doteq \frac12 A_\mu \left[ \eta^{\mu\nu}\Box -\left(1-\frac1\xi\right) \partial^\mu\partial^\nu \right]A_\nu ,$$

where $\doteq$ denotes equality up to the discarded divergence. The field equation is therefore

$$\Box A^\mu -\left(1-\frac1\xi\right) \partial^\mu(\partial\cdot A)=0.$$

The gauge-fixing term does not impose $\partial\cdot A=0$ as an operator identity. It makes the quadratic operator invertible. In Feynman gauge, $\xi=1$, the same bulk action has the Fermi representative

$$\mathcal L_1 \doteq -\frac12\partial_\rho A_\mu\partial^\rho A^\mu, \qquad \Box A^\mu=0.$$

This representative will be used for the canonical construction. Changing a Lagrangian by a divergence can change boundary data and canonical surface terms, so this step depends on the falloff assumption; it is not an identity for arbitrary boundaries. The gauge-fixed quadratic operator and its inverse are developed in Schwartz 2014, § 8.5, printed pp. 128–130.

Four covariant oscillators form an auxiliary space

Treat $A^\mu$ as the coordinates of the Fermi-form Lagrangian. Their canonical momenta and equal-time algebra are

$$\begin{aligned} \Pi_\mu&=-\dot A_\mu, \\ [A^\mu(t,\mathbf x),\Pi_\nu(t,\mathbf y)] &=i\delta^\mu{}_\nu \delta^{(3)}(\mathbf x-\mathbf y), \end{aligned}$$

and hence

$$[A^\mu(t,\mathbf x),\dot A^\nu(t,\mathbf y)] =-i\eta^{\mu\nu}\delta^{(3)}(\mathbf x-\mathbf y).$$

For $\omega_{\mathbf k}=|\mathbf k|$ and

$$\int_{\mathbf k} \equiv \int\frac{\mathrm d^3\mathbf k} {(2\pi)^3\,2\omega_{\mathbf k}},$$

expand the Hermitian field as

$$A^\mu(x) =\int_{\mathbf k} \left[ a^\mu(\mathbf k)e^{-ik\cdot x} +a^{\mu\dagger}(\mathbf k)e^{ik\cdot x} \right].$$

The canonical commutator fixes

$$\begin{aligned} [a^\mu(\mathbf k),a^{\nu\dagger}(\mathbf q)] &=-(2\pi)^3\,2\omega_{\mathbf k}\, \eta^{\mu\nu} \delta^{(3)}(\mathbf k-\mathbf q), \\ [a^\mu,a^\nu]&=[a^{\mu\dagger},a^{\nu\dagger}]=0. \end{aligned}$$

The sign is forced rather than conventional. At equal times, the two mixed terms in $[A^\mu,\dot A^\nu]$ each supply half of $-i\eta^{\mu\nu}\delta^{(3)}$. Replacing the right-hand side of the oscillator algebra by $+\eta^{\mu\nu}$ would reverse the canonical commutator.

Let $a^\mu(f)|0\rangle=0$ define the auxiliary vacuum representation. Here $\dagger$ denotes the adjoint for the auxiliary indefinite-metric construction; the resulting sesquilinear form is not a Hilbert norm. Indeed, for any nonzero common wave packet $f$,

$$\begin{aligned} \bigl( a^{0\dagger}(f)|0\rangle, a^{0\dagger}(f)|0\rangle \bigr)_{\mathrm{aux}}&<0, \\ \bigl( a^{i\dagger}(f)|0\rangle, a^{i\dagger}(f)|0\rangle \bigr)_{\mathrm{aux}}&>0. \end{aligned}$$

The timelike oscillator is therefore negative-metric, whereas the three spatial oscillators are positive-metric. This auxiliary space is useful because $A^\mu$ transforms covariantly; it is not yet the physical state space.

The subsidiary condition selects physical representatives

In Feynman gauge the divergence is itself a free massless field. Its positive-frequency part is

$$(\partial_\mu A^\mu)^{(+)}(x) =-i\int_{\mathbf k} k_\mu a^\mu(\mathbf k)e^{-ik\cdot x}.$$

Define the Gupta–Bleuler subspace by the smeared condition

$$\mathcal V_{\mathrm{GB}} = \left\{ |\psi\rangle: (\partial\cdot A)^{(+)}(f)|\psi\rangle=0 \ \text{for every test function }f \right\}.$$

Equivalently,

$$k_\mu a^\mu(\mathbf k)|\psi\rangle=0$$

as a momentum-space distribution. Only the annihilation part is imposed. The stronger equation $\partial\cdot A(x)|\psi\rangle=0$ would let the creation part act on the vacuum and would remove even $|0\rangle$. Free evolution preserves the subsidiary condition because $\partial\cdot A$ obeys the wave equation. Moreover, if both bra and ket are Gupta–Bleuler states, then

$$\langle\phi|\partial\cdot A(x)|\psi\rangle=0:$$

the positive-frequency part kills the ket, and the adjoint negative-frequency part kills the bra.

The condition is not yet a quotient and does not say that every vector in $\mathcal V_{\mathrm{GB}}$ has strictly positive norm. Its precise effect is visible already in the one-particle sector.

Null states are quotiented, not kept as photons

Consider a wave-packet state

$$|f\rangle =\int_{\mathbf k} f_\mu(\mathbf k)a^{\mu\dagger}(\mathbf k)|0\rangle.$$

The auxiliary form and subsidiary condition reduce to

$$\begin{aligned} (f,g)_{\mathrm{aux}} &=-\int_{\mathbf k} f_\mu^*(\mathbf k)\eta^{\mu\nu}g_\nu(\mathbf k), \\ k^\mu f_\mu(\mathbf k)&=0. \end{aligned}$$

At a fixed momentum choose

$$k^\mu=(\omega,0,0,\omega), \qquad k_\mu=(\omega,0,0,-\omega).$$

Then $k^\mu f_\mu=0$ gives $f_3=-f_0$, and consequently

$$(f,f)_{\mathrm{aux}} =|f_1|^2+|f_2|^2\ge0.$$

The time–longitudinal combination has become null rather than negative. Its covector is proportional to $k_\mu$, so the one-particle physical space is

$$\mathcal H_{\mathrm{phys}}(\mathbf k) \simeq \frac{k^\perp}{\operatorname{span}\{k\}},$$

a positive two-dimensional quotient. Representatives related by $f_\mu\sim f_\mu+c\,k_\mu$ define the same class, and the field-strength polarization $k_\mu f_\nu-k_\nu f_\mu$ is unchanged. Circular combinations of the two spacelike transverse classes have helicities $+1$ and $-1$.

The same mechanism controls the full free Fock construction. At the displayed momentum, suppressing the common delta-function normalization, set

$$b=\frac{a^0-a^3}{\sqrt2}, \qquad c=\frac{a^0+a^3}{\sqrt2}.$$

Then

$$[b,b^\dagger]=[c,c^\dagger]=0, \qquad [b,c^\dagger]\neq0,$$

and the subsidiary condition is $b|\psi\rangle=0$. Transverse creators and $b^\dagger$ preserve the condition, while a $c^\dagger$ excitation does not. Every excitation containing $b^\dagger$ but no compensating nonphysical partner lies in the radical of the physical form. Thus one must define

$$\begin{aligned} \mathcal N &=\mathcal V_{\mathrm{GB}} \cap\mathcal V_{\mathrm{GB}}^\perp, \\ \mathcal H_{\mathrm{phys}} &=\overline{\mathcal V_{\mathrm{GB}}/\mathcal N}. \end{aligned}$$

The first line—not “all zero-norm vectors in an arbitrary indefinite space”—is the linear null subspace being removed. The form on $\mathcal V_{\mathrm{GB}}$ is positive semidefinite in this free theory and becomes positive definite on the completed quotient. Steinmann states this indefinite-space, subsidiary-subspace, null-quotient sequence explicitly in Steinmann 1989, p. 300.

With the same vacuum, boundary, zero-mode, and completion choices, the quotient is naturally isomorphic to the symmetric Fock space over the two transverse polarizations constructed on the preceding page. The covariant potential $A_\mu$ itself does not generally descend to an operator on equivalence classes: it can move a state by an unphysical direction. Gauge-invariant quantities such as $F_{\mu\nu}$ do descend.

The Feynman-gauge propagator

Time ordering the mode expansion gives the Minkowski-vacuum Feynman distribution

$$\begin{aligned} D_{F,1}^{\mu\nu}(x-y) &\equiv \langle0|\mathrm T A^\mu(x)A^\nu(y)|0\rangle \\ &= \int\frac{\mathrm d^4p}{(2\pi)^4} \frac{-i\eta^{\mu\nu}}{p^2+i0} e^{-ip\cdot(x-y)}. \end{aligned}$$

The numerator and its minus sign come from the four-component auxiliary oscillator algebra. They are not a sum over two positive physical polarizations. As an inverse-kernel check,

$$\begin{aligned} K_{\mu\nu}^{(1)}(p) D_{F,1}^{\nu\rho}(p) &=i\delta_\mu{}^\rho, \\ K_{\mu\nu}^{(1)}(p)&=-p^2\eta_{\mu\nu}. \end{aligned}$$

The $+i0$ selects the time-ordered boundary value. The same algebra without time ordering produces a Wightman function, while retarded, advanced, and commutator kernels require different support or boundary prescriptions. The covariant gauge-fixed propagator is developed in Schwartz 2014, § 8.5, printed pp. 128–130.

The propagator passes the conserved-current check

For $p^2\neq0$, introduce the algebraic transverse and longitudinal projectors

$$\begin{aligned} T^\mu{}_\nu &=\delta^\mu{}_\nu-\frac{p^\mu p_\nu}{p^2}, & L^\mu{}_\nu &=\frac{p^\mu p_\nu}{p^2}. \end{aligned}$$

The momentum-space quadratic operator is

$$K^\mu{}_\nu =-p^2T^\mu{}_\nu -\frac{p^2}{\xi}L^\mu{}_\nu.$$

Inverting the two orthogonal sectors and taking the conventional Feynman boundary-value extension gives

$$D_{\mu\nu}^{(\xi)}(p) =\frac{-i}{p^2+i0} \left[ \eta_{\mu\nu} -(1-\xi) \frac{p_\mu p_\nu}{p^2+i0} \right].$$

For $\xi\neq1$, the longitudinal double pole is a prescribed distribution, not an ordinary fraction that may be manipulated without its boundary prescription.

Now let $J^\mu$ and $J'^\nu$ be conserved test currents: $p_\mu J^\mu=p_\nu J'^\nu=0$. The difference between two gauge parameters is

$$D_{\mu\nu}^{(\xi)}(p) -D_{\mu\nu}^{(\xi')}(p) =-i(\xi-\xi') \frac{p_\mu p_\nu}{(p^2+i0)^2}.$$

Both longitudinal factors vanish upon contraction, so

$$J^\mu(-p)D_{\mu\nu}^{(\xi)}(p)J'^\nu(p) =-\frac{i\,J(-p)\cdot J'(p)}{p^2+i0},$$

independently of $\xi$. This is a bounded free-field check: it neither makes the gauge-variant correlator $\langle A_\mu A_\nu\rangle$ parameter independent nor proves the Ward identities of interacting QED. The quadratic inverse and the corresponding conserved-current diagnostic appear in Schwartz 2014, § 8.5.1, printed p. 130.

Two formulations, one physical photon sector

The reduced and covariant descriptions organize the same free radiative content differently.

Feature	Transverse quantization	Covariant Gupta–Bleuler quantization
Variables	Two spatial transverse modes	Four components of $A^\mu$
State form before constraints	Positive definite	Indefinite
Physical-state operation	Constraints and quotient solved first	Subsidiary condition, then null quotient
Covariance and locality	Not manifest; spatial projector is nonlocal	Local covariant potential and propagator are manifest
Physical result	Two positive photon helicities	The same two positive photon helicities

The equivalence concerns physical classes and gauge-invariant observables, not the gauge-fixed potentials term by term. Zinn-Justin compares canonical reduced gauges with the covariant gauge-fixed functional integral and gives the formal conserved-source equivalence for gauge-invariant quantities in Zinn-Justin 2021, §§ 21.5–21.6, printed pp. 516–520. That functional-integral comparison does not establish the Hilbert-space quotient; the subsidiary-condition calculation above does.

This page has used a free Abelian field, the Minkowski vacuum, nonzero plane-wave momenta, and no boundary or global sectors. Gauge Fixing, BRST, and BV develops the general gauge-fixed framework; Ghosts, Auxiliary Fields, and Gauge-Parameter Dependence explains structures deliberately absent from this free Abelian calculation. The QED Action, Charges, and Observables introduces interacting charged matter, while Dynamical Gauge Fields and Matter opens the non-Abelian generalization. Gupta–Bleuler alone should not be silently promoted to a proof for any of those settings.

Common pitfalls

Reading $-\eta_{\mu\nu}$ as a physical polarization sum. It is the numerator of the auxiliary covariant propagator and includes nonphysical directions. The physical one-particle space is $k^\perp/\operatorname{span}\{k\}$ and has only two positive directions.

Imposing the full Lorenz condition on every ket. The creation part of $\partial\cdot A$ does not annihilate the vacuum. Gupta–Bleuler imposes only the positive-frequency part and recovers the full condition in matrix elements between physical representatives.

Discarding time and longitudinal oscillators separately. The subsidiary condition combines them into a null direction and a forbidden partner. The null direction must then be quotiented; deleting two named components before that calculation obscures covariance and can give the wrong state form.

Calling every zero-norm auxiliary vector a removable state. In a general indefinite space, zero-norm vectors need not form a linear subspace. The relevant object is the radical $\mathcal V_{\mathrm{GB}}\cap\mathcal V_{\mathrm{GB}}^\perp$ of the positive-semidefinite subsidiary subspace.

Using the conserved-current check as an interacting theorem. The displayed cancellation is algebraic for free conserved test currents. Interacting gauge-parameter independence requires the appropriate Ward or BRST identities and a careful specification of observables.

Check your understanding

1. Recover the sign and normalization of the covariant oscillator commutator from the equal-time field algebra.

   Solution

   Insert the two mode expansions into $[A^\mu,\dot A^\nu]$. The annihilator–creator term contributes $-i\eta^{\mu\nu}/2$ times the Fourier delta, and the creator–annihilator term gives the same result after $\mathbf k\mapsto-\mathbf k$. Their sum is $-i\eta^{\mu\nu}\delta^{(3)}(\mathbf x-\mathbf y)$, which fixes $[a^\mu,a^{\nu\dagger}]$ to the displayed negative-metric normalization.

2. Prove positivity of the one-particle Gupta–Bleuler quotient at $k^\mu=(\omega,0,0,\omega)$.

   Solution

   The subsidiary condition gives $f_3=-f_0$. Substitution into $-f_\mu^*\eta^{\mu\nu}f_\nu$ cancels the time and longitudinal terms, leaving $|f_1|^2+|f_2|^2$. It vanishes precisely along $f_\mu\propto k_\mu$; quotienting that line leaves the two positive transverse components.

3. Show directly why the free conserved-current contraction is independent of $\xi$.

   Solution

   The difference of two propagators is proportional to $p_\mu p_\nu$. Contracting the first momentum with $J^\mu(-p)$ or the second with $J'^\nu(p)$ gives zero by current conservation. Only the $\eta_{\mu\nu}$ part remains, so the result is independent of the longitudinal coefficient and hence of $\xi$.

References

• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge: Cambridge University Press, 2014. DOI.
• Steinmann, Othmar. “On the Characterization of Physical States in Gauge Theories.” Annales de l’Institut Henri Poincaré. Physique Théorique 51, no. 3 (1989): 299–321. NUMDAM.
• Zinn-Justin, Jean. Quantum Field Theory and Critical Phenomena. 5th ed. Oxford: Oxford University Press, 2021. DOI.