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4.2. HARTREE-FOCK 43 44 CHAPTER 4. TWO ELECTRONS
1.4
exact atom E HF E E X E C
1.2 approx
H − -0.486 -0.528 -0.381 -0.042
1 1d He atom in HF He -2.862 -2.904 -1.025 -0.042
φ(x) 0.8 Be ++ -13.612 -13.656 -2.277 -0.044
Table 4.1: Energies for two-electron ions.
0.6
0.4
4.3 Correlation
0.2
0 The approximate solution of the Hartree-Fock equations is not exact, because the true wave-
0 0.5 1 1.5 2 2.5 3 3.5 4
function is not a product of two orbitals, but is rather a complicated function of both vari-
x
ables simultaneously. The true wavefunction satisfies the exact Schr¨odinger equation, and
Figure 4.1: HF orbital for 1d He atom, both exact and simple exponential with effective charge. also minimizes the ground-state energy functional for the given external potential. In tra-
ditional quantum chemistry, the correlation energy is defined as the difference between the
Exercise 16 Analytic solution to HF for 1d He
Hartree-Fock energy and the exact ground-state energy, i.e.,
This exercise is best done with a table of integrals, a mathematical symbolic program, or E trad = E − E HF . (4.20)
C
simply numerically on a grid.
We illustrate the effects of correlation on the simplest system we have, He. We can
1. Show that the orbital of Eq. (4.17) is normalized. adjust the relative importance of correlation by considering various values of Z. Perhaps
counterintuitively, for large Z, the external potential dominates over the repulsion, and so the
2. Show that the orbital of Eq. (4.17) satisfies the HF equation with the correct eigenvalue.
non-interacting solution becomes highly accurate. As Z is reduced, the interaction becomes
3. Check that the HF solution satisfies the virial theorem (for power -1). more important. For larger Z values, Hartree and exchange dominate over correlation. But
when Z is of order 1, correlation becomes important. Eventually, if Z is made too small, the
Exercise 17 Ionization energy of 1-d He system becomes unstable to ejecting one of the electrons, i.e., its energy becomes less than
The ionization energy of He is defined as that of the one-electron ion, i.e., its ionization potential passes through zero. For Z just a
little greater than that, the system has significant correlation. We illustrate these results in
I = E 1 − E 2 (4.19) 2
Table 4.3. For large Z, the total energy grows as −Z (the Hydrogenic result), while the
where E N is the ground-state energy with N electrons in the potential. exchange energy grows as −5Z/8. Thus the magnitude of the exchange energy grows with
Z, but its relative size diminishes. Similarly, clearly E C is almost independent of Z. So E C
1. What is the estimate of the ionization energy given by the crude Hydrogenic orbital becomes an ever smaller fraction of E X as Z grows. Finally, notice that, for H , the exchange
−
approximation? energy is so small that correlation is 11% of it. The two-electron ion unbinds at the critical
2. What is the ionization energy given by the HF solution? value of Z = 0.911.
Returning to the 1d world, all the same remarks apply qualitatively. The ground-state
3. Can you think of another way to estimate the ionization energy from the HF solution? energy for 1d He is -3.154. Since the HF energy is exactly -3 1/12, the correlation energy is
71 mH. Thus we need to estimate correlation energies to within about 10% for useful accuracy
Exercise 18 He in 3-d
Repeat the approximate HF calculation for He in three dimensions. What is the orbital energy, in quantum chemistry. This may seem daunting, but we’ve made great strides so far: Even
our crudest method was good to within 5%. This is a gift of the variational principle, which
the effective nuclear charge, and the total physical energy? Make a rigorous statement about
both the exact HF energy and the true ground-state energy. Plot the effective potential implies that all errors in the ground-state energy are at least second-order in the error in the
wavefunction (don’t expect other expectation values to be as good). Its just that we need
(external plus half Hartree) using your approximate solution. (For help, this calculation is
done in many basic quantum books). about 0.2%!