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

The setup is $\mathcal{\mathcal{N}}=2$ supersymmetric when $m=\mu =0$. When we let $\left|m\right|\to \infty $, we decouple the chiral superﬁeld $\Phi $, and we end up with $\mathcal{\mathcal{N}}=1$ 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.

First let us consider the $\mathcal{\mathcal{N}}=1$ system. When $m$ is very small, the term $mtr\phantom{\rule{0.3em}{0ex}}{\Phi}^{2}$ can be considered as a perturbation to the $\mathcal{\mathcal{N}}=2$ solution we just obtained. In terms of the variable $u$, the term $\int {d}^{2}\mathit{\theta}mtr\phantom{\rule{0.3em}{0ex}}{\Phi}^{2}$ is $\sim \int {d}^{2}\mathit{\theta}mu$, 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

$$\int {d}^{2}\mathit{\theta}Q{a}^{\prime}\stackrel{\u0303}{Q}=\int {d}^{2}\mathit{\theta}c\left(u-{u}_{0}\right)Q\stackrel{\u0303}{Q}$$ | (4.4.2) |

where the constant $c$ was introduced in (4.3.21). Together with the term $\int {d}^{2}\mathit{\theta}mu$, the F-term equations with respect to $u$, $Q$ and $\stackrel{\u0303}{Q}$ are given respectively by

$$m=cQ\stackrel{\u0303}{Q},\phantom{\rule{1em}{0ex}}\left(u-{u}_{0}\right)\stackrel{\u0303}{Q}=0,\phantom{\rule{1em}{0ex}}\left(u-{u}_{0}\right)Q=0.$$ | (4.4.3) |

Then we ﬁnd a solution at

$$u={u}_{0},\phantom{\rule{1em}{0ex}}Q\stackrel{\u0303}{Q}=m\u2215c.$$ | (4.4.4) |

The vacuum is pinned at $u={u}_{0}$, and there is a nonzero condensate of the monopole $Q\stackrel{\u0303}{Q}=m\u2215c$. A similar argument at $u=-{u}_{0}$ says that there is another supersymmetric vacuum given by

$$u=-{u}_{0},\phantom{\rule{1em}{0ex}}{Q}^{\prime}{\stackrel{\u0303}{Q}}^{\prime}=m\u2215c$$ | (4.4.5) |

where ${Q}^{\prime}$, ${\stackrel{\u0303}{Q}}^{\prime}$ 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

$$\Phi \to {e}^{i\phi}\Phi ,$$ | (4.4.6) |

can be compensated by the

$${\mathit{\theta}}_{UV}\to {\mathit{\theta}}_{UV}+4\phi .$$ | (4.4.7) |

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

$$m\u27e8tr\phantom{\rule{0.3em}{0ex}}{\Phi}^{2}\u27e9=\frac{-i}{2\pi}\u27e8tr\phantom{\rule{0.3em}{0ex}}{W}^{\alpha}{W}_{\alpha}\u27e9$$ | (4.4.8) |

with which we ﬁnd

$$\u27e8{\lambda}_{\alpha}{\lambda}^{\alpha}\u27e9\propto \pm 2\pi im{\Lambda}^{2}=:\pm {\Lambda}_{\mathcal{\mathcal{N}}=1}^{3}.$$ | (4.4.9) |

It is important to keep in mind that the right hand side contains ${e}^{i{\mathit{\theta}}_{UV}\u22152}$ as the phase.

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

Let us now make $\mu \ne 0$, keeping $\left|\mu \right|\ll \left|{\Lambda}_{\mathcal{\mathcal{N}}=1}\right|$. 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

$$V\propto Re\left(\pm \mu {\Lambda}_{\mathcal{\mathcal{N}}=1}^{3}\right)\propto {\Lambda}_{\mathcal{\mathcal{N}}=0}^{4}Re\left(\pm {e}^{i{\mathit{\theta}}_{UV}\u22152}\right).$$ | (4.4.10) |

This was ﬁrst pointed out in [46].

We see that two degenerate vacua of the $\mathcal{\mathcal{N}}=1$ supersymmetric theory are split into two levels with diﬀerent energy density, corresponding to monopole condensation and dyon condensation, respectively. A slow change of ${\mathit{\theta}}_{UV}$ from $0$ to $2\pi $ exchanges the two levels, which cross at ${\mathit{\theta}}_{UV}=\pi $. So there is a ﬁrst-order phase transition at ${\mathit{\theta}}_{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 $\left(n,m\right)=\left(0,1\u22152\right)$.

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 $\mathit{\theta}=0$, 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 $\left(n,m\right)=\left(0,1\u22152\right)$. 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 $\left(1,1\u22152\right)$ 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:

$\left(0,1\u22152\right)$ | $\left(1,1\u22152\right)$ | |

monopole branch | screened | conﬁned |

dyon branch | conﬁned | screened |

These two behaviors are exchanged under a slow continuous change of $\mathit{\theta}$ 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.