Detailed balance condition

EXAMPLES (1) Isotropic Peripheral PGA - consider the isotropic version of the peripheral PCA defined by equations 7.61 and 7.63 i.e. take a-2 — os = 02.3- In this case, the detailed balance condition is satisfied when the 3-spin coupling constant hi23 = 0. From equation 7.96, we see that this condition translates to... 

If the two representations are equivalent then Eqs. (3.79) and (3.80) describe how A s and B s must be transformed in terms of a s and /Ts. (These identities are performed explicitly by Sanchez and Di Marzio, [49]. Frank and Tosi [105] further show that if a s and /Ts are chosen to satisfy detailed balance conditions, that is equilibrium behaviour, then the occupation numbers of the two representations are only equivalent if the nv s are in an equilibrium distribution within each stage. This is likely to be true if there is a high fold free energy barrier at the end of each stem deposition, and thus will probably be a good representation for most polymers. In particular, the rate constant for the deposition of the first stem, A0 must contain the high fold free energy term, i.e. ... 

We begin with the microstate probability i(i —> j) of making a move from configuration i to j, each characterized by a volume, number of particles, and set of coordinates q. This probability and its reverse satisfy the detailed balance condition ... 

Just as in a conventional Monte Carlo simulation, correct sampling of the transition path ensemble is enforced by requiring that the algorithm obeys the detailed balance condition. More specifically, the probability n [ZW( ) - z(n)( )]2 to move from an old path z ° 22) to a new one " (2/ ) in a Monte Carlo step must be exactly balanced by the probability of the reverse move from 22) to z<,J> 22)... 

This detailed balance condition makes sure that the path ensemble sg[z )] is stationary under the action of the Monte Carlo procedure and that therefore the correct path distribution is sampled [23, 25]. The specific form of the transition matrix tt[z(° 2 ) -> z(n, 9-) depends on how the Monte Carlo procedure is carried out. In general, each Monte Carlo step consists of two stages in the first stage a new path is generated from an old one with a certain generation probability... 

The detailed balance condition (7.12) can now be satisfied by selecting an appropriate acceptance probability. By inserting the product in (7.14) into the detailed balance condition we find that for a symmetric generation probability the acceptance probabilities for the forward and the reverse move must be related by... 

For short times, the correlation function (7(f) depends on the microscopic details of the dynamics as the system crosses from to 38. These motions take place on a molecular time scale rmoi essentially equal to the time required to move through the transition region. For times f larger than rmoi but still very small compared to the reaction time rrxn (if the crossing event is rare rrxn L> rmoi such that such an intermediate time regime exists), C(f) can be replaced by an approximation linear in time. Using the detailed balance condition k jk = h )/ h ) [33] one then obtains... 

After the momenta are selected from the distribution (8.39), the dynamics is propagated by a standard leapfrog algorithm (any symplectic and time-reversible integrator is suitable). The move is then accepted or rejected according to a criterion based on the detailed balance condition... 

The acceptance criteria for particle insertion and deletion moves are determined from the detailed balance condition applied to these probabilities. The final expressions are... 

Here x,x denote two configurations of the system (specified, for instance, by the set of coordinates of all atoms r or the position of one chain end for all chains and all bond lengths, bond angles, and torsion angles rf, If, 0a, where a = 1,... M runs over all chains and the indices i,j, k run over all internal degrees of freedom of one chain). The transition rates W(x —> x ) are chosen to fulfill the detailed balance condition... 

Using the detailed balance condition Eq. (8), this expression reduces to... 

This is just another way of stating the detailed balance condition, Eq. 5.73. 

The X and Y are thus expressed in terms of the profiles of the up and their inverse down (r v ) transitions. The spectral moments can thus be written as a simple combination of the moments of the up and down transitions, which may be computed from the induced dipole components and the interaction potential. Furthermore, the function X satisfies the (old) detailed balance condition, Eq. 6.72, and is conveniently represented by the successful BC or K0 models. A simple choice for Y could be (co/A) r (ft)) where A is a constant to be specified and r (ci)) is another model function T which satisfies Eq. 6.72. In other words, according to Eq. 6.75, the ro to vibrational profiles K can be represented by the familiar model functions whose parameters may be defined with the help of the associated moment expressions. 

That F ((o) is an even function of the frequency and consequently F (t) is an even function of the time follows from the detailed balance condition and the previous theorem. Combining the condition of detailed balance... 

It is then easy to verify that the detailed balance condition [Eq. (11)] is satisfied, if the acceptance probability is chosen as... 

However, as several authors have pointed out (5,7,82), it is incorrect to directly replace the quantum mechanical correlation function with its classical analog because the detailed balance condition will not be met. Therefore, the correct expression is... 

This relationship is a consequence of the detailed balance condition that in equilibrium each of the reactions in this scheme is in equilibrium with its forward flux equal to reverse flux [127]. Similarly,... 

The condition of detailed balancing is that each individual reaction should be at equilibrium. Consider, for instance, a system of three isomers. A, B, and C. It is conceivable that A only reacts to form B, B only to form C, and C only to form A. If the three kinetic constants are equal, the equilibrium condition would be that the concentrations of all three isomers are the same. However, this would not satisfy the detailed balancing condition, since the reaction between A and B would not by itself be at equilibrium (equilibrium is only attained through a cycle). 

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