Essentials of Computational Chemistry

432 12 EXPLICIT MODELS FOR CONDENSED PHASES
and B do not differ from one another by enormous amounts, the ratio of the two expectation
values in the last line on the r.h.s. of Eq. (12.12) (which is the inverse of the equilibrium
constant for the reaction A → B) is sufficiently close to unity that errors in the individual
ensemble averages on the order of one or two percent have a much smaller impact on the
error of the ratio than was the case for U. Moreover, one takes the natural logarithm of
the ratio to compute A, so the absolute magnitude of the error is reduced still more.
Nevertheless, converging the individual expectation values to the level of a few percent
error is a painfully slow task, since the reduced probabilities of high-energy points are
balanced by their exponentially larger contributions to the partition function. However, one
may take advantage of the relatively small difference between systems A and B to introduce
a further approximation that is extraordinarily useful.
12.2.2 Free-energy Perturbation
If the ensembles in Eq. (12.2), over which the property averages for systems A and B are
taken, were somehow to be the same, one would be able to take advantage of the properties
of exponentials to write
〈A〉B −〈A〉A = kBT ln
(EB−EA)/kBT
e A
(12.13)
where we have arbitrarily chosen to label the ensemble average as having been selected
based on system A. This formulation offers some enormous advantages over Eq. (12.12).
One of the most important is that all contributions to the energy from solvent–solvent
interactions (which are enormously dominant over solvent–solute interactions, since there
are so many more solvent molecules) cancel out in the energy difference, since the ensembles
are (somehow) identical.
What is meant by an identical ensemble for two different species? It is helpful to return
to our specific example of HCN and HNC. To determine the proper identical ensemble for
HNC based on one chosen in the usual fashion for HCN, we first stipulate that all particles
that are common to the two systems, i.e., all solvent molecules, the carbon atom, and the
nitrogen atom, have identical positions and momenta when we evaluate the energy in system
B as when we evaluate it in A. Then, the only contribution to the energy difference in
Eq. (12.13) would be the different interactions that the hydrogen atom has with all of the
other atoms, based on whether it is attached to C or N (see Figure 12.1).
So, for each snapshot of the simulation that contributes to the ensemble (by either MC
or MD evaluation), we compute the energy differential for all of the atoms interacting with
HB rather than HA. In Figure 12.1, the particular case of one of the hydrogen atoms on a
first-shell water molecule is illustrated. As this is a non-bonded interaction in each case, the
contribution from HD in a simple force field might be
EHD =
aHH
r12 −
HBHD
bHH
r6 +
HBHD
qHBqHD εrHBHD
−
aHH
r12 −
HAHD
bHH
r6 +
HAHD
qHAqHD εrHAHD
(12.14)