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I quote from Huang's Introduction to Statistical Mechanics (not the book Statistical Mechanics), page 266

Consider, for simplicity, a one-component order parameter, a scalar field $\phi(x)$ over a space of any number of dimensions. The state of the system is specified by $\phi(x)$, and the statistical ensemble describing the thermal properties of the system is a set of such fields, with assigned statistical weights given by the Boltzmann factor $$e^{-\beta E[\phi]}$$ where $\beta=(k_BT)^{-1}$. The functional $E[\phi]$, called the Landau free energy specifies the system.

Then in order to calculate the partition function $Q$ he used (in Eq. (19.4)) the following functional integral $$Q=\int D\phi e^{-\beta E[\phi]}\tag{1}$$ which appears to be the generalization of the canonical partition function formula $$Z=\sum\limits_{\rm{states}~ j} e^{-\beta E_j}\tag{2}$$ in the limit when $j$ becomes a continuous label. Therefore, as far as I can understand, $E[\phi]$ is essentially same as the energy levels of the system $E_j$. However, according to Huang's description, $E[\phi],$ is called Landau free energy.

$\bullet$ But the energy levels $E_j$ in expression (2) is independent of temperature and is determined from the classical or quantum dynamics of the system. All temperature dependencies in $Z$ enter through the factor $\beta$.

$\bullet$ However, the Landau free energy $E[\phi]$ is postulated to be of the form $$ E[\phi]=\int d^dx\Big[\frac{\epsilon^2}{2}|\nabla\phi(x)|^2+W(\phi(x))-h(x)\phi(x)\Big]$$ where $$W(\phi(x))=g_2\phi^2(x)+g_3\phi^3(x)+g_4\phi^4(x)+...$$ where the coupling constants $g_n$ are the phenomenological parameter of the theory, and may depend on the temperature. Therefore, $E[\phi]$ can be temperature-dependent.

Questions

A. Does it not mean that Landau free energy is different from the energy levels of the systems?

B. If yes, how can we use the formula (1) by generalizing (2)?

C. There is another possibility that $E[\phi]$ is defined after some coarse-graining and therefore, has temperature dependence built into it apart from $\beta$. If that is the case, (1) is not a straightforward generalization of (2). Please enlighten me.

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  • $\begingroup$ It seems that Huang's generalization refers not to the replacement of the sum by an integral. The usual canonical sum can also be written in the form of an integral $Z=\sum\limits_{\rm{states}~ j} e^{-\beta E_j}=\int \rho{(E)}e^{-\beta E}dE=\int e^{-\beta F}$, where $\rho{(E)}$ is a density of enrgy states.Free energy in this case: $F=(-ln(\rho(E))/\beta)+E$ $\endgroup$
    – Aleksey Druggist
    Commented Mar 11, 2018 at 15:35
  • $\begingroup$ @AlekseyDruggist As Thomas says, and as also I conjectured, there is a coarse-graining involved in going from (2) to (1). But I don't understand how does coarse-graining make $E[\phi]$ temperature dependent. $\endgroup$
    – SRS
    Commented Mar 11, 2018 at 15:42
  • $\begingroup$ Analogically in case of functional $E[\phi]$ we have $F[\phi]=\int dx(-s(x)/\beta)+\epsilon(x))=\int f(x, \beta) dx$, specific form of $ f (x, \beta) $ is given by the Landau theory $\endgroup$
    – Aleksey Druggist
    Commented Mar 11, 2018 at 16:03

1 Answer 1

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(1) is indeed a coarse grained version of (2). There are a number of ways of doing this, for example the renormalization group, or the mean field approximation.

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  • $\begingroup$ My question is why should $E[\phi]$ depend on temperature while $E_j$ do not. @Thomas $\endgroup$
    – SRS
    Commented Mar 11, 2018 at 14:54
  • $\begingroup$ @SRS You are performing the coarse graining at some temperature T, so the resulting functional $E[\phi]$ will naturally depend on T. Indeed, as Landau explains, this is really a free energy functional, not an energy functional. $\endgroup$
    – Thomas
    Commented Mar 11, 2018 at 15:49
  • $\begingroup$ But by coarse-graining, you mean something like the block-spin transformation of the microscopic Hamiltonian of the Ising model (for example). Right? I'm not getting how'll that introduced $T$. Well. Let me give it some more thought. @Thomas $\endgroup$
    – SRS
    Commented Mar 11, 2018 at 15:52

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