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This interaction preserves the global U(1) symmetry with relative charges $q_{\varphi }=1$, $q_{\Phi }=2$. The two conjugate interaction vertices have Feynman rule $-ig$, with either one charge-two arrow entering and two charge-one arrows leaving, or all arrows reversed. For $\varphi \varphi \to \varphi \varphi$ only the heavy $s$-channel exchange occurs at tree level. For $\varphi \bar{\varphi }\to \varphi \bar{\varphi }$, with outgoing particle labelled 3 and outgoing antiparticle labelled 4, only the heavy $u$-channel exchange occurs. For small momenta and heavy mediator mass, the amplitudes coincide but the identical-particle total scattering cross-section is smaller by the identical final-state symmetry factor $1/2$.

Divide the infinitesimal variation by its constant parameter $\alpha$, so the transformation generator is $(\delta {\varphi }_1,\delta {\varphi }_2)=({\varphi }_2,-{\varphi }_1)$. The variation of the kinetic term isand that of ${\varphi }_1^2+{\varphi }_2^2$ is $2{\varphi }_1{\varphi }_2-2{\varphi }_2{\varphi }_1=0$. Thus the Lagrangian density is invariant pointwise, without a nonzero boundary current. The finite transformation is a rotation of the two real scalar fields, giving an SO(2) global symmetry in field theory.

Noether's theorem supplies the parameter-independent Noether currentIts divergence is ${\partial }_{\mu }{j}^{\mu }={\varphi }_2\square {\varphi }_1-{\varphi }_1\square {\varphi }_2$, since the first-derivative terms cancel. Both fields obey the same Klein-Gordon equation, so this vanishes on solutions. With spatial boundary conditions giving zero flux at infinity, the Noether charge isA finite box with periodic boundary conditions works as well. The spatial integral is conserved because the local continuity equation converts its time derivative into a boundary flux; neither vanishing local current nor constant individual fields is required. Equivalently, $\Phi =({\varphi }_1+i{\varphi }_2)/\sqrt{2}$ transforms as $\Phi \mapsto e^{-i\alpha }\Phi$, exhibiting the equivalent global U(1) symmetry and fixing the sign of the generator.

Take the free squared mass $m^2\ge 0$, as for the ordinary massive free theory. The quartic part of the scalar potential can be writtenIt is nonnegative for every real field value exactly whenNecessity follows from the axis ${\varphi }_2=0$ and the diagonal ${\varphi }_1={\varphi }_2$: a negative coefficient on either ray makes $V$ tend to minus infinity under rescaling. The displayed sum proves sufficiency, since $m^2({\varphi }_1^2+{\varphi }_2^2)/2$ is also nonnegative. This is a copositive matrix condition on the quadratic form in the nonnegative variables $({\varphi }_1^2,{\varphi }_2^2)$, rather than positive definiteness on all independent signed variables.

The strict inequalities $\lambda >0$, $\kappa >-\lambda$ make the quartic part grow in every nonzero direction. They are not necessary for a stable vacuum when $m^2>0$. For example, $\lambda =\kappa =0$ gives two stable free massive fields; $\lambda >0$, $\kappa =-\lambda$ has a flat quartic diagonal but a confining quadratic term. If $m^2=0$, boundary cases have flat vacuum directions and need not have a unique isolated minimum. If one instead allows negative squared mass, a flat quartic ray makes the full potential unbounded below; the stability conditions must then be strengthened. This distinguishes boundedness from strict quartic coercivity.

Under the generator of a, the variation isIt vanishes identically if and only if $\kappa =\lambda$. In that case $V_4=\lambda ({\varphi }_1^2+{\varphi }_2^2{)}^2$ preserves the entire global U(1) symmetry. More explicitly, the interacting Euler-Lagrange equations giveHenceA special solution lying on an axis does not establish an operator conservation law. Among the stable couplings, the symmetry-preserving line is $\kappa =\lambda \ge 0$.

Let $C(u,v)=2D_F(u-v)$ be the action-consistent free propagator established in a. Write $S=S_0+S_I$ and putThe perturbative expansion isDivide numerator and denominator by the same free vacuum functional $Z_0$. For $\mathcal{O}=\Phi (y_1)\cdots \Phi (y_N)$, they becomeFor $\mathcal{O}=\varphi (y_1){\varphi }^{\ast }(y_2)$, abbreviate $C_{12}=C(y_1,y_2)$ and $C_{zz}=C(z,z)$. Since only opposite-charge fields contract, Wick's theorem givesandThe first multiplicity is $2!$ internal pairings. In the connected term, there are two choices for the vertex's ${\varphi }^{\ast }$ contracted to the external $\varphi$ and two for its $\varphi$ contracted to the external ${\varphi }^{\ast }$; the remaining pair forms a tadpole diagram. ThereforeExpanding the inverse denominator cancels the free-line times vacuum bubble term, leavingThe local quartic interaction vertex has factor $-i\lambda$, since $2!2!$ cancels its denominator 4. The oriented tadpole diagram has Feynman-diagram symmetry factor 1; the isolated two-loop vacuum vertex has factor $1/2$ in its coefficient. Coincident kernels and the integrals require a common regularization; none is being assigned a finite numerical value here.

More generally, vacuum bubbles exponentiate and factor from every numerator. Here global U(1) symmetry also makes all one-point functions zero. Thus a surviving graph with the two external insertions in separate connected components would have a component with one unbalanced charge and vanish. This proves that only connected graphs contribute to this particular two-point function to all perturbative orders. Vacuum normalization alone would not remove disconnected components containing external legs in a general four-point function. If one formally chooses a different free kernel, the displayed Wick multiplicities are unchanged; replacing $C$ by $C_{\mathrm{lit}}$ gives a formal oscillator series, not the Gaussian integral of the action as printed.

Work in the symmetric vacuum and continue with the action-consistent kernel $C(u,v)=2D_F(u-v)$. Under global U(1) symmetry, a correlator with $r$ fields $\varphi$ and $4-r$ fields ${\varphi }^{\ast }$ picks up a phase $e^{i\alpha (2r-4)}$. Invariance for every $\alpha$ requires $r=2$. Consequently ten of the sixteen ordered field-type choices vanish, and the six placements with two fields of each type can be nonzero. This is the charge selection rule for complex scalar correlators, not an assertion about a symmetry-breaking vacuum with a nonzero one-point function.

Choose external fields $\varphi (x_1)\varphi (x_2){\varphi }^{\ast }(y_1){\varphi }^{\ast }(y_2)$. Its zeroth-order full correlator is $C(x_1,y_1)C(x_2,y_2)+C(x_1,y_2)C(x_2,y_1)$ and is disconnected; its connected part starts at order $\lambda$. DefineThe tree term has Feynman-diagram symmetry factor 1 and is $-i\lambda \int d^{D}zP(z)$.

At second order, a connected graph has two quartic vertices, two internal lines and one loop. The three bubble integrals areThe oriented complex-scalar quartic graph factors distinguish the channels. In $B_s$ the two internal lines have the same orientation and can be interchanged, giving symmetry factor $S_s=2$. In each mixed channel the two orientations are opposite; interchanging them is not an oriented graph automorphism, so $S_t=S_u=1$. Equality of the numerical scalar kernels $C(z,w)=C(w,z)$ does not erase the field-slot distinction used in counting contractions.

There are also four connected external-leg tadpole diagrams. They must be included because a connected correlation function is neither amputated nor restricted to one-particle-irreducible Feynman diagrams. Representative integrals, with the other two obtained by the indicated external permutations, areEach oriented external-leg graph has symmetry factor 1: a directed self-loop has no independent reversal symmetry. Altogether,These expressions retain every external propagator. The physical normalization is explicit: each $C$ is $2D_F$, not $D_F$. Equivalently set $\chi =\varphi /\sqrt{2}$; its quadratic action has unit mixed propagator and its interaction is $-\lambda |\chi {|}^4$, giving vertex $-4i\lambda$. Multiplying a four-$\chi$ correlator by four converts it to the four-$\varphi$ correlator, confirming the powers of two.

Completeness can also be seen without a picture. With two vertices and four external legs, $I=(8-4)/2=2$. Connectivity allows either two intervertex lines with a two-and-two external split, or one bridge and one self-loop with a one-and-three split. The charge assignments give exactly the three bubble attachments and four leg choices above. Direct labelled Wick contraction counting gives respectively $16,32,32,32,32,32,32$ contractions. Dividing by the expansion denominator $2!4^2=32$ reproduces $1/2,1,1,1,1,1,1$. Vacuum factors cancel, and there are no additional connected topologies at this order.

For nonzero real $g$, the continuous internal symmetry of a classical field theory is the global U(1) symmetryThe kinetic terms and mass terms are invariant separately, while the phases in ${\varphi }^2{\Phi }^{†}$ cancel. Independent phase rotations would instead require $2{\alpha }_{\varphi }-{\alpha }_{\Phi }=0$ modulo $2\pi$, leaving one connected circle. Simultaneous complex conjugation is also a discrete symmetry, reversing the continuous charge; including it gives $U(1)⋊{\mathbb{Z}}_2\cong O(2)$ for generic parameters. At $g=0$ the continuous symmetry enlarges to $U(1)\times U(1)$.

The Noether current with particles assigned positive charges isFor example, put $A={\varphi }^2{\Phi }^{†}$ and $B={\varphi }^{†2}\Phi$. The Euler-Lagrange equations give ${\partial }_{\mu }{j}_{\varphi }^{\mu }=ig(A-B)$ and ${\partial }_{\mu }{j}_{\Phi }^{\mu }=ig(B-A)$, with the factor $2$ already included in $j_{\Phi }^{\mu }$. Their sum vanishes. The Noether charge $Q=\int d^{3}x{j}^0$ is conserved and, after normal ordering, countsThus a $\Phi$ particle carries two units of the same conserved charge carried by a $\varphi$ particle; their antiparticles carry the negatives. The interaction permits conversion of one $\Phi$ particle into two $\varphi$ particles when allowed by conservation of energy, while preserving this weighted number.

At mean-field level, positive gradient stiffness favours a spatially uniform order parameter. Writing $r=|\varphi |$, minimization of $f=a{r}^2+r^4/2$ giveswith arbitrary constant $\delta$. The minimized density is zero above the transition and $-a^2/2$ below it. The order parameter vanishes continuously as $(-a{)}^{1/2}$; the first derivative of the minimized free energy is continuous while the second changes. This is the continuous mean-field transition.

The broken symmetry is global U(1) symmetry, not merely the Ising sign symmetry. Selecting one ordered phase breaks this continuous symmetry and leaves a soft phase direction. These are minima of the supplied Landau free energy, not an exact finite-temperature long-range-order claim in one dimension: phase fluctuations invalidate that inference for a short-range one-dimensional system.^[ref]

Set $r={\varphi }_0$ and $A_k=a+c{k}^2+k^4$. The trial density is $A_k{r}^2+r^4/2$. At fixed $k$, its minimum is $r=0$ if $A_k\ge 0$ and $r^2=-A_k$ otherwise, with density $-A_k^2/2$. Minimizing $A_k$ usesFor $c\ge 0$, $k=0$ minimizes the quadratic coefficient. For $c<0$, the minima are at $k=\pm \sqrt{-c/2}$ and have $A_k=a-c^2/4$. ThusEverywhere else the minimum is $\varphi =0$; at the transition boundary the ordered amplitude tends to zero. The zero-field minimum is respectively $-a^2/2$ or $-(a-c^2/4{)}^2/2$ in the two ordered regions. For $c<0$ the helical solution is lower than a uniform ordered extremum whenever the latter exists, and lower than the disordered state whenever its amplitude is nonzero. The assumed positive-amplitude ansatz must allow its zero-amplitude boundary to describe disorder.

The phase $\delta$ remains arbitrary because the functional has global U(1) symmetry. At nonzero $k$, a translation shifts that phase; separate translation and phase operations retain a combined unbroken subgroup. The two signs of $k$ also select opposite handedness. The meeting point of the uniform, modulated and disordered mean-field regions is the Lifshitz point $(a,c)=(0,0)$.