While trying to really understanding the partition function in statistical mechanics, I tried looking at it for a 2D ising model, as that's been helpful for me for all kinds of thermodynamic values. So the partition function is

$$ Z=\sum_r \exp(-\beta E_r) $$

As I understand it, it's measure of the number of accessible energy states. When I looked for a plot of it for different temperatures, they are rare to find but usually looked like this:

enter image description here

Note the inverse temperature on the x-axis which I'll call $\beta$. Also $Z$ is named $Q$ there, and that a log is plotted.

This is exactly the opposite of my intuitive understanding. I expected a small value for low T / high $\beta$s. That's also how it's described in many text sources I found.

But the function like in the plot actually works to calculate the Energy for example via $E=-\frac{\partial}{\partial\beta} \ln(Z)$. Also I found at least two sources for that kind of function.

So I'm deeply confused about what's actually correct. As so often, there are multiple sources telling the exact opposite. Maybe it works different for the Ising model? There is a high potential for errors with all the inverses for temperature, difference of using positive or negative energies etc, but I'm mildly confident it's not a simple sign error.


The plot you have for $Z$ looks right. To get some intuition, at low $T$(large $\beta$) the spins are ordered, so the dominant term in the partition function is just $e^{2\beta N J}$, which grows like an exponential with $\beta$. (the Hamiltonian for the Ising model is $H=-\sum_{\langle ij\rangle} \sigma_i\sigma_j$, notice the important minus sign in $H$!)

And actually for 2D Ising model there is a beautiful exact duality relation between the two limits, known as the Kramers-Wannier duality: define $\beta^*$ as $e^{-2\beta^*}=\tanh \beta$ (I'm assuming the coupling in the Ising model is $1$). This maps low $T$ to high $T$. Then


So if $\beta<\beta^*$, then $Z(\beta)<Z(\beta^*)$ since $\sinh$ is a monotonically increasing function.

  • $\begingroup$ So the plot is right, as I thought because the math works out. But I still don't understand the physical meaning of Z then. Most sources are conspicuously vague about that, but for example this (people.cornellcollege.edu/aault/Chemistry/…) says it's 1 for 0K and "high" for high temperatures. That's the opposite, no? $\endgroup$ – Basti Mar 12 '15 at 21:27
  • $\begingroup$ You can easily convince yourself that at least for large $\beta$, the Ising model partition function is increasing with $\beta$: because for low $T$, only the ground state energy contributes (that is because the minus sign in $e^{-\beta E}$, and the ground state energy for the Ising model is a negative number. $\endgroup$ – Meng Cheng Mar 12 '15 at 22:39
  • $\begingroup$ Yes, but isn't that the exact opposite of what I quoted? And also the opposite to descriptions like "ratio of the total number of particles to the number of particles in the ground state" or "number of thermally accessible energy states"? (to quote two different sources) $\endgroup$ – Basti Mar 12 '15 at 22:51
  • $\begingroup$ Well, maybe these quotes have different contexts or make some assumptions. The particular values of partition functions are not very meaningful anyways. I don't know why "number of thermally accessible energy states", it is not just the numbers of states that matter in $Z$, the energy value itself is also important (maybe more important) in determining the value of $Z$. And what we saw in Ising model is that at large $\beta$, the only relevant (accessible) state is the ground state, and for small $\beta$ basically all states contribute equally. These seem perfectly sensible. $\endgroup$ – Meng Cheng Mar 12 '15 at 23:04
  • $\begingroup$ Hmm OK, thanks a lot! Still, I find it surprising that such a fundamental function is rarely even introduced with a physical meaning. And its basic properties seem to be of no interest/ignored/wrong in pretty much all sources. So much is written about entropy, yet so little about this. $\endgroup$ – Basti Mar 12 '15 at 23:13

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