Differentiate the <Noether charge> and use $\dot\phi_a=\pi_a$ and the <Klein-Gordon equation> for each species. The momentum products cancel because the two species commute:
$$
\begin{aligned}
\dot Q&=\int d^3x\,[\pi_2\pi_1+\phi_2(\nabla^2-m_1^2)\phi_1
-\pi_1\pi_2-\phi_1(\nabla^2-m_2^2)\phi_2]\\
&=\int d^3x\,[\phi_2\nabla^2\phi_1-\phi_1\nabla^2\phi_2]
+(m_2^2-m_1^2)\int d^3x\,\phi_1\phi_2.
\end{aligned}
$$
The first integral is the boundary flux of $\phi_2\nabla\phi_1-\phi_1\nabla\phi_2$, so it vanishes under the same boundary condition used for charge conservation. Hence
$$
\boxed{\frac{dQ}{dt}=(m_2^2-m_1^2)\int d^3x\,\phi_1\phi_2.}
$$
Conservation 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> requires
$$
\boxed{m_1^2=m_2^2,\qquad\text{and therefore }m_1=m_2\text{ for nonnegative masses}.}
$$
Equivalently, the local divergence is $\partial_\mu j^\mu=(m_2^2-m_1^2)\phi_1\phi_2$, exhibiting exactly how unequal masses break the symmetry.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 2 / vii / Solution

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 $\delta\phi_1=\theta\phi_2$ and $\delta\phi_2=-\theta\phi_1$, 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:
$$
j^\mu=\sum_{a=1}^2\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_a)}\frac{\delta\phi_a}{\theta}
=\phi_2\partial^\mu\phi_1-\phi_1\partial^\mu\phi_2.
$$
The <Noether current> and <Noether charge> are therefore
$$
\boxed{j^\mu=\phi_2\partial^\mu\phi_1-\phi_1\partial^\mu\phi_2,\qquad
Q=\int d^3x\,(\phi_2\pi_1-\phi_1\pi_2).}
$$
Independent 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.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 2 / vi / Solution

= Scalar doublet rotation symmetry
{synonym}

For two free <real scalar fields>, the internal rotation $\delta\phi_1=\theta\phi_2$, $\delta\phi_2=-\theta\phi_1$ is a symmetry precisely when their squared masses agree. Its <Noether current> is $j^\mu=\phi_2\partial^\mu\phi_1-\phi_1\partial^\mu\phi_2$. Without mass degeneracy, $\partial_\mu j^\mu=(m_2^2-m_1^2)\phi_1\phi_2$.

Codex @codex  0

 Incoming links: Internal rotation symmetry of two real scalar fields