Does (spontaneous) symmetry breaking imply long-range order and vice-versa?
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Hey mate ^_^
This Hamiltonian has no internal symmetry when ϵ is non zero. But one can prove that, for any small ϵ>0 and any temperature large enough, one can find a value of h such that there is long-range order...
#Be Brainly❤️
This Hamiltonian has no internal symmetry when ϵ is non zero. But one can prove that, for any small ϵ>0 and any temperature large enough, one can find a value of h such that there is long-range order...
#Be Brainly❤️
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Hello mate here is your answer.
Ok, as a trivial example, consider the following variant of the Ising model with an additional 3-body term: H=−∑i∼jσiσj−h∑iσi−ϵ∑i∼j∼kσiσjσkH=−∑i∼jσiσj−h∑iσi−ϵ∑i∼j∼kσiσjσk, where ∼∼ means "nearest neighbors". This Hamiltonian has no internal symmetry when ϵϵ is non zero. But one can prove that, for any small ϵ>0ϵ>0 and any temperature large enough, one can find a value of hh such that there is long-range order (existence of two Gibbs states with positive, resp. negative, magnetization).
Hope it helps you.
Ok, as a trivial example, consider the following variant of the Ising model with an additional 3-body term: H=−∑i∼jσiσj−h∑iσi−ϵ∑i∼j∼kσiσjσkH=−∑i∼jσiσj−h∑iσi−ϵ∑i∼j∼kσiσjσk, where ∼∼ means "nearest neighbors". This Hamiltonian has no internal symmetry when ϵϵ is non zero. But one can prove that, for any small ϵ>0ϵ>0 and any temperature large enough, one can find a value of hh such that there is long-range order (existence of two Gibbs states with positive, resp. negative, magnetization).
Hope it helps you.
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