Monopoles, dyons, and a physical BPS bound

The classical Coulomb branch leaves a massless photon. Does its magnetic field have to come from a point singularity? In the full non-Abelian theory, the fields can instead make a smooth core. The core costs energy, and a surface integral gives a lower bound on that energy.

We will calculate both the bound and a configuration that attains it. The connection and flux lesson supplies the patching language. At the end we will connect the result to the supersymmetry central charge; the general algebraic argument is in extended four-dimensional supersymmetry.

Fix the energy and the vacuum normalization

Work in four-dimensional Lorentzian pure N=2 $SU(2)$ Yang–Mills theory: eight real supercharges, four manifest in N=1 notation, and gauge rank one. Space is $\mathbb R^3$, the theta angle is zero, and $g>0$ is a fixed classical coupling. Set fermions to zero. Use

$$t^a=\frac{\sigma_a}{2},\qquad \operatorname{Tr}_2(t^at^b)=\frac12\delta^{ab},\qquad \mathscr D_\mu=\partial_\mu-ig[A_\mu,\,\cdot\,], \qquad F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu-ig[A_\mu,A_\nu].$$

Define canonically normalized Hermitian real scalars by $\phi=(X+iY)/\sqrt2$. Thus $X,Y$ here are $\sqrt2$ times the temporary real and imaginary Hermitian parts used to solve the preceding lesson’s vacuum equation. In components, $\mathscr D_\mu X=\partial_\mu X+gA_\mu\times X$, where $(A\times X)^a=\epsilon^{abc}A^bX^c$ and $\epsilon^{123}=+1$.

The kinetic term $2\operatorname{Tr}_2|\mathscr D\phi|^2$ becomes $\frac12(\mathscr D X)^2+\frac12(\mathscr D Y)^2$, and $g^2\operatorname{Tr}_2[\phi,\phi^\dagger]^2=\frac{g^2}{2}|X\times Y|^2$. With $E_i^a=F_{0i}^a$ and $B_i^a=\frac12\epsilon_{ijk}F_{jk}^a$, the bosonic energy is therefore

$$\mathcal E=\frac12\int_{\mathbb R^3}d^3x\, \left[E_i^2+B_i^2+(\mathscr D_iX)^2+(\mathscr D_iY)^2 +(\mathscr D_0X)^2+(\mathscr D_0Y)^2+g^2|X\times Y|^2\right]. \tag{SW03.1}$$

Squares include color contraction. The factor $1/2$ follows from the fundamental trace; there is no extra overall $1/g^2$ in these canonical fields.

Choose the real vacuum with $|X|\to v>0$, $Y\to0$. In a local gauge that diagonalizes the scalar at infinity,

$$X_\infty=vt^3,\qquad \phi_\infty=\frac v{\sqrt2}t^3,\qquad a_{\rm cl}=\frac{gv}{\sqrt2},\qquad M_W=gv=\sqrt2\,a_{\rm cl}. \tag{SW03.2}$$

The last equality uses precisely the scalar rescaling already established on the Coulomb branch.

Complete the magnetic energy

First take static fields with $A_0=0$, so $E_i=\mathscr D_0X=\mathscr D_0Y=0$. For either sign $\eta=\pm1$, expand a square in (SW03.1):

$$\mathcal E= \frac12\int d^3x\left[ (B_i-\eta\mathscr D_iX)^2+(\mathscr D_iY)^2+g^2|X\times Y|^2 \right] +\eta\int d^3x\,B_i^a\mathscr D_iX^a . \tag{SW03.3}$$

The non-Abelian Bianchi identity is $\mathscr D_iB_i=0$. In the invariant color contraction, the connection terms cancel:

$$B_i^a\mathscr D_iX^a =\partial_i(X^aB_i^a)-X^a\mathscr D_iB_i^a =\partial_i(X^aB_i^a).$$

Let $\widehat X=X/|X|$ on the sphere at infinity, and define the magnetic flux $q_m=\lim_{R\to\infty}\int_{S_R^2}\widehat X^aB_i^a\,dS_i$. For localized configurations with the usual Coulomb falloff, the surface term is $vq_m$. Choosing $\eta$ to make it positive gives

$$\boxed{\mathcal E\ge v|q_m|.} \qquad B_i=\eta\mathscr D_iX,\quad \mathscr D_iY=0,\quad [X,Y]=0 \quad\text{at equality}. \tag{SW03.4}$$

The second real scalar has not been discarded from the energy. Its square must also vanish. A covariantly constant $Y$ has constant norm, so $Y_\infty=0$ then implies $Y=0$ everywhere.

This is the classical Bogomolny bound. In the supersymmetric theory its saturated solitons are BPS states, named for Bogomolny, Prasad and Sommerfield. For comparison with Tachikawa’s (1.3.9)–(1.3.13), his connection and real scalar are $gA$ and $gX$. His diagonal scalar parameter is $a_T=gv/2$; his bound $(4\pi/g^2)\,2a_T|m|$ becomes $4\pi v|m|/g$. This agrees with $a_T=a_{\rm cl}/\sqrt2$ in the preceding lesson. Neither source’s parameter called “$a$” is the canonical real magnitude $v$.

Why the magnetic charge is an integer

At infinity, $X$ chooses a direction in the adjoint orbit $SU(2)/U(1)\simeq S^2$. Thus $\widehat X$ defines a map from the spatial sphere to a sphere of scalar directions. Its degree cannot change continuously while $|X|$ stays nonzero on the boundary. A nonzero degree also prevents extending $\widehat X$ throughout the ball: the full scalar must vanish somewhere in a smooth core.

In a diagonal gauge, write the asymptotic connection as $A^3t^3$. For flux $q_m=4\pi n_m/g$, regular north and south potentials are

$$A_N^3=\frac{n_m}{g}(1-\cos\theta)\,d\varphi,\qquad A_S^3=-\frac{n_m}{g}(1+\cos\theta)\,d\varphi . \tag{SW03.5}$$

Their overlap is the $SU(2)$ transformation

$$U_{NS}(\varphi)=e^{\,2in_m\varphi t^3}. \tag{SW03.6}$$

Since $\varphi$ is periodic with period $2\pi$, closure in $SU(2)$ requires $n_m\in\mathbb Z$. The fundamental probe sees phases $e^{\pm in_m\varphi}$; an adjoint charged component sees $e^{\pm2in_m\varphi}$. Hence

$$q_m=\frac{4\pi}{g}n_m,\qquad |q_{\rm fundamental}|=\frac g2,\qquad |q_W|=g. \tag{SW03.7}$$

There is no dynamical fundamental matter in this theory. The allowed fundamental Wilson probe still tests the global group. A transition ending at $-\mathbf1$ is not a closed $SU(2)$ transition.

The overall sign of degree depends on the field-strength convention. With our $\mathscr D=\partial+gA\times$ and $B_i=\epsilon_{ijk}F_{jk}/2$, it is $n_m=-\deg(\widehat X)$. One way to check the sign is to solve $\mathscr D\widehat X=0$ at infinity: $A_i=-\widehat X\times\partial_i\widehat X/g+a_i\widehat X$. Then

$$\widehat X\cdot F_{\theta\varphi} =\partial_\theta a_\varphi-\partial_\varphi a_\theta -\frac1g\widehat X\cdot (\partial_\theta\widehat X\times\partial_\varphi\widehat X).$$

In a smooth global non-Abelian gauge on this sphere, $a=\widehat X\cdot A$ is a global one-form and its curl integrates to zero. The remaining integral is $-4\pi\deg(\widehat X)/g$. Equations (SW03.5)–(SW03.7) fix the charge sign without leaving it implicit.

The full $SU(2)$ fields need no physical Dirac string. The diagonal gauge fails where the scalar direction cannot be chosen globally, and the Abelian reduction fails at the core where $X=0$. An Abelian field with this flux cannot extend smoothly across the ball; the smooth non-Abelian solution does not require that reduction there.

Worked solution: a regular unit monopole

Let $r=|\mathbf x|$, $n_i=x_i/r$, $\rho=gvr$, and identify spatial and color axes for this ansatz. Take

$$X^a=-vH(\rho)n_a,\qquad A_i^a=\frac{1-K(\rho)}{gr}\epsilon_{aij}n_j,\qquad Y=A_0=0 . \tag{SW03.8}$$

The minus sign in $X$ makes the solution below have $n_m=+1$ in our stated convention. Direct differentiation, including the commutator in $F$, gives

$$\begin{aligned} \mathscr D_iX^a &=-gv^2H' n_in_a-\frac{vHK}{r}(\delta_{ia}-n_in_a),\\ B_i^a &=-\frac{1-K^2}{gr^2}n_in_a +\frac{vK'}r(\delta_{ia}-n_in_a). \end{aligned} \tag{SW03.9}$$

Primes now mean $d/d\rho$. Radial and transverse tensors are independent, so $B_i=\mathscr D_iX$ reduces to

$$\rho^2H'=1-K^2,\qquad K'=-KH . \tag{SW03.10}$$

To solve these equations, put $K=\rho/f$. The second equation gives $H=f'/f-1/\rho$, and the first becomes $ff''-(f')^2=-1$. The function $f=\sinh\rho$ satisfies it and the required boundary conditions. We obtain

$$\boxed{K(\rho)=\frac{\rho}{\sinh\rho},\qquad H(\rho)=\coth\rho-\frac1\rho .} \tag{SW03.11}$$

The expressions with $1/r$ are not evidence of a singular core. Their expansions are $K=1-\rho^2/6+O(\rho^4)$ and $H=\rho/3+O(\rho^3)$, so

$$X^a=-\frac{gv^2}{3}x_a+O(r^2x_a),\qquad A_i^a=\frac{gv^2}{6}\epsilon_{aij}x_j+O(r^2x_j).$$

Both have smooth Cartesian power series. In particular, $B_i^a\to-gv^2\delta_{ia}/3$. At large radius, $K\to0$, $H=1-1/\rho+O(e^{-2\rho})$ and $B_i^a\sim-n_in_a/(gr^2)$. Since $\widehat X=-\mathbf n$, the projected flux is $+4\pi/g$.

There is an independent energy check. Angular integration of the original energy, before using its bound, gives

$$\mathcal E=\frac{4\pi v}{g}\int_0^\infty d\rho \left[(K')^2+\frac{(1-K^2)^2}{2\rho^2} +\frac{\rho^2(H')^2}{2}+K^2H^2\right]. \tag{SW03.12}$$

The integrand is

$$(K'+KH)^2+ \frac12\left(\rho H'-\frac{1-K^2}{\rho}\right)^2 +\frac{d}{d\rho}\left[H(1-K^2)\right].$$

For (SW03.11), the squares vanish and the boundary values of the last expression are zero and one. Thus $\mathcal E=4\pi v/g$, with a core radius of order $(gv)^{-1}$. A pointlike Abelian $B\sim1/(gr^2)$ continued to $r=0$ would instead give an energy proportional to $\int_0 dr/r^2$.

Add electric charge without changing the charge basis

A dyon carries both electric and magnetic flux. Retain static fields and the consistent sector $Y=0$, but allow $A_0\ne0$. Define $q_e=\int_{S_\infty^2}\widehat X^aE_i^a\,dS_i$. For real $c,s$ with $c^2+s^2=1$, the same energy is

$$\mathcal E=\frac12\int d^3x\left[ (B_i-c\mathscr D_iX)^2+(E_i-s\mathscr D_iX)^2 +(\mathscr D_0X)^2\right]+v(cq_m+sq_e). \tag{SW03.13}$$

The electric cross term is a surface term on the Gauss constraint: $\mathscr D_iE_i+gX\times\mathscr D_0X=0$ implies $X\cdot\mathscr D_iE_i=0$. No electric source at a singular core has been inserted. Maximizing over $(c,s)$ yields

$$\mathcal E\ge v\sqrt{q_m^2+q_e^2},\qquad c=\frac{q_m}{\sqrt{q_m^2+q_e^2}},\quad s=\frac{q_e}{\sqrt{q_m^2+q_e^2}}. \tag{SW03.14}$$

For the unit magnetic sector $q_m=4\pi/g>0$, replace $\rho$ in (SW03.8) by $gvc\,r$ and set $A_0=-sX$. The monopole equations then give $B=c\mathscr D X$ and $E=-\mathscr D A_0=s\mathscr D X$, while $\mathscr D_0X=0$. Bianchi also gives $\mathscr D_iE_i=0$, so Gauss is satisfied. The fluxes obey $q_e=q_m s/c$ and the energy attains the bound.

This family has a continuous classical parameter. Quantum electric labels arise on quantizing the compact gauge-rotation collective coordinate; they are not quantized by square completion alone. At theta zero the first Seiberg–Witten basis assigns

$$q_e=gn_e,\qquad q_m=\frac{4\pi}{g}n_m,\qquad Z_{\rm cl}=a_{\rm cl}\left(n_e+\frac{4\pi i}{g^2}n_m\right). \tag{SW03.15}$$

Here the W charges have $n_e=\pm1$, not $\pm2$. A fundamental probe would have $n_e=\pm1/2$. This is the first paper’s (4.2). Tachikawa’s collective-coordinate discussion uses an electric label twice this one; his even integers become our integers. Section 1.3.2, (1.3.14)–(1.3.16)

Choose the common phase of the N=2 supercharges so that their algebraic central charge is

$$\mathcal Z_{\rm cl}=\sqrt2\,Z_{\rm cl} =v(q_e+iq_m). \tag{SW03.16}$$

The algebraic bound $M\ge|\mathcal Z|$ is now precisely (SW03.14). Electric charge is the real part in this phase choice; for $q_m>0$, the square-completion angle $c=\cos\beta$, $s=\sin\beta$ gives $\arg\mathcal Z=\pi/2-\beta$. Saturation leaves four of the eight real supercharges unbroken. The algebra supplies this statement; the classical energy calculation supplies an actual saturating soliton.

What becomes exact in the quantum theory?

The classical profile is useful semiclassically at large Higgs scale, where the coupling is small. It is not an exact quantum field configuration. The protected statement is instead the central-charge formula

$$Z_\gamma(u)=n_ea(u)+n_ma_D(u),\qquad M_\gamma\ge\sqrt2\,|Z_\gamma(u)|, \tag{SW03.17}$$

with equality for an existing BPS state. Here $a,a_D$ are the electric and magnetic special coordinates, later constructed as periods. Seiberg–Witten, (4.3)–(4.5); Bilal, section 4.3, (4.14).

At tree level $a_D=(4\pi i/g^2)a_{\rm cl}$ in our theta-zero example. Quantum mechanically $a_D$ is not obtained by replacing that constant with the effective coupling and multiplying by $a$: the effective coupling is a derivative, $\tau_{\rm eff}=da_D/da$. The first paper explicitly distinguishes these statements after (4.6). Nor does (SW03.17) assert that every integer pair labels a stable particle everywhere. Determining the periods and determining which states exist are further dynamical questions.

Exercises

1. Diagnose a proposed half monopole

Keep the specified $SU(2)$ theory, but try $n_m=1/2$ in (SW03.5). Compute the endpoint of its transition loop and the phase seen by a fundamental probe. Why can the adjoint fields alone miss the obstruction? Does changing the global group to $SO(3)$ automatically make this a smooth soliton on all of $\mathbb R^3$?

Hint 1

Evaluate $e^{4\pi i n_mt^3}$, not only its conjugation action.

Hint 2

A loop can be closed in $SO(3)$ but fail to extend through a disk. A smooth configuration on the ball must admit the extension in the full gauge group.

Solution

The endpoint is $e^{2\pi it^3}=-\mathbf1$. A fundamental doublet changes sign, while conjugation on every adjoint field is unchanged. Thus the proposed overlap is not a closed $SU(2)$ transition even though the adjoint-only local action cannot detect the central sign.

In $SO(3)$ it becomes closed, but represents the nontrivial element of $\pi_1(SO(3))=\mathbb Z_2$. It cannot be filled in by a disk in that group. A bundle on the surrounding sphere with this transition therefore cannot be the restriction of a smooth bundle on a three-ball. It describes allowed magnetic defect data for an appropriate global theory, not automatically a smooth monopole core without a defect. The doubled loop is contractible in the full group and has our smooth-monopole flux $4\pi/g$. Closure, extendibility, and an actual finite-energy solution are separate tests.

2. A smooth deformation that loses saturation

Keep $K$ in (SW03.11) and change $H$ to $H_\varepsilon=H+\varepsilon h$, where $h(\rho)=\rho e^{-\rho^2}$ and $\varepsilon$ is real. Show that the core remains smooth and the magnetic charge stays one. Calculate the energy above the bound and decide whether a nonzero sufficiently small $\varepsilon$ still saturates it.

Hint 1

The change in $X^a$ is proportional to $x_a e^{-(gvr)^2}$ near the core. Use the completed radial energy rather than expanding four long terms independently.

Hint 2

The two nonzero square residuals are $\varepsilon Kh$ and $\varepsilon\rho h'$. The surface term has the same endpoints as before.

Solution

The added scalar is $-\varepsilon gv^2 x_a e^{-(gvr)^2}$, a smooth Cartesian function, and decays at infinity. It changes neither the asymptotic scalar direction nor the flux. Also $H_\varepsilon(0)=0$, so the boundary term still runs from zero to one.

Using the original profile equations,

$$\mathcal E_\varepsilon-\frac{4\pi v}{g} =\frac{4\pi v}{g}\varepsilon^2 \int_0^\infty d\rho\, e^{-2\rho^2} \left[K^2\rho^2+\frac12\rho^2(1-2\rho^2)^2\right]. \tag{SW03.18}$$

This finite integral is strictly positive: its first term is positive for $\rho>0$. Thus every nonzero $\varepsilon$ loses saturation, including arbitrarily small deformations. Magnetic charge fixes a bound, not the energy of every configuration in that sector. The absence of a linear term checks that the original solution is stationary under this allowed deformation; positivity gives an additional check on the signs in the square completion.

3. Match one dyon in both languages

At theta zero take $(n_m,n_e)=(1,1)$ and $a_{\rm cl}=gv/\sqrt2>0$. Find $s/c$, its classical mass, and its ratio to the purely magnetic mass. Check the central-charge formula, and explain why this calculation alone does not establish a stable particle at every quantum value of $u$.

Hint 1

The ratio in the first-order equations is the ratio of physical fluxes, not the ratio of the two integer labels.

Hint 2

Compare $v\sqrt{q_m^2+q_e^2}$ with $\sqrt2|a_{\rm cl}(1+4\pi i/g^2)|$. Keep the square root before expanding at small $g$.

Solution

The fluxes give

$$\frac sc=\frac{g^2}{4\pi},\qquad M_{(1,1)}=\frac{4\pi v}{g} \sqrt{1+\frac{g^4}{16\pi^2}},\qquad \frac{M_{(1,1)}}{M_{(1,0)}}= \sqrt{1+\frac{g^4}{16\pi^2}}.$$

The square root is greater than one. At weak coupling the fractional excess is $g^4/(32\pi^2)+O(g^8)$, consistent with a small electric component compared with the magnetic flux. Substituting $a_{\rm cl}=gv/\sqrt2$ into the central charge gives the identical mass.

This identifies a semiclassical dyon and the normalization of its charge label. Away from weak coupling one needs the quantum periods to compute the bound, and separate information about the spectrum and possible decay channels to determine whether it is saturated by a stable particle. A charge lattice and a mass bound alone do not supply that information.

The surface integral has now become a physical central charge with a fixed electric and magnetic normalization. Next, derive the running coupling and its dynamical scale. The logarithm identifies weakly coupled vacua, while threshold matching separates their low-energy description from the microscopic theory.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading the first Seiberg–Witten paper