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The total number of these states is proportional to N, moreover, the energy required to pass to the completely ordered state is

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Statistical Physics 3 October 13th, 2010

Solution 2

With the mean field approximation we find a phase transition in 1D, and that’s not correct (sim- ulations and analytical solution). To see that a qualitative argument can be used: we know that the completely ordered case minimize the energy and we are interested in the states that have the closer possible energy with respect to this minimum. These states are composed of a cluster of spins down (−) and a cluster of spins up (+), as shown in the figure.

The total number of these states is proportional to N, moreover, the energy required to pass to the completely ordered state is

∆E = −2J and the entropy variation is

∆S = −k

B

ln(N − 1)

where J is the interaction energy between two nearest neighbours. Therefore the free energy differ- ence is

∆F ∼ −2J + T k

B

ln(N − 1) and we can conclude that

∆F > 0 ⇐⇒ T > 2J k

B

ln(N − 1) ,

so in the thermodynamic limit (N → ∞) the disordered state is always more favourable than the or- dered one at T > 0.

Considering only the states that minimize the energy frustrations, the magnetization of the system takes the value m = ∑

i

hs

i

i/N. If N → ∞ the magnetization is 0. This result has been obtained considering only a small number of the disordered configurations, but the more favourable ones on the energetic level, for this reason the conclusions can be generalized.

We can use a similar construction introducing a cluster of spins down among the spins up. In this case the difference in energy is 4J.

It’s crucial to understand that this argument works only because the frustration is the same for each size of the cluster of spins down and this is true only in 1D.

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