= Solution
Use <Minkowski spacetime> with $g_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$ and units $\hbar=c=1$. The transformation rotates the two components in their internal plane. In particular,
$$
\delta(\phi_1^2+\phi_2^2)=2\alpha(\phi_1\phi_2-\phi_2\phi_1)=0,
\qquad
\delta\bigl(\partial_\mu\phi_1\partial^\mu\phi_1+\partial_\mu\phi_2\partial^\mu\phi_2\bigr)=0.
$$
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 current>
$$
j^\mu=\sum_{k=1}^2\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_k)}\frac{\delta\phi_k}{\alpha}
=\phi_2\partial^\mu\phi_1-\phi_1\partial^\mu\phi_2.
$$
Indeed, the cross terms in its divergence cancel, and the two <Klein-Gordon equations> imply
$$
\partial_\mu j^\mu=\phi_2\Box\phi_1-\phi_1\Box\phi_2
=-m^2\phi_2\phi_1+m^2\phi_1\phi_2=0.
$$
For fields decaying sufficiently at spatial infinity, there is no outward current flux. The conserved <Noether charge> is therefore
$$
\boxed{Q=\int d^3x\,(\phi_2\dot\phi_1-\phi_1\dot\phi_2).}
$$
In <canonical quantization>, the <canonical momenta> are $\pi_k=\dot\phi_k$, with equal-time <canonical commutation relations>
$$
[\phi_k(\mathbf x),\pi_l(\mathbf y)]=i\delta_{kl}\delta^3(\mathbf x-\mathbf y),
\qquad [\phi_k,\phi_l]=[\pi_k,\pi_l]=0.
$$
Consequently $Q=\int d^3x\,(\pi_1\phi_2-\pi_2\phi_1)$. Operators of different components commute, so this expression is <Hermitian> without an ordering correction. Write $\int_{\mathbf p}=\int d^3p/(2\pi)^3$, $E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}$ and $a_{k\mathbf p}=a^k_{\mathbf p}$. Substituting the oscillator expansions and using the <Dirac delta function> from the spatial integral gives
$$
Q=-\frac i2\int_{\mathbf p}\left[
(a_{1\mathbf p}-a_{1,-\mathbf p}^{\dagger})(a_{2,-\mathbf p}+a_{2\mathbf p}^{\dagger})
-(a_{2\mathbf p}-a_{2,-\mathbf p}^{\dagger})(a_{1,-\mathbf p}+a_{1\mathbf p}^{\dagger})
\right].
$$
The 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 use
$$
[a_{k\mathbf p},a_{l\mathbf q}^{\dagger}]=(2\pi)^3\delta_{kl}\delta^3(\mathbf p-\mathbf q).
$$
There is no <commutator> constant between the distinct components. The result is already in <normal ordering>:
$$
\boxed{Q=-i\int_{\mathbf p}\bigl(a_{2\mathbf p}^{\dagger}a_{1\mathbf p}-a_{1\mathbf p}^{\dagger}a_{2\mathbf p}\bigr).}
$$
For the <charged oscillator basis of a scalar doublet>, compute
$$
[Q,a_{1\mathbf p}^{\dagger}]=-ia_{2\mathbf p}^{\dagger},\qquad
[Q,a_{2\mathbf p}^{\dagger}]=ia_{1\mathbf p}^{\dagger},\qquad
b_{\pm\mathbf p}^{\dagger}=\frac{a_{1\mathbf p}^{\dagger}\mp ia_{2\mathbf p}^{\dagger}}{\sqrt2}.
$$
It follows that $[Q,b_{\pm\mathbf p}^{\dagger}]=\pm b_{\pm\mathbf p}^{\dagger}$. The <Fock vacuum> satisfies $Q|0\rangle=0$, so \b[$b_{+\mathbf p}^{\dagger}|0\rangle$ has charge $+1$, and $b_{-\mathbf p}^{\dagger}|0\rangle$ has charge $-1$]. To obtain a normalized <one-particle state>, replace the sharp momentum by $\int_{\mathbf p}f(\mathbf p)b_{\pm\mathbf p}^{\dagger}|0\rangle$, where $\int_{\mathbf p}|f(\mathbf p)|^2=1$. The charge is unchanged. Equivalently,
$$
Q=\int_{\mathbf p}(b_{+\mathbf p}^{\dagger}b_{+\mathbf p}-b_{-\mathbf p}^{\dagger}b_{-\mathbf p}).
$$
The generator convention is $\delta\phi_k=i\alpha[Q,\phi_k]$, which reproduces the original rotation signs.
For <stability of a two-scalar quartic potential>, the relevant <potential energy> density is
$$
V=\frac{m^2}{2}(\phi_1^2+\phi_2^2)+V_4,\qquad
V_4=\lambda(\phi_1^2-\phi_2^2)^2+2(\lambda+\mu)\phi_1^2\phi_2^2.
$$
Along either axis boundedness requires $\lambda\ge0$; along $\phi_1=\phi_2$ it requires $\lambda+\mu\ge0$. 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 are
$$
\boxed{\lambda\ge0,\qquad\mu\ge-\lambda.}
$$
Strict 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> by
$$
\delta V_4=4\alpha(\lambda-\mu)\phi_1\phi_2(\phi_1^2-\phi_2^2).
$$
The interacting <Euler-Lagrange equations> consequently give
$$
\partial_\mu j^\mu=-4(\lambda-\mu)\phi_1\phi_2(\phi_1^2-\phi_2^2).
$$
Thus the same <Noether charge> is conserved for all field configurations precisely when \b[$\mu=\lambda$]. Then $V_4=\lambda(\phi_1^2+\phi_2^2)^2$ preserves the <internal rotation symmetry of two real scalar fields>; imposing stability additionally requires $\lambda\ge0$.
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