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Consider a 1d Ising model with no external magnetic field $(h=0)$ and adopt a decimation transformation in which every other spin is traced out.

So the Hamiltonian $H$ is given by $$H = -J\sum_{(i,j)} s_i s_j$$ where $(i,j)$ corresponds to the nearest neighbors of $i$.

I am trying to sketch the RG flow of this system, and I am struggling. How do you start a problem like this? I know you coarse grain a system by some spin-block tranformation $\tau$, but I am unsure of how to use it.

The end goal is to derive the RG equation, so the map $K' = R_l[K]$ for this transformation and find the fixed points in the flow.

Any advice would be appreciated!

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Take the partition function $$ Z = \sum_{\{ S_{j} \}} \exp\left( \beta J \sum_{( i,j )} s_i s_j \right) = \sum_{\{ s_{j} \}} \prod_{(i,j)} \exp\left( \beta J s_i s_j \right) \ , $$ and sum over every other spin (so explicitly evaluate $\{s_{2n+1}\} \in \pm 1$ in the sum, but don't sum over $s_{2n}$).

This should be equal to a similar partition function over the coarse-grained lattice (with a new coupling $J'$ and new spin variables $s_i'$) $$ Z = \sum_{\{ S_{j} \}} \exp\left( \beta J' \sum_{( i,j )} s'_i s'_j \right) = \sum_{\{ s_{j} \}} \prod_{(i,j)} \exp\left( \beta J' s'_i s'_j \right) \ . $$ By comparing term-by-term, you can relate $J$ to $J'$.

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  • $\begingroup$ Thank you for your answer @QuantumEyeda! Why did you elect to sum over every other spin? $\endgroup$
    – megamence
    Commented Nov 23, 2020 at 22:14
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    $\begingroup$ @megamence Summing over every other spin has precisely the same meaning as "every other spin is traced out". $\endgroup$ Commented Nov 23, 2020 at 23:01

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