3
$\begingroup$

Perform real space renormalization on a discrete lattice model with a finite energy gap. Is it always true that under the flow of coarse-graining, the energy gap will only increase?

I think the argument is intuitively true, since correlation length, which is inverse proportional to the gap, decrease in such a process. But is there any rigorous proof or counterexample that support or deny the argument in general?

Any reference will be appreciated. And feel free to edit my question if it is needed.

$\endgroup$
  • $\begingroup$ Energy of what? $\endgroup$ – GiorgioP Sep 21 '19 at 8:47
  • 1
    $\begingroup$ @GiorgioP Of course the lattice model is equipped with some Hamiltonian. When I mention about energy gap, I am referring to the gap between its ground state and first excited state. $\endgroup$ – SSSSiwei Sep 21 '19 at 17:41
0
$\begingroup$

Let's say you start with some gapped lattice model, which is equipped with two length/energy scales: i) a correlation length $\xi$, and ii) an energy gap $E_G$. Next, you initiate an indefinite coarse graining process. How the scales evolve along this renormalization trajectory depends on which terms in the Hamiltonian are important (in renormalization group language these are the relevant operators). Nevertheless, I suspect, but don't have a rigorous proof for it, that you're guaranteed toeventually hit a fixed point of the renormalization group flow. That is, a point where attempted coarse graining results in an invariant action/Hamiltonian. In other words the system is scale invariant at this point, and there are no scales left in the problem. There are two options (see e.g. Fradkin's Condensed Matter Field Theory book):

  1. $\xi\rightarrow 0$ and $E_G\rightarrow \infty$ (stable fixed point)
  2. $\xi\rightarrow \infty$ and $E_G\rightarrow 0$ (unstable fixed point)

The first type of fixed point represents stable phases of matter, and the second type represents (quantum or thermal) critical systems, i.e. it represents phase transitions. The first type is stable to all local perturbations, but the second one is unstable to at least one local perturbation.

Hence, a common counterexample to your argument would be an initial system that lies in the basin of attraction of an unstable fixed point. This will make the energy gap decrease as the fixed point is approached. If the system is fine tuned, it will also remain at the fixed point under further similar renormalization transformations. If it isn't fine tuned, the presence of a perturbation under which the fixed point is unstable would proceed to open up a gap.

$\endgroup$

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.