Hermite functions

SPECIAL FUNCTIONS
where, in the second equality, we have expanded the RHS using the binomial theorem. On
equating coefficients of hn , we immediately obtain
Lmn (0) =
(n + m)!
.
n!m!
Recurrence relations
The various recurrence relations satisfied by the associated Laguerre polynomials
may be derived by differentiating the generating function (18.123) with respect to
either or both of x and h, or by differentiating with respect to x the recurrence
relations obeyed by the ordinary Laguerre polynomials, discussed in section 18.7.1.
Of the many recurrence relations satisfied by the associated Laguerre polynomials,
two of the most useful are as follows:
m
m
(n + 1)Lm
n+1 (x) = (2n + m + 1 − x)Ln (x) − (n + m)Ln−1 (x),
x(Lm
n ) (x)
=
nLm
n (x)
− (n +
m)Lm
n−1 (x).
(18.124)
(18.125)
For proofs of these relations the reader is referred to exercise 18.7.
18.9 Hermite functions
Hermite’s equation has the form
y − 2xy + 2νy = 0,
(18.126)
and has an essential singularity at x = ∞. The parameter ν is a given real
number, although it nearly always takes an integer value in physical applications.
The Hermite equation appears in the description of the wavefunction of the
harmonic oscillator. Any solution of (18.126) is called a Hermite function.
Since x = 0 is an ordinary point of the equation, we may find two linearly
independent solutions in the form of a power series (see section 16.2):
y(x) =
∞
am xm .
(18.127)
m=0
Substituting this series into (18.107) yields
∞
[(m + 1)(m + 2)am+2 + 2(ν − m)am ] xm = 0.
m=0
Demanding that the coefficient of each power of x vanishes, we obtain the
recurrence relation
2(ν − m)
am .
am+2 = −
(m + 1)(m + 2)
As mentioned above, in nearly all physical applications, the parameter ν takes
integer values. Therefore, if ν = n, where n is a non-negative integer, we see that
624
18.9 HERMITE FUNCTIONS
10
H2
5
H0
−1.5
−1
−0.5
0.5
1
1.5
x
H1
−5
H3
−10
Figure 18.8 The first four Hermite polynomials.
an+2 = an+4 = · · · = 0, and so one solution of Hermite’s equation is a polynomial
of order n. For even n, it is conventional to choose a0 = (−1)n/2 n!/(n/2)!, whereas
for odd n one takes a1 = (−1)(n−1)/2 2n!/[ 12 (n − 1)]!. These choices allow a general
solution to be written as
Hn (x) = (2x)n − n(n − 1)(2x)n−1 +
[n/2]
=
m=0
(−1)m
n(n − 1)(n − 2)(n − 3)
(2x)n−4 − · · ·(18.128)
2!
n!
(2x)n−2m ,
m!(n − 2m)!
(18.129)
where Hn (x) is called the nth Hermite polynomial and the notation [n/2] denotes
the integer part of n/2. We note in particular that Hn (−x) = (−1)n Hn (x). The
first few Hermite polynomials are given by
H0 (x) = 1,
H3 (x) = 8x2 − 12x,
H1 (x) = 2x,
H4 (x) = 16x4 − 48x2 + 12,
H2 (x) = 4x2 − 2,
H5 (x) = 32x5 − 160x3 + 120x.
The functions H0 (x), H1 (x), H2 (x) and H3 (x) are plotted in figure 18.8.
625
SPECIAL FUNCTIONS
18.9.1 Properties of Hermite polynomials
The Hermite polynomials and functions derived from them are important in the
analysis of the quantum mechanical behaviour of some physical systems. We
therefore briefly outline their useful properties in this section.
Rodrigues’ formula
The Rodrigues’ formula for the Hermite polynomials is given by
dn −x2
(e ).
dxn
This can be proved using Leibnitz’ theorem.
2
Hn (x) = (−1)n ex
(18.130)
Prove the Rodrigues’ formula (18.130) for the Hermite polynomials.
Letting u = e−x and differentiating with respect to x, we quickly find that
2
u + 2xu = 0.
Differentiating this equation n + 1 times using Leibnitz’ theorem then gives
u(n+2) + 2xu(n+1) + 2(n + 1)u(n) = 0,
which, on introducing the new variable v = (−1)n u(n) , reduces to
v + 2xv + 2(n + 1)v = 0.
(18.131)
x2
Now letting y = e v, we may write the derivatives of v as
v = e−x (y − 2xy),
2
−x2
v =e
(y − 4xy + 4x2 y − 2y).
Substituting these expressions into (18.131), and dividing through by e−x , finally yields
Hermite’s equation,
y − 2xy + 2ny = 0,
2
thus demonstrating that y = (−1)n ex dn (e−x )/dxn is indeed a solution. Moreover, since
this solution is clearly a polynomial of order n, it must be some multiple of Hn (x). The
normalisation is easily checked by noting that, from (18.130), the highest-order term is
(2x)n , which agrees with the expression (18.128). 2
2
Mutual orthogonality
We saw in section 17.4 that Hermite’s equation could be cast in Sturm–Liouville
2
2
form with p = e−x , q = 0, λ = 2n and ρ = e−x , and its natural interval is thus
[−∞, ∞]. Since the Hermite polynomials Hn (x) are solutions of the equation and
are regular at the end-points, they must be mutually orthogonal over this interval
2
with respect to the weight function ρ = e−x , i.e.
∞
2
Hn (x)Hk (x)e−x dx = 0
if n = k.
−∞
This result may also be proved directly using the Rodrigues’ formula (18.130).
Indeed, the normalisation, when k = n, is most easily found in this way.
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18.9 HERMITE FUNCTIONS
Show that
I≡
∞
−∞
√
2
Hn (x)Hn (x)e−x dx = 2n n! π.
(18.132)
Using the Rodrigues’ formula (18.130), we may write
∞
∞ n
dn
d Hn −x2
2
I = (−1)n
Hn (x) n (e−x ) dx =
e dx,
n
dx
0
−∞ dx
where, in the second equality, we have integrated by parts n times and used the fact that
the boundary terms all vanish. From (18.128) we see that dn Hn /dxn = 2n n!. Thus we have
∞
√
2
I = 2n n!
e−x dx = 2n n! π,
−∞
where, in the second equality, we use the standard result for the area under a Gaussian
(see section 6.4.2). The above orthogonality and normalisation conditions allow any (reasonable)
function in the interval −∞ ≤ x < ∞ to be expanded in a series of the form
f(x) =
∞
an Hn (x),
n=0
in which the coefficients an are given by
∞
1
2
f(x)Hn (x)e−x dx.
an = n √
2 n! π −∞
We note that it is sometimes convenient to define the orthogonal Hermite functions
2
φn (x) = e−x /2 Hn (x); they also may be used to produce a series expansion of a
function in the interval −∞ ≤ x < ∞. Indeed, φn (x) is proportional to the
wavefunction of a particle in the nth energy level of a quantum harmonic
oscillator.
Generating function
The generating function equation for the Hermite polynomials reads
2
G(x, h) = e2hx−h =
∞
Hn (x)
n=0
n!
hn ,
a result that may be proved using the Rodrigues’ formula (18.130).
Show that the functions Hn (x) in (18.133) are the Hermite polynomials.
It is often more convenient to write the generating function (18.133) as
G(x, h) = ex e−(x−h) =
2
2
627
∞
Hn (x) n
h.
n!
n=0
(18.133)