# Definition of phase transitions in statistical mechanics

Phase transitions in statistical mechanics are usually taught by working through a bunch of examples. I decided to try and think about them from a more "fundamental" point of view - but I've run into a weird little snag, in that I've come up with a system that seems to satisfy the most common definition of a (first order) phase transition, yet it's too trivially simple to really be interesting. Because of this, I'm wondering whether there's a way to define phase transitions in such a way that they include interesting examples such as the two dimensional Ising model, yet exclude the one I present below.

My reason for asking is not that I want to be finicky about definitions for their own sake, but rather because I want to get a handle on what special feature(s) less trivial models have that mine doesn't, which cause them to exhibit interesting effects such as scale free behaviour etc. around the transition point.

My trivial model is as follows. Consider a two-state (classical) system, which can either be in state 0 with energy $E_0 = 0$, or state 1 with energy $E_1 = \lambda \varepsilon$. Here $\varepsilon$ is a fixed value (with dimensions of energy) and $\lambda$ is a dimensionless parameter that I will vary later, representing the size of the system.

We assume the system is in a Boltzmann distribution with inverse temperature $\beta$. The probabilities $p_i$ of the system being in state $i$ are then as follows: $$p_0 = 1/Z \quad\text{and}\quad p_1 = e^{-\beta \varepsilon \lambda}/Z,$$ with the normalisation factor ("partition function") being given by $Z=1+e^{-\beta \varepsilon\lambda}$.

We now consider changing the "scale" of the system, $\lambda$. The expected energy density in the system is given by $$u(\beta) = \frac{\langle E \rangle}{\lambda} = \frac{\varepsilon e^{-\beta\varepsilon\lambda}}{1+e^{-\beta\varepsilon\lambda}}.$$ It is not hard to see that in the limit of infinite $\lambda$ this becomes a step function, with $u=0$ if $\beta>0$ and $u=\varepsilon$ if $\beta<0$.

Similarly, the density of the log partition function (or dimensionless "free energy"), given by $$\frac{\log Z}{\lambda} = \frac{\log(1+e^{-\beta \varepsilon \lambda})}{\lambda},$$ becomes piecewise linear in the limit of large $\lambda$, exhibiting a discontinuity in its first derivative at $\beta=0$. In the limit, $\frac{\log Z}{\lambda} = -\beta \epsilon$ if $\beta<0$, and $0$ otherwise. One can also note that at the transition point, $\beta=0$, the variance of $E$ diverges.

So this system seems to exhibit the same discontinuities as nontrivial models of phase transitions --- yet it's just a simple two-state system. Intuitively this phenomenon is too simple to be called a phase change, so I would like to understand what the defining difference is between this type of behaviour and the non-trivial phase changes that can take place in Ising type models and in physical systems.

Alternatively, I suppose it's possible that some of the "interesting" features of phase transitions are present in my example, but it just takes some clever interpretation to get at them. If this is the case I would greatly appreciate an explanation.

Addendum. The model as described above has the advantage of being the simplest I could come up with, but it has the disadvantage that the transition point is at $\beta=0$, which corresponds to infinite temperature, since $\beta=1/k_BT$. This makes the model seem somewhat pathological. However, a small change to the model can make the transition take place at a finite, positive temperature. Making this change turns out to be quite instructive.

To achieve this we give the higher energy level some degeneracy. That is, there is still only one state with $E_0=0$, but there are now $g$ states with $E_i=\varepsilon$. (If the single state with $E=0$ is aesthetically displeasing, one can also give it a degeneracy; I have chosen not to do this for simplicity.) In this case, as before, $p_0=1/Z$ and $p_i = e^{-\beta\varepsilon\lambda}/Z$ for $i>0$, but now $Z=1+ge^{-\beta\varepsilon\lambda}$. In order to get a transition at a non-zero temperature we must make the additional assumption that the degeneracy scales exponentially with $\lambda$, i.e. $g=e^{a\lambda}$ for some dimensionless constant parameter $a>0$. In this case (as gatsu deduced in his answer), $$u(\beta) = \frac{\varepsilon e^{(a-\beta\varepsilon)\lambda}}{1+e^{(a-\beta\varepsilon)\lambda}}.$$ In the limit of infinite $\lambda$ this has a transition point at $\beta = a/\varepsilon$, with $u$ being zero if $\beta$ is greater than this value, and $\varepsilon$ if it is below it. the "free energy" $\log Z(\beta)$ changes in a similar way, becoming piecewise linear with a discontinuity in its first derivative at $\beta = a/\varepsilon$.

As well as being less pathologically-behaved, this model also feels a bit less trivial, since it has many microstates rather than just two and, as gatsu points out, the transition now arises from a competition between the low energy of state $0$ and the entropy of the degenerate excited states. But nevertheless, as far as I can see, this model lacks such features as power law behaviour and symmetry breaking that are associated with phase transitions in other models. I'm interested in the essential difference between my model and those ones, which gives rise to such non-trivial behaviour around the critical point.

• The problem I see with your model is that the transition inverse temperature is $\beta=0$. Therefore, the transition temperature would be infinite. – Jonas Greitemann Oct 7 '13 at 8:37
• @Jonas well spotted - but that can be changed by making the states degenerate. If there are several states with $E=0$ and fewer (or just one) with $E=\varepsilon\lambda$, then the transition temperature is positive. I didn't explain that in the question because I wanted to present the simplest example. – Nathaniel Oct 7 '13 at 8:42
• @Jonas though it's worth noting that in order for the transition temperature to be independent of the scale parameter, the degeneracy of the states has to increase exponentially with $\lambda$. This might actually provide some kind of clue about how this model relates to less trivial ones. (I'll edit these observations into the question later.) – Nathaniel Oct 7 '13 at 8:45
• For a finite (quantum) system with #microstates <<Avogadro's number, partition function loses its thermodynamical interpretation. So even though the partition function of your system has discontinuity in limit of large $\lambda$, this discontinuity can not be interpreted to mean any "physical" phase transition in the system. Statistical mechanical formulation works well only for "large" systems. – user10001 Oct 7 '13 at 12:47
• @gatsu: I was thinking of first-order phase transitions without symmetry breaking. It is very easy to construct models with such a property. There is actually a whole mathematical apparatus aimed precisely at establishing rigorously the existence of first-order phase transition in the absence of symmetry (by considerably extending the Peierls argument): the Pirogov-Sinai theory. – Yvan Velenik Oct 8 '13 at 13:08

My first comment was more to react to the pathologies arising from the $\beta \rightarrow 0$ limit.

I kind of understand that if instead one considers a case where there is a degeneracy $g = e^{\lambda a}$ with $a > 0$ for a state of energy $\epsilon$ then the rescaled energy becomes:

$u(\beta) = \frac{\varepsilon \: e^{\lambda(a-\beta \varepsilon)}}{1+e^{\lambda(a-\beta \varepsilon)}}$

Taking the limit when $\lambda \rightarrow + \infty$ leads indeed to a case where

$u(\beta) = \varepsilon$ if $\beta \varepsilon < a$

$u(\beta) = 0$ if $\beta \varepsilon > a$

In this particular case, I understand the origin of the transition as there is a balance between entropy gain and energy loss. In the case mentioned in the original question however, there is no reason for the system to be in the high energy state even at infinite temperature.

This kind of transition reminds me of superselective particles one can encounter in drug design for instance.

As an additional note, although the complete symmetry breaking is only observed in the thermodynamic limit, the latter is not required for the high energy state to be more favoured than the low energy one.

In fact, the free energy of the state $u=0$ is simply $F(u=0)=E-TS=0$ while the free energy of the high energy state is simply $F(u=\varepsilon)= \lambda \varepsilon -T k_B \ln e^{\lambda a} = \lambda(\varepsilon- k_B T a)$

Hence, requiring that $F(u=\varepsilon) < F(u=0)$ requires only that $\beta \varepsilon = a$ irrespective of the value of $\lambda$ and this defines the transition for a finite size system.

Of course, as $\lambda$ grows bigger and bigger, the spread between the two macrostate free energies increases and the high energy state becomes overwhelmly more likely than the low energy one.

If one follows the Ehrenfest classification of phase transitions, the above transition is first order as the mean energy is discontinuous at the transition (let's not focus about definition subtleties involving divergences here which are discussed in the previous link if needed).

Now, if one is more interested in the criticality involved in some phase transitions, then he won't probably find it here as they are principally found in second order phase transitions.

One can understand why this is the case as basically a second order phase transition does not display a jump in the order parameter at the transition. Hence, that is the reason why, at the transition, one can use indefinitely the renormalization group (in the sense of Wilson) on the so called critical manifold.

For a first order phase transition however, the order parameter jumps at the transition and hence, using a renormalization scheme will end up in the dominating phase ($u = 0$ or $u=\varepsilon$) however close we are to the transition.

• Many thanks for your helpful comments. I've updated my question with a part about the positive-$T$ transition case, although it's mostly just what you already wrote here. Could you say a little bit more about the relationship to "superselective particles"? I'm not familiar with those. – Nathaniel Oct 8 '13 at 2:13
• Superselective particles are only remotely related to your model but the idea is to design multivalent binders that bind a target cell if the latter expresses a certain concentration of ligand on its surface. Eventually one can get a step-like function for the probability to bind as a function of the coverage in the limit of small binding energy per ligand but with many of them. I don't know if you have access to journals but it is described here. – gatsu Oct 8 '13 at 8:58
• @Nathaniel : It looks like a fermion (with only 2 possible states), in a grand canonical ensemble (assuming you scale the chemical potential too). BTW, You made sign errors in your question ($u(\beta)=0$ if $\beta > \beta_0$) – Trimok Oct 9 '13 at 7:40