Subject: Yet another WHAM question...

From: [email-masked]
Date: Tue, 29 Feb 2000 22:10:55 -0500 (EST)

Dear CCL'ers:

There's a line in Kumar et al.'s paper [1] on the WHAM method that
makes no sense to me.  It concerns the histogram's required
dimensionality.  Kumar considers a modified Hamiltonian of the form

       L
      ___
      \
  H = /   \lambda_i V_i = V_0 + S
      ---
      i=0

That is, the Hamiltonian is a linear combination of L+1 terms with
coefficients \lambda_i.  Moreover, \lambda_0 is assumed to be
identically 1, and the corresponding term V_0 to represent the
unmodified Hamiltonian.  The remaining terms V_i are restraining
potentials, with coupling parameters \lambda_i.  (I denote the sum of
these last L terms by S; hence the potential energy becomes V_0 + S.)
Furthermore, the authors assume that we are interested in determining
the density of states as a function of a reaction coordinate \xi.

In the appendix to the paper (p. 1020), they write: "When the
restraining potential is a function of the coordinate \xi only the
dimensionality of the histogram reduces from L + 2 to 2..."

It's the "L + 2" that makes no sense to me.  I don't see why, in this
situation, the dimensionality of the histogram needs to be any larger
than 3, even when L > 1.  At most we need to histogram the values of
V_0, S, and \xi.  (If S is entirely a function of \xi, then, as the
authors say, we can get away with only two dimensions, namely V_0 and
\xi.)  To histogram the values of each of the restraining potentials
only increases the amount of data that must be collected without
providing any obvious benefit (only the sum of these values enters in
the WHAM equations).  Besides, the restraining potentials are
presumably of little intrinsic interest, since they are merely a
technical trick to bias the sampling.  If all this is true, what's the
rationale for histogramming the values of the V_i (0 < i <= L)?

Regards,

Kynn Jones

[1] Kumar, S. et al., J. Comp. Chem. 13(8):1011-1021, 1992