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

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The ground state of the one-dimensional spin-1/2 Ising model is characterized by an ordered phase, where all spins align either up or down. Deriving this from the Hamiltonian involves minimizing the energy function, which can be expressed as the sum of interactions between neighboring spins. The initial approach of taking the derivative with respect to the spin variable leads to a trivial solution, highlighting the need for a different method. The correct approach focuses on minimizing the Hamiltonian directly, leading to the conclusion that the lowest energy configuration occurs when all spins are aligned. This confirms that the ground state is indeed the fully ordered phase.
William Crawford
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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
$$ \min_{\lbrace s_n\rbrace}\mathcal{H} = \sum_n\min\left(-Js_ns_{n+1}\right). $$