Through the RG method, one can obtain the hyperscaling relation between the critical exponents of classical second-order phase transitions: \begin{equation} 2-\alpha=\nu d \end{equation} In the case of a quantum phase transition, however, the dimensionality $d$ is substituted by an effective one $d_{eff}=d+z$, with $z$ being the dynamic critical exponent. In many cases $z=1$, consistently with the idea that quantum, $d$-dimensional systems can be mapped onto classical, $(d+1)$-dimensional ones.

What I'm looking for is a justification for the presence of the effective dimensionality $d_{eff}$ inside the RG theory. I could only find one reference that treats this, which is the first chapter of Mucio Continentino's book "Quantum Scaling in Many-Body Systems", but I found it a bit sloppy. From reading that chapter it seems that the reason why $z$ comes out is that at $T=0$ the free energy is equal to the Hamiltonian and thus $J$ can be "factored out". Since $J'=b^{-z}J$ under an RG step then an extra $z$ is added to $d$ (eqs (1.4) to (1.7)). This explanation, however, doesn't give much physical insight in my opinion and I don't understand what would happen for quantum systems at non-zero temperature.

Please note that my background in quantum field theory and path integral is limited, what I know of RG comes from studying Goldenfeld's treatment and chapter 3 of Ortiz-Nishimori's.


1 Answer 1


I've thought about this for a while more and came up with what may be an answer.

The fact that $J$ can be "factored out" reflects the fact that at $T=0$ there is only one energy scale and it is determined by $J$. Because of this, we expect that the free energy per degree of freedom scales not only with the dimensionality but also with the dynamic exponent $z$. As soon as $T\neq0$ there is a competition between energy scales, namely $J$ and $k_B T$ and we cannot say that the free energy scales with $z$ as well as with $d$.

Of what follows I'm not sure, but I would say that as soon as $T\neq 0$ the system is governed by a finite-temperature fixed point with critical exponent different than those related to the $T=0$ fixed point and for which a "classical" hyperscaling relation holds.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.