Problem Set 9 Solutions

9
6. A Hartree-Fock SCF calculation on the ground state of the H 2 molecule was carried out at
a fixed internuclear distance of 1.40 a.u. A minimal basis was used, which consisted of a
1s atomic orbital on each hydrogen atom.
The calculation yielded two MO energies,
ε 1 = − 0.619 a.u.
ε 2 = − 0.401 a.u.
The two-electron Coulomb and Exchange integrals that arise in the calculation are listed
below.
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φ 1 (1) φ 1 (2)
1
r 12
φ 1 (1)φ 1 (2) = 0.566 a.u.
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φ 1 (1) φ 2 (2)
1
r 12
φ 1 (1)φ 2 (2) = 0.558 a.u.
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φ 1 (1) φ 2 (2)
1
r 12
φ 2 (1)φ 1 (2) = 0.140 a.u.
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φ 2 (1) φ 2 (2)
1
r 12
φ 2 (1)φ 2 (2) = 0.582 a.u.
In these integrals, the functions φ 1 and φ 2 represent the molecular orbitals.
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a.) Write down a Slater determinant for the ground electronic state of H 2 .
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H 2 has two electrons so only one spatial MO will be occupied. For the ground state of H 2 , both electrons
will occupy MO1, φ 1 . The possible spin orbitals are φ 1 α and φ 1 β . Since the spins have to be paired if
the electrons are in the same spatial orbital, the Slater determinant must be:
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( ) =
Ψ H 2
1,2
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1
2
φ 1 € ( 1)α 1
φ 1 1
( ) φ 1 ( 2)α( 2)
( ) β( 1) φ 1 ( 2) β( 2) .
b.) Using the molecular € orbital energies and the two-electron integrals given above,
compute the electronic energy of H 2 at an internuclear separation of 1.40 a.u.
The electronic energy is given by
E el = E =
n
n
n
∑ 2ε i − ∑ ∑ ( 2J ij − K ij ) .
i=1 i=1 j=1
For H 2 , there are two electrons so N=2. This means that the number of occupied spatial orbitals, n, equals
N/2 = 1. Writing out € the summations above for H 2 with n=1 yields
E el ( H 2 ) = 2ε 1 − ( 2J 11 − K 11 ).
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