4.4 Less supersymmetric cases

Before continuing the study of $\mathcal{𝒩}=2$ systems, let us pause here and see what we can learn about less supersymmetric theories from the solution of the pure $\mathcal{𝒩}=2$ $SU\left(2\right)$ theory. A general Lagrangian we consider in this section is given by

The setup is $\mathcal{𝒩}=2$ supersymmetric when $m=\mu =0$. When we let $\left|m\right|\to \infty$, we decouple the chiral superfield $\Phi$, and we end up with $\mathcal{𝒩}=1$ pure $SU\left(2\right)$ theory which we discussed in Sec. 3.3. Next, by letting $\left|\mu \right|\to \infty$, we decouple the gaugino $\lambda$ and recover pure bosonic Yang-Mills.

4.4.1 $\mathcal{𝒩}=1$ system

First let us consider the $\mathcal{𝒩}=1$ system. When $m$ is very small, the term $mtr{\Phi }^2$ can be considered as a perturbation to the $\mathcal{𝒩}=2$ solution we just obtained. In terms of the variable $u$, the term $\int \; <!--nolimits-->d^2𝜃mtr{\Phi }^2$ is $\sim \int \; <!--nolimits-->d^2𝜃mu$, and therefore the F-term equation with respect to $u$ cannot be satisfied unless $u$ is at the singularity. There is no supersymmetric vacuum at generic value of $u$.

When $u$ is close to $u_0=2{\Lambda }^2$, there are additional terms in the superpotential given by

where the constant $c$ was introduced in (4.3.21). Together with the term $\int \; <!--nolimits-->d^2𝜃mu$, the F-term equations with respect to $u$, $Q$ and $\tilde{Q}$ are given respectively by

Then we find a solution at

The vacuum is pinned at $u=u_0$, and there is a nonzero condensate of the monopole $Q\tilde{Q}=m∕c$. A similar argument at $u=-u_0$ says that there is another supersymmetric vacuum given by

where $Q^{\prime }$, ${\tilde{Q}}^{\prime }$ are the dyon fields.

Summarizing, we found two supersymmetric vacua at $u=\pm u_0$, where monopoles or dyons condense, concretely realizing the idea that the confinement is given by condensation of magnetically-charged objects, see Fig. 4.10.

Recall that the anomalously broken continuous R-symmetry

can be compensated by the

Applying it to the Lagrangian (4.4.1), we see that

with which we find

It is important to keep in mind that the right hand side contains $e^{i{𝜃}_{UV}∕2}$ as the phase.

We now take the limit $m\to \infty$ keeping ${\Lambda }_{\mathcal{𝒩}=1}$ fixed. This should give the pure $\mathcal{𝒩}=1$ $SU\left(2\right)$ Yang-Mills theory. It is reassuring to find that we also see two vacua here, as in Sec. 3.3.

4.4.2 Pure bosonic system

Let us now make $\mu \ne 0$, keeping $\left|\mu \right|\ll \left|{\Lambda }_{\mathcal{𝒩}=1}\right|$. In this limit, the effect of the gaugino mass term $\mu {\lambda }_{\alpha }{\lambda }^{\alpha }$ is given by the first order perturbation theory, and the vacuum energy is given by

This was first pointed out in [46].

We see that two degenerate vacua of the $\mathcal{𝒩}=1$ supersymmetric theory are split into two levels with different energy density, corresponding to monopole condensation and dyon condensation, respectively. A slow change of ${𝜃}_{UV}$ from $0$ to $2\pi$ exchanges the two levels, which cross at ${𝜃}_{UV}=\pi$. So there is a first-order phase transition at ${𝜃}_{UV}=\pi$, at least when $\left|\mu \right|$ is sufficiently small.

It is an interesting question to ask if this first order phase transition persists in the limit $\left|\mu \right|\to \infty$, i.e. in the pure bosonic Yang-Mills theory. Let us give an argument for the persistence. The idea is to use the behavior of the potential between two external particles which are magnetically or dyonically charged as the order parameter [47].

First let us consider the dynamics more carefully. Two branches differ in the types of particles which condense: we can call the branches the monopole branch and the dyon branch, accordingly. In our convention, the charges of the particles are $\left(n,m\right)=\left(0,1\right)$ and $\left(2,1\right)$, respectively. The charge of the $SU\left(2\right)$ adjoint fields, under the unbroken $U\left(1\right)$ symmetry, is $\left(2,0\right)$ in our normalization. As there are no dynamical particles of charge $\left(1,0\right)$, the charge $\left(0,1\right)$ of the monopole is twice that of a minimally allowed one. The charge of this external monopole can then be written as $\left(n,m\right)=\left(0,1∕2\right)$.

Consider first introducing two external electric particles with charge $\left(n,m\right)=\left(1,0\right)$. In both branches, the electric field is made into a flux tube by the condensed monopoles or dyons. The flux tube has constant tension, and cannot pair-create dynamical particles, since all the dynamical particles have charge $\left(\pm 2,0\right)$. Therefore the flux tube does not break, and the potential is linear. The electric particles with charge $\left(1,0\right)$ are confined.

Instead, let us consider introducing external monopoles into the system, and measure the potential between the two. At $𝜃=0$, we can assume, without loss of generality, that the monopole branch has lower energy. There are dynamical monopole particles with charge $\left(n,m\right)=\left(0,1\right)$ condensing in the background. Let us introduce two external monopoles of charge $\left(n,m\right)=\left(0,1∕2\right)$. The magnetic field produced by the external particles with charges $\left(n,m\right)=\left(0,1\right)$ is screened and damped exponentially. The potential between them is then basically constant.

Instead, consider introducing two external particles with charge $\left(1,1∕2\right)$ into the monopole branch. The dynamical monopole cannot screen the electric charge, which is then confined into a flux tube. The potential between them is linear and they are confined.

We can repeat the analysis in the dyon branch. The behavior of the potential between external particles can be summarized as follows:

These two behaviors are exchanged under a slow continuous change of $𝜃$ from $0$ to $2\pi$. Therefore, there should be at least one phase transition. It would be interesting to confirm this analysis by a lattice strong-coupling expansion, or by a computer simulation.