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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 $\lambda \ge 0$.

Interpret the forecasts as best linear prediction from a finite past. For a causal autoregressive process, the innovations are orthogonal to previous observations. If they also have zero conditional mean given the past, the same forecasts are conditional means; white noise by itself guarantees only the linear-prediction interpretation.

Let $m$ be the process mean and write its centered autoregression as $X_t-m={\sum }_{j=1}^p{\varphi }_j(X_{t-j}-m)+{ϵ}_t$. Set ${\hat{X}}_{T,h}=X_{T+h}$ for $h\le 0$, and recursively putThe forecast is a linear combination of the observed last $p$ values. Subtracting the recursion from the future autoregression shows that the prediction error is a linear combination of future innovations, hence orthogonal to every observed value. This proves the optimal projection property, rather than merely suggesting that future noise should be replaced by zero. In particular,The empty sum for $p=1$ is zero. Taking $m=0$ gives the centered convention. This is the recursive forecasts of an autoregressive process construction.

For the stable order-one equation, use its stationary causal solutionThe series converges in mean square because $\sum {\varphi }^{2j}<\infty$ and the white noise innovations have a common finite variance ${\sigma }^2$. Orthogonality of their distinct time indices givesAdding the deterministic mean therefore yields weak stationarity for $Z$, withThe autocorrelation function assumes ${\sigma }^2>0$; a zero-variance process has no normalized correlation. A recursion from an arbitrary nonstationary initial value would have transient moments, so the stationary-solution interpretation is important here.

Iteration gives $X_{T+k}={\varphi }^k{X}_T+{\sum }_{j=1}^k{\varphi }^{k-j}{ϵ}_{T+j}$. The innovation sum is orthogonal to the observed past, givingFor the integrated process, the backshift operator subtracts consecutive levels, so $(I-B)W_t=\mu +(I-B)Y_t=\mu +X_t=Z_t$. Consequently $W_{T+k}=W_T+{\sum }_{j=1}^k{Z}_{T+j}$. If the last increment is observed, substitute its forecast and sum the geometric progression:This is the forecasts of an integrated AR(1) process formula. Its final term is bounded as $k\to \infty$, even for negative $\varphi$, hence the forecast's long-run slope is the drift:With an initial level orthogonal to future innovations, the level formula is also the best linear forecast from the observed levels. Without that initial-level condition, it is the increment-based recursive forecast, and the levels could contain additional predictive information. The level formula uses $T\ge 2$, or a known $W_0$ when $T=1$. With just $W_1$ and no model for the initial level, the last increment cannot be recovered from the supplied observations; a one-observation level forecast would need that additional information.