We can now compute the energy of these states.
〈H 0 〉± =
1
2[1±S(R)]
〈ψA±ψB|H 0 |ψA±ψB〉=
1
2[1±S(R)]
[〈ψA|H 0 |ψA〉+〈ψB|H 0 |ψB〉±〈ψA|H 0 |ψB〉±〈ψB|H 0 |ψA〉]=
〈ψA|H 0 |ψA〉±〈ψA|H 0 |ψB〉
1 ±S(R)We can compute the integrals needed.
〈ψA|H 0 |ψA〉 = E 1 +e^2
R(
1 +
R
a 0)
e−^2 R/a^0〈ψA|H 0 |ψB〉 =(
E 1 +
e^2
R)
S(R)−
e^2
a 0(
1 +
R
a 0)
e−R/a^0We have reused the calculation ofS(R) in the above. Now, we plug these in and rewrite things in
terms ofy=R/a 0 , the distance between the atoms in units of the Bohr radius.
〈H 0 〉± =
E 1 +e2
R(1 +R/a^0 )e− 2 R/a (^0) ±
(
E 1 +e2
R)
S(R)−e2
a 0 (1 +R/a^0 )e−R/a 01 ±S(R)〈H 0 〉± = E 11 −(2/y)(1 +y)e−^2 y±[
(1− 2 /y)(1 +y+y^2 /3)e−y−2(1 +y)e−y]
1 ±(1 +y+y^2 /3)e−yThe symmetric (bonding) state has a large probability for the electron to be found between nuclei.
The antisymmetric (antibonding) state has a small probability there, and hence, a much larger
energy.
The graph below shows the energies from our calculation for the space symmetric (Eg) and antisym-
metric (Eu) states as well as the result of a more complete calculation (ExactEg) as a function of the
distance between the protonsR. Our calculation for the symmetric state shows a minimum arount
1.3 Angstroms between the nuclei and a Binding Energy of 1.76 eV. Wecould get a better estimate
by introduction some parameters in our trial wave function and using the variational method.
The antisymmetric state shows no minimum and never goes below -13.6eV so there is no binding
in this state.