# 1D Ising model as indpendent coin tosses : Partition function calculation

My query refers to the Wikipedia article on the one-dimensional Ising model, namely the section "Comments", immediately after "Ising's exact solution".

An Ising model with Hamiltonian $$H = -J \sum \sigma_i \sigma_j$$ is considered. The point is made that the lowest energy state has energy $-JN$ (all spins aligned). It is then added that the energy added by $k$ spin sign changes is $2kJ$. Then it is stated that “as the energy is additive in the number of flips, the probability $p$ of having a spin-flip at each position is independent. The ratio of the probability of finding a flip to the probability of not finding one is the Boltzmann factor” $$\frac{p}{1-p} = e^{-2 \beta J}$$ FIRST QUESTION: I did not find this statement obvious and I wonder how it was arrived at. In the above statement I do not see how the energy is additive in the number of flips (the energy difference from the minimum energy state perhaps).

I can arrive at the same result in two ways, via a direct calculation, see below, but I wonder if there is a quicker way using the independence and energy additivity.

I confirmed the results as follows. First approach: using the spin independence, I write down the partition function for a specific site, $$Z_i = e^{-2\beta kJ} + 1$$ Then, the probabaility of haivng a flip is $p = e^{- \beta 2kJ} / Z_j$, which matches the expression for $p$ sought for. I wonder though why it would be given as a ratio, $p / (1-p)$.

Second approach: The average energy can be found to be $$U = - (N-1) J tanh(\beta J)$$ And this has to be equal to $$- (N-1) J + 2kJ$$ (lowest energy plus energy diffeence linear inhe number of flips) So the average number of flips $k$ equals $$-\frac{N-1}{2N} [tanh (\beta J) + 1]$$ Divinding by $N$ in the large $N$ limit one gets the chance of getting a flip at a specific location $$p = \frac{-tanh (\beta J) + 1}{2 }$$ Which matches the previously given expression for $p$, via the identity $(1-tanh(x))/2 = e^{-2x} / (1 + e^{-2x} )$

SECOND QUESTION Later in the same section the partition function is calculated. The expression is given $$Z = \sum_{configs} e^{\sum_k S_k} = \prod_k (1+p) = (1+p)^N$$ Incidentally, I do not see the equivalence between the last two terms ($N$ is different from $k$). Most importantly, I do not see how this expression could be the partition function. I tried to calculate the partition function myself following this route (focusing on total energy and not energy differences from the lowest energy state, so to be able to compare it with the known partition function expression). I neglect a factor of 2 (coming from summing over both the first spin possiiblities) as it goes to zero when divided by $N$ to get the free energy density. So $$Z = \sum_{configs} e^{-\beta J ( -N + \sum_k S_k)} = e^{J \beta N} \sum_{configs} e^{ -\beta J \sum_k S_k} = e^{\beta J N} \sum_k \binom{N}{k} e^{- 2 \beta J k} = e^{\beta J N } (1 + e^{- 2 \beta J })^N$$

The free energy density $$A = -\frac{1}{\beta N} ln Z = -\frac{1}{\beta } (\beta J + ln (1 + e^{-2 \beta J}))$$ Which matches with the expression

$$A =-\frac{1}{\beta } (ln (e^{ \beta J} + e^{- \beta J}))$$ which is often presented. But the one reported in the quoted paragraph, I cannot make sense of, it seems to give numerically different results. Many thanks as usual

I think the "comments" section is just really, really badly written with inconsistent notation. All the ideas are correct and with the right definitions each equation can be made correct (except for the final equation for free energy, which is missing a sign), but together they're totally inconsistent. What $p$ means changes, what $L$ means changes, the ground state is shifted in the equation for the partition function, which in turn screws up the free energy. It's a mess, but I will try to answer your questions and straighten it out.

# Question 1

## "Energy is additive in what exactly?"

"Spin flip" is an unclear way for the article to refer to it.

Indeed, if "spin flip" refers to sending a single spin to its negative, then the energy above the ground state is NOT proportional to the number of spin flips. But the text reads:

the extra energy is equal to the number of sign changes as you scan the configuration from left to right. If we designate the number of sign changes in a configuration as k, the difference in energy from the lowest energy state is 2k.

and concludes "the energy is additive in the number of flips". This is referring to a different kind of flip. If we consider the ground state (-,-,-,-,-,-,-,-), and send this to (-,-,-,-,+,+,+,+), we only have one sign "flip", and so $k=1$ (I am considering non-looping boundary conditions). You could say that "there is only one flip".

Section 3.1 of this document does a similar computation, but it refers to a boundary of a segment as a "kink" rather than a flip. The following diagram consists of two kinks, and so it is an energy $4J=2kJ$ above the ground state. The computation in that section may be helpful for you too, and it only takes a line or three.

## "Why p/(1-p)?"

I think the author of the section brings it up because it's an easy consequence of a two state system. The partition function in a two state system (ground state 0 and excited state $E$) is $Z=1+e^{-\beta E}$, so the two probabilities are $e^{-\beta E}/Z$ and $1/Z$, so their ratio is clearly $e^{-\beta E}$.

# Question 2

This is where the inconsistencies are the worst. Looking at the wikipedia talk page on this section, one person who authored part of the section writes "So that if you define p=e^{J}, the sum is (1+p)^L, where L is the number of sites." J, L, and P are all inconsistent with the definitions on this page, the sign of the free energy is wrong too. It's totally mutilated. You can piece together what the author meant like this:

First, it is an important fact of statistical mechanics that the partition function of two noninteracting statistically independent systems is the product of their two partition functions.

It is easy to confirm that this rule gives the correct probabilities: if $E_1$ and $E_2$ are the energies of two configurations with partition functions $Z_1$ and $Z_2$, the probability of choosing system one with energy $E_1$ and system two with energy $E_2$ is the product of the probabilities: $$\frac{e^{-\beta E_1}}{Z_2}\frac{e^{-\beta E_1}}{Z_2}=\frac{e^{-\beta(E_1+E_2)}}{(Z_1Z_2)}=\frac{e^{-\beta E_{\text{total}}}}{Z}$$

Alternatively, you can prove this general rule by expanding with binomial coefficients and counting arguments and all that, like you did in your question. That is extra work, though.

Shifting to the point of view where to the left of any site there can either be a kink or not, we have a system of $N$ completely independent systems. If there is no kink, it contributes an energy of $-J$. If there is a kink, it contributes an energy of $+J$. Then the partition functions of the systems are each equal to $e^{-\beta J}+e^{\beta J}$, so the total partition function for $N$ sites is $(e^{-\beta J}+e^{\beta J})^N$, so the free energy density is $-\frac{1}{\beta N}\log(Z)=-\frac{1}{\beta}\log(e^{-\beta J}+e^{\beta J})$. Easy as pie!