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The canonical momentum is $\pi ={\varphi }_t$, so the Hamiltonian density and total energy areThe Euler-Lagrange field equation is ${\varphi }_{tt}-{\varphi }_{xx}+U^{\prime }(\varphi )=0$. Multiplication by ${\varphi }_t$ gives the local conservation of energy identityThus $E$ is conserved when the energy flux vanishes at spatial infinity. For finite energy with asymptotically constant fields, the endpoints ${\varphi }_{\pm }$ must be scalar-field vacua, where $U=0$.

Choose an antiderivative $G^{\prime }(\varphi )=\sqrt{2U(\varphi )}$. For either sign $s=\pm 1$, completing the square givesChoosing $s$ to make the boundary term nonnegative proves the Bogomolny boundThese Bogomolny equations obtained by square completion for a one-dimensional kink describe minima within a fixed endpoint sector when a saturating configuration exists. Differentiating the spatial equation gives ${\varphi }_{xx}=U^{\prime }(\varphi )$ wherever $U>0$, with smooth extension to its limiting scalar-field vacua. The absolute minimum across all sectors is a constant zero-potential scalar-field vacuum when one exists; the nontrivial minima are sector-specific kinks.

For $\beta \ne 0$, write $b=|\beta |$. The scalar-field vacua are $-b,0,b$. A static finite-energy solution has the first integralbecause differentiating its left side gives ${\varphi }_x({\varphi }_{xx}-U^{\prime })=0$, and its value is zero at infinity. In either adjacent interval, $\sqrt{2U}=|\varphi |(b^2-{\varphi }^2)$. Set $y={\varphi }^2/b^2$. The first-order equation becomes $y_x=\eta 2b^{2}y(1-y)$, with $\eta =\pm 1$, and separation gives $\log [y/(1-y)]=2\eta b^2(x-x_0)$. HenceFor $\eta =1$ the endpoints are $0\to \sigma b$; for $\eta =-1$ they are $\sigma b\to 0$. Thus there are four oriented static kink families, up to translation, namely two increasing kinks and two decreasing antikinks. Each has rest energyA static field joining $-b$ directly to $b$ would have to pass through $\varphi =0$ at finite $x$. The first integral then gives ${\varphi }_x=0$, and the Picard-Lindelöf theorem applied to the smooth second-order equation with those initial conditions forces $\varphi \equiv 0$. Therefore there is no such additional kink; this is the intermediate-vacuum obstruction to a kink. The Bogomolny classification of a rescaled phi-six kink also shows that if $\beta =0$, only the scalar-field vacuum $0$ remains and there is no nontrivial finite-energy static kink. Indeed the first integral gives a monotone field wherever it is nonzero, precluding a return to the same scalar-field vacuum.

A Lorentz boost of the $0\to b$ profile gives a moving solution with arbitrary $|v|<1$:If $K$ denotes the static profile and $\xi =\gamma (x-vt-x_0)$, then ${\varphi }_{tt}-{\varphi }_{xx}={\gamma }^2(v^2-1)K^{\prime \prime }=-K^{\prime \prime }$, proving the Euler-Lagrange field equation directly. Its energy is $\gamma b^4/4$: since $K^{\prime 2}=2U(K)$, changing variable to $\xi$ in the energy integral gives the factor $\gamma$. Reversing $\sigma$ or $\eta$ yields the other moving elementary profiles.

The nonexistence conclusion concerns nonconstant fields: a constant scalar-field vacuum with $U=0$ is a zero-energy critical point of an energy functional in every dimension. Assume the potential is smooth, so the classical Euler-Lagrange equation is well defined. Write the two nonnegative energy terms asFor the admissible Derrick scaling ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$,Stationarity at $\lambda =1$ gives the Derrick virial identityFor $D>2$, both terms have nonpositive sign, forcing $T=V=0$ and hence a constant vacuum. For $D=2$, scaling alone only proves $V=0$; the gradient term is scale invariant, so it must not simply be declared zero.

The two-dimensional flat-target Derrick obstruction supplies that remaining step. Since $U\ge 0$ and its integral is zero, continuity gives $U(\varphi (x))=0$ everywhere; smooth nonnegative $U$ has $U^{\prime }(\varphi (x))=0$ at those values. The field equation $\Delta \varphi =U^{\prime }(\varphi )$ therefore becomes $\Delta \varphi =0$. Each ${\partial }_i\varphi$ is an entire harmonic function in $L^2({\mathbb{R}}^2)$. Its mean value property and the Cauchy-Schwarz inequality give, for every radius $R$,Let $R\to \infty$ to get ${\partial }_i\varphi =0$. Thus $D=2$ also has only constant stationary field configurations of finite energy. The unconstrained real scalar target is essential in this argument; the spherical target in Question 2 has different field equations.

In $D=1$ the field equation is ${\varphi }^{\prime \prime }=U^{\prime }(\varphi )$. Multiplication by ${\varphi }^{\prime }$ gives the first integralFinite energy makes $\frac{1}{2}({\varphi }^{\prime }{)}^2+U(\varphi )$ an integrable nonnegative function, so along some sequence $x_n\to +\infty$ it tends to zero. Evaluating the constant first integral on that sequence gives $K=0$. ChooseConsequently $({\varphi }^{\prime }{)}^2=2U(\varphi )$ andFor a nonconstant entire solution the sign is fixed. Indeed a finite-$x$ zero of ${\varphi }^{\prime }$ would have $U(\varphi )=U^{\prime }(\varphi )=0$; uniqueness of the smooth second-order equation with initial data $(\varphi ,{\varphi }^{\prime })=(v,0)$ would make the solution the constant vacuum. Thus ${\varphi }^{\prime }$ never vanishes in a nonconstant solution. Conversely a smooth solution of the displayed first-order equation satisfies ${\varphi }^{\prime \prime }=U^{\prime }(\varphi )$ wherever it is nonconstant.

Under the usual soliton boundary conditions $\varphi (x)\to v_{\pm }$ as $x\to \pm \infty$, with $v_{\pm }$ in the vacuum manifold $\{U=0\}$, the two endpoint components label the topological sector. For isolated scalar-field vacua this is an ordered pair of vacuum labels, not necessarily one universally normalized integer. More explicitly, for any smooth $F$, the topological currentis identically conserved, taking ${ϵ}^{01}=1$ and boundary conditions fixed during evolution. The choices $F(u)=u$ and $F(u)=W(u)$ give the field-difference charge and the kink energy charge. The square completion for a one-dimensional kink givesThe first-order critical solution saturates this Bogomolny bound. A nonzero difference between endpoint vacuum components prevents continuous deformation to a vacuum while those boundary conditions remain fixed. Finite energy alone need not give finite endpoint values for arbitrary potentials, so these topological labels use the stated vacuum boundary conditions.

To evade the scalar-only obstruction, introduce a gauge field and a charged, generally multicomponent Higgs field. For a Yang-Mills-Higgs energy write $E=T_H+V+E_B$, where $T_H$ contains $|D_i\Phi {|}^2$ and $E_B$ contains $|F_{ij}{|}^2$. Rescale ${\Phi }_{\lambda }(x)=\Phi (\lambda x)$ and $A_{i,\lambda }(x)=\lambda A_i(\lambda x)$. Then $D_i\Phi$ scales by $\lambda$, while $F_{ij}$ scales by ${\lambda }^2$, soThe magnetic term supplies the missing opposing scaling power. In $D=2$ the relation is $E_B=V$, permitting Abelian Higgs vortices; in $D=3$ it is $E_B=T_H+3V$, permitting 't Hooft-Polyakov monopoles. In the monopole Bogomolny limit $V=0$, $B_i=\pm D_i\Phi$ achieves $E_B=T_H$. Also, asymptotic ordinary derivatives need not vanish when a Higgs phase winds: the relevant vanishing quantity is the gauge covariant derivative. This allows nontrivial boundary winding with finite energy. Merely appending a gauge potential to a neutral one-component real field would not by itself furnish that Higgs mechanism.

There is a sign inconsistency in the printed model. For the ordinary cross product let $J(a)z=a\times z$. The vector triple-product identity gives $[J(a),J(b)]=J(a\times b)$. Consequently the printed $D_{\mu }={\partial }_{\mu }+eJ(A_{\mu })$ has gauge curvatureIts gauge curvature cannot instead contain the printed negative quadratic term. For a noncommuting pure gauge potential for this gauge covariant derivative, $F^{+}=0$ whereas $F^{-}=-2e{A}_{\mu }\times A_{\nu }\ne 0$. Thus the printed gauge kinetic term assigns a nonzero value to a connection obtained by applying the gauge-field transformation law to the zero connection. Literally the printed action has global rotations but not the requested local triplet gauge symmetry. The cross-product convention for an adjoint covariant derivative makes the minimal repair explicit: change the gauge field strength quadratic term to a plus sign, retaining the printed gauge covariant derivative. Changing the sign in the gauge covariant derivative instead would be another consistent convention. The remaining calculation uses the first repair.

The corrected theory has local SO(3) gauge symmetry, equivalently the adjoint-field realization of the SU(2) Lie algebra; its adjoint scalar fields and gauge fields are insensitive to the center of a group of SU(2). Explicitly, $\Phi \mapsto R(x)\Phi$ andBoth $D_{\mu }\Phi$ and $F_{\mu \nu }$ transform as triplets, so their dot products and the scalar potential are invariant. Assume $\lambda >0$, $e\ne 0$ and take $v\ge 0$ without loss of generality. For $v>0$ the minima have ${\Phi }^2=v^2$. Choose the scalar-field vacuum ${\Phi }_0=v(0,0,1)$, whose stabilizer consists of rotations around the third axis: the residual gauge symmetry is SO(2), locally U(1).

In unitary gauge write $\Phi =(0,0,v+h)$, $A_{\mu }=A_{\mu }^3$ and $W_{\mu }^{\pm }=(A_{\mu }^1\mp i{A}_{\mu }^2)/\sqrt{2}$. These are the neutral massless gauge field, two charged massive gauge fields and a real radial scalar field. DefineSubstitution gives $F_{\mu \nu }^3=f_{\mu \nu }-ie{C}_{\mu \nu }$ and $(F_{\mu \nu }^1\mp i{F}_{\mu \nu }^2)/\sqrt{2}=G_{\mu \nu }^{\pm }$. The entire physical-field Lagrangian density is thereforeIn particular $D_{\mu }\Phi =(e(v+h)A_{\mu }^2,-e(v+h)A_{\mu }^1,{\partial }_{\mu }h)$; this directly verifies the scalar field kinetic and vector mass terms. Expanding the scalar potential gives $\lambda v^2{h}^2/2+\lambda v{h}^3/2+\lambda h^4/8$. Comparing the quadratic terms with standard real scalar field and charged-vector mass terms yields the adjoint triplet Higgs spectrum,This also rewrites the purely vector interactions, which are already contained in the displayed gauge field strengths rather than omitted from the theory.

The original symmetry remains unbroken at $v=0$. The unique minimum is then $\Phi =0$, with three massless gauge fields and three massless real scalar field components; the quartic scalar field interaction remains. The physical degrees of freedom number $3\times 2+3=9$. At $v>0$ the two broken generators supply two Goldstone bosons, eaten to give longitudinal gauge-boson polarizations of the charged vectors. The physical count is $2\times 3+2+1=9$: two massive vectors, one massless vector and one radial scalar field. The scalar components eaten in unitary gauge are not additional physical particles.

For the broken phase all interactions containing the physical scalar field areMultiplying interaction coefficients by $i$ and differentiating with respect to the external fields includes the identical-field factorials. The scalar vertices of an adjoint triplet Higgs model areFor example, the $h^3$ rule contains $3!$, whereas the $hh{W}^{+}{W}^{-}$ rule contains only $2!$ from its two identical scalar fields. There are no $hAA$ or $hhAA$ vertices: the radial scalar field is neutral under the residual U(1). There is no derivative scalar-vector vertex in this physical spectrum in unitary gauge.

For completeness, if the unbroken phase is chosen, its physical scalar fields are the three components ${\varphi }_a$, not a single radial $h$. Expanding its gauge-covariant kinetic term givesWith all momenta incoming and Fourier transform convention $e^{-ipx}$, the three unbroken-phase rules areThe first rule follows by differentiating once on each possible scalar: for $a=3,b=1,c=2$, the interaction is $e{A}_{\mu }^3({\varphi }_1{\partial }^{\mu }{\varphi }_2-{\varphi }_2{\partial }^{\mu }{\varphi }_1)$, which fixes the momentum sign. The remaining rules follow from the quadratic gauge interaction and $-\lambda ({\varphi }_a{\varphi }_a{)}^2/8$. There are no unbroken-phase cubic scalar field Feynman vertices.

Use units $\hslash =c=1$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. Treat the complex scalar field and its adjoint as independent variables when varying the action. Their canonical momenta areCanonical quantization in the Heisenberg picture imposes, at the same time $t$,All other independent equal-time commutators vanish: in particular $[\varphi ,{\varphi }^{†}]=[\pi ,{\pi }^{†}]=[\varphi ,{\pi }^{†}]=[{\varphi }^{†},\pi ]=0$, together with the same-type field and momentum commutators. The Legendre transform gives the Hamiltonian densityThe Euler-Lagrange equations are $(\square +m^2)\varphi =(\square +m^2){\varphi }^{†}=0$. The positive- and negative-frequency plane waves form a complete basis of solutions of the Klein-Gordon equation. Since the field is complex, their operator coefficients are independent rather than being adjoints of each other. Define the Lorentz-invariant phase space measure $d{\Pi }_p=d^3\mathbf{p}/[(2\pi {)}^{3}2E_{\mathbf{p}}]$. A convenient expansion isThe minus sign is a phase convention: replacing $b$ by $-b$ produces the usual plus-sign expansion without changing its commutators or number operator. The normalization is fixed by the canonical commutation relations. For example,when $[a(\mathbf{p}),a^{†}(\mathbf{q})]=[b(\mathbf{p}),b^{†}(\mathbf{q})]=(2\pi {)}^{3}2E_{\mathbf{p}}{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$. The cross commutators vanish and the other canonical relation follows by taking adjoints with the operator order reversed. Conversely these mode commutators can be obtained by Fourier inversion of the equal-time relations. This is canonical quantization of a complex scalar field.

Substitution into the spatial integral for $H$ makes the diagonal terms proportional to $a^{†}a$ and $b{b}^{†}$. Terms creating or annihilating a pair have $\mathbf{q}=-\mathbf{p}$ and a coefficient $E_{\mathbf{p}}^2-{\mathbf{p}}^2-m^2=0$, so they cancel. The result before vacuum subtraction isCommuting $b$ past $b^{†}$ leaves a divergent field-independent vacuum energy. The displayed zero-vacuum-energy form must therefore be understood as a normal-ordered Hamiltonian:A regulator can be introduced before subtracting the constant. This is the normal-ordered Hamiltonian of a free complex scalar field; the unrenormalized Hamiltonian is not literally equal to this expression without that convention.

The scalar-field vacuum is annihilated by both annihilation operators. Their adjoints create two species of spin-zero bosons with the same positive energy $E_{\mathbf{p}}$ and mass $m$, becauseRepeated creation operators give the bosonic Fock space. The negative-frequency part does not create a negative-energy physical state; it creates the antiparticle of positive energy. The charge distinguishes the two species.

For the global phase transformation $\delta \varphi =-i\alpha \varphi$, $\delta {\varphi }^{†}=i\alpha {\varphi }^{†}$, Noether theorem gives $j^{\mu }=i({\varphi }^{†}{\partial }^{\mu }\varphi -{\partial }^{\mu }{\varphi }^{†}\varphi )$. Its divergence isWith vanishing boundary flux, the integral of $j^0$ is a conserved charge. Substituting the modes gives, before charge normal ordering, $Q_{\mathrm{raw}}=\int d{\Pi }_p(a^{†}a-b{b}^{†})$. The pair terms cancel because their frequency difference is zero when their spatial momenta sum to zero. Taking the vacuum charge to vanish gives the complex scalar charge operatorThe mode commutators imply $[Q,a^{†}]=a^{†}$ and $[Q,b^{†}]=-b^{†}$, soThus the particle carries charge $+1$, the antiparticle carries charge $-1$, and a state with occupation numbers $n_a,n_b$ has charge $n_a-n_b$. Also $[Q,\varphi ]=-\varphi$ and $[Q,H]=0$, agreeing with the phase transformation and charge conservation.