How to Derive the Hermite Polynomial Derivative Relation

For the physicists’ Hermite polynomials $H_n(x)$, the derivative relation is

$$H_n'(x)=2nH_{n-1}(x), \qquad n\geq 1.$$

You can derive it directly from the Hermite differential equation. The key is that differentiating the equation shows $H_n'$ satisfies the equation for the next lower degree. The differential equation identifies the polynomial up to a constant; the leading coefficient determines that constant.

Start with the physicists’ equation

The physicists’ Hermite polynomials satisfy

$$H_n''(x)-2xH_n'(x)+2nH_n(x)=0.$$

Differentiate both sides with respect to $x$. Applying the product rule to $-2xH_n'(x)$ gives

$$H_n'''(x)-2H_n'(x)-2xH_n''(x)+2nH_n'(x)=0.$$

Collect the terms involving $H_n'$:

$$H_n'''(x)-2xH_n''(x)+2(n-1)H_n'(x)=0.$$

Now set $y(x)=H_n'(x)$. Since $y'=H_n''$ and $y''=H_n'''$, this becomes

$$y''(x)-2xy'(x)+2(n-1)y(x)=0.$$

That is the Hermite differential equation with degree parameter $n-1$. Also, because $H_n$ is a degree-$n$ polynomial, $y=H_n'$ is a degree-$(n-1)$ polynomial when $n\geq1$.

Identify the polynomial and its constant

For a fixed nonnegative integer $m$, the polynomial solution of

$$u''-2xu'+2mu=0$$

is unique up to an overall constant. One way to see this is to substitute a polynomial $u(x)=\sum_k a_kx^k$. Equating the coefficient of $x^k$ gives

$$(k+2)(k+1)a_{k+2}+2(m-k)a_k=0.$$

The coefficients are therefore determined by the leading coefficient, so there is only one polynomial solution up to scaling. Since $H_n'$ and $H_{n-1}$ are both degree-$(n-1)$ polynomial solutions, there is a constant $C_n$ such that

$$H_n'(x)=C_nH_{n-1}(x).$$

For the physicists’ convention, the leading term of $H_n(x)$ is $2^n x^n$. Therefore the leading term of its derivative is $n2^n x^{n-1}$. The leading term of $H_{n-1}(x)$ is $2^{n-1}x^{n-1}$. Comparing these coefficients gives

$$C_n=\frac{n2^n}{2^{n-1}}=2n,$$

so

$$\boxed{H_n'(x)=2nH_{n-1}(x)}.$$

The equation gives the proportionality; the chosen normalization of the Hermite polynomials gives the factor $2n$.

Worked example: n = 3

The first relevant physicists’ Hermite polynomials are

$$H_3(x)=8x^3-12x, \qquad H_2(x)=4x^2-2.$$

Direct differentiation gives

$$H_3'(x)=24x^2-12.$$

The derivative relation predicts $H_3'=2(3)H_2$, and indeed

$$6H_2(x)=6(4x^2-2)=24x^2-12=H_3'(x).$$

This example also illustrates why the leading-coefficient comparison is necessary: the differential equation alone tells you that the derivative is proportional to $H_2$, not what the proportionality constant is.

Probabilists’ convention

A second common convention uses Hermite polynomials $\operatorname{He}_n(x)$ satisfying

$$\operatorname{He}_n''(x)-x\operatorname{He}_n'(x)+n\operatorname{He}_n(x)=0.$$

Differentiating this equation shows that $\operatorname{He}_n'$ satisfies the equation with degree parameter $n-1$. These polynomials are monic, so comparing leading coefficients yields

$$\operatorname{He}_n'(x)=n\operatorname{He}_{n-1}(x).$$

Thus the factor differs because the two conventions use different normalizations. Check which convention is being used before applying the formula.

Common mistakes

• Mixing conventions: $H_n'=2nH_{n-1}$ is for physicists’ Hermite polynomials; the probabilists’ version has factor $n$.
• Skipping the normalization step: The differential equation proves proportionality to the lower-degree polynomial, but the leading coefficient fixes the constant.
• Using the formula at $n=0$ without defining $H_{-1}$: The derivative of $H_0=1$ is zero. The stated relation is normally written for $n\geq1$.
• Confusing this identity with a three-term recurrence: This is a derivative relation; it is not the recurrence that expresses $H_{n+1}$ using $xH_n$ and $H_{n-1}$.