I am trying to understand the derivation of the variational principle as presented in Bob Evan's 1979 work.[1]. The part that is tripping me up is when he presents the following result. He starts by defining a functional of the form

$$ \Omega [f_N] = \mathrm{Tr_{cl}} \left[ f_N \left( H - \mu N + \beta^{-1}\ln{f_N} \right) \right] \, .\tag{1}\label{functional}$$ where $f_N$ is an $N$-particle probability distribution and $H$ is the Hamiltonian. Note that $H$ here is a phase space function, and not equivalent to the thermodynamic energy $E$.

This functional reduces to the grand potential $\Omega$ when $f_N$ is the equilibrium distribution $\rho^{(N)}$, which is given by

$$ \rho^{(N)} = \frac{\exp-\beta(H - \mu N)}{\Xi} \tag{2}\label{rho}$$

An important step in deriving the variational principle is showing that

$$ \Omega [f_N] = \Omega[\rho^{(N)}] + \beta^{-1}(\mathrm{Tr_{cl}}f_N\ln f_N -\mathrm{Tr_{cl}}f_N\ln \rho^{(N)}) \tag{3}\label{var}$$

This is accomplished by using \eqref{rho} to show that $$ -\beta^{-1}\ln\rho^{(N)} = H - \mu N + \Omega[\rho^{(N)}]$$ or $$ \Omega = H - \mu N + \beta^{-1}\ln\rho^{(N)} \tag{4} \label{omega}$$ and substituting \eqref{omega} into \eqref{functional}. This is where I am hung up; it seems very odd that \eqref{omega} is true. From the thermodynamic perspective, the grand potential should be given by the expectation values for the energy $\langle E\rangle$ and average particle number $\langle N \rangle$. This gives $$\Omega = \langle E \rangle - \mu \langle N \rangle - TS $$ Replacing $S$ with the Gibbs entropy formula, and using the fact that the expectation values can be calculated from the equilibrium probability distribution by $\langle A \rangle = \mathrm{Tr_{cl}}\left[\rho^{(N)}A\right]$ one now has for the grand potential

$$\Omega = \mathrm{Tr_{cl}}[\rho^{(N)}H] - \mu \mathrm{Tr_{cl}}[\rho^{(N)}N] + \beta^{-1}\mathrm{Tr_{cl}}[\rho^{(N)}\ln\rho^{(N)}] \tag{5}\label{omegaTrue}$$

Comparing equations \eqref{omega} and \eqref{omegaTrue}, it seems that the trace is unnecessary; but from a statistical mechanics perspective, it seems odd that one doesn't need to average over phase space to find the value of the grand potential, and that simply knowing the value of $H$ and $\rho^{(N)}$ at a single phase space point is enough to specify $\Omega$, which is what \eqref{omega} seems to imply. Is there some assumption that I am missing that makes this work. Or is \eqref{omega} really valid?

1 Evans, R. (1979). The nature of the liquid-vapour interface and other topics in the statistical mechanics of non-uniform, classical fluids. Advances in Physics, 28(2), 143–200.


1 Answer 1


eq(4) and eq(5) are equivalent. First I rewrite eq(5) as the result of $f_N = \rho^{(N)}$ in eq(1): $$\Omega [\rho^{(N)}] = \mathrm{Tr_{cl}} \left[ \rho^{(N)} \left( H - \mu N + \beta^{-1}\ln{\rho^{(N)}} \right) \right] \tag{1}$$

$$ \rho^{(N)} = \frac{\exp-\beta(H - \mu N)}{\Xi} \tag{2}\label{2}$$ From the above equation \eqref{2} we can get: $$ H - \mu N + \beta^{-1} \ln {\rho^{(N)} } = -\beta^{-1} \ln{ \Xi } \tag{3}\label{3} $$ Substituting 3 into 1, we get: $$ \begin{aligned} \Omega [\rho^{(N)}] =&\mathrm{Tr_{cl}} \left[ \rho^{(N)} \left( H - \mu N + \beta^{-1}\ln{\rho^{(N)}} \right) \right] \\ =& \mathrm{Tr_{cl}} \left[ \rho^{(N)} \left(-\beta^{-1} \ln{ \Xi } \right)\right] \\ =&\sum_{N=0}^{\infty} \dfrac{1}{h^{3N}N!} \int d\mathbf{r}^N d\mathbf{p}^N \frac{\exp-\beta(H - \mu N)}{\Xi} \left(-\beta^{-1} \ln{ \Xi} \right) \\ =&-\beta^{-1} \ln{ \Xi} \dfrac{1}{\Xi}\sum_{N=0}^{\infty} \dfrac{1}{h^{3N}N!} \int d\mathbf{r}^N d\mathbf{p}^N \exp-\beta(H - \mu N) \\ =&-\beta^{-1} \ln{ \Xi} \\ =&H - \mu N + \beta^{-1} \ln {\rho^{(N)} } \end{aligned} $$ So the two are equivalent.

  • $\begingroup$ Equations 4 and 5 are not equivalent. The value of H, N, and $\rho^{(N)}$ at a point in phase space are not the same as their expectation values. The algebraic manipulations aren't where I am having a problem; the problem I am having is that Eq. (3) in your answer implies that the entire partition function, and by extension the grand potential, can be determined from knowing $H$, $N$, and $\rho^{(N)}$ at a single point in phase space. $\endgroup$
    – scmartin
    May 20, 2022 at 14:14
  • $\begingroup$ The value of $H$,$N$and $\rho^{(N)}$ at a point in phase are not the same as their expectation values. but their sum $-\beta^{-1} \ln \Xi$ is a constant. So eq(4) is equivalent to eq(5), you should look at my derivation again. $\endgroup$ May 21, 2022 at 0:08
  • $\begingroup$ I have proved that eq(4) is equal to eq(5). If you think the two are not equivalent, then you should point out the error in my derivation. $\endgroup$ May 21, 2022 at 0:14

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