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Average the equations in part b and make exactly the prescribed scale-separation approximation: the retained perturbations vary on lengths much greater than $\ell$, and perturbations at the tangle scale are omitted. Products of two retained perturbations can then be neglected. Writing $\mathbf{v}=\delta \mathbf{u}$, $m=\delta M$ and $\pi =\delta p$, the linearized equations areNo unmodelled small-scale response has been inserted into this closure.

Introduce the fluid displacement $\mathbf{\xi }$ by $\mathbf{v}={\partial }_t\mathbf{\xi }$ and choose $\nabla \cdot \mathbf{\xi }=0$. Integrating the magnetic second-moment tensor equation giveswhere $s$ is a time-independent integration tensor set by the initial stress. For a pure oscillatory normal mode, $s=0$. Its restoring force is therefore $v_A^2{\nabla }^2\mathbf{\xi }$: the isotropic magnetic tangle has the same transverse response as an incompressible elastic medium with shear modulus ${\rho }_0{v}_A^2$.

For a plane wave proportional to $e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, with $\omega \ne 0$, incompressibility gives $\mathbf{k}\cdot \hat{\mathbf{v}}=0$. The tensor equation and magnetohydrodynamic momentum equation becomeDotting the second equation with $\mathbf{k}$ yields $\hat{\pi }=0$ for $k\ne 0$, since $k_i{k}_j{\hat{m}}_{ij}=0$. Substitution then givesThere are two independent transverse polarizations. The phase speed is $v_A$ in every direction, and the two signs of $\omega$ give group velocities $\pm v_A\hat{\mathbf{k}}$. These magnetoelastic waves in an isotropic magnetic tangle exist because a shear deformation distorts the second moment of the tangled magnetic field, producing a restoring magnetic tension even though its mean is zero.

For comparison, a uniform magnetic field with Alfvén velocity ${\mathbf{B}}_0$ gives the incompressible Alfvén wave relation ${\omega }^2=(\mathbf{k}\cdot {\mathbf{B}}_0{)}^2$. Its frequency depends on the direction of $\mathbf{k}$, and energy propagates along ${\mathbf{B}}_0$. Here the averaged restoring response is isotropic: there is no preferred mean-field direction. Both systems resist transverse deformation through magnetic tension, rather than through compressive sound.

The propagating branch is not the entire perturbation space. For example, $\mathbf{v}=0$, $m_{ij}=f(\mathbf{r}){\delta }_{ij}$ and $\pi =f(\mathbf{r})$ solve the linearized equations without propagation, for any small, slowly varying $f$. Such a static stress perturbation in magnetoelastic wave equations balances a pressure gradient. Thus the requested wave conclusion applies to the oscillatory branch; it must not exclude stationary initial stress perturbations.

There are two chemical atoms in the primitive unit cell, hence six displacement coordinates and six phonon branches at each wavevector. Three rigid translations cost no restoring potential energy in the long-wavelength limit. ConsequentlyFor propagation along $[100]$, one acoustic phonon has longitudinal polarization along $x$ and two have transverse polarization along $y,z$. The cubic crystal symmetry makes those two transverse directions equivalent. In terms of the elastic stiffness tensor and the mass density,Thus there are two distinct acoustic velocities in general, one longitudinal and a doubly degenerate transverse value. This degeneracy of normal modes concerns the two transverse polarizations; it does not imply equality with the longitudinal speed of sound.

Use $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ throughout. Vary the Maxwell action with respect to the electromagnetic four-potential. Antisymmetry of the electromagnetic field tensor gives $\delta \mathcal{L}=-F^{\mu \nu }{\partial }_{\mu }\delta A_{\nu }$. After integration by parts, the Euler-Lagrange field equation isA gauge transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$ leaves $F_{\mu \nu }$ unchanged because partial derivatives commute. It therefore leaves the Maxwell action and observable electromagnetic fields unchanged: different potentials represent the same physical field. This redundancy matters for canonical quantization: before gauge fixing, the momentum conjugate to $A_0$ vanishes and the four potential components cannot be treated as four independent physical oscillators.

Choose the Lorenz gauge $D={\partial }_{\mu }{A}^{\mu }=0$, attainable by solving $\square \chi =-D$ with suitable boundary data. The Euler-Lagrange field equation then reduces toResidual Lorenz gauge symmetry consists of transformations with $\square \chi =0$. The gauge condition and this residual redundancy ultimately leave two transverse polarizations of a free photon.

For the specified kinetic density ${\mathcal{L}}_C=-\frac{1}{2}{\partial }_{\mu }{A}_{\nu }{\partial }^{\mu }{A}^{\nu }$, one has $\partial {\mathcal{L}}_C/\partial ({\partial }_{\mu }{A}_{\nu })=-{\partial }^{\mu }{A}^{\nu }$. Its Euler-Lagrange field equation is therefore $-\square A^{\nu }=0$. This density is the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction, sinceThe divergence follows by expanding both sides and commuting the second derivatives. Thus the wave equation follows from a Feynman gauge action; the additional physical-state condition will implement the Lorenz gauge. One must not impose $D=0$ inside the original action before varying it.

Take the covariant components $A_{\nu }$ as canonical coordinates. The canonical momentum computed from ${\mathcal{L}}_C$ isThe equal-time canonical Poisson brackets are $\{A_{\mu }(t,\mathbf{x}),{\Pi }^{\nu }(t,\mathbf{y})\}={\delta }_{\mu }{}^{\nu }{\delta }^3(\mathbf{x}-\mathbf{y})$, with coordinate-coordinate and momentum-momentum brackets zero. Applying canonical quantization givesIn particular $[A_{\mu },{\dot{A}}_{\nu }]=-i{g}_{\mu \nu }{\delta }^3$. The negative temporal sign is the source of the indefinite state-space norm in covariant photon quantization.

The spatial Fourier transform of the wave equation gives an oscillator of frequency $\omega =|\mathbf{k}|$ for each component. The Hermitian adjoint condition on the electromagnetic four-potential then yields the mode expansion of a free fieldThese are massless on shell modes. Consistently with the preceding equal-time canonical commutation relation,The normalization is the same as for the scalar modes, with $-g_{\mu \nu }$ replacing the species identity matrix.

Gupta-Bleuler quantization starts with the Fock vacuum annihilated by every $a_{\mu }$ and imposes only the annihilation, or positive-frequency part of a quantum field, of the Lorenz gauge condition:Imposing the entire Hermitian field divergence as a strong operator identity would be incompatible with the four-component canonical commutation relation. The weaker condition suffices: between two physical states its positive-frequency part vanishes on the ket, and its negative-frequency part vanishes on the bra. Hence $⟨{\mathrm{phys}}^{\prime }|{\partial }_{\mu }{A}^{\mu }|\mathrm{phys}⟩=0$.

For a one-photon state $|\mathbf{k},ϵ⟩={ϵ}^{\nu }{a}_{\nu }^{†}(\mathbf{k})|0⟩$, commuting the constraint through the creator givesThus the allowed photon polarization vectors satisfy $\boxed{k\cdot ϵ=0}$. Their state inner product, with the common momentum-normalization factor suppressed, is $⟨{ϵ}^{\prime }|ϵ⟩=-{ϵ}^{\prime \ast }\cdot ϵ$. Momentum eigenstates are delta-normalized; the same statements hold with finite norms for normalizable wave packets.

Rotate the spatial axes so that $k=\omega (1,0,0,1)$. The constraint is ${ϵ}^0={ϵ}^3$, so the complete allowed family isThe first vector is the allowed longitudinal gauge polarization, proportional to the null four-momentum $k$. Its state has zero norm because $k^2=0$; it is also orthogonal to every constrained state because $k\cdot ϵ=0$. A purely spatial vector $(0,0,0,1)$ alone is not an allowed physical polarization: its time component must accompany it to satisfy the constraint. The other two vectors are transverse polarizations, perpendicular to $\mathbf{k}$, with positive unit norms. For the displayed family,The timelike negative-norm excitation has been excluded by the constraint, and the remaining form is positive semidefinite. The Gupta-Bleuler null-state quotient identifies $ϵ$ and $ϵ+\lambda k$, removing the null direction. It leaves exactly two transverse physical photon polarizations with a positive inner product; their bosonic Fock space is the physical photon state space.

Finally use Feynman slash notation, $/a={\gamma }^{\mu }{a}_{\mu }$, and impose four-momentum conservation $p+q=k+k^{\prime }$. The external Dirac spinors obeySet $r=p-k$ and $r^{\prime }=p-k^{\prime }$. Replacing $ϵ$ by $k$ in the first term of the electron-positron annihilation into two photons amplitude, the rightmost contracted vertex obeysSince the Clifford algebra implies $(/r+m)(/r-m)=(r^2-m^2)I_4$, that term becomesFor the second term, four-momentum conservation gives $k=q+r^{\prime }$. The contracted vertex is now adjacent to the outgoing positron spinor on the left, soConsequentlyThe two terms cancel, proving the two-photon fermion Ward identityExchanging the two photon labels proves the same result for ${ϵ}^{\prime }=k^{\prime }$. By linearity the amplitude is invariant under $ϵ\mapsto ϵ+\lambda k$ on either external photon leg. Therefore the null longitudinal gauge polarization has zero production amplitude: physical scattering produces only the two transverse photon polarizations. This cancellation also shows why both photon orderings in the tree-level Feynman diagrams are essential; either term alone fails the Ward identity.

Take $0\le \sigma \le \pi$ and split the coordinates into $a=0,\dots ,p$ along the branes and $I=p+1,\dots ,25$ transverse to them. In conformal gauge the wave equation permits a separated mode $e^{-i\omega \tau }f(\sigma )$ with $f^{\prime \prime }+{\omega }^{2}f=0$. For Neumann boundary conditions at both ends, $f^{\prime }(0)=f^{\prime }(\pi )=0$ selects $\cos m\sigma$ and integer $m$. The zero-frequency solution is independent of $\sigma$, with a constant position and a term linear in time. Therefore the open-string mode expansion isThe canonical momentum density integrated over this interval is $k^a$.

For Dirichlet boundary conditions, subtract the static interpolation between fixed endpoint positions $y_0^I$ and $y_{\pi }^I$. The remaining field vanishes at both ends, selecting $\sin m\sigma$ with integer $m$ and excluding a momentum worldsheet zero mode. The expansion isHere ${\alpha }_m^{I†}={\alpha }_{-m}^I$. There is no factor $i$ in the sine series with this reality convention: changing $m$ to $-m$ reverses both the sine and the denominator. These are the integer-mode expansion with parallel D-brane endpoints; they are not the half-integer modes of a coordinate with a Neumann condition at one endpoint and a Dirichlet condition at the other.

The endpoints can move on $(p+1)$-dimensional worldvolumes but are fixed in the transverse directions. Thus these are open strings attached to D-branes, or stretched between parallel D-branes at $y_0$ and $y_{\pi }$. The tangential momenta describe propagation on the brane; the transverse positions describe its embedding. Let $L=|y_{\pi }-y_0|$. The linear spatial worldsheet zero mode contributes $L^2/(4{\pi }^2{\alpha }^{\prime })$ to $L_0$. Quantization with string intercept one givesThe first term is also the square of the classical stretched-string energy $TL$, since the string tension is $T=1/(2\pi {\alpha }^{\prime })$.

For two coincident D3-branes, label the ends of an oriented open string by $i,j\in \{1,2\}$. The four endpoint combinations give Chan-Paton factors ${\lambda }_{ij}$, which form the full algebra of $2\times 2$ matrices. Under a change of brane basis these transform as $\lambda \mapsto U\lambda U^{-1}$, the adjoint action of $U(2)$. At $L=0$, the first oscillator level is massless. Its tangential polarization supplies a four-dimensional vector $A_a$, while the 22 transverse polarizations supply real adjoint scalar fields ${\Phi }^I$. This is the bosonic D3-brane low-energy field content.

The endpoint matrices multiply in their boundary order in disk amplitudes. The difference between the two orders in the three-vector interaction gives the matrix commutator; the kinematic factor is the one-derivative Yang-Mills theory cubic vertex. Factorization and the vector gauge redundancy then supply the quartic interaction with the same commutator. Equivalently, the leading low-energy action is the dimensional reduction of 26-dimensional Yang-Mills theory, withFor Hermitian scalar fields the last displayed term corresponds to a nonnegative potential, because their commutators are anti-Hermitian. Higher-derivative terms are suppressed at energies well below $1/\sqrt{{\alpha }^{\prime }}$. Ignoring the tachyon, the gauge sector is four-dimensional $U(2)$ Yang-Mills theory, accompanied by 22 adjoint scalar fields. It is not pure Yang-Mills theory, nor the supersymmetric six-scalar D3-brane theory of a ten-dimensional superstring.

Separate the two branes and use the D-brane scalar-position normalizationA Yang-Mills gauge transformation preserves this expectation value only when it is diagonal if $y_1\ne y_2$. Thus the unbroken group is $U(1)\times U(1)$. The scalar field kinetic term supplies a mass for the off-diagonal vectors, sinceSumming over $I$ gives $m_{12}^2=L^2/(4{\pi }^2{\alpha }^{\prime 2})$, exactly the mass of the stretched $N=1$ string. The two diagonal gauge vectors remain massless. This identifies brane separation with an adjoint Higgs mechanism, rather than an explicit deletion of the off-diagonal strings.

For the ground state of an open string joining the two different branes, $N=0$, so the bosonic stretched-string tachyon threshold isAt equality it is massless, and above it the stretched ground state has positive mass squared. The same-brane ground strings still have $M_0^2=-1/{\alpha }^{\prime }$; this calculation removes only the stretched-string instability, not the original instability of the bosonic branes.

The starting point is a physical massless closed-string stateIts symmetric traceless part is a spacetime spin-two field, the graviton; the antisymmetric part is the Kalb–Ramond field, and the scalar field part gives the dilaton. The physical-state conditions impose transverse polarizations, and longitudinal null states identifyfor the symmetric component. In position space this is the linearized diffeomorphism $\delta h_{\mu \nu }={\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }$. Thus the massless spin-two state has the gauge redundancy required of a gravitational field, rather than an arbitrary massive tensor interaction.

Closed-string amplitudes factorize on this graviton pole. Its leading soft coupling is universal and couples to the energy-momentum tensor of the other string states. Decoupling longitudinal polarizations enforces the gravitational Ward identities. The two-derivative cubic graviton vertex extracted from the low-momentum three-point amplitude is the cubic vertex of the Einstein-Hilbert action; continuing these Ward identities requires the nonlinear diffeomorphism-invariant completion. A complementary way to derive the same dynamics is to demand quantum consistency of propagation in a curved background.

Replace the flat target metric in the worldsheet action by $G_{\mu \nu }(X)$, and allow the two-form and dilaton backgrounds. After Euclidean continuation, their couplings have the schematic normalizationThese backgrounds are couplings of a string nonlinear sigma model. Quantum Weyl invariance requires their Weyl anomaly coefficients to vanish. To first order in ${\alpha }^{\prime }$, the metric equation isThe dilaton and two-form equations accompany it. In a critical-dimensional background with $H=0$ and constant $\Phi$, the leading metric equation is $R_{\mu \nu }=0$, precisely the vacuum Einstein field equations. Matter and other massless string fields supply the corresponding stress-energy when retained. This argument also explains why arbitrary curved target metrics are not consistent string backgrounds.

The equations arise from the leading string-frame massless effective actionA constant-dilaton expansion of its curvature term directly gives the Einstein-Hilbert kinetic and self-interaction terms. For a varying dilaton, introduce the Einstein-frame metric $G_{\mu \nu }^E=e^{-4\Phi /(D-2)}{G}_{\mu \nu }$. The Weyl rescaling of the metric and the corresponding transformation of the Ricci scalar and an integration by parts giveThis displays ordinary Einstein gravity coupled to the dilaton and two-form. Gravity need not be the only low-energy field: those fields, compactification moduli and gauge fields have to be retained or stabilized as appropriate. Four-dimensional gravity additionally requires compactification from the critical target dimension, with its Newton constant depending on the compactification volume.

The approximation is controlled when external energies obey $E\sqrt{{\alpha }^{\prime }}\ll 1$ and curvatures obey ${\alpha }^{\prime }|R_{\mu \nu \rho \sigma }|\ll 1$. Massive string modes then produce suppressed higher-derivative interactions, such as higher-curvature corrections. Small string coupling $g_s=e^{{\Phi }_0}$ also suppresses higher-genus quantum corrections; dimensionally $G_D$ scales as $g_s^2{\alpha }^{\prime (D-2)/2}$. Thus Einstein gravity is the universal leading two-derivative dynamics of the massless closed-string spin-two field, with additional massless matter, string-length corrections and quantum corrections systematically controlled.

For the bosonic theory this is the massless-sector effective description in 26 dimensions; the tachyon means that it is not by itself a proof of a stable vacuum. Tachyon-free supersymmetric string theories have the same massless graviton mechanism in ten dimensions before compactification. Removing the tachyon and obtaining a realistic four-dimensional model require extra physical input beyond the emergence of the Einstein action.

The oscillator commutator givesThe temporal mode $\lambda =0$ has negative norm. More precisely, the nonzero wave packet $|f,0⟩=\int d^{3}pf(\mathbf{p})a_{\mathbf{p}}^{0†}|0⟩/(2\pi {)}^3$ has norm $-\int d^{3}p|f(\mathbf{p}){|}^2/(2\pi {)}^3<0$. The difficulty is an indefinite Hermitian form, not the divergent normalization of an unsmeared momentum eigenstate. The covariant photon Fock space is therefore not a positive physical Hilbert space.

Gupta-Bleuler quantization imposes only the annihilation, or positive-frequency part of a quantum field, of the divergence:A strong operator identity $\partial \cdot A=0$ is incompatible with the four unconstrained canonical commutation relations; in the literal momentum convention it would set ${\pi }^0=0$. The weaker condition instead implies $⟨\Phi |\partial \cdot A|\Psi ⟩=0$ between physical states: the positive-frequency part annihilates the ket, and its adjoint annihilates the bra.

For a nonzero momentum choose upper-index polarization vectorswhere ${\mathbf{e}}_1,{\mathbf{e}}_2$ are an orthonormal pair perpendicular to $\mathbf{p}$. Labels 1 and 2 have transverse polarization, label 3 has spatial longitudinal polarization, and label 0 has timelike photon polarization. With $p^0=|\mathbf{p}|=\omega$, the divergence of the annihilation part is proportional to $-i\omega (a_{\mathbf{p}}^0-a_{\mathbf{p}}^3)$. The physical condition is thereforeFor a one-mode state $|\chi ⟩={\sum }_{\lambda }{c}_{\lambda }{a}^{\lambda †}|0⟩$, the condition reads $-c_0-c_3=0$. ThusThe last direction has zero norm and is orthogonal to every constrained state. Its field wavefunction is proportional to ${ϵ}^{\mu 0}+{ϵ}^{\mu 3}=p^{\mu }/\omega$, hence it is a pure gauge symmetry direction. A purely longitudinal creator $a^{3†}$ alone is not physical, despite having positive norm.

For the multiparticle sketch, put $C=a^0-a^3$ and $C^{†}=a^{0†}-a^{3†}$. The commutator $[C,C^{†}]=0$ allows polynomials in $C^{†}$ times transverse creator states to obey the constraint. Conversely, on creator polynomials the constraint is $-{\partial }_{a^{0†}}-{\partial }_{a^{3†}}$, so these are all the constrained polynomials. Every term containing $C^{†}$ is orthogonal to every constrained bra, since $⟨\Phi |C^{†}=⟨C\Phi |=0$. Quotienting those null vectors and completing the remaining positive space is the Gupta-Bleuler null-state quotient. It givesThe one-particle instance is the transverse one-photon physical quotient.

A stable symmetry-breaking minimum requires $\lambda >0$ and $m^2<0$. Put $v^2=-m^2/\lambda >0$. The common scalar potential can then be written, up to a constant, asIts minima form the sphere $\Phi \cdot \Phi =v^2$. Choose the vacuum direction ${\Phi }_0=(0,0,v{)}^T$. If instead $m^2>0$ with positive $\lambda$, the minimum is at the origin and the assumed nonzero vacuum expectation value is not a stable vacuum of these models.

In the ungauged theory, the internal symmetry is global $O(3)$; its connected part is $SO(3)$. Constant orthogonal transformations preserve both the kinetic term and the potential. The chosen vacuum is invariant under $O(2)$ acting on its first two components, so the continuous breaking is $SO(3)\to SO(2)$ with two broken generators. The vacuum orientations are genuinely different degenerate global vacua.

The global triplet scalar symmetry breaking is visible directly in the fluctuation masses. Write $\Phi =({\pi }_1,{\pi }_2,v+h{)}^T$. The potential Hessian isThe quadratic Lagrangian is thereforeThere is one radial scalar of mass $m_h^2=2\lambda v^2=-2m^2$ and two massless Goldstone bosons. The two Goldstone modes describe motion tangential to the vacuum sphere. They remain physical, since a global rotation has only constant parameters and cannot remove arbitrary spacetime-dependent fluctuations. All three original real-scalar degrees of freedom remain. Interactions persist: the shifted potential is $V-V_{\min }=\frac{\lambda }{4}(2vh+h^2+{\pi }_1^2+{\pi }_2^2{)}^2$, containing cubic and quartic terms. Masslessness of the angular modes is protected by the exact continuous symmetry through Goldstone's theorem.

In the gauged theory, the three-component real field is the vector representation of local $SO(3)$, equivalently the Adjoint representation of $SU(2)$. The center of $SU(2)$ acts trivially on this field, so both descriptions have the same local field content and perturbative spectrum. Take $(t^a{)}_{bc}=-i{ϵ}^{abc}$, for which $[t^a,t^b]=i{ϵ}^{abc}{t}^c$. The covariant kinetic term and the gauge kinetic term are invariant under local transformations with the associated connection transformation. The scalar inversion $\Phi \mapsto -\Phi$, with the gauge field unchanged, is also a discrete invariance of the displayed action.

The vacuum leaves rotations about its third axis unbroken. The residual gauge group is $SO(2)$, or the corresponding $U(1)$ subgroup in the $SU(2)$ description. Expanding the covariant kinetic term about ${\Phi }_0$ givesThus the gauge-boson mass matrix is $g^2{v}^2\mathrm{diag}(1,1,0)$. The third generator annihilates the vacuum, while the first two do not. The masses are $m_{B^1}=m_{B^2}=gv$, $m_{B^3}=0$, and $m_h=\sqrt{2\lambda }v$.

The quadratic derivative mixing displays what happens to the Goldstone fields. With the generator convention just chosen,Local rotations can set the two angular fields to zero in unitary gauge. Then $\Phi =(0,0,v+h{)}^T$ andThe two Goldstone modes supply the longitudinal polarizations of the two massive vector bosons. They do not survive as additional physical massless scalars. The radial scalar is still physical, and $B^3$ retains only two transverse polarizations. This adjoint triplet Higgs spectrum realizes the Higgs mechanism; the gauge group is only partially broken, so one massless gauge boson remains.

The degree-of-freedom check isIn contrast, the global theory has just the three scalar modes, including its two physical Goldstone bosons. A vacuum expectation value in the local theory is a choice of gauge-fixed description: the direction can be changed by gauge transformations and is not itself an observable. Gauge-independent masses and the physical degree count express the actual effect. Both theories retain Lorentz symmetry in the constant scalar vacuum.

The oscillator-generated covariant photon Fock space has an indefinite Hermitian form, not a positive Hilbert space inner product. For a normalizable one-photon wave packet of polarization $\lambda$, its squared norm is proportional to $-{\eta }^{\lambda \lambda }\int d^{3}p|f(\mathbf{p}){|}^2$. Thus $\lambda =0$ gives a negative-norm photon state. The divergent ${\delta }^3(0)$ of an unsmeared momentum eigenstate is a separate normalization issue, avoided by the wave packet.

Choose contravariant polarization vectors ${ϵ}^{0\mu }=(1,0)$, ${ϵ}^{3\mu }=(0,\hat{\mathbf{p}})$, and two ${ϵ}^{r\mu }=(0,{\mathbf{e}}_r)$ with ${\mathbf{e}}_r\cdot \mathbf{p}=0$ for $r=1,2$. The third spatial polarization is longitudinal polarization; the first two are transverse polarization. The timelike photon polarization is distinct from the longitudinal one.

The Gupta-Bleuler quantization condition sets the divergence of the positive-frequency part of a quantum field to zero on physical states:Restore the factors $e^{-i|\mathbf{p}|t}$ to the annihilation terms. Since $p_{\mu }{ϵ}^{0\mu }=|\mathbf{p}|$ and $p_{\mu }{ϵ}^{3\mu }=-|\mathbf{p}|$, Fourier transform gives the equivalent conditionIts sign depends on the chosen sign of the longitudinal polarization vector; the covariant condition does not.

To see its content for a general Fock state, temporarily discretize momentum and decompose one unphysical oscillator sector as $|\Psi ⟩={\sum }_{n,m\ge 0}|n,m{⟩}_{0,3}\otimes |{\chi }_{nm}{⟩}_{\perp }$, where $|n,m⟩=(a^{0†}{)}^n(a^{3†}{)}^m|0⟩/\sqrt{n!m!}$. The temporal oscillator obeys $a^0|n,m⟩=-\sqrt{n}|n-1,m⟩$, while $a^3|n,m⟩=\sqrt{m}|n,m-1⟩$. The condition therefore becomesEquivalently, the allowed finite-particle states use the transverse creation operators and only $b^{†}=a^{0†}-a^{3†}$ in the unphysical sector. Indeed the constraint acts on a polynomial of the two unphysical creation operators as $-{\partial }_{a^{0†}}-{\partial }_{a^{3†}}$, whose kernel consists of polynomials in their difference. Also $[b,b^{†}]=-1+1=0$, and a state containing $b^{†}$ is orthogonal to every constrained state because $b$ annihilates every such state. This argument applies mode by mode and extends by smearing to continuum momentum.

For one photon, $c_0{a}^{0†}|0⟩+c_3{a}^{3†}|0⟩$ is constrained only when $c_3=-c_0$, producing a null state. The Gupta-Bleuler null-state quotient removes these null directions. The condition excludes negative-norm physical states; quotienting its null states leaves the two positive-norm transverse photon polarizations. The condition alone gives a positive semidefinite Hermitian form, not yet a positive definite Hilbert space.

In vacuum the Maxwell equations areSubstitution of the complex plane electromagnetic wave amplitudes into these equations givesThus the electric field must be transverse to $k$; taking the positive-frequency branch gives $\omega =c|k|$.

For $E_0=E(1,i,0)$, $k=k\hat{z}$, and $\phi =kz-\omega t$, the real fields areThe electric vector rotates with constant magnitude, so this is circular polarization. Its Poynting vector isThe Poynting vector is the electromagnetic energy flux: its direction is the direction of propagation and its magnitude is the power crossing unit area.

In a vacuum with no charge or current, the Maxwell equations areFor the proposed plane electromagnetic wave, $\nabla \cdot B=0$ givesThe Ampère-Maxwell equation givesand substituting this into Faraday's law gives the dispersion relationThus both fields have transverse polarization relative to $k$, and the corresponding real electric field is

For incidence in the positive $x$-direction, choose the incident fieldsThe perfect conductor requires the tangential electric field to vanish at $x=0$. The reflected wave therefore hasThis is normal reflection of an electromagnetic wave from a perfect conductor. The magnetic field in $x\le 0$ isso its tangential value at the surface is

For a fixed propagation direction and spatial mode, a photon has two independent transverse polarization amplitudes. Orthogonal horizontal and vertical states span its two-dimensional Hilbert space: $\alpha |H⟩+\beta |V⟩$, with $|\alpha {|}^2+|\beta {|}^2=1$. Polarization analyzers measure bases in this space, and phase plates implement transformations of its relative amplitudes. Additional spatial or frequency modes enlarge the space and must not be silently discarded.