Yet another WHAM question...

[kynn \\at// panix.com]

 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