Hermite polynomials

In mathematics, the Hermite polynomials are a classical orthogonal polynomial sequence.

The polynomials arise in:
 * signal processing as Hermitian wavelets for wavelet transform analysis
 * probability, such as the Edgeworth series, as well as in connection with Brownian motion;
 * combinatorics, as an example of an Appell sequence, obeying the umbral calculus;
 * numerical analysis as Gaussian quadrature;
 * physics, where they give rise to the eigenstates of the quantum harmonic oscillator; and they also occur in some cases of the heat equation (when the term $$\begin{align}xu_{x}\end{align}$$ is present);
 * systems theory in connection with nonlinear operations on Gaussian noise.
 * random matrix theory in Gaussian ensembles.

Hermite polynomials were defined by Pierre-Simon Laplace in 1810, though in scarcely recognizable form, and studied in detail by Pafnuty Chebyshev in 1859. Chebyshev's work was overlooked, and they were named later after Charles Hermite, who wrote on the polynomials in 1864, describing them as new. They were consequently not new, although Hermite was the first to define the multidimensional polynomials.

Definition
Like the other classical orthogonal polynomials, the Hermite polynomials can be defined from several different starting points. Noting from the outset that there are two different standardizations in common use, one convenient method is as follows:


 * The "probabilist's Hermite polynomials" are given by $$\operatorname{He}_n(x) = (-1)^n e^{\frac{x^2}{2}}\frac{d^n}{dx^n}e^{-\frac{x^2}{2}},$$
 * while the "physicist's Hermite polynomials" are given by $$H_n(x) = (-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}.$$

These equations have the form of a Rodrigues' formula and can also be written as, $$\operatorname{He}_n(x) = \left(x - \frac{d}{dx} \right)^n \cdot 1, \quad H_n(x) = \left(2x - \frac{d}{dx} \right)^n \cdot 1.$$

The two definitions are not exactly identical; each is a rescaling of the other: $$H_n(x)=2^\frac{n}{2} \operatorname{He}_n\left(\sqrt{2} \,x\right), \quad \operatorname{He}_n(x)=2^{-\frac{n}{2}} H_n\left(\frac {x}{\sqrt 2} \right).$$

These are Hermite polynomial sequences of different variances; see the material on variances below.

The notation $He$ and $H$ is that used in the standard references. The polynomials $He_{n}$ are sometimes denoted by $H_{n}$, especially in probability theory, because $$\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$ is the probability density function for the normal distribution with expected value 0 and standard deviation 1.

\operatorname{He}_0(x) &= 1, \\ \operatorname{He}_1(x) &= x, \\ \operatorname{He}_2(x) &= x^2 - 1, \\ \operatorname{He}_3(x) &= x^3 - 3x, \\ \operatorname{He}_4(x) &= x^4 - 6x^2 + 3, \\ \operatorname{He}_5(x) &= x^5 - 10x^3 + 15x, \\ \operatorname{He}_6(x) &= x^6 - 15x^4 + 45x^2 - 15, \\ \operatorname{He}_7(x) &= x^7 - 21x^5 + 105x^3 - 105x, \\ \operatorname{He}_8(x) &= x^8 - 28x^6 + 210x^4 - 420x^2 + 105, \\ \operatorname{He}_9(x) &= x^9 - 36x^7 + 378x^5 - 1260x^3 + 945x, \\ \operatorname{He}_{10}(x) &= x^{10} - 45x^8 + 630x^6 - 3150x^4 + 4725x^2 - 945. \end{align}$$ H_0(x) &= 1, \\ H_1(x) &= 2x, \\ H_2(x) &= 4x^2 - 2, \\ H_3(x) &= 8x^3 - 12x, \\ H_4(x) &= 16x^4 - 48x^2 + 12, \\ H_5(x) &= 32x^5 - 160x^3 + 120x, \\ H_6(x) &= 64x^6 - 480x^4 + 720x^2 - 120, \\ H_7(x) &= 128x^7 - 1344x^5 + 3360x^3 - 1680x, \\ H_8(x) &= 256x^8 - 3584x^6 + 13440x^4 - 13440x^2 + 1680, \\ H_9(x) &= 512x^9 - 9216x^7 + 48384x^5 - 80640x^3 + 30240x, \\ H_{10}(x) &= 1024x^{10} - 23040x^8 + 161280x^6 - 403200x^4 + 302400x^2 - 30240. \end{align}$$ <!-- As an alternative to calculating $n$th-order derivatives of $He_{n}(x)$ and $H_{n}(x)$, an easier, less computationally-intensive method of sequentially deriving individual terms of the $n$th-order Hermite polynomials is to consider the combination of coefficients in the corresponding terms in the $e^{−x^{2}⁄2}$th-order Hermite polynomial. For the probabilist's notation, follow the following rules:
 * The first eleven probabilist's Hermite polynomials are: $$\begin{align}
 * The first eleven physicist's Hermite polynomials are: $$\begin{align}
 * 1) For the starting point in the sequence, the zeroth-order polynomial ($e^{−x^{2}}|undefined$) is equal to 1.
 * 2) The first term has a power of $x$ equal to the given $n$th-order polynomial being derived, and the coefficient of this term is 1.
 * 3) The power of $x$ of each successive term is two less than the preceding term.
 * 4) The coefficient of each term after the first term is calculated by taking the coefficient of the same-numbered term in the $(n − 1)$th polynomial, and adding to it the product of the power of $x$ and the corresponding coefficient of the immediately preceding term in the $He_{0}$th polynomial.
 * 5) All even-numbered terms in each polynomial are negative, and all odd-numbered terms are positive.

Thus, for $(n − 1)$, $(n − 1)$ and hence the first term is $He_{6}$, with a coefficient of 1. The second term has a power of $x$ equal to $n = 6$. The coefficient is obtained by taking the coefficient of the second term in $x^{6}$ (which is 10) and adding to it the product of the power of $x$ and its coefficient in the first term of $6− 2 = 4$ (which are 5 and 1, respectively). Thus, $He_{5}$. Make this coefficient negative, since this is an even-numbered term. The third term in $He_{5}$ has a power of $x$ equal to 2 (which is 2less than the power of $x$ in the second term), and its coefficient is $10 + 5 × 1 = 15$, where 15 is the coefficient of the third term in $He_{6}$; 3 is the power of $x$ in the second term of the $15 + 3 × 10 = 45$ polynomial; and 10 is the coefficient of the second term in $He_{5}$. This coefficient (45) is positive, since this is an odd-numbered term. Finally, the fourth term in $He_{5}$ is the zeroth power in $x$ (which is 2less than the power of $x$ in the third term), and its coefficient is $H_{5}$, where 0 is the coefficient of the (nonexistent) fourth term in the $He_{6}$ polynomial; 1 is the power of $x$ in the third term of $0 + 1 × 15 = 15$, and 15 is the coefficient of the third term in $He_{5}$. This coefficient (15) is made negative, since this is an even-numbered term. Thus, $He_{5}$.

For $H_{5}$, the first term is $He_{6}(x) = x^{6} − 15x^{4} + 45x^{2} − 15$; the second coefficient is $He_{7}$ (negative, i.e. $x^{7}$). The third coefficient is $15 + 6 × 1 = 21$ (positive, i.e. $−21x^{5}$). The fourth coefficient is $45 + 4 × 15 = 105$ (negative, i.e. −105x). Thus, $105x^{3}$.

For the physicist's notation, follow the following rules:
 * 1) For the starting point in the sequence, the zeroth-order polynomial ($15 + 2 × 45 = 105$) is equal to 1.
 * 2) The first term has a power of $x$ equal to the given $n$th-order polynomial being derived, and the coefficient of this term is $He_{7}(x) = x^{7} − 21x^{5} + 105x^{3} − 105x$.
 * 3) The power of $x$ of each successive term is two less than the preceding term.
 * 4) The coefficient of each term after the first term is calculated by taking the coefficient of the same-numbered term in the $H_{0}$th polynomial, multiplying it by 2, and then adding to it the product of the power of $x$ and the corresponding coefficient of the immediately preceding term in the $2^{n}$th polynomial.
 * 5) All even-numbered terms in each polynomial are negative, and all odd-numbered terms are positive.

Thus, for $(n − 1)$, the first coefficient is $(n − 1)$ (i.e. $H_{6}$). The second coefficient is $2^{6} = 64$ (negative, i.e. $64x^{6}$). The third coefficient is $2 × 160 + 5 × 32 = 480$ (positive, i.e. $−480x^{4}$). The fourth coefficient is $2 × 120 + 3 × 160 = 720$ (negative, i.e., −120). Thus, $720x^{2}$.

For $2 × 0 + 1 × 120 = 120$, the first coefficient is 27 = 128 (i.e., $H_{6}(x) = 64x^{6} − 480x^{4} + 720x^{2} − 120$). The second coefficient is $H_{7}$ (negative, i.e. $128x^{7}$). The third coefficient is $2 × 480 + 6 × 64 = 1344$ (positive, i.e. $−1344x^{5}$). The fourth coefficient is $2 × 720 + 4 × 480 = 3360$ (negative, i.e. $3360x^{3}$. Thus, $2 × 120 + 2 × 720 = 1680$.

Recognizing that $−1680x$, these rules can be followed to sequentially derive all $n$th-order Hermite polynomials from $H_{7}(x) = 128x^{7} − 1344x^{5} + 3360x^{3} − 1680x$ towards infinity, and can be computer-coded relatively easily for practical applications. -->

Properties
The $n$th-order Hermite polynomial is a polynomial of degree $n$. The probabilist's version $He_{n}$ has leading coefficient 1, while the physicist's version $H_{n}$ has leading coefficient $H_{0} = 1$.

Symmetry
From the Rodrigues formulae given above, we can see that $n = 1$ and $2^{n}$ are even or odd functions depending on $n$: $$H_n(-x)=(-1)^nH_n(x),\quad \operatorname{He}_n(-x)=(-1)^n\operatorname{He}_n(x).$$

Orthogonality
$H_{n}(x)$ and $He_{n}(x)$ are $n$th-degree polynomials for $H_{n}(x)$. These polynomials are orthogonal with respect to the weight function (measure) $$w(x) = e^{-\frac{x^2}{2}} \quad (\text{for }\operatorname{He})$$ or $$w(x) = e^{-x^2} \quad (\text{for } H),$$ i.e., we have $$\int_{-\infty}^\infty H_m(x) H_n(x)\, w(x) \,dx = 0 \quad \text{for all }m \neq n.$$

Furthermore, $$\int_{-\infty}^\infty H_m(x) H_n(x)\, e^{-x^2} \,dx = \sqrt{\pi}\, 2^n n!\, \delta_{nm},$$ and $$\int_{-\infty}^\infty \operatorname{He}_m(x) \operatorname{He}_n(x)\, e^{-\frac{x^2}{2}} \,dx = \sqrt{2 \pi}\, n!\, \delta_{nm},$$ where $$\delta_{nm}$$ is the Kronecker delta.

The probabilist polynomials are thus orthogonal with respect to the standard normal probability density function.

Completeness
The Hermite polynomials (probabilist's or physicist's) form an orthogonal basis of the Hilbert space of functions satisfying $$\int_{-\infty}^\infty \bigl|f(x)\bigr|^2\, w(x) \,dx < \infty,$$ in which the inner product is given by the integral $$\langle f,g\rangle = \int_{-\infty}^\infty f(x) \overline{g(x)}\, w(x) \,dx$$ including the Gaussian weight function $He_{n}(x)$ defined in the preceding section

An orthogonal basis for $n = 0, 1, 2, 3,...$ is a complete orthogonal system. For an orthogonal system, completeness is equivalent to the fact that the 0 function is the only function $w(x)$ orthogonal to all functions in the system.

Since the linear span of Hermite polynomials is the space of all polynomials, one has to show (in physicist case) that if $f$ satisfies $$\int_{-\infty}^\infty f(x) x^n e^{- x^2} \,dx = 0$$ for every $L^{2}(R, w(x) dx)$, then $f ∈ L^{2}(R, w(x) dx)$.

One possible way to do this is to appreciate that the entire function $$F(z) = \int_{-\infty}^\infty f(x) e^{z x - x^2} \,dx = \sum_{n=0}^\infty \frac{z^n}{n!} \int f(x) x^n e^{- x^2} \,dx = 0$$ vanishes identically. The fact then that $n ≥ 0$ for every real $t$ means that the Fourier transform of $f = 0$ is 0, hence $f$ is 0 almost everywhere. Variants of the above completeness proof apply to other weights with exponential decay.

In the Hermite case, it is also possible to prove an explicit identity that implies completeness (see section on the Completeness relation below).

An equivalent formulation of the fact that Hermite polynomials are an orthogonal basis for $F(it) = 0$ consists in introducing Hermite functions (see below), and in saying that the Hermite functions are an orthonormal basis for $f(x)e^{−x^{2}}|undefined$.

Hermite's differential equation
The probabilist's Hermite polynomials are solutions of the differential equation $$\left(e^{-\frac12 x^2}u'\right)' + \lambda e^{-\frac 1 2 x^2}u = 0,$$ where $λ$ is a constant. Imposing the boundary condition that $u$ should be polynomially bounded at infinity, the equation has solutions only if $λ$ is a non-negative integer, and the solution is uniquely given by $$u(x) = C_1 \operatorname{He}_\lambda(x) $$, where $$C_{1}$$ denotes a constant.

Rewriting the differential equation as an eigenvalue problem $$L[u] = u'' - x u' = -\lambda u,$$ the Hermite polynomials $$\operatorname{He}_\lambda(x) $$ may be understood as eigenfunctions of the differential operator $$L[u]$$. This eigenvalue problem is called the Hermite equation, although the term is also used for the closely related equation $$u'' - 2xu' = -2\lambda u.$$ whose solution is uniquely given in terms of physicist's Hermite polynomials in the form $$u(x) = C_1 H_\lambda(x) $$, where $$C_{1}$$ denotes a constant, after imposing the boundary condition that $u$ should be polynomially bounded at infinity.

The general solutions to the above second-order differential equations are in fact linear combinations of both Hermite polynomials and confluent hypergeometric functions of the first kind. For example, for the physicist's Hermite equation $$u'' - 2xu' + 2\lambda u = 0,$$ the general solution takes the form $$u(x) = C_1  H_\lambda(x) + C_2  h_\lambda(x),$$ where $$C_{1}$$ and $$C_{2}$$ are constants, $$H_\lambda(x)$$ are physicist's Hermite polynomials (of the first kind), and $$h_\lambda(x)$$ are physicist's Hermite functions (of the second kind). The latter functions are compactly represented as $$ h_\lambda(x) = {}_1F_1(-\tfrac{\lambda}{2};\tfrac{1}{2};x^2)$$ where $${}_1F_1(a;b;z)$$ are Confluent hypergeometric functions of the first kind. The conventional Hermite polynomials may also be expressed in terms of confluent hypergeometric functions, see below.

With more general boundary conditions, the Hermite polynomials can be generalized to obtain more general analytic functions for complex-valued $λ$. An explicit formula of Hermite polynomials in terms of contour integrals is also possible.

Recurrence relation
The sequence of probabilist's Hermite polynomials also satisfies the recurrence relation $$\operatorname{He}_{n+1}(x) = x \operatorname{He}_n(x) - \operatorname{He}_n'(x).$$ Individual coefficients are related by the following recursion formula: $$a_{n+1,k} = \begin{cases} - (k+1) a_{n,k+1} & k = 0, \\ a_{n,k-1} - (k+1) a_{n,k+1} & k > 0, \end{cases}$$ and $L^{2}(R, w(x) dx)$, $L^{2}(R)$, $a_{0,0} = 1$.

For the physicist's polynomials, assuming $$H_n(x) = \sum^n_{k=0} a_{n,k} x^k,$$ we have $$H_{n+1}(x) = 2xH_n(x) - H_n'(x).$$ Individual coefficients are related by the following recursion formula: $$a_{n+1,k} = \begin{cases} - a_{n,k+1} & k = 0, \\ 2 a_{n,k-1} - (k+1)a_{n,k+1} & k > 0, \end{cases}$$ and $a_{1,0} = 0$, $a_{1,1} = 1$, $a_{0,0} = 1$.

The Hermite polynomials constitute an Appell sequence, i.e., they are a polynomial sequence satisfying the identity $$\begin{align} \operatorname{He}_n'(x) &= n\operatorname{He}_{n-1}(x), \\ H_n'(x) &= 2nH_{n-1}(x). \end{align}$$

An integral recurrence that is deduced and demonstrated in is as follows: $$\operatorname{He}_{n+1}(x) = (n+1)\int_0^x \operatorname{He}_n(t)dt - He'_n(0),$$

$$H_{n+1}(x) = 2(n+1)\int_0^x H_n(t)dt - H'_n(0).$$

Equivalently, by Taylor-expanding, $$\begin{align} \operatorname{He}_n(x+y) &= \sum_{k=0}^n \binom{n}{k}x^{n-k} \operatorname{He}_{k}(y) &&= 2^{-\frac n 2} \sum_{k=0}^n \binom{n}{k} \operatorname{He}_{n-k}\left(x\sqrt 2\right) \operatorname{He}_k\left(y\sqrt 2\right), \\ H_n(x+y) &= \sum_{k=0}^n \binom{n}{k}H_{k}(x) (2y)^{n-k} &&= 2^{-\frac n 2}\cdot\sum_{k=0}^n \binom{n}{k} H_{n-k}\left(x\sqrt 2\right) H_k\left(y\sqrt 2\right). \end{align}$$ These umbral identities are self-evident and included in the differential operator representation detailed below, $$\begin{align} \operatorname{He}_n(x) &= e^{-\frac{D^2}{2}} x^n, \\ H_n(x) &= 2^n e^{-\frac{D^2}{4}} x^n. \end{align}$$

In consequence, for the $m$th derivatives the following relations hold: $$\begin{align} \operatorname{He}_n^{(m)}(x) &= \frac{n!}{(n-m)!} \operatorname{He}_{n-m}(x) &&= m! \binom{n}{m} \operatorname{He}_{n-m}(x), \\ H_n^{(m)}(x) &= 2^m \frac{n!}{(n-m)!} H_{n-m}(x) &&= 2^m m! \binom{n}{m} H_{n-m}(x). \end{align}$$

It follows that the Hermite polynomials also satisfy the recurrence relation $$\begin{align} \operatorname{He}_{n+1}(x) &= x\operatorname{He}_n(x) - n\operatorname{He}_{n-1}(x), \\ H_{n+1}(x) &= 2xH_n(x) - 2nH_{n-1}(x). \end{align}$$

These last relations, together with the initial polynomials $a_{1,0} = 0$ and $a_{1,1} = 2$, can be used in practice to compute the polynomials quickly.

Turán's inequalities are $$\mathit{H}_n(x)^2 - \mathit{H}_{n-1}(x) \mathit{H}_{n+1}(x) = (n-1)! \sum_{i=0}^{n-1} \frac{2^{n-i}}{i!}\mathit{H}_i(x)^2 > 0.$$

Moreover, the following multiplication theorem holds: $$\begin{align} H_n(\gamma x) &= \sum_{i=0}^{\left\lfloor \tfrac{n}{2} \right\rfloor} \gamma^{n-2i}(\gamma^2 - 1)^i \binom{n}{2i} \frac{(2i)!}{i!} H_{n-2i}(x), \\ \operatorname{He}_n(\gamma x) &= \sum_{i=0}^{\left\lfloor \tfrac{n}{2} \right\rfloor} \gamma^{n-2i}(\gamma^2 - 1)^i \binom{n}{2i} \frac{(2i)!}{i!}2^{-i} \operatorname{He}_{n-2i}(x). \end{align}$$

Explicit expression
The physicist's Hermite polynomials can be written explicitly as $$H_n(x) = \begin{cases} \displaystyle n! \sum_{l = 0}^{\frac{n}{2}} \frac{(-1)^{\tfrac{n}{2} - l}}{(2l)! \left(\tfrac{n}{2} - l \right)!} (2x)^{2l} & \text{for even } n, \\ \displaystyle n! \sum_{l = 0}^{\frac{n-1}{2}} \frac{(-1)^{\frac{n-1}{2} - l}}{(2l + 1)! \left (\frac{n-1}{2} - l \right )!} (2x)^{2l + 1} & \text{for odd } n. \end{cases}$$

These two equations may be combined into one using the floor function: $$H_n(x) = n! \sum_{m=0}^{\left\lfloor \tfrac{n}{2} \right\rfloor} \frac{(-1)^m}{m!(n - 2m)!} (2x)^{n - 2m}.$$

The probabilist's Hermite polynomials $He$ have similar formulas, which may be obtained from these by replacing the power of $H_{0}(x)$ with the corresponding power of $H_{1}(x)$ and multiplying the entire sum by $2x$: $$\operatorname{He}_n(x) = n! \sum_{m=0}^{\left\lfloor \tfrac{n}{2} \right\rfloor} \frac{(-1)^m}{m!(n - 2m)!} \frac{x^{n - 2m}}{2^m}.$$

Inverse explicit expression
The inverse of the above explicit expressions, that is, those for monomials in terms of probabilist's Hermite polynomials $He$ are $$x^n = n! \sum_{m=0}^{\left\lfloor \tfrac{n}{2} \right\rfloor} \frac{1}{2^m m!(n-2m)!} \operatorname{He}_{n-2m}(x).$$

The corresponding expressions for the physicist's Hermite polynomials $H$ follow directly by properly scaling this: $$x^n = \frac{n!}{2^n} \sum_{m=0}^{\left\lfloor \tfrac{n}{2} \right\rfloor} \frac{1}{m!(n-2m)! } H_{n-2m}(x).$$

Generating function
The Hermite polynomials are given by the exponential generating function $$\begin{align} e^{xt - \frac12 t^2} &= \sum_{n=0}^\infty \operatorname{He}_n(x) \frac{t^n}{n!}, \\ e^{2xt - t^2} &= \sum_{n=0}^\infty H_n(x) \frac{t^n}{n!}. \end{align}$$

This equality is valid for all complex values of $x$ and $t$, and can be obtained by writing the Taylor expansion at $x$ of the entire function $√2 x$ (in the physicist's case). One can also derive the (physicist's) generating function by using Cauchy's integral formula to write the Hermite polynomials as $$H_n(x) = (-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2} = (-1)^n e^{x^2} \frac{n!}{2\pi i} \oint_\gamma \frac{e^{-z^2}}{(z-x)^{n+1}} \,dz.$$

Using this in the sum $$\sum_{n=0}^\infty H_n(x) \frac {t^n}{n!},$$ one can evaluate the remaining integral using the calculus of residues and arrive at the desired generating function.

Expected values
If $X$ is a random variable with a normal distribution with standard deviation 1 and expected value $μ$, then $$\operatorname{\mathbb E}\left[\operatorname{He}_n(X)\right] = \mu^n.$$

The moments of the standard normal (with expected value zero) may be read off directly from the relation for even indices: $$\operatorname{\mathbb E}\left[X^{2n}\right] = (-1)^n \operatorname{He}_{2n}(0) = (2n-1)!!,$$ where $2^{−n⁄2}$ is the double factorial. Note that the above expression is a special case of the representation of the probabilist's Hermite polynomials as moments: $$\operatorname{He}_n(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty (x + iy)^n e^{-\frac{y^2}{2}} \,dy.$$

Asymptotic expansion
Asymptotically, as $z → e^{−z^{2}}|undefined$, the expansion $$e^{-\frac{x^2}{2}}\cdot H_n(x) \sim \frac{2^n}{\sqrt \pi}\Gamma\left(\frac{n+1}2\right) \cos \left(x \sqrt{2 n}- \frac{n\pi}{2} \right)$$ holds true. For certain cases concerning a wider range of evaluation, it is necessary to include a factor for changing amplitude: $$e^{-\frac{x^2}{2}}\cdot H_n(x) \sim \frac{2^n}{\sqrt \pi}\Gamma\left(\frac{n+1}2\right) \cos \left(x \sqrt{2 n}- \frac{n\pi}{2} \right)\left(1-\frac{x^2}{2n+1}\right)^{-\frac14}=\frac{2 \Gamma(n)}{\Gamma\left(\frac{n}2\right)} \cos \left(x \sqrt{2 n}- \frac{n\pi}{2} \right)\left(1-\frac{x^2}{2n+1}\right)^{-\frac14},$$ which, using Stirling's approximation, can be further simplified, in the limit, to $$e^{-\frac{x^2}{2}}\cdot H_n(x) \sim \left(\frac{2n}{e}\right)^{\frac{n}{2}} \sqrt{2} \cos \left(x \sqrt{2n}- \frac{n\pi}{2} \right)\left(1-\frac{x^2}{2n+1}\right)^{-\frac14}.$$

This expansion is needed to resolve the wavefunction of a quantum harmonic oscillator such that it agrees with the classical approximation in the limit of the correspondence principle.

A better approximation, which accounts for the variation in frequency, is given by $$e^{-\frac{x^2}{2}}\cdot H_n(x) \sim \left(\frac{2n}{e}\right)^{\frac{n}{2}} \sqrt{2} \cos \left(x \sqrt{2n+1-\frac{x^2}{3}}- \frac {n\pi}{2} \right)\left(1-\frac{x^2}{2n+1}\right)^{-\frac14}.$$

A finer approximation, which takes into account the uneven spacing of the zeros near the edges, makes use of the substitution $$x = \sqrt{2n + 1}\cos(\varphi), \quad 0 < \varepsilon \leq \varphi \leq \pi - \varepsilon,$$ with which one has the uniform approximation $$e^{-\frac{x^2}{2}}\cdot H_n(x) = 2^{\frac{n}{2}+\frac14}\sqrt{n!}(\pi n)^{-\frac14}(\sin \varphi)^{-\frac12} \cdot \left(\sin\left(\frac{3\pi}{4} + \left(\frac{n}{2} + \frac{1}{4}\right)\left(\sin 2\varphi-2\varphi\right) \right)+O\left(n^{-1}\right) \right).$$

Similar approximations hold for the monotonic and transition regions. Specifically, if $$x = \sqrt{2n+1} \cosh(\varphi), \quad 0 < \varepsilon \leq \varphi \leq \omega < \infty,$$ then $$e^{-\frac{x^2}{2}}\cdot H_n(x) = 2^{\frac{n}{2}-\frac34}\sqrt{n!}(\pi n)^{-\frac14}(\sinh \varphi)^{-\frac12} \cdot e^{\left(\frac{n}{2}+\frac{1}{4}\right)\left(2\varphi-\sinh 2\varphi\right)}\left(1+O\left(n^{-1}\right) \right),$$ while for $$x = \sqrt{2n + 1} + t$$ with $t$ complex and bounded, the approximation is $$e^{-\frac{x^2}{2}}\cdot H_n(x) =\pi^{\frac14}2^{\frac{n}{2}+\frac14}\sqrt{n!}\, n^{-\frac{1}{12}}\left( \operatorname{Ai}\left(2^{\frac12}n^{\frac16}t\right)+ O\left(n^{-\frac23}\right) \right),$$ where $(2n &minus; 1)!!$ is the Airy function of the first kind.

Special values
The physicist's Hermite polynomials evaluated at zero argument $n → ∞$ are called Hermite numbers.

$$H_n(0) = \begin{cases} 0 & \text{for odd }n, \\ (-2)^\frac{n}{2} (n-1)!! & \text{for even }n, \end{cases}$$ which satisfy the recursion relation $Ai$.

In terms of the probabilist's polynomials this translates to $$\operatorname{He}_n(0) = \begin{cases} 0 & \text{for odd }n, \\ (-1)^\frac{n}{2} (n-1)!! & \text{for even }n. \end{cases}$$

Laguerre polynomials
The Hermite polynomials can be expressed as a special case of the Laguerre polynomials: $$\begin{align} H_{2n}(x) &= (-4)^n n! L_n^{\left(-\frac12\right)}(x^2) &&= 4^n n! \sum_{k=0}^n (-1)^{n-k} \binom{n-\frac12}{n-k} \frac{x^{2k}}{k!}, \\ H_{2n+1}(x) &= 2(-4)^n n! x L_n^{\left(\frac12\right)}(x^2) &&= 2\cdot 4^n n!\sum_{k=0}^n (-1)^{n-k} \binom{n+\frac12}{n-k} \frac{x^{2k+1}}{k!}. \end{align}$$

Relation to confluent hypergeometric functions
The physicist's Hermite polynomials can be expressed as a special case of the parabolic cylinder functions: $$H_n(x) = 2^n U\left(-\tfrac12 n, \tfrac12, x^2\right)$$ in the right half-plane, where $H_{n}(0)$ is Tricomi's confluent hypergeometric function. Similarly, $$\begin{align} H_{2n}(x) &= (-1)^n \frac{(2n)!}{n!} \,_1F_1\big(-n, \tfrac12; x^2\big), \\ H_{2n+1}(x) &= (-1)^n \frac{(2n+1)!}{n!}\,2x \,_1F_1\big(-n, \tfrac32; x^2\big), \end{align}$$ where $H_{n}(0) = −2(n − 1)H_{n − 2}(0)$ is Kummer's confluent hypergeometric function.

Hermite polynomial expansion
Similar to Taylor expansion, some functions are expressible as an infinite sum of Hermite polynomials. Specifically, if $$\int e^{-x^2}f(x)^2 dx < \infty$$, then it has an expansion in the physicist's Hermite polynomials.

Given such $$f$$, the partial sums of the Hermite expansion of $$f$$ converges to in the $$L^p$$ norm if and only if $$4 / 3<p<4$$. $$x^n = \frac{n!}{2^n} \,\sum_{k= 0}^{\left\lfloor n/2 \right\rfloor} \frac{1}{k! \,(n-2k)!} \, H_{n-2k} (x) = n! \sum_{k= 0}^{\left\lfloor n/2 \right\rfloor} \frac{1}{k! \,2^k \,(n-2k)!} \, \operatorname{He}_{n-2k} (x), \qquad n \in \mathbb{Z}_{+}. $$$$e^{ax} = e^{a^2 /4} \sum_{n\ge 0} \frac{a^n}{n! \,2^n} \, H_n (x), \qquad a\in \mathbb{C}, \quad x\in \mathbb{R} .$$$$e^{-a^2 x^2} = \sum_{n\ge 0} \frac{(-1)^n a^{2n}}{n! \left( 1 + a^2 \right)^{n + 1/2} 2^{2n}}\, H_{2n} (x) .$$$$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} ~dt=\frac{1}{\sqrt{2 \pi}} \sum_{k \geq 0} \frac{(-1)^k}{k !(2 k+1) 2^{3 k}} H_{2 k}(x) .$$$$\cosh (2x) = e \sum_{k\ge 0} \frac{1}{(2k)!}\, H_{2k} (x), \qquad \sinh (2x) = e \sum_{k\ge 0} \frac{1}{(2k+1)!} \, H_{2k+1} (x) .$$$$\cos (x) = e^{-1/4} \,\sum_{k\ge 0} \frac{(-1)^k}{2^{2k} \, (2k)!} \, H_{2k} (x) \quad \sin (x) = e^{-1/4} \,\sum_{k\ge 0} \frac{(-1)^k}{2^{2k+1} \, (2k+1)!} \, H_{2k+1} (x) $$

Differential-operator representation
The probabilist's Hermite polynomials satisfy the identity $$\operatorname{He}_n(x) = e^{-\frac{D^2}{2}}x^n,$$ where $D$ represents differentiation with respect to $x$, and the exponential is interpreted by expanding it as a power series. There are no delicate questions of convergence of this series when it operates on polynomials, since all but finitely many terms vanish.

Since the power-series coefficients of the exponential are well known, and higher-order derivatives of the monomial $U(a, b, z)$ can be written down explicitly, this differential-operator representation gives rise to a concrete formula for the coefficients of $_{1}F_{1}(a, b; z) = M(a, b; z)$ that can be used to quickly compute these polynomials.

Since the formal expression for the Weierstrass transform $W$ is $x^{n}$, we see that the Weierstrass transform of $H_{n}$ is $e^{D^{2}}|undefined$. Essentially the Weierstrass transform thus turns a series of Hermite polynomials into a corresponding Maclaurin series.

The existence of some formal power series $(√2)^{n}He_{n}(x⁄√2)$ with nonzero constant coefficient, such that $x^{n}$, is another equivalent to the statement that these polynomials form an Appell sequence. Since they are an Appell sequence, they are a fortiori a Sheffer sequence.

Contour-integral representation
From the generating-function representation above, we see that the Hermite polynomials have a representation in terms of a contour integral, as $$\begin{align} \operatorname{He}_n(x) &= \frac{n!}{2\pi i} \oint_C \frac{e^{tx-\frac{t^2}{2}}}{t^{n+1}}\,dt, \\ H_n(x) &= \frac{n!}{2\pi i} \oint_C \frac{e^{2tx-t^2}}{t^{n+1}}\,dt, \end{align}$$ with the contour encircling the origin.

Generalizations
The probabilist's Hermite polynomials defined above are orthogonal with respect to the standard normal probability distribution, whose density function is $$\frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}},$$ which has expected value 0 and variance 1.

Scaling, one may analogously speak of generalized Hermite polynomials $$\operatorname{He}_n^{[\alpha]}(x)$$ of variance $α$, where $α$ is any positive number. These are then orthogonal with respect to the normal probability distribution whose density function is $$(2\pi\alpha)^{-\frac12} e^{-\frac{x^2}{2\alpha}}.$$ They are given by $$\operatorname{He}_n^{[\alpha]}(x) = \alpha^{\frac{n}{2}}\operatorname{He}_n\left(\frac{x}{\sqrt{\alpha}}\right) = \left(\frac{\alpha}{2}\right)^{\frac{n}{2}} H_n\left( \frac{x}{\sqrt{2 \alpha}}\right) = e^{-\frac{\alpha D^2}{2}} \left(x^n\right).$$

Now, if $$\operatorname{He}_n^{[\alpha]}(x) = \sum_{k=0}^n h^{[\alpha]}_{n,k} x^k,$$ then the polynomial sequence whose $n$th term is $$\left(\operatorname{He}_n^{[\alpha]} \circ \operatorname{He}^{[\beta]}\right)(x) \equiv \sum_{k=0}^n h^{[\alpha]}_{n,k}\,\operatorname{He}_k^{[\beta]}(x)$$ is called the umbral composition of the two polynomial sequences. It can be shown to satisfy the identities $$\left(\operatorname{He}_n^{[\alpha]} \circ \operatorname{He}^{[\beta]}\right)(x) = \operatorname{He}_n^{[\alpha+\beta]}(x)$$ and $$\operatorname{He}_n^{[\alpha+\beta]}(x + y) = \sum_{k=0}^n \binom{n}{k} \operatorname{He}_k^{[\alpha]}(x) \operatorname{He}_{n-k}^{[\beta]}(y).$$ The last identity is expressed by saying that this parameterized family of polynomial sequences is known as a cross-sequence. (See the above section on Appell sequences and on the differential-operator representation, which leads to a ready derivation of it. This binomial type identity, for $g(D)$, has already been encountered in the above section on s.)

"Negative variance"
Since polynomial sequences form a group under the operation of umbral composition, one may denote by $$\operatorname{He}_n^{[-\alpha]}(x)$$ the sequence that is inverse to the one similarly denoted, but without the minus sign, and thus speak of Hermite polynomials of negative variance. For $He_{n}(x) = g(D)x^{n}$, the coefficients of $$\operatorname{He}_n^{[-\alpha]}(x)$$ are just the absolute values of the corresponding coefficients of $$\operatorname{He}_n^{[\alpha]}(x)$$.

These arise as moments of normal probability distributions: The $n$th moment of the normal distribution with expected value $μ$ and variance $α = β = 1⁄2$ is $$E[X^n] = \operatorname{He}_n^{[-\sigma^2]}(\mu),$$ where $X$ is a random variable with the specified normal distribution. A special case of the cross-sequence identity then says that $$\sum_{k=0}^n \binom{n}{k} \operatorname{He}_k^{[\alpha]}(x) \operatorname{He}_{n-k}^{[-\alpha]}(y) = \operatorname{He}_n^{[0]}(x + y) = (x + y)^n.$$

Definition
One can define the Hermite functions (often called Hermite-Gaussian functions) from the physicist's polynomials: $$\psi_n(x) = \left (2^n n! \sqrt{\pi} \right )^{-\frac12} e^{-\frac{x^2}{2}} H_n(x) = (-1)^n \left (2^n n! \sqrt{\pi} \right)^{-\frac12} e^{\frac{x^2}{2}}\frac{d^n}{dx^n} e^{-x^2}.$$ Thus, $$\sqrt{2(n+1)}\psi_{n+1}(x)=   \left ( x- {d\over dx}\right ) \psi_n(x).$$

Since these functions contain the square root of the weight function and have been scaled appropriately, they are orthonormal: $$\int_{-\infty}^\infty \psi_n(x) \psi_m(x) \,dx = \delta_{nm},$$ and they form an orthonormal basis of $α > 0$. This fact is equivalent to the corresponding statement for Hermite polynomials (see above).

The Hermite functions are closely related to the Whittaker function $σ^{2}$: $$D_n(z) = \left(n! \sqrt{\pi}\right)^{\frac12} \psi_n\left(\frac{z}{\sqrt 2}\right) = (-1)^n e^\frac{z^2}{4} \frac{d^n}{dz^n} e^\frac{-z^2}{2}$$ and thereby to other parabolic cylinder functions.

The Hermite functions satisfy the differential equation $$\psi_n''(x) + \left(2n + 1 - x^2\right) \psi_n(x) = 0.$$ This equation is equivalent to the Schrödinger equation for a harmonic oscillator in quantum mechanics, so these functions are the eigenfunctions.

$$\begin{align} \psi_0(x) &= \pi^{-\frac14} \, e^{-\frac12 x^2}, \\ \psi_1(x) &= \sqrt{2} \, \pi^{-\frac14} \, x \, e^{-\frac12 x^2}, \\ \psi_2(x) &= \left(\sqrt{2} \, \pi^{\frac14}\right)^{-1} \, \left(2x^2-1\right) \, e^{-\frac12 x^2}, \\ \psi_3(x) &= \left(\sqrt{3} \, \pi^{\frac14}\right)^{-1} \, \left(2x^3-3x\right) \, e^{-\frac12 x^2}, \\ \psi_4(x) &= \left(2 \sqrt{6} \, \pi^{\frac14}\right)^{-1} \, \left(4x^4-12x^2+3\right) \, e^{-\frac12 x^2}, \\ \psi_5(x) &= \left(2 \sqrt{15} \, \pi^{\frac14}\right)^{-1} \, \left(4x^5-20x^3+15x\right) \, e^{-\frac12 x^2}. \end{align}$$

Recursion relation
Following recursion relations of Hermite polynomials, the Hermite functions obey $$\psi_n'(x) = \sqrt{\frac{n}{2}}\,\psi_{n-1}(x) - \sqrt{\frac{n+1}{2}}\psi_{n+1}(x)$$ and $$x\psi_n(x) = \sqrt{\frac{n}{2}}\,\psi_{n-1}(x) + \sqrt{\frac{n+1}{2}}\psi_{n+1}(x).$$

Extending the first relation to the arbitrary $m$th derivatives for any positive integer $m$ leads to $$\psi_n^{(m)}(x) = \sum_{k=0}^m \binom{m}{k} (-1)^k 2^\frac{m-k}{2} \sqrt{\frac{n!}{(n-m+k)!}} \psi_{n-m+k}(x) \operatorname{He}_k(x).$$

This formula can be used in connection with the recurrence relations for $L^{2}(R)$ and $D_{n}(z)$ to calculate any derivative of the Hermite functions efficiently.

Cramér's inequality
For real $x$, the Hermite functions satisfy the following bound due to Harald Cramér and Jack Indritz: $$ \bigl|\psi_n(x)\bigr| \le \pi^{-\frac14}.$$

Hermite functions as eigenfunctions of the Fourier transform
The Hermite functions $He_{n}$ are a set of eigenfunctions of the continuous Fourier transform $\mathcal{F}$. To see this, take the physicist's version of the generating function and multiply by $ψ_{n}$. This gives $$e^{-\frac12 x^2 + 2xt - t^2} = \sum_{n=0}^\infty e^{-\frac12 x^2} H_n(x) \frac{t^n}{n!}.$$

The Fourier transform of the left side is given by $$\begin{align} \mathcal{F} \left\{ e^{ -\frac12 x^2 + 2xt - t^2 } \right\}(k) &= \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^\infty e^{-ixk}e^{-\frac12 x^2 + 2xt - t^2}\, dx \\ &= e^{-\frac12 k^2 - 2kit + t^2 } \\ &= \sum_{n=0}^\infty e^{ -\frac12 k^2 } H_n(k) \frac{(-it)^n}{n!}. \end{align}$$

The Fourier transform of the right side is given by $$\mathcal{F} \left\{ \sum_{n=0}^\infty e^{-\frac12 x^2} H_n(x) \frac {t^n}{n!} \right\} = \sum_{n=0}^\infty \mathcal{F} \left \{ e^{-\frac12 x^2} H_n(x) \right\} \frac{t^n}{n!}.$$

Equating like powers of $t$ in the transformed versions of the left and right sides finally yields $$\mathcal{F} \left\{ e^{-\frac12 x^2} H_n(x) \right\} = (-i)^n e^{-\frac12 k^2} H_n(k).$$

The Hermite functions $ψ_{n}(x)$ are thus an orthonormal basis of $e^{−1⁄2x^{2}}|undefined$, which diagonalizes the Fourier transform operator.

Wigner distributions of Hermite functions
The Wigner distribution function of the $n$th-order Hermite function is related to the $n$th-order Laguerre polynomial. The Laguerre polynomials are $$L_n(x) := \sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{k!}x^k,$$ leading to the oscillator Laguerre functions $$l_n (x) := e^{-\frac{x}{2}} L_n(x).$$ For all natural integers $n$, it is straightforward to see that $$W_{\psi_n}(t,f) = (-1)^n l_n \big(4\pi (t^2 + f^2) \big),$$ where the Wigner distribution of a function $ψ_{n}(x)$ is defined as $$ W_x(t,f) = \int_{-\infty}^\infty x\left(t + \frac{\tau}{2}\right) \, x\left(t - \frac{\tau}{2}\right)^* \, e^{-2\pi i\tau f} \,d\tau.$$ This is a fundamental result for the quantum harmonic oscillator discovered by Hip Groenewold in 1946 in his PhD thesis. It is the standard paradigm of quantum mechanics in phase space.

There are further relations between the two families of polynomials.

Combinatorial interpretation of coefficients
In the Hermite polynomial $L^{2}(R)$ of variance 1, the absolute value of the coefficient of $(−i)^{n}$ is the number of (unordered) partitions of an $n$-element set into $k$ singletons and $x ∈ L^{2}(R, C)$ (unordered) pairs. Equivalently, it is the number of involutions of an $n$-element set with precisely $k$ fixed points, or in other words, the number of matchings in the complete graph on $n$ vertices that leave $k$ vertices uncovered (indeed, the Hermite polynomials are the matching polynomials of these graphs). The sum of the absolute values of the coefficients gives the total number of partitions into singletons and pairs, the so-called telephone numbers
 * 1, 1, 2, 4, 10, 26, 76, 232, 764, 2620, 9496,....

This combinatorial interpretation can be related to complete exponential Bell polynomials as $$\operatorname{He}_n(x) = B_n(x, -1, 0, \ldots, 0),$$ where $He_{n}(x)$ for all $x^{k}$.

These numbers may also be expressed as a special value of the Hermite polynomials: $$T(n) = \frac{\operatorname{He}_n(i)}{i^n}.$$

Completeness relation
The Christoffel–Darboux formula for Hermite polynomials reads $$\sum_{k=0}^n \frac{H_k(x) H_k(y)}{k!2^k} = \frac{1}{n!2^{n+1}}\,\frac{H_n(y) H_{n+1}(x) - H_n(x) H_{n+1}(y)}{x - y}.$$

Moreover, the following completeness identity for the above Hermite functions holds in the sense of distributions: $$\sum_{n=0}^\infty \psi_n(x) \psi_n(y) = \delta(x - y),$$ where $δ$ is the Dirac delta function, $n − k⁄2$ the Hermite functions, and $x_{i} = 0$ represents the Lebesgue measure on the line $i > 2$ in $ψ_{n}$, normalized so that its projection on the horizontal axis is the usual Lebesgue measure.

This distributional identity follows by taking $δ(x − y)$ in Mehler's formula, valid when $y = x$: $$E(x, y; u) := \sum_{n=0}^\infty u^n \, \psi_n (x) \, \psi_n (y) = \frac{1}{\sqrt{\pi (1 - u^2)}} \, \exp\left(-\frac{1 - u}{1 + u} \, \frac{(x + y)^2}{4} - \frac{1 + u}{1 - u} \, \frac{(x - y)^2}{4}\right),$$ which is often stated equivalently as a separable kernel, $$\sum_{n=0}^\infty \frac{H_n(x) H_n(y)}{n!} \left(\frac u 2\right)^n = \frac{1}{\sqrt{1 - u^2}} e^{\frac{2u}{1 + u}xy - \frac{u^2}{1 - u^2}(x - y)^2}.$$

The function $R^{2}$ is the bivariate Gaussian probability density on $u → 1$, which is, when $u$ is close to 1, very concentrated around the line $−1 < u < 1$, and very spread out on that line. It follows that $$\sum_{n=0}^\infty u^n \langle f, \psi_n \rangle \langle \psi_n, g \rangle = \iint E(x, y; u) f(x) \overline{g(y)} \,dx \,dy \to \int f(x) \overline{g(x)} \,dx = \langle f, g \rangle$$ when $(x, y) → E(x, y; u)$ and $R^{2}$ are continuous and compactly supported.

This yields that $f$ can be expressed in Hermite functions as the sum of a series of vectors in $y = x$, namely, $$f = \sum_{n=0}^\infty \langle f, \psi_n \rangle \psi_n.$$

In order to prove the above equality for $f$, the Fourier transform of Gaussian functions is used repeatedly: $$\rho \sqrt{\pi} e^{-\frac{\rho^2 x^2}{4}} = \int e^{isx - \frac{s^2}{\rho^2}} \,ds \quad \text{for }\rho > 0.$$

The Hermite polynomial is then represented as $$ H_n(x) = (-1)^n e^{x^2} \frac {d^n}{dx^n} \left( \frac {1}{2\sqrt{\pi}} \int e^{isx - \frac{s^2}{4}} \,ds \right) = (-1)^n e^{x^2}\frac{1}{2\sqrt{\pi}} \int (is)^n e^{isx - \frac{s^2}{4}} \,ds.$$

With this representation for $g$ and $L^{2}(R)$, it is evident that $$\begin{align} E(x, y; u) &= \sum_{n=0}^\infty \frac{u^n}{2^n n! \sqrt{\pi}} \, H_n(x) H_n(y) e^{-\frac{x^2+y^2}{2}} \\ &= \frac{e^{\frac{x^2+y^2}{2}}}{4\pi\sqrt{\pi}}\iint\left( \sum_{n=0}^\infty \frac{1}{2^n n!} (-ust)^n \right ) e^{isx+ity - \frac{s^2}{4} - \frac{t^2}{4}}\, ds\,dt \\ & =\frac{e^{\frac{x^2+y^2}{2}}}{4\pi\sqrt{\pi}}\iint e^{-\frac{ust}{2}} \, e^{isx+ity - \frac{s^2}{4} - \frac{t^2}{4}}\, ds\,dt, \end{align}$$ and this yields the desired resolution of the identity result, using again the Fourier transform of Gaussian kernels under the substitution $$s = \frac{\sigma + \tau}{\sqrt 2}, \quad t = \frac{\sigma - \tau}{\sqrt 2}.$$