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

Gold Member

## Homework Statement

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.

## Homework 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.

## 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:

Orodruin
Staff Emeritus
Homework Helper
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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?

• fluidistic
Gold Member
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:
Orodruin
Staff Emeritus
$$\sum_{\tau_1, \sigma_1 = \pm} \sum_{{\rm states}\ > 2} \tau_1 \sigma_1 \prod_{k = 1}^N \exp(\beta J \sigma_k \tau_k) = \sum_{\tau_1, \sigma_1 = \pm} \tau_1 \sigma_1 \exp(\beta J \sigma_1 \tau_1) \sum_{{\rm states}\ i > 2}\prod_{k = 2}^N \exp(\beta J \sigma_k \tau_k).$$