In many places in statistical physics we use the partition function. To me, the explanations of their use are clear, but I wonder what their physical significance is. Can anyone please explain with a good example without too many mathematical complications?
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$\begingroup$ Aside from being a normalization factor, many of its significant features for calculations arise from its likeness to Z and Laplace transforms, thanks to the exponential-with-energy Boltzmann distribution, which is kind of a "co-indidence" in that they wouldn't work with a different distribution. $\endgroup$– Selene RoutleyCommented Sep 1, 2015 at 10:14
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3$\begingroup$ Did you read the "meaning" section in the Wikipedia article? If yes, what doesn't satisfy you about "it encodes how the probabilities are partitioned among the different microstates"? $\endgroup$– ACuriousMind ♦Commented Sep 1, 2015 at 10:40
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1$\begingroup$ Possible duplicate of The unreasonable effectiveness of the partition function $\endgroup$– tparkerCommented Feb 2, 2017 at 20:53
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$\begingroup$ @tparker NOT a duplicate imo, but "complementary": the way that question has been posted is much more precise and attracted more interesting answers, especially this one physics.stackexchange.com/a/174180/226902 $\endgroup$– QuilloCommented Mar 2, 2022 at 10:10
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$\begingroup$ @ACuriousMind, the only interesting bit on Wiki is "the partition function is also equivalent to performing a Laplace transform of the density of states function from the energy domain to the $\beta$ domain, and the inverse Laplace transform of the partition function reclaims the state density function of energies." This says it all, but it may not be intuitively (or physically) super clear. $\endgroup$– QuilloCommented Mar 2, 2022 at 10:13
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3 Answers
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The partition function is a measure of the volume occupied by the system in phase space. Basically, it tells you how many microstates are accessible to your system in a given ensemble. This can be easily seen starting from the microcanonical ensemble.
In the microcanonical ensemble, where every microstate with energy between $E$ and $E+\Delta E$ is equally probable, the partition function is
$$Z_{mc}(N,V,E)= \frac 1 {N! h^{3N}}\int_{E<\mathcal H(\{p,q\})<E+\Delta E} d^{3N}p \ d^{3N} q \tag{1}$$
where the integral is just the hypervolume of the region of phase space where the energy (hamiltonian) $\mathcal H$ of the system is between $E$ and $E+\Delta E$, normalized by $h^{3N}$ to make it dimensionless. The factor $N!^{-1}$ takes into account the fact that by exchanging the "label" on two particles the microstate does not change.
$$S=k_B \log(Z_{mc})\tag{2}$$
tells you that the entropy is proportional to the logarithm of the total number of microstates corresponding to the macrostate of your system, and this number is just $Z_{mc}$.
In the canonical and grand-canonical ensembles the meaning of the partition function remains the same, but since energy is not anymore fixed the expression is going to change.
The canonical partition function is
$$Z_c(N,V,T)= \frac 1 {N! h^{3N}}\int e^{-\beta \mathcal H(\{p,q\})} d^{3N}p \ d^{3N} q\tag{3}$$
In this case, we integrate over all the phase space, but we assign to every point $\{p,q\}=(\mathbf p_1, \dots \mathbf p_N, \mathbf q_1, \dots \mathbf q_N)$ a weight $\exp(-\beta \mathcal H)$, where $\beta=(k_B T)^{-1}$, so that those states with energy much higher than $k_B T$ are less probable. In this case, the connection with thermodynamics is given by
$$-\frac{F}{T}=k_B \log(Z_c)\tag{4}$$
where $F$ is the Helmholtz free energy.
The grand canonical partition function is
$$Z_{gc}(\mu,V,T)=\sum_{N=0}^\infty e^{\beta \mu N} Z_c(N,V,T)\tag{5}$$
where this time we are also summing over all the possible values of the number of particles $N$, weighting each term by $\exp(\beta \mu N)$, where $\mu$ is the chemical potential.
The connection with thermodynamics is given by
$$\frac{PV}{ T} = k_B \log (Z_{gc}\tag{6})$$
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$\begingroup$ How can you relate phase space volume to a normalization factor which appeared in probabilities of canonical distribution? Because partition function in literature is generally defined like this. $\endgroup$– PrabhatCommented Apr 24, 2021 at 11:25
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$\begingroup$ @Prabhat It is both the integral over phase space volume and the normalization factor when you want to calculate averages. For example in the canonical ensemble the probability density is $\rho(p,q)\propto e^{-\beta H}/(h^{3N}N!)$, and in order to insure normalization you want that the integral over phase space of $\rho(p,q)$ is unity. In order to do that, you divide $\rho(p,q)$ by $Z$. $\endgroup$– valerioCommented May 1, 2021 at 14:44
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It's $e^{-F/T}$, where $F/T$ is the free energy normalized by the relevant thermodynamic energy scale, the temperature. The exponential is just a monotonic reparameterization, so morally speaking, the partition function is just the free energy that's available to do useful work.
Another interpretation: if you normalize it so that $E = 0$ is the ground state, then roughly speaking, it's the reciprocal of the "fraction of the system that's in the ground state." Extremely heuristically, let $g$ be the total amount of the system that's in the ground state, $e$ be the total amount of the system that's in an exited state, and $s = g + e$ be the total amount of the system. Then $g/s$ is the fraction of the system that's in the ground state, and its reciprocal is $s/g = (g + e)/g = 1 + e/g$. The Boltzmann weight gives that the relative weight (or "amount") of each excited state $i$ with energy $E_i$ relative to the weight of the ground state is $e^{-\beta E_i}$. Summing over all the excited states $i$, we get the partition function $s/g = 1 + e^{-\beta E_1} + e^{-\beta E_2} + \dots$.
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Partition function physical meaning is the following: It expresses the number of thermally accesible states that a system provides to carriers (e.g. electrons).
