Special functions · Orthogonal polynomials

Hermite Polynomials

The third classical family completes the set: Legendre polynomials live on \([-1, 1]\), Laguerre polynomials on \([0, \infty)\), and Hermite polynomials on the whole real line, with the Gaussian bell \(e^{-x^2}\) as weight. They are the polynomials of probability theory (the normal distribution) and of quantum mechanics, where they describe the stationary states of the harmonic oscillator. As before, everything flows from a generating function.

1. Definition by a generating function

1° Hermite polynomials The Hermite polynomials \(H_n(x)\) are defined by \[ e^{2tx - t^2} = \sum_{n=0}^\infty\frac{H_n(x)}{n!}\,t^n. \]

(The book numbers this point as 2° as well; it is the first point, 1°.) The generating function is an entire function of \(t\), so the series converges for all \(t\) and \(x\). Writing \( e^{2tx - t^2} = e^{x^2}e^{-(t - x)^2} \) shows where it comes from: shifting a Gaussian bell. Expanding:

\[ H_0 = 1,\quad H_1 = 2x,\quad H_2 = 4x^2 - 2,\quad H_3 = 8x^3 - 12x,\quad H_4 = 16x^4 - 48x^2 + 12. \]

The leading coefficient of \(H_n\) is \(2^n\), and since \( e^{2t(-x) - t^2} = e^{2(-t)x - (-t)^2} \), \( H_n(-x) = (-1)^nH_n(x) \): \(H_n\) is even or odd with \(n\).

2. The explicit formula and Rodrigues' formula

2° 1) Explicit formula \[ H_n(x) = \sum_{k=0}^{[n/2]}\frac{(-1)^kn!}{k!\,(n - 2k)!}(2x)^{n-2k}. \]

Why: multiply the series \( e^{2tx} = \sum_j\frac{(2x)^j}{j!}t^j \) and \( e^{-t^2} = \sum_k\frac{(-1)^k}{k!}t^{2k} \). The coefficient of \(t^n\) collects \(j + 2k = n\), i.e. \( \sum_k\frac{(-1)^k(2x)^{n-2k}}{k!(n - 2k)!} \); multiplying by \(n!\) gives the formula.

Example 1 \( n = 4 \): \( k = 0 \): \( (2x)^4 = 16x^4 \); \( k = 1 \): \( -\frac{24}{1\cdot2}(2x)^2 = -48x^2 \); \( k = 2 \): \( \frac{24}{2\cdot1} = 12 \). So \( H_4 = 16x^4 - 48x^2 + 12 \). ✓

Not in the book, but the most compact description, is Rodrigues' formula:

\[ H_n(x) = (-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}. \]

It follows from the generating function: \( H_n = \frac{\partial^n}{\partial t^n}\Big[e^{x^2}e^{-(x - t)^2}\Big]_{t=0} \), and differentiating \(e^{-(x - t)^2}\) in \(t\) is the same as differentiating in \(x\) with a minus sign. Check: \( (e^{-x^2})' = -2xe^{-x^2} \), so \( H_1 = 2x \). ✓

3. Recurrence and derivative

2° 2) Recurrence \[ H_{n+1}(x) - 2x\,H_n(x) + 2n\,H_{n-1}(x) = 0. \]

Derivation: let \(G = e^{2tx - t^2}\). Then \( \frac{\partial G}{\partial t} = (2x - 2t)G \). Comparing coefficients of \(\frac{t^n}{n!}\): \( H_{n+1} = 2xH_n - 2nH_{n-1} \).

Differentiating \(G\) in \(x\) instead, \( \frac{\partial G}{\partial x} = 2tG \), gives the derivative rule

\[ H_n'(x) = 2n\,H_{n-1}(x), \]

so, like Bernoulli polynomials, Hermite polynomials form an Appell-type sequence.

Example 2 \( H_3 = 2xH_2 - 4H_1 = 2x(4x^2 - 2) - 8x = 8x^3 - 12x \). ✓ And \( H_3' = 24x^2 - 12 = 6(4x^2 - 2) = 6H_2 \). ✓

4. Orthogonality on the whole line

2° 3) Orthogonality \[ \int_{-\infty}^\infty e^{-x^2}H_m(x)\,H_n(x)\,dx = 2^n\,n!\,\sqrt\pi\;\delta_{mn}. \]

On the whole line, the Gaussian weight \(e^{-x^2}\) makes all integrals of polynomials finite, with \( \int_{-\infty}^\infty e^{-x^2}dx = \sqrt\pi \) (the Gaussian integral).

Why (the same pattern as for Legendre and Laguerre): for \(m < n\), use Rodrigues' formula, \( e^{-x^2}H_n = (-1)^n\frac{d^n}{dx^n}e^{-x^2} \), and integrate by parts \(n\) times against \(H_m\). The boundary terms vanish because \(e^{-x^2}\) decays faster than any polynomial grows, and \( H_m^{(n)} = 0 \). For \(m = n\), only the leading term \(2^nx^n\) contributes after \(n\) differentiations: \( 2^n\,n!\int e^{-x^2}dx = 2^nn!\sqrt\pi \).

Example 3 \( \int_{-\infty}^\infty e^{-x^2}H_1^2\,dx = 4\int x^2e^{-x^2}dx = 4\cdot\frac{\sqrt\pi}{2} = 2\sqrt\pi = 2^1\cdot1!\cdot\sqrt\pi \). ✓ And \( \int e^{-x^2}H_0H_2\,dx = \int e^{-x^2}(4x^2 - 2)dx = 2\sqrt\pi - 2\sqrt\pi = 0 \). ✓

5. Hermite's differential equation

2° 4) Hermite's equation \(H_n(x)\) is a particular solution of \[ y'' - 2x\,y' + 2n\,y = 0. \]

Why: from the two rules of Section 3, \( H_n' = 2nH_{n-1} \) and \( H_n'' = 2nH_{n-1}' = 2n\cdot2(n - 1)H_{n-2} \). Rewrite the recurrence with \(n - 1\): \( H_n - 2xH_{n-1} + 2(n - 1)H_{n-2} = 0 \). Multiply by \(2n\) and substitute: \( 2nH_n - 2xH_n' + H_n'' = 0 \).

For non-integer \(n\), the equation still has solutions, but none of them is a polynomial, and they grow like \(e^{x^2}\) at infinity. Only for \(n = 0, 1, 2, \dots\) is there a solution (the polynomial \(H_n\)) with \( e^{-x^2/2}y \) decaying at \(\pm\infty\). This is exactly the mechanism of energy quantization in the next section.

6. The quantum harmonic oscillator

A particle in the potential \( V = \frac12m\omega^2x^2 \) (a mass on a spring, a vibrating molecule) obeys the Schrödinger equation. In suitable units it reads

\[ -\psi'' + x^2\psi = 2\varepsilon\,\psi. \]

Substituting \( \psi = e^{-x^2/2}y(x) \) turns it into Hermite's equation \( y'' - 2xy' + (2\varepsilon - 1)y = 0 \). A physically acceptable solution must be normalizable (\(\int|\psi|^2 < \infty\)), which by Section 5 forces \( 2\varepsilon - 1 = 2n \). Hence:

\[ \psi_n(x) = \frac{1}{\sqrt{2^nn!\sqrt\pi}}\,H_n(x)\,e^{-x^2/2}, \qquad \varepsilon_n = n + \frac12. \]

The energy comes in equal steps, \(E_n = \hbar\omega\big(n + \frac12\big)\), and the lowest energy is not zero (the zero-point energy). The normalization constant is exactly the square root of the orthogonality constant, so the \(\psi_n\) form an orthonormal family.

V = x²/2 x n = 0 n = 1 n = 2 n = 3
Figure 1. The oscillator wavefunctions \(\psi_n(x) = c_nH_n(x)e^{-x^2/2}\) for \(n = 0, 1, 2, 3\), each drawn at the height of its energy level \(n + \frac12\), inside the parabola \(V = \frac{x^2}{2}\). The state \(\psi_n\) has \(n\) zeros (the zeros of \(H_n\)).
Two normalizations in use The book's \(H_n\) (weight \(e^{-x^2}\), leading coefficient \(2^n\)) are the "physicists' Hermite polynomials". In probability and statistics one also uses the "probabilists' Hermite polynomials" \( He_n(x) = 2^{-n/2}H_n\big(\frac{x}{\sqrt2}\big) \), with weight \(e^{-x^2/2}\) (the standard normal density), generating function \( e^{tx - t^2/2} \), and leading coefficient \(1\): \( He_2 = x^2 - 1 \), \( He_3 = x^3 - 3x \).

7. Summary: the three classical families

Legendre \(P_n\)Laguerre \(L_n\) (book)Hermite \(H_n\)
Interval\([-1, 1]\)\([0, \infty)\)\((-\infty, \infty)\)
Weight\(1\)\(e^{-x}\)\(e^{-x^2}\)
Generating function\( (1 - 2xt + t^2)^{-1/2} \)\( \frac{1}{1 - t}e^{-\frac{xt}{1-t}} \) (\(\frac{t^n}{n!}\))\( e^{2tx - t^2} \) (\(\frac{t^n}{n!}\))
Rodrigues\( \frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2 - 1)^n \)\( e^x\frac{d^n}{dx^n}(x^ne^{-x}) \)\( (-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2} \)
Recurrence\( (n + 1)P_{n+1} = (2n + 1)xP_n - nP_{n-1} \)\( L_{n+1} = (2n + 1 - x)L_n - n^2L_{n-1} \)\( H_{n+1} = 2xH_n - 2nH_{n-1} \)
Norm \(\int w\,p_n^2\)\( \frac{2}{2n + 1} \)\( (n!)^2 \)\( 2^nn!\sqrt\pi \)
Equation\( (1 - x^2)y'' - 2xy' + n(n + 1)y = 0 \)\( xy'' + (1 - x)y' + ny = 0 \)\( y'' - 2xy' + 2ny = 0 \)
Physicsangular part (spheres)radial part (hydrogen)harmonic oscillator

8. Practice problems

Try each one before opening the solution. More solved problems from this topic are in Series.

Problem 1: Compute \(H_5\) from the recurrence.

\( H_5 = 2xH_4 - 8H_3 = 32x^5 - 96x^3 + 24x - 64x^3 + 96x = 32x^5 - 160x^3 + 120x \).

Problem 2: Find \(H_{2k}(0)\).

From the explicit formula only the term with \(n - 2k = 0\) survives: \( H_{2k}(0) = \frac{(-1)^k(2k)!}{k!} \): \( 1, -2, 12, -120, \dots \) (and \(H_{2k+1}(0) = 0\)).

Problem 3: Verify that \(H_2\) satisfies Hermite's equation with \(n = 2\).

\( y = 4x^2 - 2 \), \( y' = 8x \), \( y'' = 8 \): \( 8 - 16x^2 + 4(4x^2 - 2) = 0 \). ✓

Problem 4: Expand \(x^2\) in Hermite polynomials.

\( H_2 = 4x^2 - 2 \Rightarrow x^2 = \frac14H_2 + \frac12H_0 \).

Problem 5: Compute \( \int_{-\infty}^\infty e^{-x^2}H_3(x)^2\,dx \).

\( 2^3\cdot3!\cdot\sqrt\pi = 48\sqrt\pi \).

Problem 6: Where are the zeros of \(\psi_2\), and what is its energy?

Zeros of \(H_2 = 4x^2 - 2\): \( x = \pm\frac{1}{\sqrt2} \). Energy \( \varepsilon_2 = \frac52 \) (in units of \(\hbar\omega\)).

Final thoughts

Hermite polynomials complete the classical trio of orthogonal polynomials, each tied to its interval and weight: Legendre with \(1\) on \([-1, 1]\), Laguerre with \(e^{-x}\) on \([0, \infty)\), Hermite with \(e^{-x^2}\) on the whole line. The structure is the same each time: a generating function gives the explicit formula and the recurrence; a Rodrigues formula gives orthogonality through repeated integration by parts; and a second-order differential equation has the polynomial as its only well-behaved solution. In the harmonic oscillator, that last fact is exactly why energy is quantized.

Related posts: Legendre polynomials, Laguerre polynomials, and the Gamma function (\(\int e^{-x^2}dx = \sqrt\pi = \Gamma(\frac12)\)).