1.

HERMITE DIFFERENTIAL EQUATION

French Mathematician Charles Hermite (1822-1901), an inspiring teacher is
renowned for his proof of the transcendental character of e and solution of
differential equation.
The second order homogeneous differential equation of the form
𝒅𝟐𝒚 𝒅𝒚
− 𝟐𝒙 + (𝝀 − 𝟏)𝒚 = 𝟎 … … … . (𝟏)
𝒅𝒙𝟐 𝒅𝒙
where 𝜆 is a constant is known as Hermite differential equation. When 𝜆 is an
odd integer i.e., when 𝜆 = 2𝑛 + 1; 𝑛 = 0,1,2 … ….then one of the solutions of
equation (1) becomes a polynomial. These polynomial solutions are known as
Hermite Polynomial denoted by 𝐻n(𝑥). Hermite polynomials appear in many
diverse areas, the most important being in the solutions of the simple wave
functions of hydrogen atom.

2. SOLUTION OF HERMITE DIFFERENTIAL EQUATION

Hermite differential equation does not have any singularity in the finite plane. So,
we shall use the Power Series method to solve Hermite differential equation as
given by
𝑑2𝑦 𝑑𝑦
− 2𝑥 + (𝜆 − 1)𝑦 = 0 … … … . (1)
𝑑𝑥2 𝑑𝑥
where 𝜆 is a constant given by 𝜆 = 2𝑛 + 1
Equation (1) takes the form
𝑑2𝑦 𝑑𝑦
− 2𝑥 + 2𝑛𝑦 = 0 … … … . (2)
𝑑𝑥2 𝑑𝑥
Here 𝑛 is a non-negative constant
The various steps followed to solve the above equation are discussed below:
Step-I:
Suppose the series solution of equation (1) as
∞
𝑦(𝑥) = Σ 𝑎r 𝑥k+r...............(2𝑎)
r=0
∞

→ 𝑦'(𝑥) = Σ 𝑎r (𝑘 + 𝑟)𝑥k+r–1................. (2𝑏)

r=0
∞

→ 𝑦''(𝑥) = Σ 𝑎r (𝑘 + 𝑟)(𝑘 + 𝑟 − 1)𝑥k+r–2................ (2𝑐)

r=0
Using the equations (2a) – (2c) in equation (2), we obtain;
∞ ∞ ∞
Σ 𝑎r(𝑘 + 𝑟) (𝑘 + 𝑟 − 1)𝑥k+r–2 − 2𝑥 Σ 𝑎r (𝑘 + 𝑟)𝑥k+r–1 + 2𝑛 Σ 𝑎r 𝑥k+r = 0
r=0 r=0 r=0
 ∞ ∞

→ Σ 𝑎𝑟 (𝑘 + 𝑟)(𝑘 + 𝑟 − 1)𝑥𝑘+𝑟−2 − 2 Σ 𝑎𝑟 (𝑘 + 𝑟 − 𝑛)𝑥𝑘+𝑟 = 0

𝑟=0 𝑟=0
∞

→ Σ[(𝑘 + 𝑟)(𝑘 + 𝑟 − 1)𝑥𝑘+𝑟−2 − 2(𝑘 + 𝑟 − 𝑛)𝑥𝑘+𝑟]𝑎𝑟 = 0 … … . (3)

𝑟=0
Equation (3) is an identity and that is why the coefficients of various powers of 𝑥
must be zero.
Step – II: Setting up of Recursion Relation
Equating the coefficient of the lowest power of 𝑥 i.e. 𝑥k–2 (putting 𝑟 = 0) to zero,
we get
𝑎0𝑘(𝑘 − 1) = 0
𝑖. 𝑒. , 𝒌 = 𝟎; 𝒌=𝟏
(As 𝑎0 ≠ 0, as it is the first term of the series)
Again equating the coefficient of 𝑥k–1 to zero (putting 𝑟 = 1), we get
𝑎1(𝑘 + 1)𝑘 = 0
As 𝑘 = 0 so, from the above relation, we can write, 𝑎1 = 0
Now equating the coefficient of general term 𝑥k+r to zero, we obtain,
𝑎r+2(𝑘 + 𝑟 + 2)(𝑘 + 𝑟 + 1) − 2𝑎r(𝑘 + 𝑟 − 𝑛) = 0
𝟐(𝒌 + 𝒓 − 𝒏)
→ 𝒂𝒓+𝟐 = 𝒂 … … … (𝟒)
(𝒌 + 𝒓 + 𝟐)(𝒌 + 𝒓 + 𝟏) 𝒓
This is the recursion or recurrence relation between the coefficients.

Step-III: Determination of values of coefficients

Case-A: 𝒌=𝟎
For 𝑘 = 0, the recursion relation given by equation (4) takes the form:
2(𝑟 − 𝑛)
𝑎r+2 = 𝑎 … … … (4𝑎)
(𝑟 + 2)(𝑟 + 1) r
Putting 𝑟 = 0, 1, 2, 3, ............ , we get
−2𝑛 −2𝑛
𝑎2 = 2.1 𝑎0 = 2! 𝑎0
2(1 − 𝑛) −2(𝑛 − 1)
𝑎3 = 3.2.1 𝑎1 = 𝑎1 = 0; 𝑠𝑖𝑛𝑐𝑒 𝑎1 = 0
3!
2(2 − 𝑛) −2(𝑛 − 2) −2𝑛 22 (𝑛 − 2)
𝑎4 = 𝑎2 = . 𝑎 = 4.3.2.1 𝑎0
4.3 4.3 2! 0
3(2 − 𝑛)
𝑎5 = 𝑎3 = 0
5.4
.
.
Thus the general terms of the coefficients are given by
(−2)k 𝑛(𝑛 − 2) .......... (𝑛 − 2𝑘 + 2)
𝑎2k = 𝑎0
(2𝑘)!
(−2)k 𝑛(𝑛 − 2)........... (𝑛 − 2𝑘 + 1)
𝑎2k+1 = 𝑎0
(2𝑘 + 1)!
Therefore the general solution for the case 𝑘 = 0 is given by,
∞

𝑦(𝑥) = Σ 𝑎r 𝑥k+r = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + 𝑎3𝑥3 + 𝑎4𝑥4 + ⋯

r=0
 2n 22 n(n–2) (–2)kn( n–2)…(n–2k+2)
→ 𝑦(𝑥) = 𝑎 [1 − 𝑥2 + 𝑥4 − ⋯ + 𝑥2k] +
0 2! 2(n–1)(n–3)
4! (2k)!
2(n–1) 2
𝑎 [𝑥 − 𝑥3 + 𝑥5 + ⋯ ] (for 𝑎 ≠ 0)
1 3! 5! 1

2n 2 2n(n–2) (–2) kn(n–2)…(n–2k+2)

𝑦(𝑥) = 𝑎 [1 − 𝑥2 + 𝑥4 − ⋯ + 𝑥2k]= 𝑢(𝑥) (say)….(5a)
0 2! 4! (2k)!
(for 𝑎1 = 0)
Case-B: For 𝒌 = 𝟏
In this case, the recursion relation (4) takes the form:
2(1 + 𝑟) − 2𝑛
𝑎r+2 = 𝑎 … … . (4𝑏)
(𝑟 + 3)(𝑟 + 2) r
Putting 𝑟 = 1, 3, 5 … … in the above equation, we get,
𝑎3 = 𝑎5 = 𝑎7 = ⋯ = 0 𝑠𝑖𝑛𝑐𝑒 𝑎1 = 0
Also putting, 𝑟 = 0, 2, 4 … …, we obtain
2 − 2𝑛 2(1 − 𝑛)
𝑎2 = 𝑎0 = 𝑎0
3.2 3!

6 − 2𝑛 2(3 − 𝑛) 2(1 − 𝑛) 22(𝑛 − 1)(𝑛 − 3)

𝑎4 = 𝑎2 = . 𝑎0 = 𝑎0
5.4 5.4 3! 5!
And so on.
So the general solution in this case will be,
2 k
2(n–1) 3 2 (n–1)(n–2) 5 (–1) (n–1)(n–3)…(n–2k+1) 2k+1
𝑦(𝑥) = 𝑎0 [𝑥 −
3!
𝑥 +
5!
𝑥 − ⋯+
(2k+1)!
𝑥 ]= 𝑣(𝑥)
(say)….(5b)
Here, we have seen that that the solution (5b) is a part of solution (5a). So,
𝑎1 = 0 and 𝑘 = 0; the solution of equation (1) can therefore be expressed as the
superposition of equations (5a) and (5b). So, the general solution of Hermite
differential equation is given by,
𝒚(𝒙) = 𝑨𝒖(𝒙) + 𝑩𝒗(𝒙) (𝟔)
Here A and B are two arbitrary constants.

3. HERMITE POLYNOMIAL 𝑯𝒏(𝒙)

As stated earlier, Hermite polynomials 𝐻n(𝑥) appear in diverse areas of physics,
the most important of which is the harmonic oscillator problem in quantum
mechanics.
The Hermite Polynomial 𝐻n(𝑥) of order 𝑛 can be expressed as
𝒏/𝟐
𝒓
𝒏!
(−𝟏) (𝟐𝒙) 𝒏–𝟐𝒓 , 𝒏 = 𝒆𝒗𝒆𝒏 … … (𝟏𝒂)
𝑯𝒏(𝒙) = Σ 𝒓! (𝒏 − 𝟐𝒓)!
𝒓=𝟎
𝒏
𝟐–𝟏 𝒏!
(−𝟏)𝒓 (𝟐𝒙)𝒏–𝟐𝒓, 𝒏 = 𝒆𝒗𝒆𝒏 .......... (𝟏𝒃)
𝑯𝒏 (𝒙) = Σ 𝒓! (𝒏 − 𝟐𝒓)!
𝒓=𝟎
(for proof see supplementary examples 6.3.1.)
4. GENERATING FUNCTION FOR HERMITE POLYNOMIAL
𝑯𝒏(𝒙)
The generating function for Hermite Polynomial is defined as
∞ n
2 𝑡
𝑔(𝑥, 𝑡) = 𝑒 2xt–t =Σ 𝐻n (𝑥) .......... (1)
𝑛!
n=0
Proof:
We have,
∞ ∞
2 2 (2𝑥𝑡)r (−𝑡2)s
𝑔(𝑥, 𝑡) = 𝑒2xt–t = 𝑒2x . 𝑒–t =Σ .Σ
r=0
𝑟! s=0
𝑠!
∞
(2𝑥)r r+2s
→ 𝑔(𝑥, 𝑡) = Σ .𝑡
𝑟! 𝑠!
r,s=0
Thus the coefficient of 𝑡n(for a fixed value of 𝑠) is given by,
(2𝑥)n–2s
(−1)s.
(𝑛 − 2𝑠)! 𝑠!
Putting 𝑟 + 2𝑠 = 𝑛
The total coefficients of will however be obtained by adding all values for
𝑡n
n
possible 𝑠. Since 𝑟 = 𝑛 − 2𝑠 ≥ 0 → 𝑠 ≤
2
Now, if 𝑛 is even then 𝑠 ranges from 0 to n and if 𝑛 is odd then 𝑠 ranges from 0 to
2
n −1.
2
So, the desired coefficient of 𝑡n is expressed as;
n/2
(2𝑥)n–2s 𝐻 (𝑥)
Σ(−1)s = n
(𝑛 − 2𝑠)! 𝑠! 𝑛!
s=0
Thus, we have,
∞
𝟐 𝟐–(𝒙–𝒕)𝟐 𝒕𝒏
𝒈(𝒙, 𝒕) = 𝒆𝟐𝒙𝒕–𝒕 = 𝒆𝒙 =Σ 𝑯𝒏 (𝒙) … … . (𝟐)
𝒏=𝟎
𝒏!
5. RODRIGUE’S FORMULA FOR HERMITE POLYNOMIAL 𝑯𝒏(𝒙)
The Rodrigue’s formula for Hermite Polynomial 𝐻n(𝑥) is its differential form,
which is given by,
n
2 𝑑 2
𝐻n(𝑥) = (−1) . 𝑒 . n (𝑒–x ) … … … (1)
n x
𝑑𝑥
Proof:
We know that the Hermite Polynomial 𝐻n(𝑥) is obtained from the generating
function as,
∞ n
2xt–t2 x 2–(x–t)2 𝑡
𝑔(𝑥, 𝑡) = 𝑒 =𝑒 =Σ 𝐻 (𝑥) … … . (2)
n=0
𝑛! n
For all integral values of 𝑛 and all real values of 𝑥, equation (2) can be expressed
as,
∞ n
x
2
–(x–t)
2 𝑡
𝑒 .𝑒 = Σ 𝐻n (𝑥)
𝑛!
n=0
 2 2 𝐻0(𝑥) 𝐻1(𝑥) 𝐻2(𝑥) 𝐻n(𝑥)
→ 𝑔(𝑥, 𝑡) = 𝑒x . 𝑒–(x–t) = + 𝑡+ 𝑡2 + ⋯ + 𝑡n
0! 1! 2! 𝑛!
So that
&n x2 –(x–t) 2
𝐻n(𝑥)
[𝑒 . 𝑒 ]t=0 = 𝑛! = 𝐻n(𝑥) … … … (3)
&𝑡n 𝑛!
д
Now, putting 𝑧 = 𝑡 − 𝑥 at 𝑡 = 0, 𝑧 = −𝑥, so that д ≡
дt дz
n &n 𝑑n
2 2 2 2
&𝑡n [𝑒x . 𝑒–(t–x) ]t=0 = &𝑧n (𝑒–z ) = (−1)n 𝑑𝑥n (𝑒–x )
𝒅𝒏 𝟐
∴ 𝑯𝒏 (𝒙) = (−𝟏)𝒏𝒆𝒙𝟐 (𝒆–𝒙 ) … … . . (𝟒)
𝒅𝒙𝒏
This is required differential form of Hermite Polynomials and is known as
Rodrigue’s formula.

6. VALUES OF FOR HERMITE POLYNOMIALS

The Rodrigue’s formula
n
for Hermite Polynomial 𝐻n(𝑥) is given by,
n𝑒x d (𝑒–x ) ...........(1)
2 2
𝐻 (𝑥) = (−1)
n dxn
Putting 𝑛 = 0, 1, 2, 3 … …. in relation (1) we can get the Hermite polynomials of
different orders.
0 2
𝑑0 2 2 2
∴ 𝐻0(𝑥) = (−1) 𝑒 x
0 (𝑒 –x ) = 𝑒x . 𝑒–x = 1

𝑑1𝑑𝑥
1 2 2 2 2
𝐻1(𝑥) = (−1) 𝑒x (𝑒 –x ) = −1. 𝑒x . −2𝑥. 𝑒–x = 2𝑥
𝑑𝑥21
2 2
𝑑 2 2 𝑑 2
𝐻2(𝑥) = (−1) 𝑒 x (𝑒 –x ) = 𝑒 x . [−2𝑥. 𝑒–x ]
𝑑𝑥 2 x2 𝑑𝑥 –x2 –x2 2
= 𝑒 . (−2𝑥. −2𝑥. 𝑒 − 2𝑒 ) = (4𝑥 − 2)
Similarly,
𝐻3(𝑥) = 8𝑥3 − 12𝑥
𝐻4(𝑥) = 16𝑥4 − 48𝑥2 + 12, 𝑒𝑡𝑐.

7. RECURRENCE RELATION FOR HERMITE POLYNOMIALS

The generating function can be used to develop the recurrence relations
associated with Hermite Polynomials. Here, we shall derive some of the
important recursion/recurrence relations in connection with Hermite
Polynomials.

RELATION I: 𝑯'𝒏(𝒙) = 𝟐𝒏𝑯𝒏–𝟏(𝒙)

Proof:
We know the generating function for Hermite Polynomial 𝑯𝒏(𝒙) can be
expressed as,
∞
𝟐 𝑯𝒏(𝒙) ...............
𝒈(𝒙, 𝒕) = 𝒆 𝟐𝒙𝒕–𝒕 =Σ 𝒕𝒏 (𝟏)
𝒏!
𝒏=𝟎
Differentiating bothsides with respect to 𝒙, we obtain,
∞
𝟐 𝑯'𝒏(𝒙) ...............
𝒈'(𝒙, 𝒕) = (𝟐𝒕). 𝒆𝟐𝒙𝒕–𝒕 = Σ 𝒕𝒏 (𝟐)
𝒏!
𝒏=𝟎
Now using equation (1) on the LHS of the equation (2), we get
∞ ∞
𝑯𝒏(𝒙) 𝑯'𝒏(𝒙) ..............
(𝟐𝒕). Σ 𝒕 =Σ
𝒏 𝒕𝒏 (𝟑)
𝒏! 𝒏!
𝒏=𝟎 𝒏=𝟎

∞ ∞
𝑯𝒏(𝒙) 𝑯'𝒏(𝒙)
𝟐. Σ 𝒕𝒏+𝟏 = Σ 𝒕𝒏 ...............(𝟒)
𝒏=𝟎
𝒏! 𝒏=𝟎
𝒏!
Now we shall equate the coefficients of 𝒕𝒏 on bothsides of equation (4):
𝑯𝒏–𝟏(𝒙) 𝑯'𝒏(𝒙)
𝟐. =
(𝒏 − 𝟏)! 𝒏!

∴ 𝑯'𝒏(𝒙) = 𝟐𝒏𝑯𝒏–𝟏(𝒙)

RELATION II: 𝟐𝒙𝑯𝒏(𝒙) = 𝟐𝒏𝑯𝒏–𝟏(𝒙) + 𝑯𝒏+𝟏(𝒙)

Proof:
We have,
∞
2 𝐻n(𝑥)
𝑔(𝑥, 𝑡) = 𝑒2xt– =Σ 𝑡n ............... (1)
𝑛!
n=0
Differentiating bothsides of equation (1) with respect to 𝑡, partially, we get,
∞
2 𝐻n(𝑥)
2(𝑥 − 𝑡)𝑒 2xt–t =Σ 𝑛𝑡n–1
𝑛!
n=0
∞ ∞
𝐻n(𝑥) 𝐻n(𝑥)
→ 2(𝑥 − 𝑡) Σ 𝑡n =Σ 𝑛𝑡n–1
n=0
𝑛! 𝑛!
∞ ∞ n=0 ∞
𝐻n(𝑥) 𝐻n(𝑥) 𝐻n(𝑥)
→ 2𝑥 Σ 𝑡n − 2 Σ 𝑡n+1 = Σ 𝑛𝑡n–1 … . . (2)
n=0
𝑛! n=0
𝑛! n=0
𝑛!
Now equating the coefficients of 𝑡n on bothsides of equation (2), we get
Hn(x) Hn—1(x) Hn+1(x)
2𝑥. − 2. = (𝑛 + 1)
n! (n–1)! (n+1)!
Hn(x) nHn—1(x) Hn+1(x)
→ 2𝑥. − 2. = (𝑛 + 1)
n! n! (n+1)n!
→ 2𝑥𝐻n(𝑥) − 2𝑛𝐻n–1(𝑥) = 𝐻n+1(𝑥)
∴ 𝟐𝒙𝑯𝒏(𝒙) = 𝟐𝒏𝑯𝒏–𝟏(𝒙) + 𝑯𝒏+𝟏(𝒙)

RELATION III: 𝑯'𝒏(𝒙) = 𝟐𝒙𝑯𝒏(𝒙) − 𝑯𝒏+𝟏(𝒙)

Proof:
From Recurrence Relations I and II, we have
𝐻'n(𝑥) = 2𝑛𝐻n–1(𝑥) … … . (1)

2𝑥𝐻n(𝑥) = 2𝑛𝐻n–1(𝑥) + 𝐻n+1(𝑥) … … … (2)

Substituting the value of 2𝑛𝐻n–1(𝑥) in equation (2) from equation (1), we obtain,
 2𝑥𝐻n(𝑥) = 𝐻'n(𝑥) + 𝐻n+1(𝑥)
∴ 𝑯'𝒏(𝒙) = 𝟐𝒙𝑯𝒏(𝒙) − 𝑯𝒏+𝟏(𝒙)

RELATION IV: 𝑯''𝒏(𝒙) − 𝟐𝒙𝑯'𝒏(𝒙) + 𝟐𝒏𝑯𝒏(𝒙) = 𝟎

Proof: We know, 𝐻'n(𝑥) = 2𝑥𝐻n(𝑥) − 𝐻n+1(𝑥) … … . (1)
Differentiating bothsides with respect to 𝑥, we get;
𝐻''n(𝑥) = 2𝐻n(𝑥) + 2𝑥𝐻'n(𝑥) − 𝐻'n+1(𝑥) … … (2)
Again we know,
𝐻'n(𝑥) = 2𝑛𝐻n–1(𝑥) … … . (3)
Replacing 𝑛 by (𝑛 + 1), we get
𝐻'n+1(𝑥) = 2(𝑛 + 1)𝐻n(𝑥) … … . (4)
Using equation (4) in equation (2), we get,
𝐻''n(𝑥) = 2𝐻n(𝑥) + 2𝑥𝐻'n(𝑥) − 2(𝑛 + 1)𝐻n(𝑥)
∴ 𝑯''𝒏(𝒙) − 𝟐𝒙𝑯'𝒏(𝒙) + 𝟐𝒏𝑯𝒏(𝒙) = 𝟎
This is required relation, which also indicates that, 𝑦 = 𝐻n(𝑥), i.e., the Hermite
polynomial is a solution of Hermite’s differential equation.

8. ORTHOGONALITY OF HERMITE POLYNOMIALS

A family of functions 𝑓0(𝑥), 𝑓1(𝑥), 𝑓2(𝑥) … … . , 𝑓n(𝑥) is said to be
orthogonal with respect to a weight 𝑤(𝑥) over an interval [a, b] if the following is
true:
b
ƒ 𝑓m(𝑥) 𝑓n(𝑥)𝑤(𝑥)𝑑𝑥 = 0 𝑓𝑜𝑟 𝑚 ≠ 𝑛
a
≠ 0 𝑓𝑜𝑟 𝑚 = 𝑛
Hermite polynomials form an orthogonal set of functions for the weight
2
𝑤(𝑥) = 𝑒–x over the interval (−∞, ∞). The exact relation runs as:
∞
2
ƒ 𝐻m(𝑥) 𝐻n(𝑥)𝑒–x 𝑑𝑥 = 0 𝑓𝑜𝑟 𝑚 ≠ 𝑛
–∞
= 2n𝑛! √𝜋 𝑓𝑜𝑟 𝑚 = 𝑛
Proof:

We know that 𝐻m(𝑥) is a solution of the Hermite differential equation

given by,

𝒅𝟐𝒚 𝒅𝒚
𝒅𝒙 𝟐 − 𝟐𝒙 𝒅𝒙
+ 𝟐𝒎𝒚 = 𝟎 … … … (𝟏)
So, we can write,
𝐻''m(𝑥) − 2𝑥𝐻'm(𝑥) + 2𝑚𝐻m(𝑥) = 0 … … . (2𝑎)
Similarly,
𝐻''n(𝑥) − 2𝑥𝐻'n(𝑥) + 2𝑛𝐻n(𝑥) = 0 … … . (2𝑏)
Multiplying equation (2a) by 𝐻n(𝑥) and equation (2b) by 𝐻m(𝑥) and subtracting,
we get,
[𝐻''m(𝑥)𝐻n(𝑥) − 𝐻''n(𝑥)𝐻m(𝑥)] − 2𝑥[𝐻'm(𝑥)𝐻n(𝑥) − 𝐻'n(𝑥)𝐻m(𝑥)]
+ 2(𝑚 − 𝑛)𝐻m(𝑥)𝐻n(𝑥) = 0 .......... (3)
 𝑑
→ [𝐻'm(𝑥)𝐻n(𝑥) − 𝐻'n(𝑥)𝐻m(𝑥)] − 2𝑥[𝐻'm(𝑥)𝐻n(𝑥) − 𝐻'n(𝑥)𝐻m(𝑥)]
𝑑𝑥
= 2(𝑛 − 𝑚)𝐻m(𝑥)𝐻n(𝑥)......... (4)
The above equation is linear differential equation,
2
So, Integrating factor (I.F.) = 𝑒ƒ –2xdx = 𝑒–x
Multiplying equation (4) by I.F. we get,
𝑑 2 2
[𝐻'm (𝑥)𝐻n (𝑥) − 𝐻'n (𝑥)𝐻m (𝑥)]𝑒–x = 2(𝑛 − 𝑚)𝑒–x 𝐻m (𝑥)𝐻n (𝑥) … . (5)
𝑑𝑥
Now, integrating bothsides with respect to 𝑥, from 𝑥 = −∞ to 𝑥 = ∞, we obtain:

∞ ∞
2 2
[𝐻'm(𝑥)𝐻n(𝑥) − 𝐻'n(𝑥)𝐻m(𝑥)]𝑒–x –∞
= 2(𝑛 − 𝑚) ƒ 𝑒–x 𝐻m(𝑥)𝐻n(𝑥)
–∞
∞ 2
→ 0 = 2(𝑛 − 𝑚) ƒ 𝑒–x 𝐻 (𝑥)𝐻 (𝑥)
m n
–∞
∞ 2
∴ ƒ 𝑒–x 𝐻 (𝑥)𝐻 (𝑥) = 0...........(6𝑎)
m n
–∞

We also know from the generating function of Hermite polynomial 𝐻n(𝑥) that,
∞
2 𝐻n(𝑥)
𝑔(𝑥, 𝑡) = 𝑒2xt–t =Σ 𝑡n … … (7𝑎)
𝑛!
n=0
Similarly,
∞
2 𝐻m(𝑥)
𝑔(𝑥, 𝑠) = 𝑒2xs–s =Σ 𝑠m … … (7𝑏)
𝑚!
m=0
Multiplying (7a) and (7b) we get,
∞ ∞
2xt–t2+2xs–s2
𝐻m(𝑥) m 𝐻n(𝑥) n
𝑒 = Σ 𝑠 Σ 𝑡
𝑚! 𝑛!
m=0 n=0

∞ ∞
2+2xs–s2 𝐻m(𝑥) 𝐻n(𝑥) m n................
𝑒2xt–t = ΣΣ 𝑠 .𝑡 (8)
𝑚! 𝑛!
m=0 n=0
2
Multiplying bothsides of equation (8) by weight 𝑤(𝑥) = 𝑒–x and then
integrating from 𝑥 = −∞ to 𝑥 = ∞, we obtain:

∞ ∞
m n ∞
∞
𝑑𝑥 = Σ Σ 𝑠 . 𝑡 . ƒ 𝐻 ( 𝑥). 𝐻n(𝑥 ). 𝑒–x 𝑑𝑥 … . . (9)
2–2st] 2
ƒ 𝑒–[(x+s+t)
𝑚!. 𝑛! m
–∞ m=0 n=0 –∞
LHS:
∞ ∞ ∞
2–2st] 2 2
ƒ 𝑒–[(x+s+t) 𝑑𝑥 = 𝑒2st ƒ 𝑒–[(x+s+t) ] 𝑑𝑥 = 𝑒2st ƒ 𝑒–u 𝑑𝑢 = 𝑒2st. √𝜋
–∞ –∞ –∞
∞
2m. 𝑠m. 𝑡m
= √𝜋. Σ
𝑚!
m=0
Thus from equation (9) using the above result, we get
 ∞ 2m. 𝑠m. 𝑡m ∞ ∞ 𝑠m. 𝑡n ∞
2
√𝜋. Σ =ΣΣ . ƒ 𝐻m (𝑥). 𝐻n (𝑥). 𝑒–x 𝑑𝑥 … … (10)
m=0
𝑚! m=0 n=0
𝑚!. 𝑛! –∞
Equating the coefficients of 𝑡n, (if 𝑚 = 𝑛) in bothsides of equation (10) we get:
∞
1 2
√𝜋. 2 = 𝑛!
n . ƒ 𝑒–x 𝐻2n (𝑥)𝑑𝑥
–∞
∞
2
∴ ƒ 𝑒–x 𝐻2 n(𝑥)𝑑𝑥 = √𝜋. 2n. 𝑛! … … . (11𝑎)
–∞
Combining the results of (6a) and (11a) we can easily write:
∞
𝟐
ƒ 𝑯𝒎(𝒙) 𝑯𝒏(𝒙)𝒆–𝒙 𝒅𝒙 = 𝟎 𝒇𝒐𝒓 𝒎 ≠ 𝒏
–∞
= 𝟐𝒏𝒏! √𝝅 𝒇𝒐𝒓 𝒎 = 𝒏……(12)

6.3.9 INTEGRAL REPRESENTATION OF HERMITE

POLYNOMIAL
The integral form of Hermite polynomial is given by,
𝟐𝒏(−𝒊)𝒏 𝟐 ∞ 𝟐
𝑯𝒏(𝒙) = 𝒆𝒙 ƒ 𝒕𝒏𝒆–𝒕 +𝟐𝒊𝒙𝒕𝒅𝒕 … . . (𝟏)
√𝝅 –∞
(the proof of the equation (1) is given in supplementary exercise)

6.3.10 APPLICATIONS OF HERMITE POLYNOMIALS IN

PHYSICS

A. The Linear Harmonic Oscillator Problem in Quantum mechanics

The one dimensional quantum mechanical harmonic oscillator is a state of
energy E and is governed by the equation (known as time independent
Schrodinger’s wave equation)
2
– ħ d2
[ 2m dx2 + 𝑉(𝑥)] T (𝑥) = 𝐸 T (𝑥) .......... (1)

Here, 𝑉(𝑥) is the potential energy function for the harmonic oscillator, which is
given by,
1 1
𝑉(𝑥) = 2 𝑘𝑥2 = 2 𝑚𝑤 2𝑥 2……(2)
Inserting equation (2) in equation and then rearranging we obtain,
𝑑2 T 2𝑚
+ 2 [𝐸 − 𝑉(𝑥)] T = 0
𝑑𝑥2 ħ
𝑑2 T 2𝑚 1
+ 2 [𝐸 − 𝑚𝑤2𝑥2] T = 0 … … . (3)
𝑑𝑥2 ħ 2