Renormalization of sine gordon theory

Question

So assume that we have a usual sine gordon theory in the the theory we have a term in the hamiltonian $$\frac{yu}{2\pi\alpha^2}\int dx \cos(\sqrt{8}\phi_\sigma(x))$$

where $\alpha$ is cut off parameter, and $y$ and $u$ are coupling constants, $\phi_\sigma(x)$ is a bosonic field operator.

So it says that, if $$y\rightarrow \pm \infty$$ $\phi$ locked into one of the minima of $\cos$ and for very large $y$ one can expand $\cos$ around the $\phi$ which minimizes the $\cos$ and we can only do this iff $y$ is very large.

I don't get it, how and why $\phi$ locks it self to minimum for large $y$ I mean how an operator lock it self(it is just an operator when acts on a state it changes number of bosons on particular x, it can not be equal to a number or a some function) to a specific value in the first place does it talk about classical limit that operators are actually just classical functions, and some how euler lagrange equations gives that solution? or is it something else going on?

Answer 1

I think I get the answer,

Now if $y$ goes to infinity, in the path integral representation of partition function, that cosine term will oscillate wildly, so according to saddle point approximation, only the classical path($\phi$ that obeys to Euler Lagrange equations) will have non zero contribution the other configurations will interfere destructively.

Thus we write the E-L equations and solve them, we will have several solutions as solitons or gapped phonons, these correspond to small fluctuations around the classical path. In other words, we only take into account the classical configuration and the configurations that deviate from the classical one very slightly. Because the other configurations will destructively interfere because of the high $y$.

and, I think the value that minimizes the cosine term is the classical configuration. Because it minimizes the energy.