# 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

## 3 Answers

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.

See: http://en.wikipedia.org/wiki/Characteristic_function_(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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