# Why are strongly-correlated systems treated with equilibrium conditions?

1. Dec 25, 2007

### Gerenuk

Remembering the derivation for the Boltzmann distribution it seems to me, that it assumes that energy of a system in contact with a reservoir spreads over all degrees of freedom equally likely (sort of "structure-less"). Now I imagine correlated systems have special behaviours and reactions or tendencies, so that a statistical approach like statistical mechanics under equilibrium conditions (i.e. defineable temperature) isn't justified. Or is it?

2. Dec 25, 2007

### Gokul43201

Staff Emeritus
Why do you expect that an interacting system will not be able to reach equilibrium?

What is true is that you have to deal with interacting systems differently than you deal with non-interacting systems. Introductory Stat Mech courses deal only with non-interacting systems, and develop Boltzmann, Fermi-dirac or Bose-Einstein statistics in this framework. When dealing with interacting systems, things get tougher. The many-body Hamiltonian is no longer just a product of single-particle states. The interaction terms affect the partition function, and add a level of complexity to the problem.

For many weakly or moderately interacting systems, there are perturbative, field theoretic and other approaches to extracting different statistical quantities. In some of these systems (like the Fermi liquid, for instance) interactions can be magicked away by the wave of a wand, and the system can be pictured as containing essentially non-interacting creatures (called quasiparticles).

Last edited: Dec 25, 2007
3. Dec 25, 2007

### Gerenuk

So in the general case not even the complete state functions needs to obey any sort of statmech? Then there is no justification for defining temperature?!

I was surprised when I started reading about Green functions and the stuff. They speak about strongly correlated systems, but use the concept of temperature.

Is there a special derivation for interacting Hamiltonians to admit temperature?

Would you say that strongly correlated systems still statisfy
P(TotalState=i)=exp(-beta*E_i)
???

4. Dec 26, 2007

### Modey3

Hello Gerenuk,

When systems are strongly correlated such as an electron gas (eg The Jellium Model). The state distribution of the electrons will be dependent on temperature not though Fermi-Dirac statistics, but some other distribution function due to correlation effects.

Modey3

5. Dec 28, 2007

### kanato

Temperature is a well defined quantity for virtually any macroscopic system. Take a chunk of manganese oxide and set it on your countertop; it will eventually come to thermal equlibrium with the environment.

From a microscopic stat mech point of view, the canonical partition function is calculated via
$$Z = \sum_\psi \langle \psi | e^{-\beta H} | \psi \rangle$$
where the sum is carried out over a complete set of many-body wavefunctions, regardless of whether the system is interacting or not. For a non-interacting system, the Hamiltonian is generally writable as a sum of terms which are all single-particle terms. In that case, this greatly simplifies the calculation of the partition function as it will be just the single particle partition function to the power N.

6. Dec 28, 2007

### Gerenuk

Well, that's the statement I was questioning, but it's not an explanation why it should be true for all system however general.

I can't think of a very good example, but let's take that one: A container with water in one of two compartments. If I apply statmech without thinking, I'd conclude that the water should spread into both compartments? Why doesn't it do like so? Because there is a wall in between. So what if I have a general system with an abstract wall (i.e. an interaction that not "random"). The system could remain in a non-equilibrium state forever.

7. Dec 28, 2007

### Gokul43201

Staff Emeritus
Oh why bother with two compartments? Take a container with water. If you apply stat mech without thinking, the water should spread out of the container and flow along the floor. Why doesn't it do so? Because the container has walls.

See the point?

Moral: Do not do stat mech without thinking!

8. Dec 29, 2007

### Gerenuk

I thought of a similar example, but I wasn't sure that if you include gravity it still might work.

Anyway, if you need thinking, so how can you exclude that a not understood correlated system needs some thinking and can't have statmech applied to it?

Statmech really only applies under special circumstances, which are IMHO satisfied only for "completely random" systems.

9. Dec 29, 2007

### Gokul43201

Staff Emeritus
What you are saying is that your partitioned container has two isolated sections. I had imagined it with a partition that went up only to the liquid level. But if the container is completely partitioned, then each partition becomes an isolated system and knows nothing of the other one. So there is no real problem there once you recognize this.

In any case, I don't understand the relevance of the boundaries on role of interactions. I think perhaps you chose a poor example to illustrate your question.

Now, I'm not saying that a temperature can be defined for any interacting system. I believe you are correct that this is not true in general, but not that this is untrue in the case of every interacting system. What is true is that temperature can be defined for many types of interacting sytems. It is only those systems that are specifically dealt with using tools of equilibrium thermodynamics. Interacting systems that do not equilibrate must be studied using non-equilibrium techniques.

Last edited: Dec 29, 2007
10. Dec 29, 2007

### kanato

The reason your two containers of water don't reach equilibrium is because of the wall - the containers are not allowed to exchange particles even though they have different chemical potentials. For this to work as an analogy to temperature, you would have to have the two systems thermally isolated from each other so they could not exchange energy.

Stongly correlated systems have been studied quite a bit experimentally. If they didn't reach thermal equilibrium with their environment, I would think this would have been noticed, as it would have huge implications.

There are interactions which break stat mech; specifically, a long range interaction like gravity is problematic for stat mech because the potential energy from it is not extensive. But for systems of charges, as long as the system has no net charge, then the electric interaction is not enough to break extensivity.

There are certainly systems which exhibit long range correlations that are treated just fine with stat mech. Take a ferromagnet, for instance. Certainly some care has to be taken when interpreting what the ground state and low-lying excitations are, but that does not mean stat mech is useless in this case. A magnet certainly reaches thermal equilibrium with its environment.

A diatomic molecule has very strong interactions between the two atoms that form the molecule. Yet in spite of the strong interaction, stat mech applies very well to a diatomic gas.

So I don't understand why you think thermal equilibrium wouldn't apply to strongly correlated systems? Strong correlation doesn't mean strong interactions... by "strong interaction" I mean that the potential energy of the interactions is overwhelming. Strong correlation means (rougly speaking) that the energy scales of different types of energy come in at about the same energy scale, so that different types of order or disorder can end up competing. This is why phase diagrams for strongly correlated systems can be quite rich, because minor changes in pressure or concentration affect which phase "wins" (see the phase diagram for Pu, for instance).

11. Dec 29, 2007

### Gerenuk

I didn't say (or didn't mean to say) that temperature is useless. I just meant that there is no 100% justification in applying it to any arbitrary quantum mechanical system.

But we agree on this point.

Now I wonder which other more general concepts of statmech exist to deal with "party non-equilibrium systems". I heard about extended entropies (Tsallis, ...). Any other hints welcome!

12. Dec 29, 2007

### Gerenuk

For the same reason quantum mechanical systems might not reach equilibrium because some system specific process ("wall") forbids it.

Of course some systems can have a well defined temperature. I expect that many don't. Probably even already examined ones, since theoretical predictions in this area are indeed poor (in the sense of "far from complete"). So one cannot be sure for statmech to be right.

Is there a general proof that extensivity guarantees the full correctness of statmech? That would be interesting.

Because by applying some knowledge you know where to apply statmech and where not. In correlated systems its not clear to which part, if any, statmech might have a relevance.

Maybe that's a common definition based on observations. I can't think of a detailed formulation, but I prefer one based on equal probability distribution in phase space or something along these lines.

13. Dec 30, 2007

### Gokul43201

Staff Emeritus
Gerenuk,

We can always use stat mech to study a system. If the system is in equilibrium, we use equilibrium stat mech. Else, we must use techniques of nonequilibrium stat mech.

14. Dec 30, 2007

### Gerenuk

My feeling is that for example mere contact with a reservoir doesn't need to be enough. In fact you can't know when you have equilibrium, when you don't know the final solution. Statmech doesn't come from nowhere and it has well defined assumptions, which are not given in all systems.

Maybe I should have a look at non-equi statmech. Can you recommend a good book?