# Summing over spins in Ising model.

1. Nov 23, 2013

### wotanub

"Summing over spins" in Ising model.

I don't understand this concept very well. Is it like taking the energy each possible configuration and adding them together? For example, if we had a 2D lattice gas and wanted to sum over the "white" the spins in a "checkerboard pattern":

What are we doing mathematically? Let's say the case of only nearest neighbor interactions and each spin has two states. I understand that each neighbor interaction between an i and j will contribute a term proportional to σiσj to the energy.

But I'm having trouble understanding what we are doing when we "sum over" the spins. Each "white" has 2 states and each of its neighbors have 2 states, so there are 25 different configurations! How do we know which value to take when we sum?

And it only gets more confusing when you consider that those nearest neighbor spins are connected to their neighbors!

Does anyone have a simple, lucid example of "summing over spins" in 2D? I'm studying the RG and this seems related to "block spins" but I'm having trouble with making the connection.

2. Nov 23, 2013

### Hypersphere

I guess you don't know much statistical physics? Because most of your questions are answered and best described in that language. You might want to look at some textbook in that field, specifically to look up partition functions and "thermal averages".

The point is exactly that we do not know the exact microscopic state of the system, but we (claim that we) know all possible states the system can take. (This is in fact enough to make statistical predictions for the system, that get really accurate for large systems.) Using that information we can sum over all realizations of these states, and build the partition function Z.

Then, as shown e.g. on Wikipedia, we can use Z to calculate any observable for the system (Z serves as the normalization for the resulting probability distribution). You could say, that by summing over the spins and creating the partition function, we are weighting different energy states.

EDIT:
Based on your previous posts, it seems as if you have taken quite a lot of statistical mechanics. Are you fine with the usual calculation of partition function from microstates, and using that partition function to calculate probabilities? Because, in the Ising model, the spin up and down states are all the microstates there are.

3. Nov 23, 2013

### wotanub

I understand the physics of the partition function, but I do not quite understand the math. I'm not sure how to calculate these sums. I always see expressions like

$Z = \sum _{\{\sigma\}} e^{J\sum _{<i,j>} \sigma _i \sigma _j}$

Where {σ} indicates the sum is over all spins and <i,j> indicates nearest neighbors (in 2D for example) and J is just some arbitrary coupling constant.

I don't understand the manipulation to sum over the "checkerboard" in 2D. I picked one of them, it has 4 neighbors, but it's indexed as just <i,j>. There are in fact 4 j's that will contribute to a possible 25 combinations. For the sum in the exponent I could get anything from -4 to 4 in integral steps, some multiple times. But it seems silly that I would write down $e^{4J} \cdots e^{-4J}$ for each spin when trying to simplify the sum. Obviously some things would cancel, and I might even guess that they all cancel by some symmetry... for every positive exponent there's a reciprocal that will multiply by it.

It isn't obvious to me what the partition function looks like after we do this "summing over" the spins. Am I making any sense? I feel like I don't know what to ask or how to explain where my difficulty is coming from.

Last edited: Nov 23, 2013
4. Nov 23, 2013

### atyy

{σ} does not indicate sum over all spins, but instead means sum over all configurations of spin. So for each term in the sum, you choose a particular configuration of spins.

For example, if you have two spins each of which can be either up (1) or down (-1), the possible configurations are {1 1}, {1 -1}, {-1 1}, {-1 -1}.

5. Nov 23, 2013

### Hypersphere

As atyy pointed out, {σ} means that, for each spin in the <i,j> sum, you sum over all possible spin states for both the i and j spins. And the <i,j> symbol denotes that you sum over i and j, but only include nearest neighbors.

Does the sum make sense to you in the simplest case, i.e. 1D?

6. Nov 23, 2013

### atyy

I think it's ok as long as you understand the procedure. The exact solution was found by Onsager, but I think it's basically a magic trick

http://www.nyu.edu/classes/tuckerman/stat.mech/lectures/lecture_26/node2.html

"While the one-dimensional Ising model is a relatively simple problem to solve, the two-dimensional Ising model is highly nontrivial. It was only the pure mathematical genius of Lars Onsager that was able to find an analytical solution to the two-dimensional Ising model."

"This is the nontrivial problem that is worked out in 20 pages in Huang's book." !

Using the RG to get the critical exponents one uses approximations like
http://www.lorentzcenter.nl/lc/web/2010/404/presentations/VvedenskyI_2.pdf
http://math.arizona.edu/~tgk/541/chap3.pdf

Last edited: Nov 23, 2013
7. Nov 24, 2013

### wotanub

I see, the sum is over all configurations. But don't all the spins contribute to the partition function?
It seems like I should raise the entire thing to the Nth power then where N is the number of particles. (A little fishy though since the ones on the extremities should have a slightly different partition function.)

I make a little more sense now.

Yes, 1D is fine but 2D is hard to swallow. But I think I'm starting to get is the partition function I posted only picks one spin and sumes all the configurations associated with it, then we raise to the Nth power to get the full thing. I just need to wrap my head around actually calculating the single particle ones in a sensible way.

I see. I really appreciate the fact that it is not trivial now and am starting to see why so many approximation schemes have been devised. Well, I'll keep at it because I was trying to work this Migdal-Kadanoﬀ RG problem out.