16.2 Hermite Polynomials
201
and hence the Jacobi matrix becomes
J H =
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
0
1
2
√
2
0
0 0 · · ·
1
2
√
2
0
1
0 0 · · ·
0
1
0
1
2
√
6 0 · · ·
0
0
1
2
√
6
0
√
2 · · ·
0
0
0
√
2 0 · · ·
. . .
. . .
. . .
. . .
. . .
. . .
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
.
(16.7)
The zeros of the Hermite polynomial H n (x) are given by eigenvalues of Eq. (16.7).
The derivative of the Hermite polynomials holds
d
dx
H n (x) = 2nH n−1 (x)
(16.8)
which leads together with the above recurrence formula to
H
n (x) = 2xH n (x) − H n+1 (x)
(16.9a)
and
H
n (x) − 2xH
n (x) + 2nH n (x) = 0.
(16.9b)
Thus the Hermite differential equation is given by
y
− 2xy
+ 2ny = 0.
(16.9c)
H 2n (x) are even and H 2n+1 (x) are odd polynomials in x. Therefore we obtain
for the parity of the Hermite polynomial
H n (−x) = (−1)
n H n (x),
(16.10)
and they are connected with the confluent hypergeometric function 1 F 1 by
H 2n (x) = (−1)
n (2n)!
n!
1 F 1
−n,
1
2
; x
2
and
(16.11a)
H 2n+1 (x) = (−1)
n 2
(2n + 1)!
n!
x 1 F 1
−n,
3
2
; x
2
.
(16.11b)
Because the Laguerre polynomial holds
L
α
n (x) =
n + α
n
1 F 1 (−n, α + 1; x)