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For the internal rotation symmetry of two real scalar fields with $\delta {\varphi }_1=\alpha {\varphi }_2$ and $\delta {\varphi }_2=-\alpha {\varphi }_1$, the Noether charge is $Q=i\int (a_1^{†}{a}_2-a_2^{†}{a}_1)d^{3}p/(2\pi {)}^3$. Its commutators satisfy $[Q,b_{\pm }^{†}]=\pm b_{\pm }^{†}$, so these creation operators produce charge eigenstates in the one-particle state sector. Substitution gives $Q=N_{+}-N_{-}$, converting the rotation of real components into opposite charges in a complex basis.

Use Minkowski spacetime with $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and units $\hslash =c=1$. The transformation rotates the two components in their internal plane. In particular,Thus both terms in the Lagrangian density are invariant: this is an internal rotation symmetry of two real scalar fields, rather than a transformation of spacetime. The equal masses are essential to this symmetry.

The Noether theorem gives the Noether currentIndeed, the cross terms in its divergence cancel, and the two Klein-Gordon equations implyFor fields decaying sufficiently at spatial infinity, there is no outward current flux. The conserved Noether charge is therefore

In canonical quantization, the canonical momenta are ${\pi }_k={\dot{\varphi }}_k$, with equal-time canonical commutation relationsConsequently $Q=\int d^{3}x({\pi }_1{\varphi }_2-{\pi }_2{\varphi }_1)$. Operators of different components commute, so this expression is Hermitian without an ordering correction. Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$, $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $a_{k\mathbf{p}}=a_{\mathbf{p}}^k$. Substituting the oscillator expansions and using the Dirac delta function from the spatial integral givesThe terms with two annihilation operators cancel after $\mathbf{p}\mapsto -\mathbf{p}$; the terms with two creation operators cancel in the same way. In the remaining terms useThere is no commutator constant between the distinct components. The result is already in normal ordering:

For the charged oscillator basis of a scalar doublet, computeIt follows that $[Q,b_{\pm \mathbf{p}}^{†}]=\pm b_{\pm \mathbf{p}}^{†}$. The Fock vacuum satisfies $Q|0⟩=0$, so $b_{+\mathbf{p}}^{†}|0⟩$ has charge $+1$, and $b_{-\mathbf{p}}^{†}|0⟩$ has charge $-1$. To obtain a normalized one-particle state, replace the sharp momentum by ${\int }_{\mathbf{p}}f(\mathbf{p})b_{\pm \mathbf{p}}^{†}|0⟩$, where ${\int }_{\mathbf{p}}|f(\mathbf{p}){|}^2=1$. The charge is unchanged. Equivalently,The generator convention is $\delta {\varphi }_k=i\alpha [Q,{\varphi }_k]$, which reproduces the original rotation signs.

For stability of a two-scalar quartic potential, the relevant potential energy density isAlong either axis boundedness requires $\lambda \ge 0$; along ${\varphi }_1={\varphi }_2$ it requires $\lambda +\mu \ge 0$. Conversely the displayed sum is nonnegative whenever both conditions hold. With $m^2>0$ the origin is then the unique global minimum, including the equality cases. Hence the stable-vacuum conditions areStrict positivity of the quartic term away from the origin would instead require $\lambda >0$ and $\mu >-\lambda$. That stronger condition is unnecessary for a positive mass term. If $m=0$, the same non-strict conditions give boundedness, but the equality cases have flat directions and do not give an isolated classical vacuum.

Finally, the interaction changes the variation of the potential energy byThe interacting Euler-Lagrange equations consequently giveThus the same Noether charge is conserved for all field configurations precisely when $\mu =\lambda$. Then $V_4=\lambda ({\varphi }_1^2+{\varphi }_2^2{)}^2$ preserves the internal rotation symmetry of two real scalar fields; imposing stability additionally requires $\lambda \ge 0$.

Differentiate the Noether charge and use ${\dot{\varphi }}_a={\pi }_a$ and the Klein-Gordon equation for each species. The momentum products cancel because the two species commute:The first integral is the boundary flux of ${\varphi }_2\nabla {\varphi }_1-{\varphi }_1\nabla {\varphi }_2$, so it vanishes under the same boundary condition used for charge conservation. HenceConservation must hold for arbitrary configurations, rather than only a state in which this particular integral happens to vanish. Thus the internal rotation symmetry of two real scalar fields requiresEquivalently, the local divergence is ${\partial }_{\mu }{j}^{\mu }=(m_2^2-m_1^2){\varphi }_1{\varphi }_2$, exhibiting exactly how unequal masses break the symmetry.

For the two free real scalar fields, let $\mathcal{L}={\mathcal{L}}_1+{\mathcal{L}}_2$. The infinitesimal internal rotation symmetry of two real scalar fields has $\delta {\varphi }_1=\theta {\varphi }_2$ and $\delta {\varphi }_2=-\theta {\varphi }_1$, with constant $\theta$. Its kinetic terms are invariant because this is an orthogonal rotation of the two-component field.

Under the assumed invariance of the full Lagrangian density, Noether's theorem gives the current per unit transformation parameter:The Noether current and Noether charge are thereforeIndependent species commute at equal time, so these products require no extra ordering sign. With vanishing boundary flux, $Q$ is conserved. Part (vii) checks precisely what the symmetry assumption requires of the masses.