Ground state of the one-dimensional spin-1/2 Ising model

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SUMMARY

The ground state of the one-dimensional spin-1/2 Ising model is definitively characterized by an ordered phase, where all spins are either up or down. The Hamiltonian for this model is given by $$ \mathcal{H} = -J\sum_n s_{n}s_{n+1} $$, with the spin variable $$ s_n $$ taking values of ±1. The low energy solution can be derived by minimizing the Hamiltonian directly, leading to the conclusion that the minimum energy configuration occurs when all spins are aligned, rather than alternating. This approach circumvents the need to compute the partition function.

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  • Understanding of the Ising model and its significance in statistical mechanics.
  • Familiarity with Hamiltonian mechanics and energy minimization techniques.
  • Knowledge of spin variables and their representation in quantum mechanics.
  • Basic proficiency in mathematical derivation and solving difference equations.
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William Crawford
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TL;DR
How to derive the low energy ground state for the one-dimensional spin-1/2 Ising model on either a periodic or an infinite chain.
Hi,

I know that the ground state of the spin-1/2 Ising model is the ordered phase (either all spin up or all spin down). But how do I actually go about deriving this from say the one-dimensional spin hamiltonian itself, without having to solve system i.e. finding the partition function? $$ \mathcal{H} = -J\sum_n s_{n}s_{n+1}, \qquad s_n=\pm1 $$ I've tried computing the derivative of ## \mathcal{H} ## w.r.t. the spin variable ## s_i ##, but this leaves me with the trivial difference equation ## s_n + s_{n+1} = 0 ## yielding the high energy solution ## s_n = (-1)^ns_0 ## and not the low energy solution that I was searching for (assuming ##J>0##).
 
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Never mind, I've solved it myself. Simply using that
$$ \min_{\lbrace s_n\rbrace}\mathcal{H} = \sum_n\min\left(-Js_ns_{n+1}\right). $$
 

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