# 1-D Ising's model without B field, stat. mech., correlation

1. Dec 16, 2014

### fluidistic

1. The problem statement, all variables and given/known data
Hi guys,
I'm going nuts on a problem. Out of memory the problem was a system of N pairs of atoms that interact between each others so that the Hamiltonian of the whole systeme is $H=-J \sum _{n=1}^N \sigma _i \tau _i$ where the only possible values of the $\tau _i$'s and $\sigma _i$'s are -1 and 1.
1)Calculate $\langle \sigma _i \tau _i \rangle$ (I think this is called the correlation function, since it seems to be the correlation between the spins of 2 adjacent pair of atoms or so).
2)Calculate $\langle \varepsilon \rangle$. The problem goes on but I don't remember the next questions.

2. Relevant equations
Mean value of A: $\frac{\sum _{\text{over all states}} \exp (-\beta H) \cdot A }{Z}$ where Z is the partition function of the system.

3. The attempt at a solution
I am stuck on part 1). I know the answer is $\tanh (\beta J)$ but I'm unable to show it.
First I notice that the partition function $Z=Z_1^N$, in other words it is the partition function of a single pair of spins to the N'th power.
Where Z_1=$\sum _{\text{over all states}} \exp (-\beta H)=\sum _{\text{over all states}} \exp (\beta J \sum _{i=1}^N \sigma _i \tau _i )$. Stuck here, I don't know how to write out these sums.
Now maybe the numerator of $\frac{\sum _{\text{over all states}} \exp (-\beta H) \cdot \sigma _i \tau _i }{Z}$ is worth the derivative of the denominator with respect to $\beta J$ and if Z is worth $\cosh (\beta J)$ then I would be done with the calculations but I am really not able to show that Z is indeed worth cosh(beta J). Also I am not even sure that this is true because I have that Z is Z_1 to the N'th power and I see no way how to obtain a hyperbolic cosine.

Thanks for any tip.

Edit: Ok I considered N=3, I know how to write the terms of the sum for the partition function but there are already a lot of them. I really don't see how to reach the result.

Last edited: Dec 16, 2014
2. Dec 17, 2014

### Orodruin

Staff Emeritus
Suggestion: Start by noticing that the exponential is the exponential of a sum. Also note that the summands can be factorised into factors which are dependent only on the state at one particular site at a time - what does this tell you about the sums?

3. Dec 17, 2014

### fluidistic

Ok thanks. I am not sure I am understanding exactly what you're saying, let's see if I did:
$$\langle \tau_i \sigma_i \rangle =\frac{ \sum _{\text{states}} \sigma _i \tau_i \exp (\beta J \sum _{i=1}^N \sigma_i \tau_i ) }{ \sum _{\text{states}} \exp (\beta J \sum _{i=1}^N \sigma_i \tau_i ) }$$. (75)

Where $\exp (\beta J \sum _{i=1}^N \sigma_i \tau_i )$ is worth $\prod _{i=1}^N \exp ( \beta J \sigma_i \tau_i ) = \exp ( \beta J \sigma_i \tau_i ) \prod _{i=1}^{N-1} \exp ( \beta J \sigma_i \tau_i )$.
Replacing this into the expression (75), I get that it's worth $$\langle \tau_i \sigma_i \rangle =\frac{ \sum _{\text{states}} \sigma _i \tau_i \exp (\beta J \sigma_i \tau_i ) }{ \sum _{\text{states}} \exp (\beta J \sigma_i \tau_i ) }$$.
Now I notice that the numerator is the derivative of the denominator with respect to $\beta J$ and that the denominator is indeed worth $4 \cosh (\beta J)$.
Hence $\langle \tau_i \sigma_i \rangle =\tanh (\beta J)$.
Does this seem correct?

Last edited: Dec 17, 2014
4. Dec 17, 2014

### Orodruin

Staff Emeritus
I think you essentially have the basic details cut out. The thing that can be improved is the mathematical reasoning in the middle steps.

The sum over all states can be written as several sums over the states of each individual site, i.e., $\sum_{\rm states} = \sum_{\tau_1, \sigma_1 = \pm} \ldots \sum_{\tau_1, \sigma_1 = \pm}$. Now take $\langle \tau_1 \sigma_1\rangle$ (without loss of generality, all sites are equivalent) and denote $\sum_{\rm states} \equiv \sum_{\tau_1, \sigma_1 = \pm} \sum_{{\rm states}\ i > 2}$. The numerator becomes
$$\sum_{\tau_1, \sigma_1 = \pm} \sum_{{\rm states}\ > 2} \tau_1 \sigma_1 \prod_{k = 1}^N \exp(\beta J \sigma_k \tau_k) = \sum_{\tau_1, \sigma_1 = \pm} \tau_1 \sigma_1 \exp(\beta J \sigma_1 \tau_1) \sum_{{\rm states}\ i > 2}\prod_{k = 2}^N \exp(\beta J \sigma_k \tau_k).$$
I am sure you can find out how to do a similar thing for the denominator and draw a conclusion from there.