Physical Chemistry Third Edition

F Some Mathematics Used in Quantum Mechanics 1279

The second derivative of our trial solution is

d^2 ψ

dx^2

e−ax

(^2) / 2

(

d^2 S

dx^2

− 2 ax

dS

dx

+(a^2 x^2 −a)S

)

(F-31)

Substitution of this expression into Eq. (F-26) gives (after cancellation of two terms)

e−ax

(^2) / 2

[

d^2 S

dx^2

− 2 ax

dS

dx

+(b−a)S

]

 0 (F-32)

The exponential factor does not vanish for any finite real value ofx, so the quantity in

square brackets must vanish. From Eq. (F-30), the first two derivatives ofSare

dS

dx



∑∞

n 1

ncnxn−^1 

∑∞

n 0

ncnxn−^1 (F-33)

d^2 S

dx^2



∑∞

n 2

n(n−1)cnxn−^2 

∑∞

j 0

(j+2)(j+1)cj+ 2 xj

∑∞

n 0

(n+2)(n+1)cn+ 2 xn

(F-34)

where we letjn−2. The indexnor the indexjcan be called a “dummy index.”

The symbol used for it is unimportant, since it just stands for successive integral values.

We can therefore replacejbynwithout changing the sum, even thoughnnow has

a different meaning than in the original sum. Also, we can add ann0 term to the

expression fordS/dxwithout any change, since then0 term has a factor of zero.

We substitute Eq. (F-33) and (F-34) into Eq. (F-32) and cancel a common factor of

e−ax

(^2) / 2

. The result is
∑∞

n 0

[(n+2)(n+1)cn+ 2 − 2 ancn+(b−a)cn]xn 0 (F-35)

The quantity in the square brackets must vanish for each value ofn, since every power

ofxon the right-hand side of the equation has a zero coefficient, and every power of

xmust have the same coefficient on both sides of the equation. Therefore, we have the

recursion relation of Eq. (15.4-6):

cn+ 2 

2 an+a−b

(n+2)(n+1)

cn (n0, 1, 2,...) (F-36)

This recursion relation leads to the termination of the series as discussed in Section 15.4

and to the energy eigenfunctions shown in Eqs. (15.4-10)–(15.4-13). The first three

energy eigenfunctions correspond to the first three Hermite polynomials:

H 0 (y) 1 (F-37)

H 1 (y) 2 y (F-38)

H 2 (y) 4 y^2 − 2 (F-39)

where in order to correspond with our wave functions,y

√

ax. All of the Hermite

polynomials can be generated by the formula:

Hn(y)(−1)ney

2 dn

dyn

(

e−y

2 )

(F-40)