The polynomial solutions of the Hermite differential equation, with n a non-negative integer, are
usually normed so that the highest degree term is (2z)n and called the Hermite polynomials H
n(z).
The Hermite polynomials may be defined explicitly by
Hn(z) := (−1)nez2
e−z2
, | | (1) |
since this is a polynomial having the highest degree term (2z)n and satisfying the Hermite
equation. The first six Hermite polynomials are
H0(z) ≡ 1,
H1(z) ≡ 2z,
H2(z) ≡ 4z2 − 2,
H3(z) ≡ 8z3 − 12z,
H4(z) ≡ 16z4 − 48z2 + 12,
H5(z) ≡ 32z5 − 160z3 + 120z,
and the general polynomial form is
Hn(z) ≡ (2z)n −
(2z)n−2 +
(2z)n−4 − +
.
Differentiating this termwise gives H′n(z) = 2n
,
i.e.
We shall now show that the Hermite polynomials form an orthogonal set on the interval
(−∞, ∞) with the weight factor e−x2. Let m < n; using (1) and integrating by parts we
get
The substitution portion here equals to zero because e−x2 and its derivatives vanish at ±∞. Using
then (2) we obtain
Repeating the integration by parts gives the result
whereas in the case m = n the result
(see the area under Gaussian curve). The results mean that the functions x
e−
form
an orthonormal set on (−∞, ∞).
The Hermite polynomials are used in the quantum mechanical treatment of a harmonic oscillator,
the wave functions of which have the form