Stat. Phys. : Renormalization Group and scaling hypothesis

Nohman
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Hello everyone,

I am currently studying the renorm. group in Stat. physics, more precisely how a rescaling (of space) leaves the partition function unchanged, at the price of having an infinite space of parameters due to the interaction proliferation at each rescaling.

Let K be our infinite parameters vector and R_{b} the application sending K to R_{b}K after a rescaling of a factor b.

Let K_{0} be a fixed point, i.e. such that K_{0}=R_{b}K_{0}.

Now on the problem itself: let K^{(n)}=K_{0} + δ^{(n)} be our parameters vector at the nth rescaling step, where δ^{(n)} is infinitesimal, this implies that (via a Taylor development)

δ^{(n+1)} = J_{b} δ^{(n)} where J_{b} is the jacobian matrix associated to R_{b} evaluated in K_{0} .


Let λ_{i} be the eigenvalues of J_{b} .

We then have the property that, in the associated orthogonal eigenvectors basis the i-th component of δ^{(n+1)} = λ_{i}
δ^{(n)} i-th component.

I would like to prove that λ_{i}= b^{y_i} with {y_i} independent of b (so that the Renormalization group proves the scaling hypothesis).

It is clear that performing two times a rescaling with factor b or one with factor b^2 is equivalent, and therefore δ^{(n+2)}=J_{b^2}δ^{(n)} = (J_b)^2 δ^{(n)}

However I can not manage to get rid of the fact that the eigenvectors of J_{b^2} are not the same as those of J_b. I tried performing a change of basis, but it did not help, so I hope someone out here will be able and willing to : )

Have a nice day,

Nohman
 
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Ok so the answer came to me, from

(J_{b^2} -J_b^2)\delta^{(n)}=0

we deduce, due to the arbitrariness of δ that both operators are equal and therefore share the same eigenvalues, i.e omitting an index

\lambda_{b^2}=\lambda_b^2 \Rightarrow b^{2y(b^2)}= b^{2y(b)}

which implies that y is indeed a constant function.

Thank you to anyone who has thought about this problem
 
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