Stat. Phys. : Renormalization Group and scaling hypothesis

  • Thread starter 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[itex]_{b}[/itex] the application sending K to R[itex]_{b}[/itex]K after a rescaling of a factor b.

Let K[itex]_{0}[/itex] be a fixed point, i.e. such that K[itex]_{0}[/itex]=R[itex]_{b}[/itex]K[itex]_{0}[/itex].

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

δ[itex]^{(n+1)}[/itex] = J[itex]_{b}[/itex] δ[itex]^{(n)}[/itex] where J[itex]_{b}[/itex] is the jacobian matrix associated to R[itex]_{b}[/itex] evaluated in K[itex]_{0}[/itex] .


Let λ[itex]_{i}[/itex] be the eigenvalues of J[itex]_{b}[/itex] .

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

I would like to prove that λ[itex]_{i}[/itex]= b[itex]^{y_i}[/itex] with [itex]{y_i}[/itex] 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 [itex]b^2[/itex] is equivalent, and therefore δ[itex]^{(n+2)}[/itex]=[itex]J_{b^2}δ^{(n)}[/itex] = [itex](J_b)^2 δ^{(n)}[/itex]

However I can not manage to get rid of the fact that the eigenvectors of [itex]J_{b^2}[/itex] are not the same as those of [itex]J_b[/itex]. 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
 

Answers and Replies

  • #2
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Ok so the answer came to me, from

[itex] (J_{b^2} -J_b^2)\delta^{(n)}=0 [/itex]

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

[itex] \lambda_{b^2}=\lambda_b^2 \Rightarrow b^{2y(b^2)}= b^{2y(b)}[/itex]

which implies that y is indeed a constant function.

Thank you to anyone who has thought about this problem
 

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