3
$\begingroup$

I'm going over some (molecular dynamics) related literature - specifically the derivation of the Weighted Histogram Analysis Method (WHAM).

As a quick backdrop WHAM is a method for stitching together conceptually overlapping and independent experiments done in Monte Carlo or Molecular Dynamic simulations. This isn't actually super important, as this is more a notation question.

Before the paper gets to the WHAM derivation, there's a discussion on the actual problem, where two variables are defined

$ x = \text{coordinates of atoms}$

$ \xi = \text{reaction coordinate - is a function of x} $

There's also a discussion on the notation used, where a caret above a value indicates it's a function,

"Thus, $\hat{V_i}(x)$ denotes the function and $V_i$ a particular value the function takes"

Having defined this and after some more work, the paper defines the probability density obtained from a simulation, which can be written in the following manner.

$P(\xi) = \langle\delta[\xi - \hat{\xi}(x)]\rangle$

For the record

$P(\xi) = e^{\big(-\beta W(\xi)\big)}$

Firstly, there's no explanation of what $\delta$ is at all, which makes me think it's probably some common operator. I don't think a Kronecker delta because that doesn't make sense in this context (and it lacks subscripts) but maybe?

Secondly, is there some common way to interpret a variable which can also be a function? When I initially read it I assumed that $\xi(x)$ was looking at deviation from the reaction coordinate the $x$ variable showed, but in hind-site I don't know why I thought that.

The reaction coordinate, traditionally, is some physical component of the system. It could be the distance between two residues, a specific orientation of all the atoms in the system, the distance of a specif residue to some arbitrary point etc. I'm not sure if that's the same here, or if it has some other, more specific meaning in this context.

EDIT: I'm currently reading another paper in the literature ("Use of the Weighted Histogram Analysis Method for the "Analysis of Simulated and Parallel Tempering Simulations" by Chodera et al.) which a colleague recommended and uses very different notation - hopefully I can use this to reconcile what's going on here...

$\endgroup$
4
  • $\begingroup$ The $\delta$ is the dirac delta function. You can think of it as the continuous case of the kronecker delta. $\endgroup$
    – Kitchi
    Mar 5, 2013 at 7:11
  • $\begingroup$ I don't know the answer, but I too am interested in it. $\endgroup$
    – Nick
    Mar 6, 2013 at 3:54
  • $\begingroup$ @Kitchi so that was one of my initial thoughts, but that means that if $\xi = \hat{\xi}(x)$ then $P(\xi)=1$ else $P(\xi)=0$ which doesn't seem to make too much sense in the contex $\endgroup$
    – Alex
    Mar 6, 2013 at 4:34
  • $\begingroup$ @Alex : That is definitely not what it means. You're forgetting the important ensemble average sign $\right<P(\xi)\left>$ in it $\endgroup$
    – user35952
    Jun 15, 2016 at 8:15

1 Answer 1

1
$\begingroup$

As @Kitchi writes $\delta$ is the Dirac Delta function. Your question is basically why is this true: $P(\xi) = \langle\delta[\xi - \hat{\xi}(x)]\rangle$.

As already noted (by Alex), when $ ξ=\hat ξ (x)$, we have $ \delta(ξ-\hat ξ (x)) = 1$. So the average that needs to be computed has this form: a configuration which has the reaction coordinate value $\xi$ contributes 1 and other configurations contribute 0.

So we have: $\langle\delta[\xi - \hat{\xi}(x)]\rangle = { \frac{ \text {# of conf. with }\xi}{\text{total # configs}}}$ . Clealy the RHS is the $P(\xi)$ So that this computes the ENSEMBLE average.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.