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I'm reading Shang-keng Ma's book Modern theory of critical phenomena, and I'm a bit confused as to how the real-space RG transformations are defined. Ma basically says that these transformations are defined for any integer $s\ge 1$, $\mu^{\prime} = {\rm R}_s \mu$, in the parameter space to represent the transformation of probability distribution ${\rm P} \propto e^{-\cal{H}}$ to probability distribution ${\rm P}^{\prime} \propto e^{-\cal{H}^{\prime}}$, according to the rule: $$e^{-\cal{H}^{\prime}[\sigma]} = \int e^{-\cal{H}[\sigma^{\prime\prime}]} \prod_{x^{\prime}} \delta (\lambda_s\sigma_{x^{\prime}} - s^{-d}{\sum_{y}}^{x}\sigma^{\prime\prime}_y)\prod_y d\sigma^{\prime\prime}_y$$ where $x^{\prime} = x/s$, $\lambda_s > 0$ is real, and $${\sum_{y}}^{x}$$ represents a sum over the position vectors $y$ of $s$ blocks that form a big block centered at $x$. (The system is a cubic spin lattice in $d$ dimensions.) Then, he goes on to say that $${<\sigma_x>}_{\rm P} = \lambda_s {<\sigma_{x^{\prime}}>}_{{\rm P}^{\prime}}$$ and I don't see how this relation can be proven. I would very much appreciate a detailed proof. Frankly speaking, I don't even know how to interpret the spins $\sigma_x$ and $\sigma_{x^{\prime}}$, i.e. how they correspond to each other, since $\sigma_x$ is a variable of the initial block Hamiltonian $\cal{H}$, whereas $\sigma_{x^{\prime}}$ is a variable of the renormalized Hamiltonian after one RG transformation ${\cal{H}}^{\prime}$.

Also, how can one rigorously prove that ${\rm R}_s{\rm R}_{s^{\prime}} = {\rm R}_{ss^{\prime}}$ (if $\lambda_s = s^a$)? It seems that Ma's notation is a bit misleading since $\cal{H}^{\prime}[\sigma]$ also depends on $\lambda_s$.

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It seems the author is implicitly using translation invariance so that $$ \langle \sigma_x\rangle_P=\langle s^{-d}\sum_{y}{}^x\ \sigma_y\rangle_{P}\ . $$ Then insert $1=\int \prod_{x'} \lambda_{s} \delta(\lambda_s\sigma_{x'}-s^{-d}\sum_{y}{}^x \sigma_{y}'') d\sigma_{x'}$ in the numerator and denominator of Gibbs expectations, and finally integrate over the original spin field $\sigma$ or $\sigma''$.

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  • $\begingroup$ Thank you very much! Your idea works just fine. I would very much appreciate it if you could prove the second part, namely ${\rm R}_s {\rm R}_{s^{\prime}} = {\rm R}_{ss^{\prime}}$ if ${\lambda}_s {\lambda}_{s^{\prime}} = {\lambda}_{ss^{\prime}}$ starting from the definition of ${\rm R}_s$ transformation, namely $e^{-{\cal H}^{\prime}[\lambda]} = \ldots$. $\endgroup$ Commented Dec 1, 2015 at 3:12

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