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I am studying the Ising model in 1D, in the absence of magnetic interaction but in presence of an external magnetic field. The Hamiltonian for an Ising chain with $n$ sites is hence described by $$H = \sum_{j=1}^{n} \sigma_j,$$ where $\sigma$ can take the values $\sigma = 1,0$. Once the partition function $Z$ is calculated one can answer questions on the average magnetization, etc. One instead might wonder on the probability to have all spins down ($\sigma = 0$) and only one up. Let us denominate one such state, i.e. all spins down but a certain one up, as $\hat{\sigma}$. Its energy equals unity, so $$Pr(\hat{\sigma}) = \frac{1}{Z} \exp(-\beta ).$$ If one is not interested in this state, with a specific spin up, but simply wonders about the probability of having only one spin up, regardless of its position, and given that there are $n$ such configurations should not the probability be $$Pr(any\ spin \ up, remaining \ down)= n \frac{1}{Z} \exp(-\beta )$$ ? I might just get confused by “degenerate” states, i.e. states with the same energy. Indeed all the states with any spin up and all the rest down have the same energy, but as they are specifically counted as different states in the partition function, so my reasoning should apply.

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That is exactly right. I would say "exactly one spin up" rather than "any spin up", but that is just phrasing.

Another way of looking at this is that there is an increase in entropy (of $\log n$ times Boltzman's constant) associated with flipping one spin, which will sometimes overcome the energy cost.

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