Calculation of Berry's phase due to monopole tunneling event of $O(3)$ NLSM on square lattice

Question

I am currently reading the seminal paper by Duncan Haldane: http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.61.1029

In this paper, he asserts that for a unit-vector field $\hat{\Omega}(x,y,t)$ configuration which is everywhere well defined, there will be no Hopf term. But there are monopole tunneling events which can change the number of skyrmions in the system, which are associated with certain Berry phases and can potentially affect the system.

Tunneling event will create $+4\pi$ and $-4\pi$ vortices of the $\omega(x,y)$ fields (a detailed definition can be found in his paper). Therefore it is necessary to introduce the discontinuity lines in $(x,y)$ plane across which $\omega$ jumps by $4\pi$ in order to fully describe those vortices.

Haldane states that when tunneling process occurs, the only nontrivial contribution of the Berry phase term is from the bonds cut by the lines joining a $+4\pi$ vortex and a $-4\pi$ vortex, and every such bond will contribute $\pi S\eta$.

I tried to repeat his calculation, but I find that each cut bond contributes an amount of $2\pi S\eta$, which comes from the fact that every time you do the alternating sum of $\omega$ field over a plaquette which includes a cut bond, you will get $\pi S\eta$, and every such bond is counted twice since it belongs to two different plaquettes, which finally yields $2\pi S\eta$.

As a simplest example, if I have a $+4\pi$ vortex at $(0,0)$ and a $-4\pi$ skyrmion at $(0,1)$, and the bond which is cut by the discontinuity line is $(-1,0)->(0,0)$. Therefore when I sum over the plaquette $(0,0)$ and the plaquette $(0,1)$, this bond will be counted twice and will yield a Berry phase of $2\pi S$. So in this case, the total Berry phase is $e^{i2\pi S}$. But from formula $(5)$ in Haldane's paper, we should have a Berry phase of $i^{-2S}$ given such a field configuration, which leads to the contradiction.

Can anyone explain to me what I'm doing wrong?

Answer 1

I now realize that in doing an alternating sum over a plaquette, you have an extra $1/2$ factor which results in the following:

$\sum_{i\in plaq.}\eta_{i}\omega_i=\frac{1}{2}\eta\int dx \omega(x)=\eta\frac{1}{2}\times 4\pi=2\pi\eta$.

Therefore in evaluating the Berry phase over a plaquette, each bond cut by the discontinuity line contained in the plaquette will contribute the phase $2\pi \eta S$. Divided by four to avoid double counting, it will give us $\frac{\pi}{2} \eta S$. Therefore each bond will indeed contribute $\pi \eta S$ as explained by Haldane, since it is included in two plaquettes.

Answer 2

You should be summing over plaquettes, not the links.

Monopoles occuring at the center of a given plaquette assigns a 4 pi phase to that plaquette (introduction of the 4 pi branch cut which cuts across links (which you mention) allows you to fix the gauge ambiguity).
What you should do now is to convert the site-summation to a plaquette-summation. Each site will then be counted four times, so to correct for this overcounting, the phases contributed by the monopoles (which is assigned to plaquettes) has to be divided by four, not by two.