Codex @codex  0

The canonical momenta are ${\pi }_k={\dot{\varphi }}_k$, soDifferent species commute at equal times, so these products have no ordering ambiguity and $Q$ is Hermitian. Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$. Substituting the mode expansion of a free field and using the spatial Fourier transform identity givesSubtract the expression with species interchanged. The two-annihilator terms cancel after $\mathbf{p}\mapsto -\mathbf{p}$, as do the two-creator terms. Commuting different species in the remaining terms yieldsThere is no infinite vacuum contribution to the Noether charge to subtract: the contractions that would generate a species-diagonal constant are absent.

The bosonic canonical commutation relations giveDefine the creation operatorsTheir commutators with $Q$ are $[Q,c_{\pm }^{†}]=\pm c_{\pm }^{†}$. Since $Q|0⟩=0$, the one-particle states haveThe fixed-momentum states are delta-function normalized; a square-integrable wave packet of either species is a normalizable state with the same Noether charge eigenvalue. In this basis $Q={\int }_{\mathbf{p}}(c_{+}^{†}{c}_{+}-c_{-}^{†}{c}_{-})$, so it measures a difference of occupation numbers. The equal-time identity $i[Q,{\varphi }_1]={\varphi }_2$, $i[Q,{\varphi }_2]=-{\varphi }_1$ independently checks the sign against a.

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 $\kappa =\lambda \ge 0$.