Suppose we have a classical Hamiltonian that can be divided into an “easy” part $H_0$ and a “difficult” part $\Delta H$ that depends on a parameter $g$:

\begin{equation} H = H_0 + g \Delta H ~. \end{equation}

The partition function can be written as

\begin{equation} Z=\sum_\text{states}e^{-\beta H_0}e^{-g\beta \Delta H} ~, \end{equation}

because it's a classical system so $H_0$ and $\Delta H$ commute. The partition function of the easy part,

\begin{equation} Z_0=\sum_\text{states} e^{-\beta H_0} ~, \end{equation}

is supposed to be easy to compute.

A common way to solve this problem is to work perturbatively on $g$. We find

\begin{equation} \begin{aligned} Z &= \sum_\text{states} \Big( e^{-\beta H_0}\Big\{ 1-g\beta \Delta H +\frac{1}{2}\big(g\beta \Delta H\big)^2 +\cdots\Big\} \Big) \\ &= \sum_\text{states} e^{-\beta H_0} \cdot \Big( 1 - g\beta\langle \Delta H \rangle_0 + \frac{g^2\beta^2}{2}\langle\Delta H^2\rangle_0 + \cdots \Big) \end{aligned} \end{equation}

Where $\langle A \rangle_0 \equiv \frac{1}{Z_0} \sum_\text{states} A \, e^{-\beta H_0}$ is the mean value of $A$ using $H_0$.

The problem is that $H_0 \propto N$ because it is an extensive variable, so $e^{-\beta H_0}\, \propto \ e^{N}$ while $\langle \Delta H \rangle_0$ only scales as $N$. When calculating the Helmholtz energy in the thermodynamic limit,

\begin{equation} \begin{aligned} f&= \lim_{N\rightarrow\infty}-\frac{kT}{N}\ln(Z) \\ &= \lim_{N\rightarrow\infty}-\frac{kT}{N}\ln(Z_0)-\frac{kT}{N}\ln\Big( 1 - g\beta\langle \Delta H \rangle_0 + \frac{g^2\beta^2}{2}\langle\Delta H^2\rangle_0 \Big) ~. \end{aligned} \end{equation}

The first term scales as $\frac{1}{N}\ln(e^{N})$ so it is finite, but the second term scales as $\frac{1}{N}\ln(1 + N + N^2)$, so it should vanish.

How do we use perturbation theory in the thermodynamic limit then?

  • 1
    $\begingroup$ Expand the logarithm in powers of $g$ using $\log(1+x) = x - x^2/2 + x^3/3\cdots$. $\endgroup$ Mar 24, 2018 at 2:56
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    $\begingroup$ Notice that the 2nd-order term in the logarithm scales like $N^2$, 3rd-order term like $N^3$, etc, so we cannot conclude that the perturbative correction to the free energy vanishes. In fact, we'll have to organize the series expansion more cleverly even for proving that the perturbative correction is small. $\endgroup$
    – higgsss
    Mar 24, 2018 at 2:56
  • $\begingroup$ @higgsss Could you say a bit more on what does it mean to "organize the series expansion more cleverly"? Thanks! $\endgroup$ Mar 25, 2018 at 16:52
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    $\begingroup$ You could look up "cumulant expansion". I don't know of a general proof that the Helmholtz free energy is an extensive quantity in all orders of perturbation theory, but there should be one. $\endgroup$
    – higgsss
    Mar 25, 2018 at 17:23
  • $\begingroup$ Your first formula for Z is valid only if $\Delta H$ commutes with $H_0$. $\endgroup$ Mar 26, 2018 at 15:06

4 Answers 4


As far as I know, in many-body condensed-matter theory, a simplistic finite-order Taylor expansion is (almost) never used, unless for pedagogical reasons. In fact, one always needs to “organize the series expansion more cleverly” [as higgsss commented] in form of infinite perturbative resummations or asymptotic expansions.

A simple Taylor-expansion, as in $ e^{-x} = 1 - x + \frac{1}{2}x^2 + \mathcal{O}(x^3) $, works only for some single-particle problems in physics.

In words of Ref. [1] (§5.1, p. 195):

The moral to be taken ... is that perturbative expansions should not be confused with rigorous Taylor expansions. Rather they represent asymptotic expansions, in the sense that, for weaker and weaker coupling, a partial resummation of the perturbation series leads to an ever more precise approximation to the exact result. For weak enough coupling the distinction between Taylor expansion and asymptotic expansion becomes academic (at least for physicists). However, for intermediate or strong coupling theories, the asymptotic character of perturbation theory must be kept in mind.

Consult §5.1 of Ref. [1] for an extensive discussion.

In some contexts, eg. in the perturbative renormalization group, one actually “re-exponentiates” the finite-order perturbative expansion; for instance,

$$ Z = \sum_{\text{states}} e^{-\beta (E_0 + g \Delta E)} = Z_0 \, \langle e^{-\beta \, g\Delta E} \rangle_0 \approx Z_0 \, \langle 1 - \beta g \Delta E \rangle_0 \approx Z_0 \, e^{- \beta g \langle \Delta E \rangle} ~, $$

so that the problem observed in the question does not appear at all; see, for instance, §8.2 of Ref. [1].

[1] Altland, A., and B. D. Simons. “Condensed matter field theory”. CUP (2010) [wcat].

  • $\begingroup$ Perturbation theory is not always asymptotic. There are many cases, like the high temperature expansion for the Ising model I discussed in the answer, where the expansion is convergent and one has analyticity. This issue has to do with how you decide to split the full Hamiltonian as $H=H_0+\Delta H$. $\endgroup$ Mar 29, 2018 at 15:18
  • $\begingroup$ @AbdelmalekAbdesselam : Honestly, I have not seen any finite-order perturbative method so far, in (quantum) many-body condensed matter research. Yet, I have certainly to look deeper in your proof -- the problem is merely time. $\endgroup$
    – AlQuemist
    Mar 29, 2018 at 16:50

As you noticed, the first thing to do for the thermodynamic limit is to consider quantities which have a chance of making sense in this limit. The partition function $Z$ is not good for that, but the limit of $\frac{1}{N}\log Z$ (related to the free energy density or the pressure) is indeed the right quantity. Correlation functions also work. Using your notations, one has $$ Z=Z_0 \langle e^{-g\beta\Delta H}\rangle_0 $$ so $$ \log Z=\log Z_0+\log \langle e^{-g\beta\Delta H}\rangle_0 =\log Z_0+\sum_{m=1}^{\infty}\frac{(-g\beta)^m}{m!} \langle \Delta H,\ldots,\Delta H\rangle_0^{\rm T} $$ where $\langle \Delta H,\ldots,\Delta H\rangle_0^{\rm T}$ is the $m$-th joint cumulant of the random variable $\Delta H$ with itself (or simply the $m$-th cumulant of $\Delta H$). The underlying probability measure corresponds to expectations $\langle\cdots\rangle_0$. If you expanded the logarithm as one of the comments suggested, you would have seen, e.g., that the quadratic term would involves the variance $$ \langle \Delta H,\Delta H\rangle_0^{\rm T}=\langle (\Delta H)^2\rangle_0- \langle \Delta H\rangle_0^2\ . $$ One may naively think that, since $\Delta H$ is intensive, the cumulant $$ \langle \Delta H,\ldots,\Delta H\rangle_0^{\rm T} $$ scales like $N^m$, but this is not true. It scales like $N$ for any $m$. There are several ingredients coming together for this property. The first is that a joint cumulant $\langle X_1,\ldots,X_m\rangle^{\rm T}$ is always zero if the random variables $X_1,\ldots,X_m$ can be arranged in two groups which are statistically independent. This forces a "connectedness" property. The second is that a decent $\Delta H$ should have a built-in decay property in the size and spatial extent of the support of the interaction. More concretely, take the Ising model with possibly long-range interactions. Then $H_0=0$ (if you want a less trivial $H_0$ turn on a magnetic field), and $$ \Delta H=\sum_{x,y}J_{x,y} \sigma_x\sigma_y $$ for some translation invariant couplings that satisfy a decay condition when $x,y$ get far apart. In particular, one wants a condition like $$ \sum_y |J_{x,y}|<\infty $$ for $x$ fixed, otherwise $\Delta H$ would not be "extensive". Inserting the definition of $\Delta H$ in the cumulant, you see that you have to sum over a configuration of $m$ edges thrown into the lattice. They must form a connected graph. One basically has to perform a sum $$ \sum_{x_1,y_1}\cdots\sum_{x_m,y_m} J_{x_1,y_1}\cdots J_{x_m,y_m} \langle \sigma_{x_1}\sigma_{y_1},\ldots,\sigma_{x_m}\sigma_{y_m} \rangle_0^{\rm T}\ . $$ This results in an overall factor of $N$ to pick the location of $x_1$ say.Then by translation invariance you can assume that this point is pinned at the origin say. But you have to sum over the $2m-1$ remaining points. This is where the connectedness and summability condition on the $J$'s will save you.

This is a simple case of a very general method called a cluster expansion.

To learn more about this, have a look a the notes I wrote

  1. "Notes on the cluster expansion for the polymer gas, a.k.a., the Mayer expansion."
  2. "Notes on the Brydges-Kennedy-Abdesselam-Rivasseau forest interpolation formula" needed for Lemma 1 in the previous note and for more sophisticated versions of the cluster expansion.

This was from a graduate course I taught a while ago.

  • $\begingroup$ Interesting answer, indeed. Although I am not sure if I understood the proof properly. Furthermore, I think one should avoid any assumptions about the existence or nature of an underlying lattice. The proof must be more general. $\endgroup$
    – AlQuemist
    Mar 28, 2018 at 7:47
  • $\begingroup$ @AlQuemist: This was not a detailed proof but a sketch which hopefully explains the key points. If there are points you could not understand properly, I would be happy to try to clarify them, but you would have to be specific. This kind of proof cannot really be general. The method is general, but then it has to adapted to the specific model at hand, and this requires a nontrivial amount of ingenuity. $\endgroup$ Mar 28, 2018 at 13:46
  • $\begingroup$ One of the main points which I don't see is how to extend such a proof for the systems where there is no underlying lattice, eg. an interacting fluid. $\endgroup$
    – AlQuemist
    Mar 29, 2018 at 9:20
  • $\begingroup$ @AlQuemist: the method works also when there is no lattice, e.g., when $Z$ is the grand canonical partition function of a gas of particles in a box $\Omega\subset\mathbb{R}^n$ subject to two body interactions. $\endgroup$ Mar 29, 2018 at 15:10
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    $\begingroup$ @AlQuemist: It's hard for me to expand my answer, if you don't say anything about what direction I should expand into. That's why I asked earlier about specific points you would like to see clarified. $\endgroup$ Mar 29, 2018 at 17:31

I think the problem arises because you assume that $\langle \Delta H^2\rangle_0$ is in the same order as $\langle \Delta H\rangle_0^2$, i.e in $O(N^2)$. In fact, for any canonical ensemble the energy fluctuation, $$\frac {\langle H^2\rangle}{\langle H\rangle^2}\propto \frac{1}{N},$$ (see, for example in here). Thus, following your perturbation case, $\langle \Delta H^2\rangle_0$ should be in $O(N)$; you can check this at the higher orders (notice that this has some connections to Abdelmalek's answer).

Therefore for any $\left|\frac{g\Delta H}{H_0}\right|\ll1$ the partition function should converge .

  • $\begingroup$ The result you cited is well-known. However, in the derivation, one implicitly assumes that the internal energy (or equivalently, the free energy) is an extensive quantity. OP is asking a more fundamental question: how can we see that the free energy is actually extensive in perturbation theory? $\endgroup$
    – higgsss
    Apr 1, 2018 at 15:10

Every answer was really useful so I will try to put it all together in one answer

As Abdelmalek Abdesselam said, the partition function can we written as

\begin{equation} \tag{1}\label{part} Z=\sum_{\{\sigma\}}e^{-\beta H_0}e^{-g\beta H'}=Z_0 \langle e^{-g\beta H'}\rangle_0 \end{equation}

Then, the free energy is

\begin{equation} -\beta F = \log(Z)=\log(Z_0)+\log\langle e^{-g\beta H'}\rangle_0 \end{equation}

The last term can be expanded in Taylor series as

\begin{equation} \log\langle e^{-g\beta H'}\rangle_0=\log\Big(1-g\beta \langle H'\rangle_0 + \frac{g^2 \beta ^2}{2}\langle H'^2\rangle_0+...\Big) \end{equation}

If now we expand the $\log$ in powers of $g$ we get

\begin{equation} \log\langle e^{-g\beta H'}\rangle_0=-g\beta\langle H'\rangle_0+\frac{g^2 \beta^2 }{2}\Big[ \langle H'^2 \rangle_0-\langle H' \rangle_0 ^2 \Big]+...=\sum_{n=1}^{\infty}\frac{(-g\beta)^n}{n!}C_{H'}^{\ (n)} \end{equation}

Where $C_{H'}^{\ (n)}$ is the n-th momentum of the aleatory variable $H'$. The first one is the mean value, the second one is the variance $\langle \Delta H'^2 \rangle_0$ and so on. The free energy is then

\begin{equation} \tag{2}\label{ener} F = F_0+g\langle H'\rangle_0 - \frac{\beta g^2}{2}\langle \Delta H'^2 \rangle_0 + ... \end{equation}

In equation (\ref{part}) one can easily see that $Z\ \alpha\ e^N$. So $F$ is clearly an extensive quantity. The question is if this holds at every order in perturbation theory. The answer is yes, because every momentum $C_{H'}^{\ (n)} $ is extensive so equation (\ref{ener}) holds to every order. The proof is fairly simple. If one notices

\begin{equation} \frac{1}{\beta^(n-1)} \frac{\partial^n F}{\partial g^n}|_{g=0}=C_{H'}^{\ (n)} \end{equation}

because we've already proved that $F$ is extensive, derivatives respect to $g$ (which is intensive) will remain extensive, so every moment $C_{H'}^{\ (n)}$ is extensive too.

This result make look familiar to those who have taken a course in statistical mechanics. As donnydm pointed out, recall that in the canonical ensemble the mean energy is obtained as

\begin{equation} U=-\frac{\partial}{\partial \beta} \log(Z) \end{equation}

and that the specific heat (an extensive quantity) is

\begin{equation} C_V=\frac{\partial U}{\partial T}=-\beta^2 \frac{\partial}{\partial \beta} U= \beta^2 \frac{\partial^2}{\partial \beta^2} \log(Z) \end{equation}

The derivation can be performed explicitly and the answer is

\begin{equation} C_V=\beta^2 \Big[\langle H^2\rangle -\langle H \rangle ^2 \Big]=\beta^2 \langle \Delta H^2\rangle \end{equation}

The specific heat is proportional to the variance of the energy. This is just an example of the more general result that may be familiar to everybody.


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