# Partition Function as characteristic function of energy?

I'm going through a book on statistical mechanics and there it says that the partition function $$Z = \sum_{\mu_S} e^{-\beta H(\mu_S)}$$ where $\mu_S$ denotes a microstate of the system and $H(\mu_S)$ the Hamiltonian, is proportional to the characteristic function $\hat p(\beta)$ of the energy probability distribution function. This allows us to make then the next step and conclude that $\ln Z$ is the cumulant generating function with the nice result that $$\langle H \rangle = -\frac{\partial \ln Z}{\partial \beta}$$ and $$\langle H - \langle H \rangle \rangle^2 = \frac{\partial^2 \ln Z}{\partial \beta^2}$$ but I fail to see why $Z$ is proportional to characteristic function. Also, if I imagine that $Z$ is the characteristic function to the energy, then wouldn't I have to evaluate the derivative at $\beta = 0$?

I know that the two formulas above can also be obtained by explicitly doing the calculation using the definition of $Z$ in the first line, but I'd like to generalize this result to the momenta and cumulants of all orders.

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I see the confusion. You can just do a shift of variables, to separate the temperature 1/beta from the the parameter in your characteristic function, which is taken to zero after the derivatives. – user1631 Oct 13 '12 at 1:13

They are not actually proportional, I think it's meant something else. The partition function $Z(\beta)$ is the Laplace Transform (LT) of the density $g(E)$ of states of energy $E$, i.e. $$Z(\beta)=\int_{0}^{\infty}e^{-\beta E}g(E)dE.$$ The partition function $Z$ plays the role of a characteristic function and $g(E)$ the role of probability density function (p.d.f.), that is the p.d.f. is $p(E)=\frac{g(E)}{\int g(E)dE}$.

To be more precise, the partition function $Z$ is a "probability-generating function" like it is the characteristic function ($\varphi_{X}:=\langle e^{itX} \rangle$ and in this case is the fourier transform of the p.d.f.), the moment-generating function $(M_{X}(t):=\langle e^{tX} \rangle)$ and the cumulant-generating function.

The Probability-generating function is defined as \begin{equation*} G(s)=\sum_{n=0}^{\infty}p_{n}s^{n} \end{equation*} where $p_{n}$ is the probability mass and the series is absolutely convergent for $|s|\leq1$. Then \begin{equation*} Z(\beta)=G(e^{-\beta})=\sum_{E}^{\infty}p_{E}e^{-\beta E}. \end{equation*} Note that this can be generalized to $$Z_{H}(\beta)=\int e^{-\beta H(\vec{X})}d\mu(\vec{X}),$$ where the integral is over the space where $\vec{X}$ belongs to (usually phase-space) and $d\mu(\vec{X})$ is a measure on this space (usually $d\mu(\vec{X})=g(\vec{X})d\vec{X}$). It's common to use the assumption that the density is uniform (microcanonical ensemble).

This is a nice question since these probability-generating functions appear in many different areas of physics such statistical mechanics, electromagnetism, quantum field theory and quantum mechanics. All with different names but the same mathematical structure which simplifies calculations of moments and correlations.

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I think the confusion is simple.

Let $\gamma$ be the transform parameter of the generating function.

The generating function is $G(\gamma) = <exp(-\gamma H)> \propto \sum_{\mu_S} e^{-(\beta+\gamma) H(\mu_S)}$ for the Gibbs distribution.

If we take $\gamma ->0$ we get the partition function.

Taking the derivative w.r.t. $\gamma$ is equivalent to taking derivative w.r.t $\beta$ for this particular distribution, since they appear only in their sum.

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The statement is a purely mathematical one.

Let $p(E)$ be the probability distribution function for energy. The characteristic function of this distribution will be $\hat{p}(\beta) = \mathbb{E}[e^{i \beta E}] = \sum_E e^{i \beta E} p(E)$. So if the distribution is the Gibbs one $p(E) \propto e^{-\beta E}$ then we see that $Z$ is proportional to $\hat{p}$. The rest then follows from standard probability theory.

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If it is still actual for you. The probability density $$e^{-\beta H\left( \mu\right) }$$ is the so called canonical (Gibbs) distribution.There are plenty of methods how to derive it. I can reproduce the simplest one.

Let's imagine that your system has the Hamiltonian $H\left( \mu\right)$ and you would like to study it for a certain temperature. In order to set the temperature you put your system inside a thermostat, so that your system exchanges only energy with the thermostat but the volume and the number of particles are constant. Let's suppose that the thermostat is a big tank filled by an ideal gas so that its energy: $$h=\sum_{i=1}^{N}\frac{P_{i}^{2}}{2m}.$$ The total system (your system+thermostat) is isolated thus the total energy is fixed. Therefore, the distribution with respect to the total energy is a delta-function: $$\rho\left( E\right) =\Lambda\delta\left( h+H-E\right) ,$$ where $\Lambda$ is a some normalization factor such that $$\int \rho\left( E\right)\,d\Gamma =1,\qquad\left( 1\right)$$ where $d\Gamma$ is an element of the full phase space: $$d\Gamma=d\mu\prod_{i=1}^{N}d^{3}P_{i}\,d^{3}Q_{i}.$$ Let's integrate out all degrees of freedom of the thermostat: $$\rho\left( H\right) =\int\prod_{i=1}^{N}d^{3}P_{i}\,d^{3}Q_{i} \,\Lambda\delta\left( H+\sum_{i=1}^{N}\frac{P_{i}^{2}}{2m}-E\right) =\Lambda V^{N}\int\prod_{i=1}^{N}d^{3}P_{i}\,\,\delta\left( H+\sum_{i=1}^{N} \frac{P_{i}^{2}}{2m}-E\right) ,$$ where $V$ is the volume of thermostat. The integration measure can be simplified as follows: $$\int\left[\prod_{i=1}^{N}d^{3}P_{i}\right] f(\epsilon)=\frac{2\pi^{3N/2}}{\Gamma\left( 3N/2\right) }\int\left[\epsilon^{3N-1}d\epsilon\right] f(\epsilon),$$ where $$\epsilon^{2}=\sum_{i=1}^{N}P_{i}^{2}.$$ Hence the integration can be preformed as follows: $$\rho\left( H\right) =\Lambda V^{N}\frac{2\pi^{3N/2}}{\Gamma\left( 3N/2\right) }\int\,d\epsilon\,\epsilon^{3N-1}\,\delta\left( H+\frac {\epsilon^{2}}{2m}-E\right) =\Lambda V^{N}\frac{2m\pi^{3N/2}}{\Gamma\left( 3N/2\right) }\,\left( E-H\right) ^{\frac{3N}{2}-1}.$$ Let's now consider the $N\rightarrow\infty$ limit so that $$\frac{E}{N}\approx\frac{h}{N}=\frac{3T}{2}.$$ The distribution takes the form: $$\rho\left( H\right) \sim\left( 1-\frac{H}{E}\right) ^{3N-2}\approx\left( 1-\frac{H}{\frac{3N}{2}T}\right) ^{\frac{3N}{2}-1}\approx\exp\left( -\frac{H}{T}\right) .$$ The normalization factor can be found from the normalization condition (1). Finally the probability density for the energy of your system takes the form: $$\rho\left( H\right) =\frac{e^{-\beta H\left( \mu\right) }}{Z},\quad Z=\int d\mu\,e^{-\beta H\left( \mu\right) }.$$ In fact, the result is independent of the nature of the thermostat, see e.g. L.D. Landau, E.M. Lifshitz, Volume 5 of Course of Theoretical Physics, Statistical physics Part 1 Ch. III, The Gibbs distribution.

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This is a very nice derivation of the canonical ensemble, but your answer has nothing to do with the question! – Mark Mitchison Oct 11 '13 at 13:36