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Use the unitary Fourier transform with phase $e^{-i\lambda x}$. It commutes with $-d^2/d{x}^2+x^2$, because differentiation transforms to multiplication by $i\lambda$ and multiplication by $x$ transforms to $id/d\lambda$. The Hermite differential equation has a unique polynomial solution of each degree up to scale, and multiplying by the Gaussian gives a Hermite function. Transforming a polynomial times this Gaussian again gives a polynomial times the Gaussian of the same degree; comparison of leading coefficients gives the eigenvalue $(-i{)}^n$.

A solution of $b^{\prime \prime }+(\Lambda -{\xi }^2)b=0$ decaying at both real infinities has $b=e^{-{\xi }^2/2}{H}_n(\xi )$ with a Hermite polynomial. Substitution gives $H^{\prime \prime }-2\xi H^{\prime }+(\Lambda -1)H=0$; its power-series recurrence terminates at degree $n$ when $\Lambda =2n+1$. The square-integrable Hermite functions give the complete oscillator spectrum, including the ground state $n=0$. This quantization also determines global modes of a quadratically confined Ginzburg-Landau equation after removing drift.

For $|t|<1$, this identity sums the Hermite function expansion into a Gaussian. One proof takes the Bargmann-Fock space inner product of the generating functions ${\pi }^{-1/4}\exp (-x^2/2+\sqrt{2}xz-z^2/2)$ at $\sqrt{t}z$. Writing $z=u+iv$, the two resulting real Gaussian integrals have quadratic coefficients $1+t$ and $1-t$ and linear coefficients $\sqrt{2t}(x+y)$ and $i\sqrt{2t}(x-y)$. Evaluation gives the formula. Setting $t=e^{-2s}$ and multiplying by $e^{-s}$ gives the harmonic oscillator transition kernel of $e^{-s(-d^2/d{x}^2+x^2)}$.

First assume the usual diffusion regime $\gamma >0$, with $U$ real; the complex-diffusion extension is described below. Substitution of a global normal mode givesSet $a=e^{Ux/(2\gamma )}b$ to remove the first derivative, then put $\xi =\sqrt{\epsilon }(\nu /\gamma {)}^{1/4}x$. The resulting oscillator equation isWriting $b=e^{-{\xi }^2/2}H(\xi )$ gives the Hermite differential equation, $H^{\prime \prime }-2\xi H^{\prime }+(\Lambda -1)H=0$. For a power series $H={\sum }_{j\ge 0}{c}_j{\xi }^j$, its recurrence isTermination at degree $n$ requires $\Lambda =2n+1$. If it does not terminate in the appropriate parity, the series has a growing Gaussian component at one infinity; the solution decaying at one end then fails at the other. The Hermite oscillator quantization can also be seen from the self-adjoint operator $-d^2/d{\xi }^2+{\xi }^2$, whose square-integrable eigenfunctions are the Hermite functions. Their spectrum is $2n+1$ with $n=0,1,2,\dots$. The ground state $H_0=1$ shows that the paper's phrase “positive integer” must include zero.

The global modes of a quadratically confined Ginzburg-Landau equation are thereforeTheir spatial functions are proportional to $e^{Ux/(2\gamma )}{e}^{-{\xi }^2/2}{H}_n(\xi )$; the quadratic Gaussian decay dominates the linear drift factor at both infinities. For real coefficients, the most dangerous mode is $n=0$. The system has global hydrodynamic instability precisely whenEquality is marginal and smaller ${\mu }_0$ gives decay of all these modes. The finite-width confinement correction explains why a locally absolutely unstable region need not immediately support a growing global mode.

If complex diffusion is intended, impose the parabolic condition $\mathrm{Re}\gamma >0$ and choose $\sqrt{\nu /\gamma }$ with positive real part. The same formulas continue along the rotated $\xi$ ray, with the corresponding branch of $\sqrt{\gamma \nu }$; instability is determined by the real part of the square bracket. Arbitrary constant $\gamma$, especially negative diffusion, does not justify the asserted whole-line Hermite decay or a well-posed evolution.

Use the unitary normalization of the Fourier transform,The Schwartz space $\mathcal{S}({\mathbb{R}}^n)$ consists of smooth functions for which every Schwartz seminorm ${\sup }_x|x^{\alpha }{\partial }^{\beta }f(x)|$ is finite. Such functions and all their derivatives are integrable. Differentiating under the integral and integrating by parts giveConsequentlyExpand the derivative by the product rule and insert the integrable weight $(1+|x|{)}^{-n-1}$; the right side is bounded by finitely many Schwartz seminorms of $f$. This proves continuity $\mathcal{F}:\mathcal{S}\to \mathcal{S}$. The identical argument with the opposite sign proves continuity of $\mathcal{G}$.

For surjectivity and injectivity an inversion proof is needed. If $f\in \mathcal{S}$, multiply $\mathcal{F}f(\xi )$ by $e^{-\epsilon |\xi {|}^2/2}$. Fubini's theorem and the elementary Gaussian integral giveThe Gaussian has total integral one and its mass outside any fixed ball tends to zero. The right side therefore tends to $f(x)$, by uniform continuity of $f$. On the left, $\mathcal{F}f\in \mathcal{S}\subset L^1$, so dominated convergence gives $\mathcal{GF}f(x)$. This proves $\mathcal{GF}=I$. Reversing the signs in the same calculation proves $\mathcal{FG}=I$. ThusIts inverse is continuous by the seminorm estimate already proved. This establishes the requested bijection without assuming Fourier inversion as an unexplained theorem.

Inversion gives ${\mathcal{F}}^{2}f(x)=f(-x)$ and the Parseval identity. Indeed, inserting the integral for $\mathcal{F}f$ into $\int \mathcal{F}f\overline{\mathcal{F}g}$ and using $f,\mathcal{F}g\in L^1$ gives $\int f\overline{g}$. Density of Schwartz functions in $L^2$ extends $\mathcal{F}$ to a unitary operator on $L^2$, the Plancherel theorem. Translation of $f$ becomes multiplication by a phase, modulation becomes frequency translation, andThese formulas explain why the Fourier transform diagonalizes constant-coefficient differential operators. By transpose it also defines a bijection of the tempered distributions: with the complex-linear distributional pairing, $⟨\mathcal{F}T,\varphi ⟩=⟨T,\mathcal{F}\varphi ⟩$, since the Fourier kernel is symmetric in its variables.

The connection with creation and annihilation operators is especially transparent for the harmonic oscillator. On the common invariant dense domain $\mathcal{S}$, setIntegration by parts shows that these are formal Hilbert-space adjoints. They are closable unbounded operators; commutators below are computed on $\mathcal{S}$, avoiding undefined products on arbitrary $L^2$ vectors. The identity $[{\partial }_j,x_k]={\delta }_{jk}I$ gives the canonical commutation relationsIt also givesThe normalized vacuum $h_0(x)={\pi }^{-n/4}{e}^{-|x{|}^2/2}$ satisfies $a_j{h}_0=0$. Define the Hermite functionsMoving annihilation operators past creation operators with the commutation relations provesThese same relations show that the Hermite functions are orthonormal: unequal occupations are orthogonal eigenvectors of one of the commuting symmetric number operators $a_j^{†}{a}_j$, and the ladder relations recursively give norm one.

For completeness, let $g\in L^2$ be orthogonal to every $h_{\alpha }$. The Hermite polynomials have nonzero leading monomials and triangular lower-degree terms, so these orthogonality relations say that all polynomial moments of $h(x)=g(x)e^{-|x{|}^2/2}\in L^1$ vanish. Its Fourier integral extends to an entire function of $\zeta \in {\mathbb{C}}^n$:The Cauchy-Schwarz inequality, with the Gaussian absorbing every exponential linear weight and every polynomial, justifies differentiation uniformly on compact sets of $\zeta$. Every Taylor coefficient at zero is a vanishing moment, so the entire function is zero. Fourier uniqueness on $L^1$ follows here without circularity: if an $L^1$ function has zero Fourier transform, convolving it with ${\rho }_{\epsilon }$ gives zero by the Gaussian integral and Fubini, and the approximate-identity limit in $L^1$ gives the original function zero. Thus $h=0$ and $g=0$. The Hermite functions form an orthonormal basis of $L^2$.

Finally, the elementary Gaussian integral gives $\mathcal{F}{h}_0=h_0$. The derivative and multiplication identities above implywhere the operators on the right act on the frequency variable. HenceThus creation and annihilation organize both the oscillator's spectral decomposition and the four possible Fourier eigenvalues. On Schwartz functions the oscillator powers control weighted derivatives, and Hermite coefficients decay faster than every power of $1+|\alpha |$; conversely this rapid coefficient decay implies all Schwartz seminorms are finite. The Fourier phases preserve this decay, providing another explanation of the Fourier automorphism of Schwartz space.

Use the real diffusive interpretation $\gamma >0$, $U,{\mu }_0\in \mathbb{R}$, with $\nu >0$. Without a positive real diffusion coefficient the printed ordering inequality and real oscillator coordinate would need different hypotheses. Remove the drift by $\eta =e^{Ux/(2\gamma )}v$. Direct differentiation reduces the linear complex Ginzburg-Landau equation toSet $v=b(\xi )e^{-i({\omega }_0+ϵ{\omega }_1)t}$ and $\xi =\sqrt{ϵ}fx$. The resulting equation isCanceling its order-one coefficient and normalizing the quadratic term givesThe supplied decay quantization has $\lambda =2j+1$, $j=0,1,2,\dots$. Equivalently the Hermite functions $b_j=e^{-{\xi }^2/2}{H}_j(\xi )$ solve this harmonic-oscillator problem, where $H_j$ is the physicists' Hermite polynomial. Thus ${\omega }_1=-i(2j+1)\sqrt{\nu \gamma }$ and the exact temporal growth rates areThe global modes of a quadratically confined Ginzburg-Landau equation areFor every fixed positive $ϵ$, the negative quadratic spatial exponent dominates the linear drift exponent and the polynomial at both infinities; all these normal modes are spatially bounded. The largest growth rate is $s_0$, so the least-node normal mode is temporally non-growing, and all normal modes are non-growing, whenEquality is a neutral ground-state normal mode; strict inequality gives decay. This proves the printed sufficient condition for a bounded stable global normal mode and identifies its stronger meaning: stability of the leading global normal mode. Existence of merely some stable higher-$j$ normal mode is not equivalent to the inequality, since sufficiently large $j$ gives $s_j<0$ even above the leading-mode threshold. In the quadratic-coefficient model the reduction and eigenvalues are exact, not just a first-order approximation to a more general profile.

To distinguish the two normalizations, denote the physicists' Hermite polynomial by $h_n(x)=(-1{)}^n{e}^{x^2}(d/dx{)}^n{e}^{-x^2}$. Define the normalized Hermite functions byIntroduce the differential operators $a=(x+d/dx)/\sqrt{2}$, $a^{\ast }=(x-d/dx)/\sqrt{2}$ on the Schwartz space. Direct multiplication gives $[a,a^{\ast }]=1$ andThe Gaussian $H_0={\pi }^{-1/4}{e}^{-x^2/2}$ satisfies $a{H}_0=0$, and repeated differentiation gives $H_n=(a^{\ast }{)}^n{H}_0/\sqrt{n!}$. The commutator identity implies $a{H}_n=\sqrt{n}{H}_{n-1}$ and $a^{\ast }{H}_n=\sqrt{n+1}{H}_{n+1}$. ThereforeFor normalization, integration by parts makes $a^{\ast }$ the adjoint operator of $a$ on these rapidly decreasing functions, so $\Vert (a^{\ast }{)}^n{H}_0{\Vert }^2=n!\Vert H_0{\Vert }^2=n!$. Distinct eigenvalues of the symmetric quantum harmonic oscillator give orthogonality.

The generating identity follows directly from the Rodrigues definition of the Hermite polynomials:The Taylor expansion of the entire Gaussian justifies the middle equality for every complex $w$. Substitute $w=z/\sqrt{2}$ and the Bargmann-Fock space basis $e_n(z)=z^n/\sqrt{n!}$ to obtainThe convergence is locally uniform in $z$ by the same entire generating series.

For $0<t<1$, $A_x(\sqrt{t}z)$ belongs to the Bargmann-Fock space: after writing $z=u+iv$, its squared modulus times $e^{-|z{|}^2}$ has strictly negative quadratic coefficients $-(1+t)u^2-(1-t)v^2$. Taking the Bargmann-Fock space inner product of the two generating functions and using the orthonormal basis proved in part (b) givesThe coefficient series converges absolutely by the Cauchy-Schwarz inequality. Substitution factors the right side intoUse the Gaussian integral $\int e^{-a{u}^2+bu}du=\sqrt{\pi /a}{e}^{b^2/(4a)}$ for $a>0$ and complex $b$; the complex case follows from the real case by analyticity. The resulting exponent isThus the Mehler formula for Hermite functions isAs a normalization check, its $t\downarrow 0$ limit is $H_0(x)H_0(y)$. Setting $t=e^{-2s}$ and multiplying by $e^{-s}$ yields the harmonic oscillator transition kernel with eigenvalues $2n+1$.