Physical Chemistry Third Edition

19.3 The Perturbation Method and Its Application to the Ground State of the Helium Atom 801

ψ 1 (0)

√

2

2 a

sin

(

π(x+a)

2 a

)

; E(0) 1 

h^2

8 m(2a^2 )



h^2

32 ma^2

E(1) 1 

1

a

∫a

−a

kx^2

2

sin^2

(

π(x+a)

2 a

)

dx

From a trigonometric identity,

sin

(πx

2 a

+

π

2

)

sin(πx/ 2 a)cos(π/2)+cos(πx/ 2 a)sin(π/2) 0 +cos(πx/ 2 a)

E(1) 1 

k

2 a

∫a

−a

x^2 cos^2 (πx/ 2 a)dx

k

2 a

(

2 a

π

) 3 π/∫^2

−π/ 2

y^2 cos^2 (y)dy, wherey

πx

2 a



8 k

π^3

a^2

π/∫ 2

0

y^2 cos^2 (y)dy

where we have used the fact that the integral from−π/2toπ/2 is twice as large as the integral

from 0 toπ/2, since the integrand is an even function ofy, having the same value for−yas

fory. This integral can be looked up to give

E(1) 1 

8 k

π^3

a^2

[

y^3

6

+

(

y^2

4

−

1

8

)

sin(2y)+

ycos(2y)

4

]π/ 2

0



8 k

π^3

a^2

(

π^3

48

+ 0 −

π

8

)



ka^2

6

−

ka^2

π^2

EE(0) 1 +E(1) 1 

h^2

32 ma^2

+ka^2

(

1

6

−

1

π^2

)



h^2

32 ma^2

+(0.06535)ka^2

Exercise 19.3

EvaluateE(0) 1 andE(1) 1 from the previous example in the case thatmis the mass of an electron,

9. 1094 × 10 −^31 kg,a 10 .0Å 1 .00 nm, andk 0 .0500 J m−^2.

We now apply the perturbation method to the ground state of the helium atom. The

zero-order Hamiltonian is a sum of two hydrogen-like Hamiltonians:

̂H(0)ĤHL(1)+ĤHL(2) (19.3-9)

The perturbation term is

Ĥ′ e

2

4 πε 0 r 12

(19.3-10)