Histograms and free energies

Transcription

1 Histograms and free energies ChE210D Today's lecture: basic, general methods for computing entropies and free energies from histograms taken in molecular simulation, with applications to phase equilibria. Overview of free energies Free energies drive many important processes and are one of the most challenging kinds of quantities to compute in simulation. Free energies involve sampling at constant temperature, and ultimately are tied to summations involving partition functions. There are many kinds of free energies that we might compute. Macroscopic free energies We may be concerned with the Helmholtz free energy or Gibbs free energy. We might compute changes in these as a function of their natural variables. For single-component systems: For multicomponent systems,,,,,,,,,,,,, Typically we are only interested in the dependence of these free energies along a single parameter, e.g.,,,, etc. for constant values of the other independent variables. Free energies for changes in the interaction potential It is also possible to define a free energy change associated with a change in the interaction potential. Initially the energy function is and we perturb it to. If this change happens in the canonical ensemble, we are interested in the free energy associated with this perturbation: Δ=,,,, M. S. Shell /29 last modified 5/22/2012

2 = ln!" #$ %& '!" ( % ' ) What kinds of states 1 and 0 might we use to evaluate this expression? Here is a small number of sample applications: electrostatic free energy charging of an atom or atoms in a molecule, in which state 0 has zero partial charges and state 1 has finite values dipolar free energy adding a point dipole to an atom between states 0 and 1 solvation free energy one can turn on interactions between a solvent and a solute as a way to determine the free energy of solvation free energy associated with a field states 0 and 1 correspond to the absence and presence, respectively, of a field, such as an electrostatic field restraint free energy turning on some kind of restraint, such as confining a molecule to have a particular conformation or location in space. Such restraints would correspond to energetic penalties for deviations from the restrained space in state 1. free energies of alchemical transforms we convert one kind of molecule (e.g., CH 4 ) to another kind (e.g., CF 4 ). This gives the relative free energies of these two kinds of molecules in the system of interest (e.g., solvation free energies in solution). Potentials of mean force (PMFs) Oftentimes we would like to compute the free energy along some order parameter or reaction coordinate of interest. These are broadly termed potentials of mean force, for reasons we will see shortly. This perspective enables us to understand free-energetic driving forces in many processes. For the purposes of this discussion, we will notate a PMF by *+ where + is the reaction coordinate of interest. This coordinate might be, for example: an intra- or intermolecular distance (or combination of distances) a bond or torsion angle a structural order parameter (e.g., degree of crystallinity, number of hydrogen bonds) Consider the example of a protein in aqueous solution interacting with a surface. The reaction coordinate might be the distance between the center of mass of the protein and the surface: M. S. Shell /29 last modified 5/22/2012

3 , The PMF along,, *, would give the free energy of the system as a function of the proteinsurface distance. It might look something like: *,, This curve would show us: the preferred distance at which the protein binds to the surface, from the value of, at the free energy minimum the free energy change upon binding, from the difference in free energy between the minimum and large values of, the barrier in free energy for binding and unbinding, from the height of the hump Importantly, the free energy function does not just include the direct potential energy interactions between atoms in the molecule with atoms in the surface. It also includes the effects of M. S. Shell /29 last modified 5/22/2012

4 all of the interactions in the solvent molecules. This may be crucial to the behavior of the system. For example, the direct pairwise interactions of an alkane with a silica surface will be the same regardless of whether the solvent is water or octanol. However, the net interaction of the alkane and surface will be very different in the two cases due to solvent energies and entropies, and this effect is exactly determined by the PMF. Definition Formally, a potential of mean force (the free energy) along some reaction coordinate + is given by a partial integration of the partition function. In the canonical ensemble, *+=,,,+ = ln-,,,+ = ln.!"$%& / ' Here, +1 is a function that returns the value of the order parameter for a particular configuration. The integral in this expression entails a delta function that filters for only those Boltzmann factors for configurations with the specified +. One can think of the PMF as the free energy when the system is constrained to a given value of +. Notice that we have the identity.!34 '+=!5 The potential of mean force is so-named because it s derivative gives the average force along the direction of + at equilibrium. To show this, we need the mathematical identity ' '6.78/8 6'8=.'78 '8 /8 6'8 Now, we proceed to find the derivative of the PMF: '*+ '+ = ' '+ ln.!"$%& / ' ' '+!"$%& / ' =!"% / ' = 9 ' '+!"$%& :/ '!"% / ' M. S. Shell /29 last modified 5/22/2012

5 9 ; ' '+!"$%& :/ ' =!"% / ' = < 4!"$%& / '!"% / ' Here, the term < 4 gives the force along the direction of +, < 4 = ' '+ = ' '+1 = ' '+1? The remainder of the terms in the PMF equation serve to average the force for a specified value of +. Thus, '*+ '+ 4 A+ Paths Keep in mind that free energies are state functions. That is, if we are to compute a change in any free energy between two conditions, we are free to pick an arbitrary path of interest between them. This ultimately lends flexibility to the kinds of simulation approaches that we can take to compute free energies. Overview of histograms in simulation Until now, we have focused mainly on computing property averages in simulation. Histograms, on the other hand, are concerned with computing property distributions. These distributions can be used to compute averages, but they contain much more information than that. Importantly, they relate to the fluctuations in the ensemble of interest, and can ultimately be tied to statistical-mechanical partition functions. It is through this connection that histograms enable us to compute free energies and entropies. Definitions and measurement in simulation For the purposes of illustration, we will consider a histogram in potential energy. In our simulation, we want to measure the distribution of the variable using a long simulation run and many observations of the instantaneous value of. M. S. Shell /29 last modified 5/22/2012

6 In classical systems, the potential energy is a continuously-varying variable. Therefore, the underlying is a continuous probability distribution. However, in the computer we must measure a discretized version of this distribution. We specify a minimum and maximum value of the energy that defines a range of energies in which we are interested. Let these be min and max. We define a set of G bins into which the energy range is discretized. Each bin has a bin width of /= max min G Let the variable be the bin index. It varies from 0 to G 1. The average energy of bin is then given by I = min +K+ 1 2 M/ We create a histogram along the energy bins. This is simply an array in the computer that measures counts of observations: N I =counts of observations in the range X I / 2, I +/ 2) For the sake of simplicity, we will often write the histogram array using the energy, rather than the bin index, N()=N I where =intk min / M Here the int function returns the integer part of its argument. For example, int(2.6)=2. To determine a histogram in simulation, we perform a very large number of observations \ from a long, equilibrated molecular simulation. At each observation, we update: N() N()+1 This update is only performed if min < max. Otherwise, the energy would be outside of the finite range of energies of interest. However, we still need to keep count of all energies, whether or not inside the range, in order to properly normalize our histogram. We can normalize the histogram to determine a discretized approximation to the true underlying distribution () in the energy range of interest: M. S. Shell /29 last modified 5/22/2012

7 `()/= N() \ where \ is the total number of observations, including those energies outside of the defined range. On the LHS we include the bin width so as to approximate the continuous probability differential ()'. Thus, `()= N() \/ In the limit of an infinite number of observations from an infinitely long, equilibrated simulation, this approximation converges to the true one in the following manner: " a bc" d `( I )/=. ()' " a c" d This equation simply says that we sum up all of the underlying probabilities for the continuous energies within a bin to obtain the observed, computed probabilities. As the bin width goes to zero, we have lim c",f g `()= () Notice that there are two components to this limit: We need an infinite number of observations. We need an infinite number of bins. These two limits compete with each other: as we increase the number of bins, we need more observations so that we have enough counts in each bin to have good statistical accuracy. Practically speaking, we must choose a finite bin width that enables us to balance the length of the run with statistical accuracy in each bin. Typically, for the energy example above, the bin width is chosen to be of the order of a basic energy scale in the force field. For a Lennard-Jones system, this might be h. the simulation is performed long enough to achieve on the order of ~1000 or more average counts per bin, that is, \ G Statistical considerations Keep in mind that the computation of a histogram is subject to the same statistical considerations as simple simulation averages. That is, the histogram needs to be performed for many M. S. Shell /29 last modified 5/22/2012

8 correlation times of the energy in order to reach good statistical accuracy. It can be shown that the expected squared error in the histogram bin N() goes as d l m(") N() This implies that the error goes as the square root as the number of counts. Similarly, the expected squared error in the corresponding estimate `() goes as: The relative error in `() is given by: d l `(") () \ l `(") `() 1 o\ () Notice that the relative error is higher for lower values of (), i.e., at the tails of the distribution. Multidimensional histograms In this example, we considered only a histogram of potential energy. However, it is possible to construct histograms of many kinds of simulation observables. In other ensembles, we might compute distributions of other fluctuating macroscopic quantities like in GCMC simulations and in simulations. We can also view histograms of arbitrary parameters of interest, such as the end-to-end distance of a polymer chain or that number of hydrogen bonds a water molecule makes. We can also compute joint distributions using multidimensional histogram arrays. For example, `(,)//= N(,) \ Note that any continuous variables will require discretization and specification of a bin width. Discrete variables, on the other hand, do not require such a definition because the underlying distribution itself is discrete: `()= N() \ Many kinds of distributions, however, require the specification of a minimum and maximum observable value. M. S. Shell /29 last modified 5/22/2012

9 Connection to statistical mechanics The true power of histograms is that they allow us to measure fluctuations in the simulation that can be used to extract underlying partition functions. That is, we measure () in simulation and then we post-process this discretized function to make connections to free energies and entropies. Consider the energy distribution (). Its form in the canonical ensemble is given by the expression: ()= Ω q(,,)!" -(,,) where -(,,) is the configurational partition function, -(,,)=.Ω q (,,)!" ' Here, Ω m is the configurational density of states. It is the part of the microcanonical partition function that corresponds to the potential energy and configurational coordinates. It is defined by the equation Ω m (,,)=./X( ) r' The reason we are concerned only with the configurational energy distribution is that the kinetic component can always be treated analytically, within the context of equilibrium simulations. We can rewrite both using the dimensionless configurational entropy: s m (,,) lnω m (,,) ()= u v(",w,)!" -(,,) -(,,)=. u v(",w,)!" ' Basic histogram analysis and reweighting The above formalism shows that if we were able to compute the function s m (,,), we would be able to predict the complete energy distribution () at any temperature. In fact, things are a bit simpler than this: we only need the energy-dependence of s m, since the volume and number of particles stay fixed. For example, we can compute the average potential energy at any temperature using the expression M. S. Shell /29 last modified 5/22/2012

10 xy()=. (;)' =. u v(")!" ' -() = u v(")!" ' u v(")!" ' Here, we use the notation (;) to signify a distribution at a given specified temperature. Extracting the entropy and density of states from a histogram We might extract an estimate for s m by measuring (;) from a histogram in a canonical simulation at specified temperature. Inverting the above relationship, s m ()=ln{ -() (;)!" =ln (;)+;+ln-() =ln (;)+; ; m () Here, m () is the configurational part of the Helmholtz free energy, (,,)= m (,,)+ lnx!λ() r We can measure (;) using a histogram along a set of discrete energies. Post-simulation, we can then take this measured distribution and use it to compute a discrete approximation to the dimensionless entropy function at the same energies, s m ( I )=ln ( I ;)+; I ; m () Notice that the partition function part of this expression is a temperature-dependent constant that is independent of potential energy. Thus, we can compute the entropy function s m to within an additive constant using this approach. This also says that we can compute the configurational density of states to within a multiplicative constant. This basic inversion of the probability distribution function underlies many histogram-based methods. Keep in mind that s m () is fundamentally independent of temperature since it stems from the microcanonical partition function. That is, any temperature-dependencies on the RHS of this equation should exactly cancel to leave a -independent function. Another way of putting this is that, in principle, it does not matter from which temperature we measure (;). At equilibrium, the (;) from all temperatures should give the same s m () function. M. S. Shell /29 last modified 5/22/2012

11 Free energy differences from histograms Imagine that we measure (; ) and (; d ) from two simulations at different temperatures. We can use the relationship of both with the density of states to compute a free energy difference. We construct Rearranging, ln (; d) (; ) =Xs m() ; d +; d m ( d )r Xs m () ; +; m ( )r = (; d ; )+; d m ( d ) ; m ( ) ; d m ( d ) ; m ( )=ln (; d) (; ) +(; d ; ) This expression provides us with a way to determine the configurational free energy difference between temperatures 1 and 2 using measured histograms of the energy distribution. You may find it interesting that the RHS is dependent on, whereas the LHS is not. In fact, the free energies should have no -dependence. In principle, any value of could be plugged into the RHS and the same free energy would be returned. In practice, statistical errors in the measured () mean that the estimate for the free energy can have different accuracies depending on the value of chosen. More on this below. Reweighting With a computed a discrete approximation to the underlying energy distribution, we can also use it to predict () at temperatures other than the original simulation temperature. This basic procedure is called reweighting, since we use a distribution measured at one temperature to predict that at another. The scheme is as follows. Imagine s m () is computed from (; ) in a canonical simulation at. We have s m ()=ln (; )+; ; m ( ) We want to predict (; d ) at another temperature d. We have, Plugging in the above expression for s m, (; d )= u v(")! " -( d ) (; d )= (; ) (!! # )" -( ) -( d ) M. S. Shell /29 last modified 5/22/2012

12 However, the last term on the RHS involving the ratio of partition functions is independent of. We can find it using the probability normalization condition,. (; d )'=1 Thus, (; d )= (; ) (!! # )" (; ) (!! # )" ' This equation is an important one. It states that a distribution measured at can be used to predict a distribution at a different d. Thus, in principle, we would only need to perform a single simulation, measure () once, and use this expression to examine the energy distribution at any other temperature of interest. For example, we might compute the average energy as a function of temperature using the equation xy( d )=. (; d )' = (; ) (!! # )" ' (; ) (!! # )" ' Similar expressions could be found for other moments of the potential energy, such as x d y. These could be used to compute the temperature dependence of the heat capacity, using the relationship d w =x d y xy d. Statistical analysis Unfortunately, there are practical limits to the histogram reweighting procedure just described. The main problem is the measurement of () to good statistical accuracy in the tails of its distribution. Consider that a typical () is very sharply peaked: () M. S. Shell /29 last modified 5/22/2012

13 The width of the distribution relative to the mean goes roughly as # so that, for macroscopic systems, the distribution is infinitely peaked. The implication of this result is that it is very hard to measure the distribution at it tails, where we typically only have a few counts in each bin. If we reweight a measured distribution to a temperature where the tails change to high probability, the error can be magnified to be very large. () d > Here, the mean of the distribution at the new temperature is well within the tail region of the distribution at the original temperature. If we reweight (; ) to d, errors in the tails will be magnified. The error in the new distributions can be written approximately as d l ("; ) d l ("; # ) (; d d) (; ) This formula is derived using standard error propagation rules and the reweighting expression. Notice that if d = d, the error is the same at each energy as from the original measurement. Otherwise, we must weigh the error by a ratio involving the two probability distributions. In the above picture, the ratio ( ; d ) ( ; ) is very large due to the small probability of this energy at. Therefore, the error is greatly magnified in the reweighting procedure according to the above equation. In general, Distributions are subject to statistical inaccuracies at their tails, owing to the finite number of counts in each bin. M. S. Shell /29 last modified 5/22/2012

14 If the important parts of the energy distribution at d correspond to the tails of a measured distribution at, a reweighting procedure will fail due to large statistical errors. If computed from a measured (;), the quality of an estimated s m () function is only good around the frequently-sampled energies, i.e., for energies where the histogram has many entries. These techniques limit the determination of free energies and entropies using measurements from single histograms. The solution here, as we will see next, it to incorporate multiple histogram estimates of these quantities. Multiple histogram reweighting For the remainder of this section we will drop the subscript c from the configurational properties, for simplicity. Imagine that we were able to take histograms from multiple simulations ˆ, each at a different temperature. For each, we could get a different estimate of the underlying dimensionless configurational entropy function s: s()=ln (; )+; ; ( ) We should get the same function every time. We don t know the free energies, but we do know that these are merely additive constants in the entropy functions. We can therefore shift the different estimates for the entropy function up or down in value until they all match each other: M. S. Shell /29 last modified 5/22/2012

15 () d > > d ln ()+; +const This is the basic idea behind multiple-histogram methods. Notice that by finding the amounts to shift each curve to achieve overlap, we are essentially determining free energy differences between the adjacent state points. The trick in these calculations is to determine the shift amounts. The best ways of doing this take into account the different errors for each energy in each measured (). Below we describe methods for computing free energy differences so as to maximize the statistical quality of the predicted results. For this approach to work, we must have good overlap between the histograms. That is, there must be ranges of the histograms that overlap with a reasonable number of counts in each, for statistical efficiencies. Otherwise, it will be challenging to determine the shift amounts to good statistical accuracy. Ferrenberg-Swendsen reweighting and WHAM Ferrenberg and Swendsen in 1989 proposed an optimal way to stitch together different histograms by minimizing the statistical error in the computed density of states and entropy function. It was later generalized by [Kumar et al., 1992] and was named the Weighted Histogram Analysis Method (WHAM). Although the original derivation relied on certain forms for the M. S. Shell /29 last modified 5/22/2012

16 error propagation, the same result can be derived using a more elegant maximum likelihood approach, which is what we discuss here. The maximum likelihood approach is a statistical method for parameterizing probability models. It simply says the following: posit some form of the underlying distribution function for a random process. Then, given an observed set of events, the best parameters for that distribution are those that maximize the probability (likelihood) of the observed events. In this context, imagine we are trying to determine Ω() from Š simulations at different temperatures. We know that the underlying Ω() for each should be the same, but that the measured histograms N () will be different because the simulations are performed at different temperatures. The observed events are the energies tabulated in the histograms. Say that we make \ independent observations of the energy in each temperature. Here, independent implies that we are beyond twice the correlation time. In order to make a connection with the histogram, we will discretize all of energy space into discrete values I separated by intervals /. Before we start, let s define some notation: \ number of observations at each temperature ˆ index of observations (ˆ=1,,\) Œ index of temperature (Œ=1,,Š) index of discrete energy values I discrete energy values Ω I,s I density of states and entropy at the values I energy of the ˆth observation at temperature Œ N I count of histogram entries for temperature Œ and energy bin Derivation With these notations, we construct the total probability or likelihood of making the \Š observations from the simulations in terms of a yet-unknown unknown density of states function: f =ŽŽ ( ; ) M. S. Shell /29 last modified 5/22/2012

17 f =ŽŽ Ω$ &! " - Here, the - are formally given by - =.Ω()! " ' In terms of our discrete energies, - = Ω I! " a I = u a! " a I Note that - =!5 where the are the free energies at each temperature. With all of these considerations, we can rewrite the above probability in terms of histograms: Taking the logarithm, =ŽŽ Ω I! " a - I m a =ŽŽ$ u a! " a b! 5 & m a I ln = N I 0s I ; I +; 2 I According to the maximum likelihood approach, we want to maximize this probability with respect to any adjustable parameters in our model and the given observations. The former are the yet unknown values s I while the latter are the histograms N I. Therefore, we take the derivative with respect to s I and set it equal to zero: Evaluating, 0= ln s I for all M. S. Shell /29 last modified 5/22/2012

18 0= N I +\; s I The latter derivative is ; s I = ln- s I Substituting back in: = u a! " a - 0= $N I \ u a! " a b! 5 & for all This equation can be solved for s I : u a = N I \! " a b! 5 Notice that this equation now provides us with a recipe for computing the density of states (the LHS) based on histogram data. Let s rewrite it without the index for clarity: u" = N \! "b! 5 Here, we have made the definition that N gives the total number of histogram counts of energy, at any temperature. Even though this expression gives us a way to determine the entropy function from multiple simulation results, note that the RHS involves the free energy at every temperature. This free energy depends on the entropy itself: = ; ln u"! " " Iterative solution How can we solve for s? Ferrenberg and Swendsen suggested an iterative solution: 1. An initial guess is made for the Š values of. One can simply choose =0 for all Œ. M. S. Shell /29 last modified 5/22/2012

19 2. The (discretized) entropy at every energy is computed using s()=lnn () ln\ ln! "b!5 for all 3. The free energies are recalculated using ; =ln u"! " " for all Œ 4. Steps 2 and 3 are repeated until the free energies no longer change upon each iteration. In practice, one typically checks to see if the free energies change less than some fractional tolerance. 5. According to the above equations, the entropy function and the combinations ; cannot be determined to within more than an additive constant. Thus, with each iteration, one typically sets one of the free energies equal to zero at one temperature, =0. This iterative procedure can be fairly computationally expensive since it involves sums over all temperatures and all discretized energies with each iteration. It is not uncommon to require on the order of iterations to achieve good convergence. However, once convergence has been achieved, energy averages at any temperature even temperatures not included in the original dataset can be computed with the expression u"!" ; The constant of proportionality is determined by normalizing the probability distribution. This general approach is termed multiple histogram reweighting, and it can be shown to be the statistically optimal method for reweighting data from multiple simulations. That is, it makes the maximum use of information and results in reweighted distributions for ; that have the least statistical error. Importantly, it provides statistically optimal estimates of free energy differences b. Configurational weights Using the multiple histogram technique, it is possible to define a weight associated with each configuration ˆ at each temperature Œ at the reweighting temperature : = u"!" N $ & M. S. Shell /29 last modified 5/22/2012

20 In practice, the logarithm of the weights are maintained in computer rather than the wrights themselves, so as to maintain precision. Any property depending on the configurational coordinates can be determined by a sum over all of the \Š observations, using the weights: xšy= š f Here, š is the value of the property š for the ˆth configuration observed at temperature Œ. f For example, one can compute the mean potential energy, xy= f However, keep in mind that š can be any property of interest. The multiple histogram technique provides a way to determine its average at any arbitrary reweighting temperature and provides a statistically optimal estimator. This approach also enables one to determine the distribution (š) using the weights: f (š) / œ œ f That is, for a given value of š, the probability (š) includes all of the weights from configurations ˆŒ that have š =š. Again, the constant of proportionality stems from the probability normalization condition. Binless implementation One possible drawback of this approach is that it requires us to specify a bin width / for computing histograms and tabulating energies. It is also possible to follow this derivation in the limit that / 0. In this limit, the expression for the free energies are: f ; = ln\+ln! " ž Ÿ! " ž b! 5 where and G are also indices over temperature. As with the previous approach, this equation must be iterated to solve for the free energies. Notice that it involves a quadruple loop, once M. S. Shell /29 last modified 5/22/2012

21 over observations and three times over temperatures. Therefore, convergence can be much slower than the histogram version. Once the free energies are determined using this expression, the expression for the configuration weights is: =!"! ž5 ž! ž " Precision issues It is very important to keep in mind that the combined exponentiation and logarithm functions in these expressions require careful treatment in numerical evaluation in the computer in order to avoid precision inaccuracies. More on this issue is described in the document Simulation best practices. As a quick example, consider the computation of *=ln To get around this numerical precision issues, we rearrange this expression so that the terms in the exponential are better-behaved: where *=ln K max M max = max +ln max = max +ln max max =max Other thermodynamic potentials The multiple histogram technique is not limited to determining s(). It can also be used to determine the joint functions s(,) or s(,) if histograms are taken not only in energy but in particle number of volume fluctuations, respectively. Such might be the case in grandcanonical or isothermal-isobaric MC simulations. The Ferrenberg-Swendsen equations change M. S. Shell /29 last modified 5/22/2012

22 slightly in each case, but the derivation proceeds conceptually as above. We will discuss one implementation in the context of phase equilibria, below. Alternatively, one can find other underlying thermodynamic potentials, besides the entropy s, using the reweighting approach. Imagine, for example, that we perform Š simulations in the grand-canonical ensemble, each at the same temperature but with a different chemical potential. For each, we collect a histogram in the particle number, N I where is now an index over different values of. Statistical mechanics shows that the histogram of in the grand canonical ensemble relates to the Helmholtz free energy, (; )= ª()!«Ξ( ) =!5()b!«Ξ( ) Here, the dependence of and Ξ on and has been suppressed since these variables are constant throughout. The total likelihood is therefore f =ŽŽ ( ; ) f =ŽŽ!5$ &b!«) Ξ A derivation similar to the one above shows that the reweighting equations providing the solution to () and the Š values Ξ are: ()=lnn () ln\ ln!«b!3 for all ;* =ln 5()b!«for all Œ Notice here the change in roles relative to the previous example takes the place of. takes the place of. M. S. Shell /29 last modified 5/22/2012

23 Instead of computing s(), we compute the underlying function (). That is, we find the partition function corresponding to variations in fluctuations at constant, since all of our simulations were at the same temperature. We compute * = lnξ at each chemical potential, rather than at each temperature. By statistical mechanics, * =. Thus, the differences between these weights provide information about the pressure differences. It is possible to determine many different kinds of free-energetic quantities in this way. In general, To compute a free energy or entropy * as a function of some macroscopic parameter š, the simulation must accomplish fluctuations in š. For example, () requires fluctuations in and thus would necessitate use of the isothermal-isobaric ensemble. To have good statistical accuracy, a wide range of fluctuations in š must be sampled. This can be accomplished by performing multiple simulations at different values of the parameter conjugate to š. To compute (), different pressures would be imposed for multiple simulations at the same temperature. The multiple-histogram reweighting procedure will compute *(š) as well as the relative partition functions at the different simulation conditions. In the example for (), we would find () as well as, where is the Gibbs free energy of the isothermalisobaric ensemble. Potentials of mean force (PMFs) The connection between a PMF and simulation observables stems from the probability distribution of the reaction coordinate of interest. Based upon the fact that the total partition function can be expressed as a sum over the PMF partition function, we have that (+)=!3(4) That is, the probability that one will see configurations with any value of + is proportional to a Boltzmann factor involving the potential of mean force at +. This approach enables us to measure PMFs in simulation by computing histograms along reaction coordinates. We have, *(+)= ln (+) ln- = ln (+)+const - M. S. Shell /29 last modified 5/22/2012

24 Notice that - provides an additive constant independent of the reaction coordinate. Typically, we cannot determine the PMF in absolute units, which would require the determination of the absolute free energy of the system. Instead, we can only compute *(+) curves to within an additive constant. This is not normally a problem, however, since we are usually only interested in relative free energies along +. One of the limitations of this basic approach is that we cannot compute large changes in free energies due to the statistics of the histogram. Imagine that we examine a free energy difference at two values of +, *(+ d ) *(+ ). Say that this value is equal to 4, which is a moderate energy equal to about 2 kcal/mol at room temperature. The relative probabilities at these two values would be (+ d ) (+ ) =!$3(4 ) 3(4 # )& = 0.02 If we were to measure a histogram and the bin for + contained 1000 entries, the bin for + d would only contain ~20 entries. Thus, the statistical error for the latter would be quite high. The main point here is that it is difficult to measure free energies for rarely sampled states. One way around this, which we will discuss in a later lecture, is to bias our simulation to visit these states. Though the introduction of the bias modifies the ensemble we are in, we can exactly account for this fact in the computation of thermodynamic properties. Weighted Histogram Analysis Method (WHAM) If multiple simulations are performed at different state conditions (e.g., different,, or ), the Ferrenberg-Swendsen reweighting approach can also be used to compute potentials of mean force at some set of target conditions. This is quite straightforward. As before, imagine that Š simulations are performed at different conditions. In each simulation Œ, \ observations ˆ are made. In addition to computing the energy, one tallies the value of the reaction coordinate for each observation: + Then, the PMF at some final set of conditions (the reweighting temperature, chemical potential, and/or pressure) is found by tallying the configurational weights to compute the distribution in +: M. S. Shell /29 last modified 5/22/2012

25 (+) / 4 4 f Ultimately, this distribution will need to be performed in a discretized array if + is a continuous variable. Then, the PMF simply comes from *(+)= ln (+) Histogram methods for computing phase equilibria The computation of phase equilibria saturation conditions, critical points, phase diagrams, etc. is a major enterprise in the molecular simulation community. There are many ways that one might go about this, including the Gibbs ensemble, and they are discussed in detail in [Chipot & Pohorille, Free Energy Calculations: Theory and Applications in Chemistry and Biology, Springer, 2007]. Of all of these methods, however, the current gold-standard for phase equilibrium computations is one that relies on multiple histogram reweighting. This approach is often the quickest and has the maximum accuracy. Here we will discuss these calculations in the context of liquid-vapor equilibria. Other phase behavior such as liquid-solid, vapor-solid, or liquid-liquid (in mixtures) can be treated using this approach as well, although special techniques may be needed to treat structured phases. Grand canonical perspective The basic idea of our calculations is the following: For a given temperature, we want to find the conditions at which phase equilibrium occurs. In a single component system, we only have one free parameter to specify this condition in addition to, per the Gibbs phase rule. This could be or. We will focus on instead of and perform simulations in the grand canonical ensemble. Thus, we want to find the value of at coexistence that generates an equilibrium between liquid and vapor phases. We could equally choose and perform simulations; however, particle insertions and deletions are generally much more efficient than volume fluctuations and thus we go with the former for computational reasons. If the state conditions place us at conditions of phase equilibrium (i.e., on the liquidvapor phase line), we expect the GCMC simulation to spontaneously fluctuate between a dense liquid (low, large ) and dilute gas (high, small ). These fluctuations can be observed by examining the joint distribution (,) M. S. Shell /29 last modified 5/22/2012

26 At phase equilibrium, (,) will be bimodal, and the total probability weight under the two peaks will be equal. One peak corresponds to the liquid phase, and the other to the gas phase. We will find the conditions of phase equilibrium (, ) that result in a bimodal (,). We will generate (, ;, ) by reweighting simulation data from multiple simulations, each not necessarily on the phase coexistence line. GCMC implementation We need to gather enough histogram data so that we will be able to reweight (,;, ) with good statistical accuracy. We perform Š simulations in the grand-canonical ensemble at different temperatures and chemical potentials. For each we measure the joint distribution $,;, &. The temperatures and chemical potentials should be chosen so that the energy and particle number distributions all overlap with each other and span the complete range of liquid-to-gas densities. Typically we perform the following kinds simulations: one near the (presumed) critical point that has large fluctuations in and one in the gas phase several in the liquid phase, of varying temperature The point is that we must have good histogram statistics for all the particle numbers and energies that will ultimately be relevant at phase equilibrium. gas-phase simulation near-critical simulation liquid-phase simulation lower- liquid-phase simulation M. S. Shell /29 last modified 5/22/2012

27 Once these histograms are taken, Ferrenberg-Swendsen iterations are used to compute the free energy at each state point. The relevant equations for GCMC simulations are: s,lnn,ln\ln! "b! «b! 3 for all, ;* ln u",! "b! «" for all Œ This procedure is performed until values for the discretized entropy s, and free energies * lnξ converge. Keep in mind that one value of * must be set equal to zero. Once these values are determined, the joint distribution, can be computed for an arbitrary reweighting, using u",!"b!«,;, The constant of proportionality is given by the normalization condition. Using the reweighting equation, one finds phase equilibrium by adjusting given a value of until a bimodal distribution appears and the integral under each peak (the total probability) is the same. Notice that this is a very fast operation since the Ferrenberg-Swendsen reweighting only needs to be performed once, at the beginning, to determine s,. The following figure shows what this distribution might look like, taken from a review article by Panagiotopoulos [Panagiotopoulos, J Phys: Condens Matter 12, R25 (2000)]. M. S. Shell /29 last modified 5/22/2012

28 By repeating this procedure at different reweighting temperatures, one can map out the phase diagram. In the - plane, this might look like: The average density of each phase can be determined using: x y = 1 " " ± ±,, x y ² = 1 " " ³ ³,, Here, is the value of at the minimum between the two peaks in the distribution. Pressures can be found using the fact that ; lnξ,, We cannot compute absolute pressures using this equation, because we cannot compute Ξ absolutely (one must be set to zero). However, we can compare the pressures of two different state points: ; d d ; ln Ξ d, d, Ξ,, ln u",! "b! «" u",! #"b! # «# " By letting one state point correspond to a very dilute, high temperature gas phase, we can compute its absolute pressure using the ideal gas law. This equation then can be used to relate the pressure at other state points back to the ideal gas reference. M. S. Shell /29 last modified 5/22/2012

29 Critical point reweighting and finite-size scaling How can one predict the location of the critical point? It turns out that systems have a kind of universal behavior at this point in the phase diagram. For many fluids, their behavior at the critical point follows the same trends as the three-dimensional Ising model (or, equivalently, the lattice gas); that is, they fall in the Ising universality class or the Ising criticality class. In particular, the probability distribution, assumes a universal form. We can locate the true critical point by finding reweighting values of m, m so that, maps onto the universal Ising distribution, which has been computed to high accuracy by a number of researchers. Near the critical point, fluctuations become macroscopic in size. That means that computed properties become sensitive to the size of the system being studied. In particular, the computed m, m can have a substantial dependence on the system size. Fortunately, these parameters have a scaling law behavior with. To locate the infinite system size m, m one performs multiple studies for different values of, each time computing the critical point using the procedure above. Then, one extrapolates to. An example of this taken from [Panagiotopoulos, 2000] is: M. S. Shell /29 last modified 5/22/2012