Hermite polynomial

Named after Charles Hermite.

For the Hermite polynomials in the Rodrigues normalization $H_n=(-1{)}^n{e}^{t^2}{D}^n{e}^{-t^2}$, differentiating gives $H_{n+1}=2t{H}_n-H_n^{\prime }$ with $H_0=1$. Induction givesInsert this into the exponential generating series and reindex $n=m+2j$. The result factors into ${\sum }_m(2tz{)}^m/m!$ times ${\sum }_j(-z^2{)}^j/j!$, proving the displayed identity. Equivalently Taylor expansion of $e^{-(t-z{)}^2}$ in $z$ gives the same formula. This generating function keeps the normalization distinct from the Probabilists' Hermite polynomial convention.

Multiply two copies of the generating function of the physicists' Hermite polynomials and integrate against $e^{-t^2}$. Completing the square givesComparing coefficients gives the displayed orthogonality and normalization. Equivalently differentiation in $y,z$ at zero justifies coefficient extraction by the Gaussian integrable bound on compact parameter ranges. Hence the unnormalized Hermite functions $e^{-t^2/2}{H}_n(t)$ have norm ${\pi }^{1/4}\sqrt{2^{n}n!}$. This proves their mutual orthogonality; completeness of that family is an additional assertion.

On $\mathbb{C}[x]$, equip polynomials with the inner productThe linear operatorpreserves the space $V_n$ of polynomials of polynomial degree at most $n$. Integration by parts, whose boundary terms vanish by Gaussian decay, givesThus $T$ equals its adjoint on an inner product space. If a nonzero polynomial of degree $n$ is an eigenvector, its leading coefficient forces its eigenvalue to be $-2(n+1)$.

One can construct all these eigenvectors without solving a differential equation. The one-dimensional space $L_n=V_n\cap V_{n-1}^{\perp }$ is invariant: for $u\in V_{n-1}$ and $v\in L_n$, the displayed identity gives $⟨u,Tv⟩=⟨Tu,v⟩=0$. Its unit vector of positive real leading coefficient is therefore an eigenvector of degree $n$. These choices are exactly the Gram-Schmidt process applied to $1,x,x^2,\dots$, producing normalized Hermite polynomials. Their first three members areEvery polynomial is a finite linear combination of these orthonormal eigenvectors.

For the physicists' Hermite polynomials $H_n=(-1{)}^n{e}^{x^2}{D}^n{e}^{-x^2}$, the displayed recurrence holds for $n\ge 1$, while $H_0^{\prime }=0$ separately. Put $u=e^{-x^2}$. Differentiating $u^{\prime }+2xu=0$ exactly $n$ times using the Leibniz rule yieldsDifferentiating the defining product for $H_n$ then givesThe nonnegative integer convention is essential for the derivative-based definition; no negative-order derivative or $H_{-1}$ is needed.

The normalized Hermite functions multiply the physicists' Hermite polynomials by a Gaussian. They form an orthonormal basis of $L^2(\mathbb{R})$ and satisfy $(-d^2/d{x}^2+x^2)H_n=(2n+1)H_n$. With $a=(x+d/dx)/\sqrt{2}$ and $a^{\ast }=(x-d/dx)/\sqrt{2}$, they are $(a^{\ast }{)}^n{H}_0/\sqrt{n!}$ and the quantum harmonic oscillator is $2a^{\ast }a+1$. For completeness, if $u$ is orthogonal to all of them, all moments of $u(x)e^{-x^2/2}$ vanish. Its transform $\int u(x)e^{-x^2/2}{e}^{zx}dx$ is entire by Cauchy-Schwarz inequality, has every derivative zero at zero, and hence vanishes identically. Uniqueness of the Fourier transform gives $u=0$.

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 $(-i{)}^n$.

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 $e^{-s(-d^2/d{x}^2+x^2)}$.

The probabilists' Hermite polynomials areThey are orthogonal for the standard Gaussian weight and satisfy ${\mathrm{He}}_{n+1}=x{\mathrm{He}}_n-n{\mathrm{He}}_{n-1}$.

This is the Gaussian-probability normalization of the Hermite polynomial family.

The space-time Probabilists' Hermite polynomial $H_n(x,t)=t^{n/2}{\mathrm{He}}_n(x/\sqrt{t})$ extends polynomially to $t=0$ and solves ${\partial }_t{H}_n+\frac{1}{2}{\partial }_x^2{H}_n=0$. Therefore $H_n(B_t,t)$ is a martingale for a standard Brownian motion, by the Itô formula and Gaussian moments.