What does it mean that after the theory (1-dim Ising model here, but the question is general) is renormalized one time and $g_i\rightarrow g_i'$, that the couplings are weaker, even if the theory is still the same theory, just reformulated?

It is said that it can now be viewed at a theory at high temperatures $-$ the theory flowing with the renormalization group towards a critical point. But what does that mean that it's actually still the same theory? The unrenormalized theory itself gives different results at different temperatures, right?

So how does this fit together?


It means that if you look at every other site the correlation length decreases. This is obvious, it just says that if the decay length of correlations is l, if you look at every other site, you find a decay length of l/2, just because you as skipping every other site. There is no physics here, it's the same system viewed either as a nearest neighbor, or as a next-nearest neighbor interaction.

But if you wanted to adjust the temperature to get a decay length of l/2, a priori, you might not know how to do it. By integrating out the odd-site spins, you find a new Ising model where you know in advance that the correlation length is half as big. So doing the integrating out, and writing it as an Ising model in terms of the every-other-site variables, you learn how to adjust the coupling/temperature to get the correlation length you want.

The 1 dimensional Ising model is far too trivial to be pedagogically useful. This transformation is exact in the 1d Ising model, because 1d is a tree. The analogous thing for the 2d Ising model is nearly as simple, and involves meaningful useful approximations. This is the Migdal Kadanoff method of renormalization from the 1960s, which is still not given enough attention compared with perturbative renormalization techniques, even though it is simpler and more general both.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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