Here we summarize the features of magnetic monopoles which we will repeatedly quote in the rest of
the lecture note. For a detailed exposition of topics discussed in this subsection, the readers should
consult the reviews such as [25, 26], or the textbook [27]. The review by Coleman [28] is also very
instructive.^{3}

Consider an $SU\left(2\right)$ gauge theory with a scalar in the adjoint representation, with the action

$$S=\int {d}^{4}x\frac{1}{{g}^{2}}\left[\frac{1}{2}tr\phantom{\rule{0.3em}{0ex}}{F}_{\mu \nu}{F}_{\mu \nu}+tr\phantom{\rule{0.3em}{0ex}}{D}_{\mu}\Phi {D}_{\mu}\Phi \right].$$ | (1.3.1) |

The ﬁeld $\Phi $ is a traceless Hermitean $2\times 2$ matrix.

Consider the vacuum where

$$\Phi =\left(\begin{array}{cc}\hfill a\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -a\hfill \end{array}\right).$$ | (1.3.2) |

When $a\ne 0$, the $SU\left(2\right)$ gauge symmetry is broken to $U\left(1\right)$. Indeed, the vev (1.3.2) commutes with a gauge ﬁeld strength of the form

$${F}_{\mu \nu}=\left(\begin{array}{cc}\hfill {F}_{\mu \nu}^{U\left(1\right)}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -{F}_{\mu \nu}^{U\left(1\right)}\hfill \end{array}\right)$$ | (1.3.3) |

where ${F}_{\mu \nu}^{U\left(1\right)}$ is a $U\left(1\right)$ gauge ﬁeld strength normalized as in Sec. 1.1. Note that the quanta of the scalar ﬁelds $\Phi $ has electric charge $2$ under this $U\left(1\right)$ ﬁeld, as can be found by expanding the covariant derivative.

We are considering a gauge theory; therefore the ﬁeld $\Phi $ does not have to be given exactly as in the right hand side of (1.3.2). Rather, we just need that $\Phi $ has eigenvalues $\pm a$. Then we can consider a conﬁguration of the form

$$\Phi \left(x\right)=\frac{{x}_{i}{\sigma}^{i}}{\left|x\right|}f\left(\left|x\right|\right)a$$ | (1.3.4) |

where $i=1,2,3$ and $f\left(r\right)$ is a dimensionless function such that

$$\underset{r\to 0}{lim}f\left(r\right)=0,\phantom{\rule{2em}{0ex}}\underset{r\to \infty}{lim}f\left(r\right)=1.$$ | (1.3.5) |

At the spatial inﬁnity, the vev of $\Phi $ is conjugate to (1.3.2), and therefore this conﬁguration can be thought of as an excitation of the vacuum given by (1.3.2).

The unbroken $U\left(1\right)$ within $SU\left(2\right)$ is along $\Phi $. A more general deﬁnition of the $U\left(1\right)$ ﬁeld strength ${F}_{\mu \nu}^{U\left(1\right)}$, at least when $r\gg 0$, is then the combination

$${F}_{\mu \nu}^{U\left(1\right)}:=\frac{1}{2a}tr\phantom{\rule{0.3em}{0ex}}{F}_{\mu \nu}\Phi .$$ | (1.3.6) |

In the region $r\gg 0$, let us try to bring the conﬁguration (1.3.4) to (1.3.2) by a gauge transformation. This can be done smoothly except at the south pole, by using the gauge transformation

This gives a gauge transformation around the south pole given by

$$i\left(-{\sigma}^{1}sin\mathit{\theta}+{\sigma}^{2}cos\mathit{\theta}\right)=exp\left[-i\mathit{\theta}{\sigma}^{3}\right]\cdot \left(i{\sigma}^{2}\right).$$ | (1.3.8) |

As $\mathit{\theta}$ goes from $0$ to $2\pi $, we see that the $U\left(1\right)$ ﬁeld ${F}_{\mu \nu}^{U\left(1\right)}$ has the magnetic charge $m=1$, and therefore is a monopole. This was originally found by ’t Hooft and Polyakov. Note that its Dirac pairing with the particle of the ﬁeld $\Phi $ is 2, which is twice the minimum allowed value.

Let us evaluate the energy contained in the ﬁeld conﬁguration. The kinetic energy is $1\u2215{g}^{2}$ times

$$\begin{array}{lll}\hfill \int {d}^{3}x\left[tr\phantom{\rule{0.3em}{0ex}}{B}_{i}{B}_{i}+tr\phantom{\rule{0.3em}{0ex}}{D}_{i}\Phi {D}_{i}\Phi \right]& =\int {d}^{3}x\left[tr\phantom{\rule{0.3em}{0ex}}{\left({B}_{i}\mp {D}_{i}\Phi \right)}^{2}\pm 2tr\phantom{\rule{0.3em}{0ex}}{B}_{i}{D}_{i}\Phi \right]\phantom{\rule{2em}{0ex}}& \hfill \text{(1.3.9)}\\ \hfill & \ge \pm 2\int {d}^{3}xtr\phantom{\rule{0.3em}{0ex}}{B}_{i}{D}_{i}\Phi =\pm 2\int {d}^{3}x{\partial}_{i}tr\phantom{\rule{0.3em}{0ex}}{B}_{i}\Phi \phantom{\rule{2em}{0ex}}& \hfill \text{(1.3.10)}\\ \hfill & =\pm 2{\int}_{{S}^{2}}d\overrightarrow{n}\cdot tr\phantom{\rule{0.3em}{0ex}}\overrightarrow{B}\Phi \phantom{\rule{2em}{0ex}}& \hfill \text{(1.3.11)}\end{array}$$where the ﬁnal integral is over the sphere at the spatial inﬁnity, which according to (1.3.6) evaluates to $\pm 2\left(2a\right)\left(2\pi m\right)$, where $m$ is the magnetic charge. Therefore we have the bound

$$\text{(energyofthemonopole)}\ge \frac{4\pi}{{g}^{2}}\left(2a\right)\left|m\right|$$ | (1.3.12) |

This is called the Bogomolnyi-Prasad-Sommerﬁeld (BPS) bound. The inequality is saturated if and only if

$${B}_{i}=\pm {D}_{i}\Phi ,$$ | (1.3.13) |

which is called the BPS equation. This ﬁxes the form of the function $f\left(r\right)$ in (1.3.4).

Given such an explicit monopole solution, there is a way to construct other solutions related by the symmetry. First, the conﬁguration (1.3.4) has a center at the origin of the coordinate system. We can shift the center of the monopole at an arbitrary point $\overrightarrow{y}$ of the spatial ${\mathbb{R}}^{3}$. These give three zero-modes.

Another zero mode is obtained by the gauge transformation:

$${e}^{i\alpha \Phi \u2215a}$$ | (1.3.14) |

Note that a gauge transformation which vanishes at inﬁnity is a redundancy of the physical system, but a gauge transformation which does not vanish at inﬁnity is considered to change the classical conﬁguration. For general $\alpha $, this transformation (1.3.14) changes the asymptotic behavior of ${F}_{ij}\left(x\right)$, but for $\alpha =\pi $, the transformation (1.3.14) trivially acts on the ﬁelds in the adjoint representation. Therefore $\alpha $ is an angular variable $0\le \alpha <\pi $.

The semiclassical quantization of the monopole involves the Fock space of non-zero modes, together with a wavefunction $\psi \left(\overrightarrow{y},\alpha \right)$ depending on the zero modes $\overrightarrow{y}$ and $\alpha $. The wavefunction along $\overrightarrow{y}$ represents the spatial motion of the center of mass of the monopole. The wavefunction along $\alpha $ represents the electric charge of the monopole, which can be seen as follows.

By comparing (1.3.14) with (1.3.3) and (1.3.6), we see that the unbroken $U\left(1\right)$ global gauge transformation by ${e}^{i\phi}$ shifts $\alpha $ by

$$\alpha \to \alpha +\phi .$$ | (1.3.15) |

Recall that a state $|\psi \u27e9$ with electric charge $n$ behaves under the $U\left(1\right)$ global transformation by ${e}^{i\phi}$ by

$$|\psi \u27e9\to {e}^{in\phi}|\psi \u27e9.$$ | (1.3.16) |

Now, as $\alpha $ is a variable with period $\pi $, $\psi \left(\alpha \right)$ can be expanded as a linear combination of ${e}^{i2d\alpha}$ where $d$ is an integer. Under (1.3.15) the wavefunction changes as in (1.3.16) with $n=2d$, therefore we see that the monopole state with this zero-mode wave function has the electric charge $2d$.

Summarizing, the combination of the electric charge and the magnetic charge $\left(n,m\right)$ we obtain from the semi-classical quantization has the form $\left(n,m\right)=\left(2d,1\right)$ where $d$ is an integer. This was found originally by Julia and Zee: once we quantize the ’t Hooft-Polyakov monopole, we not only have a purely-magnetic monopole but a whole tower of dyon states, with $d=-\infty $ to $+\infty $.

Finally let us consider the eﬀect of the fermionic zero modes in the ’t Hooft-Polyakov monopole (1.3.4). First let us consider two Weyl fermions $\lambda $, $\stackrel{\u0303}{\lambda}$ in the adjoint representation, with the Lagrangian

We regard both the gauge potential in the covariant derivative $D$ and the scalar ﬁeld as backgrounds, and decompose $\lambda $, $\stackrel{\u0303}{\lambda}$ into eigenstates of the angular momentum. The lower bound of the orbital angular momentum is given by the Dirac pairing, which is $\hslash $ here. The spinor ﬁelds have spin $\hslash \u22152$. Therefore the state with lowest angular momenta has spin $\hslash \u22152$. When the coeﬃcient $c$ takes a value in a certain range, it is known that there is a pair of zero modes ${b}_{\alpha}$ where $\alpha =1,2$ the spinor index of the $SO\left(3\right)$ spatial rotation. The semiclassical quantization promotes them into a pair of fermionic oscillators

$$\left\{{b}_{\alpha},{b}_{\beta}^{\u2020}\right\}={\delta}_{\alpha \beta}.$$ | (1.3.18) |

This creates four states starting from one state $|\psi \u27e9$ from the semiclassical quantization of the bosonic part:

This counts as one complex boson and one fermion.

Suppose we introduce another pair ${\lambda}^{\prime}$, ${\stackrel{\u0303}{\lambda}}^{\prime}$ of the adjoint Weyl fermions. Then we will have another pair of fermionic oscillators ${b}_{\alpha}^{\prime}$. Together, they generate ${2}^{4}=16$ states, consisting of one massive vector (with 3 states), four massive spinors (with 8 states) and ﬁve massive scalars.

Next, consider having $2N$ Weyl fermions ${\psi}_{i}^{a}$ in the doublet representation where $a=1,2$ and $i=1,\dots ,2N$, with the Lagrangian

Note that the Lagrangian has an $SO\left(2N\right)$ ﬂavor symmetry acting on the index $i$.

The electric charge of the quanta of $\psi $, $\stackrel{\u0303}{\psi}$ with respect to the unbroken $U\left(1\right)$ is now $1$. Then the Dirac pairing is $\hslash \u22152$. Tensoring with the intrinsic spin $\hslash \u22152$, we ﬁnd that the minimal orbital angular momentum is $0$. It is known that for a suitable choice of ${c}^{\prime}$, this fermion system has zero modes ${\gamma}_{i}$, $i=1,\dots ,2N$. After semiclassical quantization, it becomes a set of fermionic operators with the commutation relation

$$\left\{{\gamma}_{i},{\gamma}_{j}\right\}={\delta}_{ij}.$$ | (1.3.21) |

This is the commutation relation of the gamma matrices of $SO\left(2N\right)$. Monopole states are representations of ${\gamma}_{i}$’s, meaning that they transform as a spinor representation of the ﬂavor symmetry $SO\left(2N\right)$.

Fields in a doublet representation of the $SU\left(2\right)$ gauge symmetry has an another eﬀect. Namely, in the gauge zero mode (1.3.14), $\alpha =\pi $ gives the matrix

$$\left(\begin{array}{cc}\hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill \end{array}\right)$$ | (1.3.22) |

which acts nontrivially on the ﬁelds in the doublet representation. Then the periodicity of the gauge zero mode $\alpha $ is now $2\pi $, and the wavefunction along the $\alpha $ direction can now be ${e}^{in\alpha}$ for arbitrary integer $n$. Therefore, the electric charge $n$ can either be even or odd. The operators ${\gamma}_{i}$ come from the modes of the ﬁelds in the doublet representation, and therefore it changes the electric charge by $\pm 1$.

We can deﬁne the ﬂavor spinor chirality $\Gamma $ by

$$\Gamma ={\gamma}_{1}{\gamma}_{2}\cdots {\gamma}_{2N},$$ | (1.3.23) |

by which the spinor of $SO\left(2N\right)$ can be split into positive-chirality and negative-chirality spinors. The action of the operators ${\gamma}_{i}$ changes the chirality of the ﬂavor spinors. Combined with the behavior of the $U\left(1\right)$ electric charge we saw in the previous paragraph, we conclude that the parity of the $U\left(1\right)$ electric charges of the monopole states is correlated with the chirality of the ﬂavor spinor representation.