Physical Chemistry Third Edition

19.1 The Variation Method and Its Application to the Helium Atom 795

wherer 12 is the distance between the electrons. This function gave a variation energy

equal to− 78 .7 eV with a value ofZ′′equal to 1.849 and a value ofbequal to 0.364.

This energy is in error by 0.3 eV, or about 0.4%. More elaborate variation functions

have been used, and have given excellent agreement with experiment.^4

The presence of the factor (1+br 12 ) corresponds to a larger probability density for

larger separations of the electrons than for smaller separations. This explicit dependence

on the electron–electron distance represents the tendency of the electrons to repel each

other and is calleddynamical electron correlation. In a one-term orbital wave function

the probability density of each electron is independent of the position of the other

electrons and there is no electron correlation.

An antisymmetrized orbital wave function can also exhibit electron correlation if it

has an antisymmetric space factor like that of Eq. (18.4-2a). Consider the wave function

ΨC[ψ 1 (r 1 ,θ 1 ,φ 1 )ψ 2 (r 2 ,θ 2 ,φ 2 )−ψ 2 (r 1 ,θ 1 ,φ 1 )ψ 1 (r 2 ,θ 2 ,φ 2 )]

whereψ 1 andψ 2 are any two different space orbitals. Ifr 1 r 2 ,θ 1 θ 2 , andφ 1 φ 2

this function vanishes. Since the wave function must be continuous it must have a

small magnitude if the two electrons are close to each other. This effect is called

statistical electron correlation. Statistical correlation does not occur in wave functions

with symmetric space factors.

PROBLEMS

Section 19.1: The Variation Method and Its Application to
the Helium Atom

19.1 Prove the variation theorem. Assume that all of the

energy eigenfunctions and energy eigenvalues are known,

and write the variation function as a linear combination of

the energy eigenfunctions:

φ

∑∞

i 1

ciψi

Substitute this expression into the formula for the

variation energy and use eigenfunction and orthogonality

properties.

19.2 Calculate the variation energy of a particle in a

one-dimensional box of lengthawith the trial function

φ(x)Ax^2 (a^2 −x^2 ). Calculate the percent error.

19.3 Calculate the variation energy of a particle in a

one-dimensional box of lengthawith the trial function

tφ(x)Ax^3 (a^3 −x^3 ). Calculate the percent error.

19.4 Calculate the variation energy of a harmonic oscillator

using the trial functionφ(x)A/(b^2 +x^2 ), wherebis a

variable parameter. Minimize the energy and find the

percent error from the correct ground-state energy.

19.5Using the variation method, calculate an upper bound to

the energy of the ground state of an anharmonic oscillator

with potential energyVkx^2 / 2 +bx, wherekandbare

constants. Use the trial functionφA(ψ 0 +cψ 1 ),

whereψ 0 andψ 1 are the first energy eigenfunctions of

the harmonic oscillator with force constantk. Minimize

the energy with respect toc. Assume the numerical

values:k576Jm−^2 ,m 1. 672623 × 10 −^27 kg,

a 9. 3075 × 1021 m−^2 , and takebmg 1. 639 ×

10 −^26 Jm−^1. These values correspond to a hydrogen

molecule held in a vertical position near the surface of the

earth with the lower nucleus somehow held fixed. How

significant is the effect of gravity?

19.6Use the variation functionφAx^2 (x−a) to calculate an

upper bound to the ground state of a particle in a box of

lengtha. Give the percent error from the correct energy.

19.7Use the variation functionφAsin^2 (πx/a) to calculate

an upper bound to the ground-state energy of a particle in

a box of lengtha. Give the percent error from the correct

energy.

19.8Calculate the ground-state energy of a harmonic oscillator

using the variation function

(^4) T. Koga,J. Chem. Phys., 94 , 5530 (1991).