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Take $D_{\mu }X={\partial }_{\mu }X+[A_{\mu },X]$, signature $(+---)$ and the trace Lagrangian density $\mathrm{Tr}(F_{\mu \nu }{F}^{\mu \nu })/4-\mathrm{Tr}(D_{\mu }\Phi D^{\mu }\Phi )/2+m^2\mathrm{Tr}({\Phi }^2)/2$. The variation of the gauge curvature is $D_{\mu }\delta A_{\nu }-D_{\nu }\delta A_{\mu }$. Cyclicity of the trace and covariant integration by parts give gauge coefficient $-D_{\mu }{F}^{\mu \nu }+[D^{\nu }\Phi ,\Phi ]$ and scalar coefficient $D_{\mu }{D}^{\mu }\Phi +m^2\Phi$. Their vanishing gives the displayed Euler-Lagrange field equations. In static temporal gauge at zero mass they reduce to $D_j{F}_{ji}=[\Phi ,D_i\Phi ]$ and $D_i{D}_i\Phi =0$.

// Destination: integrable-systems.bigb

Use metric $(+--)$, $D_{\mu }={\partial }_{\mu }-i{a}_{\mu }$ and Lagrangian $-f_{\mu \nu }{f}^{\mu \nu }/4+\overline{D_{\mu }\varphi }{D}^{\mu }\varphi /2-(1-|\varphi {|}^2{)}^2/8$. The Euler-Lagrange field equations are $D_{\mu }{D}^{\mu }\varphi +(|\varphi {|}^2-1)\varphi /2=0$ and ${\partial }_{\mu }{f}^{\mu \nu }=\mathrm{Im}(\bar{\varphi }{D}^{\nu }\varphi )$. Varying $\bar{\varphi }$ gives a kinetic contribution $-D_{\mu }{D}^{\mu }\varphi /2$ and potential contribution $(1-|\varphi {|}^2)\varphi /4$; variation of $a_{\nu }$ gives ${\partial }_{\mu }{f}^{\mu \nu }-\mathrm{Im}(\bar{\varphi }{D}^{\nu }\varphi )$. For static fields in temporal gauge $a_0=0$, the positive Bogomolny vortex equations imply the scalar equation because $(D_1-i{D}_2)(D_1+i{D}_2)=D_1^2+D_2^2+B$. They imply the spatial gauge equations because $\mathrm{Im}(\bar{\varphi }{D}_1\varphi )=-{\partial }_2|\varphi {|}^2/2$ and $\mathrm{Im}(\bar{\varphi }{D}_2\varphi )={\partial }_1|\varphi {|}^2/2$, while $B=(1-|\varphi {|}^2)/2$. The Gauss equation then vanishes identically.

The identically conserved topological current $j$ couples to an external electromagnetic gauge potential through $-A_{\mu }{j}^{\mu }=-A_0{\theta }_x+A_1{\theta }_t$. A gauge transformation changes this coupling only by a divergence. The Euler-Lagrange field equation becomes ${\theta }_{tt}-{\theta }_{xx}+\sin \theta +E=0$, with $E={\partial }_t{A}_1-{\partial }_x{A}_0$. A unit Sine-Gordon kink has unnormalized charge $Q=2\pi$ and mass $8$. For spatially uniform forcing, its field momentum balance is $\dot{P}=QE$ when the boundary stress contributions cancel.

With the usual Dirac adjoint, Hermitian ${\gamma }^5$ and $\{{\gamma }^0,{\gamma }^5\}=0$, taking the adjoint gives ${\psi }^{†}{\gamma }^5{\gamma }^0\psi =-{\psi }^{†}{\gamma }^0{\gamma }^5\psi$. Consequently a real pseudoscalar with coupling $-g\varphi \bar{\psi }{\gamma }^5\psi$ has a Hermitian interaction only if $g^{\ast }=-g$. Writing $g=i{g}_P$ with real $g_P$ gives the familiar interaction $-i{g}_P\varphi \bar{\psi }{\gamma }^5\psi$. Formal Euler-Lagrange field equations obtained with independent spinor and adjoint variables can still be computed for the original symbolic coupling; it must not silently be identified with a real physical coupling.

The source-free electromagnetic Lagrangian density is quadratic in the electromagnetic field tensor. Its Euler-Lagrange field equation is ${\partial }_{\mu }{F}^{\mu \nu }=0$. Antisymmetry of $F$ makes the density invariant under gauge transformations, and contraction of its indices makes it a Lorentz scalar.

The displayed action functional contains no mass term and no $R{\varphi }^2$ curvature coupling. Its Euler-Lagrange field equation is ${\partial }_{\mu }(\sqrt{-g}{g}^{\mu \nu }{\partial }_{\nu }\varphi )=0$. In a flat FLRW metric, a Fourier mode obeys ${\varphi }_k^{\prime \prime }+2(a^{\prime }/a){\varphi }_k^{\prime }+k^2{\varphi }_k=0$ in four dimensions. The expansion can therefore freeze these modes; the curvature term of a conformally coupled scalar field changes that conclusion.

For the global phase symmetry of a complex scalar field oriented as $\delta \psi =-i\epsilon \psi$, the Noether current is $j^{\mu }=i({\psi }^{\ast }{\partial }^{\mu }\psi -\psi {\partial }^{\mu }{\psi }^{\ast })$. The Euler-Lagrange field equations $\square \psi +V^{\prime }(|\psi {|}^2)\psi =0$ and its complex conjugate give ${\partial }_{\mu }{j}^{\mu }=i({\psi }^{\ast }\square \psi -\psi \square {\psi }^{\ast })=0$ for a real differentiable potential. Integrating $j^0$ gives $Q$ when boundary flux vanishes. A charged scalar coupled to electromagnetism has electric charge $qQ$ with the chosen charge normalization; without that physical identification it is an internal charge.

A nonminimal scalar coupling includes an interaction between the field and spacetime curvature, such as $-\xi R{\Phi }^2$ in the Lagrangian density. The Euler-Lagrange field equation for $\mathcal{L}=-\frac{1}{2}(\nabla \Phi {)}^2-\xi R{\Phi }^2$ is $\square \Phi -2\xi R\Phi =0$. The coefficient is convention dependent: an action written with $-\frac{1}{2}\xi R{\Phi }^2$ has $\xi$ instead of $2\xi$.

A four-dimensional Majorana spinor gravitino has $4\times 4=16$ real components, minus four local supersymmetry gauge functions, leaving twelve before using any Euler-Lagrange field equation. A vierbein has sixteen real components, minus six local Lorentz transformation and four diffeomorphism gauge functions, leaving six. Equivalently, a symmetric metric tensor has $10-4=6$. The propagating graviton and massless gravitino each have two states on shell. Six bosonic auxiliary field components complete an off-shell supergravity multiplet.

Old-minimal four-dimensional supergravity adds a complex scalar $M$ and a real vector $b_{\mu }$ to the vierbein and gravitino. These six real bosonic auxiliary field components have no propagating kinetic terms. They balance the off-shell component count of minimal supergravity and permit the local supersymmetry algebra to close without imposing the propagating Euler-Lagrange field equations.

A field configuration or identity is on shell when the equations of motion hold.

Vary the Maxwell Lagrangian with respect to $A_{\nu }$. Since $\delta F_{\mu \nu }={\partial }_{\mu }\delta A_{\nu }-{\partial }_{\nu }\delta A_{\mu }$ and $F$ is antisymmetric,After integration by parts, the Euler-Lagrange field equation isThe gauge transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$ leaves $F_{\mu \nu }$ unchanged because partial derivatives commute. This gauge invariance means that potentials related in this way represent the same electromagnetic field; all four potential components are not independent physical photon degrees of freedom. In particular, the ungauged kinetic operator is singular and its temporal canonical momentum vanishes.

Choose the Lorenz gauge ${\partial }_{\mu }{A}^{\mu }=0$. The field equation then becomes $\boxed{\square A_{\mu }=0}$. Residual gauge transformations with $\square \chi =0$ preserve this condition. The correctly named Lorenz gauge condition concerns the divergence of the potential; Lorentz covariance is the separate transformation property of the equations.

For the alternative quadratic Lagrangian density, direct differentiation givesso its Euler-Lagrange field equation is also $\square A^{\nu }=0$. This density is the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction. Indeed, with $D={\partial }_{\mu }{A}^{\mu }$,The divergence identity follows by expanding ${\partial }_{\mu }{K}^{\mu }$ and commuting its second derivatives. The quadratic density alone describes four wave equations; the gauge constraint must still be incorporated in the physical states. It is not an unconstrained four-positive-polarization replacement for Maxwell equations.

Using this specified boundary-term convention, the canonical momentum conjugate to $A_{\nu }$ isThe canonical commutation relations are thereforewith field-field and momentum-momentum brackets zero. Equivalently $[A_{\mu },{\dot{A}}_{\nu }]=-i{g}_{\mu \nu }{\delta }^3$. The temporal component has the opposite sign to the spatial components; replacing the metric by a positive Euclidean one would destroy these brackets.

The wave equation gives massless frequencies ${\omega }_{\mathbf{k}}=|\mathbf{k}|$. Hermiticity pairs its positive- and negative-frequency solutions, so the Heisenberg picture field has the expansionThis is a real field expansion for each component, but its oscillator metric is indefinite. The assumed brackets, with the dagger's vector index restored, areThey verify the photon oscillator completeness and canonical brackets. Inserting ${\Pi }^{\nu }=-{\dot{A}}^{\nu }$ gives at equal timesThe other brackets cancel between the two frequency contributions. Thus both the $2\omega$ normalization and the sign $-g_{\mu \nu }$ are needed.

The Gupta-Bleuler formalism imposes the Lorenz gauge condition weakly on states:Here $(+)$ is the annihilation, positive-frequency part. Setting the full divergence to zero as an operator would conflict with the four-component canonical commutation relations. The subsidiary condition instead selects a physical pre-space inside the covariant photon Fock space, followed by a quotient of its null states.

Apply it to $|k,ϵ⟩={ϵ}^{\nu }{a}_{\nu }^{†}(k)|0⟩$. Commuting $k^{\mu }{a}_{\mu }$ through the creator gives a multiple of $-k_{\nu }{ϵ}^{\nu }|0⟩$. Hence the allowed photon polarization vectors obeyTake $k=(\omega ,0,0,\omega )$. The complete solution iswith complex coefficients. The last two vectors are transverse linear photon polarizations, and $(0,1,\pm i,0)/\sqrt{2}$ gives the two circular photon polarizations. The first vector is $k/\omega$, the allowed temporal-longitudinal combination. A purely spatial longitudinal vector $(0,0,0,1)$ by itself does not satisfy the subsidiary condition.

The oscillator brackets give, apart from the positive continuum normalization factor,Momentum eigenstates are distributionally normalized; a box or a wave packet makes the norm statement literal. The allowed longitudinal combination has zero norm because $k^2=0$, and it is orthogonal to every constrained state since $k\cdot ϵ=0$. It also gives a pure-gauge field wave: the field strength of $A_{\mu }\propto k_{\mu }{e}^{-ik\cdot x}$ vanishes. Thus $ϵ$ and $ϵ+ck$ represent the same physical photon polarization. The Gupta-Bleuler null-state quotient removes this null direction and leaves two positive-norm transverse photon photon polarizations. The negative-norm temporal oscillator of the unrestricted space is not an extra physical state. This is the transverse one-photon physical quotient.

Finally verify the two-photon fermion Ward identity directly for electron-positron annihilation into two photons. Define $D(r)=/r-m$ and the reduced Dirac propagatorThe equality follows from the Clifford algebra, since $(/r-m)(/r+m)=(r^2-m^2)I$. The incoming Dirac spinors obey $D(p)u(p)=0$ and $\overline{v}(q)(/q+m)=0$. On replacing one inserted photon polarization vector by $k$, the spinor factor in the two diagrams isAt the electron end, $/k=D(p)-D(p-k)$, soFor the other diagram, four-momentum conservation gives $p-k^{\prime }=k-q$. HenceThe two contributions therefore cancel exactly:Individual diagrams need not vanish; their sum implements the Ward identity. Linearity then gives invariance under $ϵ\mapsto ϵ+ck$, as required by gauge invariance. In particular, the null longitudinal photon polarization proportional to $k$ has zero physical production amplitude. Physical scattering produces the transverse photon classes, not an independent longitudinal massless photon.

Use Minkowski spacetime signature $(+,-)$ and put $\kappa =|\lambda |>0$. The Euler-Lagrange field equation for the real scalar field isIndeed, differentiating the potential $U(\varphi )={\kappa }^2({\varphi }^2-1{)}^2/2$ gives $U^{\prime }=2{\kappa }^2\varphi ({\varphi }^2-1)$, while the kinetic term contributes ${\partial }_{\mu }{\partial }^{\mu }\varphi$.

Translation invariance gives the symmetric stress-energy tensorIts divergence is $({\partial }_{\mu }{\partial }^{\mu }\varphi +U^{\prime }){\partial }^{\nu }\varphi$, which vanishes by the Euler-Lagrange field equation. Consequently the conserved energy and physical spatial momentum areThe minus sign in $P=\int T^{01}dx$ follows from ${\partial }^1=-{\partial }_x$. Conservation assumes the associated stress-energy fluxes vanish at spatial infinity, as they do for the localized fields below.

A static finite-energy field configuration approaches vacua $\varphi =\pm 1$. Multiplying ${\varphi }^{\prime \prime }=U^{\prime }(\varphi )$ by ${\varphi }^{\prime }$ giveswhere the vacuum limits fix the constant. For a kink increasing from $-1$ to $1$, ${\varphi }^{\prime }=\kappa (1-{\varphi }^2)$. Integration gives $\mathrm{artanh}\varphi =\kappa (x-X)$ and henceIts negative is the antikink. The arbitrary center $X$ is a translational collective coordinate. The static kink mass is its rest energy; using ${\varphi }^{\prime 2}=2U$ and $y=\kappa (x-X)$ givesEquivalently, completing the static energy into a square gives $E=\frac{1}{2}\int [{\varphi }^{\prime }-\kappa (1-{\varphi }^2){]}^{2}dx+\kappa [\varphi -{\varphi }^3/3{]}_{-\infty }^{\infty }$, so this kink saturates its Bogomolny bound.

A Lorentz boost with velocity $v$ givesWriting $z=\gamma (x-X-vt)$, the derivatives are ${\varphi }_t=-\gamma v{\varphi }_{\mathrm{K}}^{\prime }(z)$ and ${\varphi }_x=\gamma {\varphi }_{\mathrm{K}}^{\prime }(z)$. The momentum integral therefore becomesThe boosted energy isbecause ${\gamma }^{-1}=\gamma (1-v^2)$. Thus $E^2-P^2=M^2$ and $P/E=v$, exactly the energy–momentum relation of a relativistic particle of rest mass $4\kappa /3$. This is the relativistic energy and momentum of a phi-four kink. If $\lambda =0$, the double-well potential disappears and there is no localized static kink connecting these vacua.

All three fields have mass $m$. In the interaction picture the quantum fields evolve with the free Hamiltonian operator, so their equations areThe interactions evolve the states instead. For $\chi \in \{\varphi ,\psi ,\Phi \}$ the appropriate mode expansion of a free field isIndependent species have commuting creation and annihilation operators. The full set of canonical commutation relations isThese follow from the separate equal-time canonical commutation relations for the three real scalar fields. For comparison, in the Heisenberg picture the full Euler-Lagrange field equations contain interactions: $(\square +m^2){\varphi }_H=-g{\varphi }_H{\Phi }_H$, $(\square +m^2){\psi }_H=-g{\psi }_H{\Phi }_H$, and $(\square +m^2){\Phi }_H=-\frac{g}{2}({\varphi }_H^2+{\psi }_H^2)$. These are distinct from the free equations used in the interaction picture.

There are no derivative couplings, so the interaction Hamiltonian is $H_I(t)=-\int d^{3}x{\mathcal{L}}_I(x)$. The interaction picture evolution operator obeysIterating its integral equation gives the Dyson seriesThe ordered integration region is equivalent to the unrestricted region with time ordering and the factor $1/n!$. Taking the scattering limit $t_0\to -\infty$, $t\to \infty$, with the usual adiabatic prescription for perturbation theory, therefore yields the S-matrixAll fields inside this expression are interaction picture fields.

Use relativistic normalization of a one-particle state and take $|i⟩=a_{\varphi }^{†}(\mathbf{p})a_{\psi }^{†}(\mathbf{q})|0⟩$, $|f⟩=a_{\varphi }^{†}({\mathbf{p}}^{\prime })a_{\psi }^{†}({\mathbf{q}}^{\prime })|0⟩$. Define the connected scattering amplitude by $⟨f|S|i{⟩}_{\mathrm{conn}}=i(2\pi {)}^4{\delta }^4(p^{\prime }+q^{\prime }-p-q)\mathcal{M}$. The identity contribution describes no scattering; the term linear in $g$ cannot connect these four external particles. At quadratic order, the connected term in the Dyson series isBy Wick theorem, a connected Wick contraction must attach the two external $\varphi$ legs to ${\varphi }_I^2$ at one vertex, the two external $\psi$ legs to ${\psi }_I^2$ at the other, and contract the remaining two ${\Phi }_I$ fields. The two assignments of species to $x,y$ cancel the Dyson factor $1/2!$. At either vertex, the $2!$ assignments of its identical fields to the incoming and outgoing particle cancel the interaction factor $1/2$. Equivalently, the coefficient $-g^2/8$ is multiplied by $2\times 2\times 2$, giving $-g^2$. Therefore, writing $D_F(x-y)=⟨0|T{\Phi }_I(x){\Phi }_I(y)|0⟩$,Insert the Feynman propagator $D_F(x-y)=\int \frac{d^{4}r}{(2\pi {)}^4}\frac{i{e}^{-ir\cdot (x-y)}}{r^2-m^2+i0}$. The two vertex integrals impose $r=p^{\prime }-p$ and $p^{\prime }+q^{\prime }=p+q$. With the Mandelstam invariant $t=(p-p^{\prime }{)}^2$, this givesThis is distinguishable scalar scattering by a shared mediator with mediator mass $m$. In physical elastic kinematics $t\le 0$, so the denominator has no pole for $m>0$. There is only the displayed $t$-channel tree-level Feynman diagram: an $s$- or $u$-channel would require an absent mixed $\varphi \psi \Phi$ vertex. Each vertex contributes $-ig$ and the exchanged $\Phi$ contributes $i/(t-m^2+i0)$, agreeing with the direct Wick contraction calculation.

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.

For a compactly supported variation $\varphi \mapsto \varphi +\epsilon \eta$, the first variation of the energy functional isIntegration by parts has no endpoint term. Since $\eta$ is arbitrary, the static Euler-Lagrange field equation isIntegrating the specified polynomial derivative gives $U(\varphi )={\varphi }^2/2-{\varphi }^4+{\varphi }^6/2+C$. The polynomial factors as a nonnegative square times ${\varphi }^2$; its global minimum is zero when $C=0$. ThereforeThese are the three constant scalar-field vacua. For a localized finite-energy kink, the two ends approach vacua and the derivative tends to zero. Multiplying the field equation by ${\varphi }^{\prime }$ gives the first integralA nonconstant kink must stay within one interval between neighboring vacua. A proposed direct connection from $-1$ to $1$ would reach $\varphi =0$ at some finite point, where the first integral forces ${\varphi }^{\prime }=0$. Uniqueness for the smooth second-order differential equation then forces the identically zero solution, a contradiction. This is the intermediate-vacuum obstruction to a kink; two elementary kinks can span the outer vacua only in an infinite-separation limit, not as one finite-width static kink.

There are consequently four oriented elementary sectors: $-1\to 0$, $0\to -1$, $0\to 1$ and $1\to 0$. If opposite orientations are grouped together, there are two kink families and their antikinks. Spatial reflection reverses orientation, and $\varphi \mapsto -\varphi$ exchanges the two adjacent intervals; both preserve the energy. All four therefore have the same rest mass, even though the central and outer vacua have different fluctuation curvatures.

For the sector $0\to 1$, define $W(\varphi )={\varphi }^2/2-{\varphi }^4/4$, so $U=(W^{\prime }{)}^2/2$ and $W^{\prime }=\varphi (1-{\varphi }^2)$. The Bogomolny bound follows by completing the square:It is saturated by the Bogomolny equationLet $y={\varphi }^2$ on this positive branch. Then $y^{\prime }=2y(1-y)$, so $\log [y/(1-y)]=2(x-x_0)$. Thus the explicit kink in a phi-six model isIt has the required limits and satisfies the first integral. The translated center $x_0$ is arbitrary. All oriented elementary solutions areFor $\eta =1$ the endpoints are $0\to \sigma$, and for $\eta =-1$ they are $\sigma \to 0$. Direct evaluation ${\int }_0^1\sqrt{2U(\varphi )}d\varphi ={\int }_0^1\varphi (1-{\varphi }^2)d\varphi =1/4$ independently gives their common mass.