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Instanton in sine-Gordon equation

+ 3 like - 0 dislike
1963 views

This is a statement from Giamarchi's book on Quantum Physics in 1D:

"For a single-particle in a cosine potential, the slightest amount of tunneling between two cosine minima leads to conduction bands, for example, and restores the translational symmetry. However, our sine-Gordon problem is a two-dimensional (one space one time) problem. In that case it is well-known that instantons with a finite action (instanton) that would connect two cosine minima cannot exist (Rajaraman, 1982). There is thus no restoring of symmetry and the field is truly locked in one of the minima. This is of course related to the Mermin-Wagner theorem stating that in two (classical) dimensions it is impossible to break a continuous symmetry but one can break a discrete one."

The potential concerns us here of a scalar field $\Phi$ is: $$ g \cos(\beta \;\Phi) $$

I wonder how to show this statement:

"It is well-known that instantons with a finite action (instanton) that would connect two cosine minima cannot exist (Rajaraman, 1982) ... the field is truly locked in one of the minima."

Question:

(1) Are "there" any criteria or conditions when the field will be locked in one of the minima? Such as $g>g_c$ and certain values of $\beta$?

(2) how to show this statement? How is the instanton analysis done here?

ps. I read Rajaraman book and S Coleman on instantons. So please do not post an answer for recommending just the Refs. Thanks. :)

NEW Edit NOTE: I suppose we are talking about this kind of 1+1D bosonic action: $$ \frac{1}{4\pi} \int_{ \mathcal{M}^2} dt \; dx \; k\, \partial_t \Phi \partial_x \Phi - v \,\partial_x \Phi \partial_x \Phi + g \cos(\beta_{}^{} \cdot\Phi_{}) $$

Ref:

(1) Rajaraman 1982, Solitons and Instantons, Volume 15: An Introduction to Solitons and Instantons in Quantum Field Theory (North-Holland Personal Library)

(2) Thierry Giamarchi, Quantum Physics in One Dimension

(3) S Coleman, Aspects of Symmetry


This post imported from StackExchange Physics at 2014-06-15 18:05 (UCT), posted by SE-user Idear

asked Jun 14, 2014 in Theoretical Physics by wonderich (1,400 points) [ revision history ]
edited Jun 18, 2014 by wonderich
Most voted comments show all comments
"This is of course related to the Mermin-Wagner theorem stating that in two (classical) dimensions it is impossible to break a continuous symmetry but one can break a discrete one", this is the key

This post imported from StackExchange Physics at 2014-06-15 18:05 (UCT), posted by SE-user Nikos M.
@ Nikos, thanks, I know Coleman theorem. Even if there is an instanton transporting between two vacuums(vev) at minimum of cosine, it only creates a discrete symmetry (not continuous symmetry) between the minimum. In that case, one can either break the discrete symmetry by trapping at a single minimum, or does NOT break the discrete symmetry by allowing instanton transporting the vev. Importantly, there is no contradiction in either cases. So my understanding is that Coleman-Mermin-Wagner theorem does NOT directly prove anything. Do you agree?

This post imported from StackExchange Physics at 2014-06-15 18:05 (UCT), posted by SE-user Idear
@ Nikos, so my point is that Coleman-Mermin-Wagner theorem does NOT directly help to prove the cited statement.

This post imported from StackExchange Physics at 2014-06-15 18:05 (UCT), posted by SE-user Idear
hmm, thinking about it, it should be a part in non-trivial way (plus some topology), will have to take a look.. thanks

This post imported from StackExchange Physics at 2014-06-15 18:05 (UCT), posted by SE-user Nikos M.
yes, CMW theorem is relevant, but does not prove it. CMW theorem only says that the outcomes has NO contradiction. :)

This post imported from StackExchange Physics at 2014-06-15 18:05 (UCT), posted by SE-user Idear
does the particle have mass? can you give the full sine-gordon eq?

This post imported from StackExchange Physics at 2014-06-15 18:05 (UCT), posted by SE-user Nikos M.
no the particle has no explicit mass, it is a free boson plus cosine potential. But the consensus is that at large coupling $g > g_c$, the boson gain masses. I like to understand this $g_c$ more carefully. Better non-perturbatively, than using RG.

This post imported from StackExchange Physics at 2014-06-15 18:05 (UCT), posted by SE-user Idear

1 Answer

+ 5 like - 0 dislike

The argument that the vacua are stable to tunneling is obvious if you think about simulating the lattice version of the theory. In order to move from one vacuum at t=-infinity to another vacuum at t=+infinity, you need to have a transition between the two at every separate value of x, and this means you end up with a finite action density per unit x, which is an infinite action overall. It's easier for an Ising model below the phase-transition point. Can the Ising model flip from a plus-state to a minus-state in 2d? No, because this requires conspiracy over the entire volume. In 1d, it can, because it can make a local patch of vacuum, and the surface area action cost is constant.

This is not at all the Mermin-Wagner theorem, because the symmetry is discrete, and it is in fact broken. The Mermin-Wagner theorem tells you that continuous symmetries can't be broken in 2d, but this is a discrete symmetry, and it is broken in 2d.

The argument I said isn't rigorous at all. It's clear it can't be done, but it's not completely trivial to prove. The rigorous proof in the Ising model is the Pieirls droplet argument. The analog of the Pieirls argument here is to consider a region in vacuum 1 inside vacuum 2, and the total action of the configuration is the boundary energy, which always goes up with the boundary length of the droplet. The number of these droplets is bounded by the number of paths on the lattice, and the assumption that the action is going up means that the droplets can't congeal, that you are below the phase transition point. It is possible to convert this into a proof, but the best way to understand it is "proof by analogy", since the argument is identical for the Ising model.

There are solitons in this model, they interpolate between the two vacua, but the do so at all time. These have finite energy, but they are solitons, they sit there forever, with a finite energy at each unit time, they are infinite action over all time.

answered Jun 16, 2014 by Ron Maimon (7,535 points) [ no revision ]

Hi Ron, +1, thanks so much for the nice answer. I wonder what is your Ref for Pieirls droplet argument? 

@Ron, p.s. for non-perturbative aspect, I myself still so far only knows on the Thirring model. That is also what Prof. R Jackiw told me. It is good to hear something different from you.

No ref, it's not in the literature, it's an analogy that is good, and it for sure can be made rigorous appropriately, but that's the devil with these QFT things, the technology for rigorous proofs is so backwards, that you really have to understand it by intuition. If you simulate sine-gordon on a computer, you'll understand, I guarantee it, even simulating Ising helps. I am sorry to give a "proof by assertion", but it's really clear, despite the difficulty in rigorous argument. Honest. Cross my heart. (more proof by assertion).

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