3
$\begingroup$

This is the sine-Gordon action: $$ \frac{1}{4\pi} \int_{ \mathcal{M}^2} dt \; dx \; k\, \partial_t \Phi \partial_x \Phi - v \,\partial_x \Phi \partial_x \Phi + g \cos(\beta_{}^{} \cdot\Phi_{}) $$ Here $\mathcal{M}^2$ is a 1+1 dimensional spacetime manifold, where 1D space is a $S^1$ circle of length $L$.

At $g=0$ : it is a chiral boson theory with zero mass, gapless scalar boson $\Phi$.

At large $g$ : It seems to be well-known that at large coupling $g$ of the sine-Gordon equation, the scalar boson $\Phi$ will have a mass gap.

Q1: What is the original Ref which states and proves this statement about the nonzero (or large) mass gap for large $g$?

-

Q2: What does the mass gap $m$ scale like in terms of other quantities (like $L$, $g$, etc)?

-

NOTE: I find S Coleman has discussion in

(1)"Aspects of Symmetry: Selected Erice Lectures" by Sidney Coleman

and this paper

(2)Quantum sine-Gordon equation as the massive Thirring model - Phys. Rev. D 11, 2088 by Sidney Coleman

But I am not convinced that Coleman shows it explicitly. I read these, but could someone point out explicitly and explain it, how does he(or someone else) rigorously prove this mass gap?

Here Eq.(17) of this reference does a quadratic expansion to show the mass gap $m \simeq \sqrt{\Delta^2+\#(\frac{g}{L})^2}$ with $\Delta \simeq \sqrt{ \# g k^2 v}/(\# k)$, perhaps there are even more mathematical rigorous way to prove the mass gap with a full cosine term?

$\endgroup$
  • 2
    $\begingroup$ Via bosonization technique you can prove that the sine Gordon model is equivalent to the neutral sector of the massive thirring model. With this technique you can also calculate exactly the correlation functions and see directly that they fall exponentially with distance so that the theory has mass gap. See also this paper by Mandelstam journals.aps.org/prd/abstract/10.1103/PhysRevD.11.3026 $\endgroup$ – TwoBs Jun 4 '14 at 6:08
  • $\begingroup$ @ TwoBs, does that correlator argument have any β dependence of $\cos(\beta \Phi)$? In S Coleman, there is some discussion when it is Thirring model, when it is energy unbounded from below, when it is just a free massive Fermion model. $\endgroup$ – wonderich Jun 4 '14 at 18:07
  • $\begingroup$ One of the easiest ways to see this is to say that for large $g$, the field is localized in the cosine potential. So, expanding around a minimum of the cosine, you immediately pick up a mass term. Of course, this is only a handwaving argument of sorts. $\endgroup$ – surajshankar Sep 21 '14 at 4:14

Your Answer

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

Browse other questions tagged or ask your own question.