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  Peierls Argument for Absence of Long Range Order

+ 1 like - 0 dislike

I'm really confused about the argument in Cardy's book for why there can't be long range order in 1D for discrete models. Let me just copy it out, and hopefully someone can explain it to me.

He takes an Ising-like system as an example. We start with the ground state with all spins up, and we want to see if this state is stable against flipping the spins in some chain of length $l$. This chain has two domain walls at the endpoints, so we get an energy change of $4J$. Then the claim is that there is an entropy of $\log l$ associated with this chain, since "each wall may occupy $O(l)$ positions." If this were true, we would get a free energy change of $4J-\beta^{-1} \log l$, and this would imply that the ground state is unstable to flipping very long chains.

The only part I'm not on board with is the claim about the entropy. I would say that if $L$ is the length of the system, then we have $L$ places to put the chain, so we get an entropy of $\log L$. Certainly as $L\to \infty$ this gives no long range order, as expected.

So, is the entropy $\log l$ or $\log L$?

(Incidentally, I'm perfectly happy with his argument in 2D...)

This post imported from StackExchange Physics at 2015-11-08 10:09 (UTC), posted by SE-user Matthew

asked Aug 4, 2013 in Theoretical Physics by Matthew (320 points) [ revision history ]
edited Nov 8, 2015 by Dilaton
Because the spin chain itself is the entire system that we study, so the length of the chain = the length of the system, i.e. $l=L$, hence $\log l$ and $\log L$ make no difference. The argument is to put domain walls (kinks) on the spin chain, but not to put a spin chain in some 1D space.

This post imported from StackExchange Physics at 2015-11-08 10:09 (UTC), posted by SE-user Everett You

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