MATH3383. Quantum Mechanics. Appendix D:Hermite Equation; Orthogonal Polynomials

1. Hermite Equation

In the study of the eigenvalue problem of the Hamiltonian for the quantum harmonicoscillator we have encountered the Hermite equation which reads

d2H

dξ2− 2ξ

dH

dξ+

(2E

~ω− 1

)H = 0 , (1)

which is a well-known homogeneous second-order linear differential equation. Note thatthis equation has nonconstant coefficients – otherwise we could readily solve the equationwith elementary methods– and it is amenable to methods encountered e.g. in the moduleMATH2370, using series expansions. We will show here that the physically importantsolutions are the polynomial ones leading to the well-known Hermite polynomials. In thenext section we will study the latter polynomials in their own right.

The emergence of polynomial solutions from the equation (1) is seen as follows. Let ustry first a power series expansion for the solution, given by

H(ξ) =∞∑

m=0

amξm ,

where the coefficients am need to be determined. Inserting this expansion in the equation(1) we obtain:

∞∑m=2

m(m− 1)amξm−2 −∞∑

m=1

2mamξm + β

∞∑m=0

amξm = 0 ,

where we have abbreviated β = 2E~ω − 1 . Changing the summation variable (by a shift

over 2 units) in the first term we obtain:

∞∑m=0

(m + 2)(m + 1)am+2ξm −

∞∑m=0

2mamξm + β

∞∑m=0

amξm = 0 ,

so that all powers of ξ within the sums are of the same form, and we can immediatelyread off a recursion relation for the coefficients (since all powers of ξ are independent):

(m + 2)(m + 1)am+2 = (2m− β)am , m = 0, 1, 2, . . . .

Since the recursion relation increases in steps of 2 units, the implementation of it will leadto two independent solutions (as one expects!). In fact, the two independent solutions ofthe Hermite equation can be obtained by the following choices of initial conditions on the

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recursion relation: given β we will have one solution starting with a0 = 0, a1 6= 0, forwhich the solution series will only have odd powers of ξ, and another solution starting witha1 = 0, a0 6= 0 for which the series will have only even powers of ξ. Clearly, the resultingtwo solutions are independent. Thus, upon iteration we obtain from the recursion relationthe following explicit expressions for the even coefficients:

a2k =[(4(k − 1)− β][4(k − 2)− β] . . . [−β]

(2k)!a0

while for the odd coefficients we have

a2k+1 =[(2(2k − 1)− β][2(2k − 3)− β] . . . [2− β]

(2k + 1)!a1 .

From these expansions one thing is immediately clear: if β = 2n for some positive integern, then either one of the sequences of coefficients will terminate:

if β = 2n = 4ν − 2 ⇒ a2ν+1 = 0 odd sequence terminatesif β = 2n = 4ν − 4 ⇒ a2ν = 0 even sequence terminates

and when these sequences terminate, the corresponding series for H(ξ) will truncate (re-sulting in only a finite number of terms in the sum) and, hence, we have a polynomialsolution.

If β 6= 2n for any n ∈ Z, then the series for H(ξ) does not truncate and we will havean infinite series. We will now show that these solutions have to be rejected on the groundthat an infinite series will lead to solutions that do not obey the integrability assumptionassociated with the normlisation of the quantum eigenfunctions. In fact, for large enoughm the recursion relation gives

am

am−2∼ 2

m, as m→∞ ,

implying that for large m

am ∼1

(m/2)!a0 (m even) and am ∼

1((m− 1)/2)!

a1 (m odd) as m→∞ .

However, the series for N large enough can then be approximated as

H(ξ) ∼ polynomial +∞∑

m=N

1(m/2)!

ξ2(m/2) ∼ eξ2.

Thus, if the series for H(ξ) does not truncate to a finite polynomial the correspondinginfinite series will roughly behave like H(ξ) ∼ eξ2

, and inserting this into the formulafor the eigenfunction φ(x) of Handout # 5 we will obtain a solution that is not only notsquare integrable, but actually unbounded as x becomes large! Clearly such solutions haveto be rejected. We conclude that the only quantum mechanically acceptable solutions ofthe Hermite equations are the polynomial ones occurrring for the values β = 2n.

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2. Properties of the Hermite Polynomials

We will now study some of the properties of the Hermite polynomials. Most of theseproperties are characteristic not only of the Hermite polynomials but of other classes ofso-called orthogonal polynomials as well.

The simplest way to study the Hermite polynomials is through its generating function.This is a function of ξ as well as of an additional “dummy” variable t (not to be confusedwith the time variable) of the form:

S(ξ, t) = e2tξ−t2 =∞∑

n=0

Hn(ξ)n!

tn . (2)

This function encodes all the Hermite polynomials in one formula. In fact, by expandingthe l.h.s. in powers of t and comparing the coefficients with the terms in the sum on ther.h.s. we get all Hermite polynomials in explicit form.

Let us now show how to derive the various properties from the generating function.

i) Orthogonality: Evaluating the integral∫ ∞

−∞S(ξ, t)S(ξ, t′)e−ξ2

dξ =∫ ∞

−∞e2tξ−t2e2t′ξ−t′2e−ξ2

dξ

=∫ ∞

−∞e−[ξ−(t+t′)]2e(t+t′)2−t2−t′2 dξ = e2tt′√π =

∞∑n=0

2n√π

n!(tt′)n

=∞∑

n=0

∞∑m=0

tnt′m

n!m!

∫ ∞

−∞Hn(ξ)Hm(ξ)e−ξ2

dξ

where as we see the insertion of both sides of (2) leads to a power series in tt′ only.Comparing both sides we are led to the orthogonality relation:∫ ∞

−∞Hn(ξ)Hm(ξ)e−ξ2

dξ = 2n n!√

π δn,m , (3)

expressing the fact that the collection of all Hermite polynomials forms a family oforthogonal polynomials, and the function w(x) = e−ξ2

, that appears in the integrandof (3) is called the weight function of the family. The weight function together withthe integration region (in our case the infinite interval: (−∞,∞) ) fully determinethe family of orthogonal polynomials up to a common constant factor.

ii) Recursion relation: Taking derivatives with respect to t on both sides of (2) we get:

∂S

∂t= 2(ξ − t)S =

∞∑n=0

2(ξ − t)Hn

n!tn

=∞∑

n=0

2ξHn

n!tn −

∞∑n=1

2Hn−1

(n− 1)!tn

=∞∑

n=1

Hn

(n− 1)!tn−1 =

∞∑n=0

Hn+1

n!tn

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which, upon comparison of both sides power by power in t, yields the relation:

Hn+1 = 2ξHn − 2nHn−1 , n = 1, 2, . . . (4)

together with H1 = 2ξH0 (corresponding to the term n = 0 in the sum). Eq.(4) is a three-point recursion relation (i.e. involving Hn+1, Hn and Hn−1, which istypical for orthogonal polynomials (all families of orthogonal polynomials obey sucha recursion relation, albeit with different coefficients).

iii) Differential relation: Taking derivatives with respect to ξ of the generating function,we get:

∂S

∂ξ= 2tS =

∞∑n=0

2Hn

n!tn+1 =

∞∑n=1

2Hn−1

(n− 1)!tn

=∞∑

n=0

H ′n

n!tn

leading, by comparison of the coefficients of the same power of t, to the relation:

H ′n = 2nHn−1 , n = 1, 2, . . . (5)

together with H ′0 = 0 (which implies H0(ξ) = constant) for the lowest-order

term. Combining the recursion relation (4) with the differential relation (5) we canactually write down a second-order differential equation in terms of one single Hn

(so eliminating shifts in n). In fact, combining (4) and (5) we get

Hn+1 = 2ξHn −H ′n ⇒ (2ξHn −H ′

n)′ = H ′n+1 = 2(n + 1)Hn ,

which after some easy manipulation yields the Hermite equation:

H ′′n − 2ξH ′

n + 2nHn = 0 . (6)

iv) Rodrigues’ formula: The nth order coeffient in the series on the r.h.s. of (2) is obtainedby n differentiations as follows:

Hn(ξ) =(

∂

∂t

)n

S(ξ, t)∣∣∣∣t=0

=(

∂

∂t

)n

e2ξt−t2∣∣∣∣t=0

=(

∂

∂t

)n

eξ2−(t−ξ)2∣∣∣∣t=0

= eξ2

(∂

∂t

)n

e−(t−ξ)2∣∣∣∣t=0

= eξ2

(− ∂

∂ξ

)n

e−(t−ξ)2∣∣∣∣t=0

= eξ2

(− d

dξ

)n

e−ξ2

This is Rodrigues’ formula, which can also be cast in the form of the recursionrelation:

Hn+1(ξ) = eξ2

(− d

dξ

)e−ξ2

Hn(ξ) . (7)

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v) Mehler’s formula: From the Gaussian integral formula one has immediately the follow-ing result:

e−ξ2=

1√π

∫ ∞

−∞e−x2+2ixξ dx

and inserting this into Rodrigues’ formula we get:

Hn(ξ) = eξ2

(− d

dξ

)n 1√π

∫ ∞

−∞e−x2+2ixξ dx =

1√π

eξ2

∫ ∞

−∞(−2ix)ne−x2+2ixξ dx

=1√π

∫ ∞

−∞(−2ix)ne−(x−iξ)2 dx

which provides us with an integral representation for the Hermite polynomials. Toobtain Mehler’s formula we use this representation as follows:

∞∑n=0

tn

2nn!Hn(ξ)Hn(η) =

∞∑n=0

tn

2nn!1π

∫ ∞

−∞dx

∫ ∞

−∞dy (−2ix)n(−2iy)ne−(x−iξ)2−(y−iη)2

=1π

∫ ∞

−∞dx

∫ ∞

−∞dy e−(x−iξ)2−(y−iη)2

∞∑n=0

(−2txy)n

n!

=1π

∫ ∞

−∞dx

∫ ∞

−∞dy exp

[−(x− iξ)2 − (y − iη)2 − 2txy

]=

1π

∫ ∞

−∞dx e−(x−iξ)2−2itxη+t2x2

∫ ∞

−∞dy e−(y−iη+tx)2

=1√π

∫ ∞

−∞dx e−(1−t2)x2+ξ2+2ix(ξ−tη) =

1√1− t2

exp[ξ2 − (ξ − tη)2

1− t2

]from which Mehler’s formula follows immediately:

∞∑n=0

tn

2nn!Hn(ξ)Hn(η) =

1√1− t2

exp(

2tξη − t2(ξ2 + η2)1− t2

). (8)

The completeness relation for the quantum oscillator eigenfunctions of Handout #5 follows by setting 1− t2 = ε and taking the limit ε→ 0 , using the representationof the Dirac δ-function:

δ(x) = limε→0

1√πε

e−x2/ε ,

(see Appendix A).

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