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If $U(s)=e^{-isQ/\hslash }$ and $U(s{)}^{†}HU(s)=H$, differentiation at $s=0$ givesIn the Heisenberg picture,

For an operator $O_H=U^{†}{O}_{S}U$ in the Heisenberg picture, differentiating the evolution operators gives $d{O}_H/dt=i[H_H,O_H]+U^{†}(\partial O_S/\partial t)U$ in units $\hslash =1$. With no explicit operator time dependence, the second term vanishes. Applying this commutator identity to the canonical momentum and real scalar field gives their operator field equations.

Use units $\hslash =c=1$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. Treat the complex scalar field and its adjoint as independent variables when varying the action. Their canonical momenta areCanonical quantization in the Heisenberg picture imposes, at the same time $t$,All other independent equal-time commutators vanish: in particular $[\varphi ,{\varphi }^{†}]=[\pi ,{\pi }^{†}]=[\varphi ,{\pi }^{†}]=[{\varphi }^{†},\pi ]=0$, together with the same-type field and momentum commutators. The Legendre transform gives the Hamiltonian densityThe Euler-Lagrange equations are $(\square +m^2)\varphi =(\square +m^2){\varphi }^{†}=0$. The positive- and negative-frequency plane waves form a complete basis of solutions of the Klein-Gordon equation. Since the field is complex, their operator coefficients are independent rather than being adjoints of each other. Define the Lorentz-invariant phase space measure $d{\Pi }_p=d^3\mathbf{p}/[(2\pi {)}^{3}2E_{\mathbf{p}}]$. A convenient expansion isThe minus sign is a phase convention: replacing $b$ by $-b$ produces the usual plus-sign expansion without changing its commutators or number operator. The normalization is fixed by the canonical commutation relations. For example,when $[a(\mathbf{p}),a^{†}(\mathbf{q})]=[b(\mathbf{p}),b^{†}(\mathbf{q})]=(2\pi {)}^{3}2E_{\mathbf{p}}{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$. The cross commutators vanish and the other canonical relation follows by taking adjoints with the operator order reversed. Conversely these mode commutators can be obtained by Fourier inversion of the equal-time relations. This is canonical quantization of a complex scalar field.

Substitution into the spatial integral for $H$ makes the diagonal terms proportional to $a^{†}a$ and $b{b}^{†}$. Terms creating or annihilating a pair have $\mathbf{q}=-\mathbf{p}$ and a coefficient $E_{\mathbf{p}}^2-{\mathbf{p}}^2-m^2=0$, so they cancel. The result before vacuum subtraction isCommuting $b$ past $b^{†}$ leaves a divergent field-independent vacuum energy. The displayed zero-vacuum-energy form must therefore be understood as a normal-ordered Hamiltonian:A regulator can be introduced before subtracting the constant. This is the normal-ordered Hamiltonian of a free complex scalar field; the unrenormalized Hamiltonian is not literally equal to this expression without that convention.

The scalar-field vacuum is annihilated by both annihilation operators. Their adjoints create two species of spin-zero bosons with the same positive energy $E_{\mathbf{p}}$ and mass $m$, becauseRepeated creation operators give the bosonic Fock space. The negative-frequency part does not create a negative-energy physical state; it creates the antiparticle of positive energy. The charge distinguishes the two species.

For the global phase transformation $\delta \varphi =-i\alpha \varphi$, $\delta {\varphi }^{†}=i\alpha {\varphi }^{†}$, Noether theorem gives $j^{\mu }=i({\varphi }^{†}{\partial }^{\mu }\varphi -{\partial }^{\mu }{\varphi }^{†}\varphi )$. Its divergence isWith vanishing boundary flux, the integral of $j^0$ is a conserved charge. Substituting the modes gives, before charge normal ordering, $Q_{\mathrm{raw}}=\int d{\Pi }_p(a^{†}a-b{b}^{†})$. The pair terms cancel because their frequency difference is zero when their spatial momenta sum to zero. Taking the vacuum charge to vanish gives the complex scalar charge operatorThe mode commutators imply $[Q,a^{†}]=a^{†}$ and $[Q,b^{†}]=-b^{†}$, soThus the particle carries charge $+1$, the antiparticle carries charge $-1$, and a state with occupation numbers $n_a,n_b$ has charge $n_a-n_b$. Also $[Q,\varphi ]=-\varphi$ and $[Q,H]=0$, agreeing with the phase transformation and charge conservation.

Write the final topological quantum order constants as $a>0$ and $\epsilon <1$, to avoid confusing the allowed support diameter with the small time coefficient. For any initial operator $A_X$ of operator norm at most one,and similarly for the partner state. The minus sign follows from $|{\psi }_0⟩=e^{itH}|{\psi }_1⟩$ and the Heisenberg picture convention $A_X(t)=e^{itH}{A}_X{e}^{-itH}$. The Lieb-Robinson bound and its localization corollary apply to either time direction, using $|t|$.

Choose $\mathrm{diam}(X)\le aL/4$, $t=\tau L$ with $v\tau \le a/8$, and localization buffer $l=aL/8$. The enlarged support $X^{\prime }$ obeysThe Lieb-Robinson localization by Haar twirling corollary supplies an operator $B_{X^{\prime }}$ approximating $A_X(-t)$ withDo not assume that the approximate operator has norm at most one: it only has $\Vert B_{X^{\prime }}\Vert \le 1+{\delta }_L$. Apply final-state local indistinguishability to $B_{X^{\prime }}/(1+{\delta }_L)$. The two approximation errors then giveIf the lattice has polynomially many sites in its diameter, $N(L)=O(L^p)$, the error tends to zero uniformly in these supports. For sufficiently large $L$, take ${\delta }_L\le (1-\epsilon )/[2(\epsilon +2)]$. The initial pair is then indistinguishable within ${\epsilon }^{\prime }=(1+\epsilon )/2<1$ on supports of diameter at most $aL/4$. Together with the exact orthogonality from condition (i), this proves the initial state is topologically ordered for sufficiently small linear-time coefficient. The constants for initial and final order need not be identical.

The backward preservation of local indistinguishability has an explicit finite-system qualification: it holds whenever the displayed $L|X|e^{-\mu aL/16}$ error is small enough. The conventional bounded-density, fixed-dimensional lattice interpretation supplies that condition. The question leaves growth control implicit; for an arbitrary collection of qudits one cannot discard the $|X|$ prefactor merely because $L$ is large. This identifies exactly the assumption needed by the supplied localization proof.

Use the Fourier transform convention $\hat{w}(E)={\int }_{\mathbb{R}}w(t)e^{itE}dt$ and define the spectral filtering of Hamiltonian terms byThe normalization $\hat{w}(0)=1$ means $\int w=1$. Since $w$ is real and nonnegative, $\hat{w}(-E)=\overline{\hat{w}(E)}$. Thus the stated positive-frequency cutoff also implies $\hat{w}(E)=0$ for $E\le -\Delta$. The integral exists in operator norm, because its integrand has norm at most $w(t)\Vert h_Z\Vert$.

Let $H|{\varphi }_i⟩=E_i|{\varphi }_i⟩$, with $E_i-E_0\ge \Delta$ for $i>0$. In this eigenbasis,Every matrix element coupling the unique ground state to an excited state vanishes, in either direction. HenceMoreover, linearity and conservation of $H$ under its own Heisenberg picture evolution giveEach filtered term is Hermitian, but it need not remain supported on its original set $Z$. The tail assumption supplies stronger long-time control than the integrability used here; no additional tail estimate is needed for the requested identities.

Evaluate the field variance in the zero-occupation Fock vacuum, ${\hat{a}}_{\mathbf{k}}|0⟩=0$, with $[{\hat{a}}_{\mathbf{k}},{\hat{a}}_{{\mathbf{k}}^{\prime }}^{†}]={\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })$. The canonical commutation relation alone does not specify a quantum state; this vacuum assumption is needed for the requested expression without occupation factors. In the Heisenberg picture, use the position-space mode expansion of a free fieldThis is the position-space form of the shorthand oscillator expansion in the question; equivalently its Fourier operator uses the creation operator with opposite wavevector. The annihilation operators give $⟨{\hat{a}}_{\mathbf{k}}{\hat{a}}_{{\mathbf{k}}^{\prime }}^{†}⟩={\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })$, while all other vacuum pairings vanish. The mean field fluctuation is zero, so the two-point function is its coincident-point vacuum field variance. Assuming spatial isotropy, angular integration gives the variance per logarithmic wavenumber interval:Here the dimensionless power spectrum is denoted ${\Delta }_v$, as in the question, rather than the alternative notation ${\Delta }_v^2$. A coincident vacuum field product needs smearing or ultraviolet regulation: the unregulated full vacuum variance is generally divergent. With a specified finite wavenumber band, the same expression holds with those integration limits. This does not affect the mode-by-mode dimensionless power spectrum.

For the leading nonzero slower branch, put $x=-k\tau =k/(aH)>0$ and $v_k=c{x}^{1/2-\nu }/\sqrt{2k}$. Since the physical fluctuation is $\delta \chi =v/a$, its dimensionless power spectrum isFor a $k$-independent coefficient $c$, this is a blue spectral tilt, because $3-2\nu >0$ in the specified mass range. At fixed $k$ it decays as $a^{-(3-2\nu )}$, consistently with the field-amplitude decay found above.

To include the other branch explicitly, the general superhorizon expression isIf $c_{-}\ne 0$, it approaches the displayed target spectrum with $c=c_{-}$. A pure faster classical branch would instead have power proportional to $x^{3+2\nu }$. For a normalized quantum mode, however, the Wronskian normalization $v_k{v}_k^{\ast \prime }-v_k^{\ast }{v}_k^{\prime }=i$ requires both independent real solutions; its slower coefficient cannot vanish. Thus the target is the leading superhorizon quantum spectrum, while the faster classical solution remains part of the complete mode analysis.

For a time-independent Hamiltonian operator and a Schrödinger picture observable with no explicit time dependence, the Heisenberg picture operator isDifferentiation gives ${\dot{O}}_H=iH{O}_H-i{O}_{H}H=i[H,O_H]$, soIf $O_S$ itself depends explicitly on time, one must add $e^{iHt}({\partial }_t{O}_S)e^{-iHt}$ to ${\dot{O}}_H$; the displayed commutator relation uses the usual convention of no explicit time dependence for the observable.

Use the free Hamiltonian operator and the canonical commutation relations. The kinetic momentum term givesThe Heisenberg picture equation therefore yields ${\dot{\varphi }}_1={\pi }_1$.

For the momentum commutator, the mass term gives $i{m}_1^2{\varphi }_1(\mathbf{x})$. The gradient term giveswhere integration by parts differentiates the field instead of the Dirac delta distribution. ThusCombining the two first-order operator equations gives the Klein-Gordon equation:Suitable decay or spatial smearing removes the boundary term in the integration by parts.

The Born rule depends only on transition probabilitiesbetween rays. A physical transformation must preserve these probabilities. By Wigner theorem, such a ray transformation is induced by a unitary or antiunitary operator. For ordinary continuously connected transformations one chooses the unitary branch, so

The states $U(g_1)U(g_2)|\psi ⟩$ and $U(g_1{g}_2)|\psi ⟩$ represent the same ray. They may therefore differ by a state-independent quantum phase:Thus quantum transformations form a projective unitary representation.

If $g$ is a symmetry of the Hamiltonian operator $H$, it also obeysequivalently $[U(g),H]=0$. For a differentiable one-parameter subgroup, writewith a self-adjoint generator $Q$. Differentiating the symmetry equation at $s=0$ yieldsFor a time-independent $Q$, the Heisenberg picture equation isHence the generator is a conserved observable. Conversely, a self-adjoint conserved $Q$ commutes with $H$ and its unitary group generates symmetries. This is the conserved generator of a continuous quantum symmetry.

Letwhere the two signs describe the Heisenberg antiferromagnet and Heisenberg ferromagnet. A standard Lieb-Robinson bound is obtained by iterating the Heisenberg picture equation and bounding nested commutators by operator norms. Its constants depend on the interaction only through quantities such asChanging $J$ to $-J$ changes neither the supports nor the norms $\Vert \Phi (X)\Vert$. Every term in the nested-commutator estimate acquires at most an irrelevant sign before its absolute value is taken. Therefore both chains obey exactly the same estimatewith the same $C,\mu$, and Lieb-Robinson velocity $v_{\mathrm{LR}}$. No unitary equivalence of the two Hamiltonians is required.

In the Schrödinger picture, states evolve with the Hamiltonian operator while observables without explicit time dependence are fixed. For time-independent $H$, the corresponding Heisenberg picture operator is $O_H(t)=e^{iHt/\hslash }{O}_S{e}^{-iHt/\hslash }$.