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I am reading Free Energy Calculations: Theory and Applications in Chemistry and Biology by Chipot and Pohorille. At the beginning of the text (page 19, for example), the authors define the Helmholtz free energy $A$ and changes $\Delta A$ in it for two states (states 0 and 1):

$$\begin{align} A&=-\beta^{-1} \ln Q(N, V, T) \;\;\;\textbf{(1.14)} \\ \Delta A&=-\beta^{-1} \ln \frac{Q_1}{Q_0} \;\;\;\textbf{(1.15)} \\ \Delta A&=-\beta^{-1} \ln \frac{Z_1}{Z_0} \;\;\;\textbf{(1.16)} \end{align}$$

where $Q_i$ is a partition function:

$$Q_i = \frac{1}{N! h^{3N}} \int_{\Gamma_i} \exp[-\beta \mathcal{H}(d\vec{x}, d\vec{p})] d\vec{x} d\vec{p}_x$$

and $Z_i$ is a configurational integral:

$$Z_i = \int_{\Gamma_i} \exp[-\beta U(\vec{x})] d\vec{x}$$

where $U$ is the potential energy.

Next (page 20), the authors write a derivation for $\Delta A$, in the framework of free energy perturbation (FEP):

Equation (1.15) indicates that our ultimate focus in calculating $\Delta A$ is on determining the ratio $Q_1/Q_0$, or equivilently $Z_1/Z_0$, rather than on individual partition functions. On the basis of computer simulations, this can be done in several ways. One approach consists of transforming (1.16) as follows:

$$\begin{align} \Delta A&=-\beta^{-1} \ln \frac{\int \exp[-\beta U_1(\vec{x})] d\vec{x}}{\int \exp[-\beta U_0(\vec{x})] d\vec{x}} \\ &=-\beta^{-1} \ln \exp\left\{-\beta [U_1(\vec{x}) - U_0(\vec{x})]\right\} P_0(\vec{x}) \\ &=-\beta^{-1} \ln \langle \exp\left\{-\beta [U_1(\vec{x}) - U_0(\vec{x})]\right\} \rangle_{0} \end{align}$$

Here, the systems 0 and 1 are described by the potential energy functions, $U_0(\vec{x})$ and $U_1(\vec{x})$, respectively. Generalization to conditions in which systems 0 and 1 are at two different temperatures is straight forward. $\beta_0$ and $\beta_1$ are equal to $(k_B T_0)^{-1}$ and $(k_B T_1)^{-1}$, respectively. $P_0(\vec{x})$ is the probability density function of finding system 0 in the microstate defined by positions $\vec{x}$ of the particles:

$$P_0(\vec{x}) = \frac{\exp[-\beta_0 U_0(\vec{x})]}{Z_0} \;\;\;\textbf{(1.19)}$$

My question is, what happened to the integrals in going from (1.18) to the next step? $P_0(\vec{x})$ only contains an integral in its denominator (i.e., in $Z_0$), and not also in the numerator. Could you please perhaps provide one more step between equation (1.18) and the equation that immediately follows it? Thanks for your time.

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1 Answer 1

There has to be an integral missing in your book citation, line 2 of $\Delta A =\ldots$ should read $\Delta A = \beta^{-1} \ln \left[\int \exp{-\beta[U_1(\vec x) - U_0(\vec x)]} P_0(\vec x) d \vec x \right]$. Other than that, the calculation is as follows:

$$\begin{align*} \Delta A &=-\beta^{-1} \ln \frac{\int \exp[-\beta U_1(\vec{x})] d\vec{x}}{\int exp[-\beta U_0(\vec{x})] d\vec{x}} &= -\beta^{-1} \ln \frac{\int \exp[-\beta U_1(\vec{x})] \frac{\exp(\beta U_0(\vec x))}{\exp(\beta U_0(\vec x))} d\vec{x}}{\int \exp[-\beta U_0(\vec{x})] d\vec{x}} \\ &= -\beta^{-1} \ln \frac{\int \exp[-\beta [U_1(\vec{x}) - U_0(\vec x)]] \exp[-\beta U_0(\vec x)] d\vec{x}}{\int \exp[-\beta U_0(\vec{x})] d\vec{x}} \\ &= -\beta^{-1} \ln \frac{\int \exp[-\beta [U_1(\vec{x}) - U_0(\vec x)]] \exp[-\beta U_0(\vec x)] d\vec{x}}{Z_0} \\ &= -\beta^{-1} \ln \int \exp[-\beta [U_1(\vec{x}) - U_0(\vec x)]] p_0(\vec x) d\vec{x} \\ &= -\beta^{-1} \ln \langle\exp[-\beta [U_1(\vec{x}) - U_0(\vec x)]] \rangle \end{align*}$$

Long, faulty latex story short: multiply the integrand in the nominator by $1=\frac{\exp[\beta U_0(\vec x)]}{\exp[\beta U_0(\vec x)]}$, unite the two "top" exponentials, convert $\frac{1}{\exp[\beta U_0(\vec x)]} = \exp[-\beta U_0(\vec x)]$ which , with $Z_0$ from the big denominator, gives $P_0(\vec x)$ and write the integral as the ensemble average.

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Welcome to PhysicsSE! To fix your beautiful TeX I removed '\left{' delimiters in first statement. There were two left/right pairs. In the next math block, I just removed the blank lines and changed '\begin{align}' to '\begin{align*}'. However the second expression still is not aligned correcly by me. –  Stefan Bischof Apr 17 '13 at 20:00
    
@StefanBischof: Thanks a lot for your help (and the welcome)! –  Simeon Carstens Apr 18 '13 at 7:39

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