# Stat. Phys. : Renormalization Group and scaling hypothesis

1. Jan 29, 2012

### Nohman

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

2. Jan 30, 2012

### Nohman

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.