3
$\begingroup$

I am reading this paper(pdf) and on page 11, the chiral boson theory on a cylinder is studied when both edges of the cylinder are brought in close proximity so that electron is allowed.

Why is it that for large tunneling amplitudes, $g$, the field $\phi$ can be treated 'classically' in the sense that it gets pinned to the classical minima of the cosine to obtain the ground state energy?

Could someone give a more detailed explanation?

$\endgroup$

1 Answer 1

2
$\begingroup$

It is simply the fact that if the $g$ is large, $$ g \cos(\hat{\phi}) \to g(1- \frac{1}{2}\hat{\phi}^2 +\dots) $$ the ground state will be trapped in one of the vacuum sectors, at the minimum of the potential energy. (Please check S Coleman's book $\it{Aspects\; of\; symmetry}$ or his Phys Rev D paper on quantum sine-Gordon eq. It may offer differ views than what I said. If so, please report to me. :-))

It is NOT the only way to treat in the $\phi$ space and counting the minimum vacuum energy, and further count (topological) ground state degeneracy, and module out the equivalence classes. You can also treat in the dual conjugate momentum of the field $\phi$ variable. In a compact 1+1D boundary, the mode expansion of $\phi$ is

$$ \hat{\Phi}_I(x) ={\hat{\phi}_{0}}_{I}+K^{-1}_{IJ} \hat{P}_{\phi_J} \frac{2\pi}{L}x+i \sum_{n\neq 0} \frac{1}{n} \hat{\alpha}_{I,n} e^{-in x \frac{2\pi}{L}} $$

In the sense of large coupling limit, from the view points of the mode expansion of $\phi$, the topological features is controlled by the nontrivial zero modes: $\phi_0$. The dual conjugate variable of $\phi_0$ variable is the nontrivial winding mode, $p_\phi$.

Please see this paper arxiv-1212.4863. You can construct the Hilbert space of either zero modes $\phi_0$, or the Hilbert space of winding mode, $p_\phi$. Both approaches should give the same consistent result, though constructing the Hilbert space of winding modes $p_\phi$ has some better advantages than $\phi_0$ in the case of modding out equivalence classes. See Appendix B of this paper arxiv-1212.4863 for some highly relevant discussions.

ps. everything listed above can be treated as operator levels, not just classical fields. i.e. $\phi \to \hat{\phi}$

$\endgroup$
5
  • $\begingroup$ Thanks for the nice reply. I think the problem that I am/was having was that I did understand the intuitive picture (meaning that for large g the field is trapped in the minima of the cosine and cannot escape by any 'flunctuations'). The problem was rather on a formal level, in the sense that one is threating $\phi$ as if wasn't an operator but a normal complex function (i.e. a classical field). Maybe you can elaborate on this. $\endgroup$
    – MrLee
    Commented Dec 23, 2013 at 23:25
  • $\begingroup$ My response: every fields listed above can be treated as quantum operators, not just classical fields. i.e. $\phi \to \hat{\phi}$. e.g. construct a Hilbert space of $\hat{\phi}_0$ or $\hat{P}_\phi$ certainly means the Hilbert space acted by operators. But honestly, the first part, "how can I consider $\phi$ treated at the minimum at large $g$?" - this is simply based on my physics intuition. (Doing physics relies on some intuition/approximation?) I have no further proof for this. Perhaps let me know if Coleman or others offers some formal answer for this? Thanks a lot. $\endgroup$
    – wonderich
    Commented Dec 23, 2013 at 23:34
  • $\begingroup$ For formal proof, I suggest to compute the correlator $\langle \hat{\phi}(0) \hat{\phi}(x) \rangle$ where $0$ and $x$ are two different minimum (vacuum expectation values). I suppose this value has exponential decay $\exp[-g \cdot\; \text{const} \cdot \text{volume}]$(?), with an exponent of some factor proportional to the volume of the spacetime of fields. (yes?) $\endgroup$
    – wonderich
    Commented Dec 23, 2013 at 23:45
  • $\begingroup$ Did you set t=0 in your mode expansion above or does "x" correspond to some "lightcone-like" coordinate? $\endgroup$
    – MrLee
    Commented Apr 1, 2014 at 14:13
  • $\begingroup$ I set t=0. It just a matter of Schrodinger picture of Heisenberg picture. You can add $e^{iHt}$ and $e^{-iHt}$ on two sides of $\Phi$. $\endgroup$
    – wonderich
    Commented Apr 1, 2014 at 15:42

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.