How an N=1 deformation produces confinement

The undeformed Coulomb vacuum has a massless photon. Giving a neutral scalar a mass cannot, by itself, remove that photon. A charged field must enter the mechanism. Near a Seiberg–Witten singularity the required field is already available: the light monopole.

Read BPS masses and particle stability first. We will add a small superpotential, solve for the new vacuum, and show how an ordinary Higgs mechanism in magnetic variables produces confinement of electric probes. The calculation uses flat four-dimensional spacetime and the microscopic global group $SU(2)$, with no dynamical fundamental matter. Keep $\Lambda>0$ and the established periods: the monopole point is $u=\Lambda^2$, where $a_D=0$.

Keep the light monopole in the action

Add the microscopic N=1 superpotential $\mu\,\operatorname{Tr}\Phi^2$, where $[\mu]=1$. Here $\mu$ is the coefficient of the operator whose expectation value is $u$; another normalization of a canonical adjoint mass would require a corresponding conversion. Work with $0<|\mu|/\Lambda\ll1$.

In the local magnetic description, write $b=a_D$. The neutral chiral superfield is $A_D$; the two independent charged chirals $M,\widetilde M$ have dual electric charges $+1,-1$. Denote their scalar components by $m,\widetilde m$. A tilde is not conjugation. The local superpotential is

$$W=\sqrt2\,A_DM\widetilde M+\mu\,u(A_D). \tag{SW13.1}$$

The first term is required by the undeformed N=2 interaction. The second represents the microscopic perturbation in the exact Coulomb coordinate. Seiberg–Witten (5.17) gives this expression and the following page explains its holomorphic protection. That protection concerns the superpotential, not an exact canonical Kähler metric or every massive observable.

Use the locally normalized hypermultiplet kinetic terms of the light-monopole theory. At a finite Wilsonian scale $E$, write $g_D=g_D(E)>0$ and

$$\mathcal L_{\rm kin} =-\frac{1}{4g_D^2}f_{D\mu\nu}f_D^{\mu\nu} +\frac{1}{g_D^2}|\partial_\mu b|^2 +|\mathscr D_\mu m|^2+|\mathscr D_\mu\widetilde m|^2, \quad \mathscr D_\mu m=(\partial_\mu-i v_{D\mu})m. \tag{SW13.2}$$

The sign in the derivative is reversed for $\widetilde m$. The coupling is in the kinetic coefficient; $v_D$ is not the canonical vector. A theta term does not affect the static vacuum calculation. Equation (SW13.2) is the leading local two-derivative approximation, with the monopole fields retained.

Eliminating the chiral auxiliaries uses the inverse metric $K^{b\bar b}=g_D^2$, while the charged metrics equal one. Eliminating $D_D$ uses $D_D^2/(2g_D^2)+D_D(|m|^2-|\widetilde m|^2)$. The resulting potential is

$$\begin{aligned} U={}&g_D^2\left|\sqrt2\,m\widetilde m+\mu u'(b)\right|^2 +2|b|^2\left(|m|^2+|\widetilde m|^2\right)\\ &+\frac{g_D^2}{2}\left(|m|^2-|\widetilde m|^2\right)^2 . \end{aligned} \tag{SW13.3}$$

Each term has dimension four and is nonnegative. In particular, the first coefficient is $g_D^2$, rather than one: forgetting the neutral scalar’s kinetic normalization would spoil the mass comparison below.

Solve the vacuum equations, including the phase

The equations for zero auxiliaries are

$$\sqrt2\,m\widetilde m+\mu u'(b)=0,\qquad b\,m=b\,\widetilde m=0,\qquad |m|=|\widetilde m|. \tag{SW13.4}$$

We need $u'(0)$ in the same branch as our physical periods. The near-monopole integral from the period calculation, with $\delta=u-\Lambda^2$, gives

$$b=\frac{i\delta}{2\Lambda} \left[1-\frac{\delta}{16\Lambda^2}+\cdots\right], \qquad u(b)=\Lambda^2-2i\Lambda b-\frac14b^2 +O(b^3/\Lambda). \tag{SW13.5}$$

For example, put $w=-2i\Lambda b$. Then $w=\delta-\delta^2/(16\Lambda^2)+\cdots$, so $\delta=w+w^2/(16\Lambda^2)+\cdots$. This recovers both the phase and the quadratic coefficient. The leading derivative agrees with the scale-restored original (6.27).

Since $u'(0)=-2i\Lambda\ne0$, the first equation requires a nonzero product. The other two then force

$$\boxed{\ b=0,\qquad m\widetilde m=i\sqrt2\,\mu\Lambda,\qquad |m|=|\widetilde m|=v,\quad v^2=\sqrt2\,|\mu|\Lambda.\ } \tag{SW13.6}$$

Let $\vartheta=\arg(i\mu)$. A convenient gauge representative is $m=\widetilde m=v e^{i\vartheta/2}$. The gauge rotation acts oppositely on their phases, so it removes their phase difference while their sum is fixed by (SW13.6). Changing both square-root signs is the gauge rotation with angle $\pi$; it does not create another vacuum. This explicitly solves the source’s (5.18)–(5.21).

The charged expectation values describe a gauge-fixed Higgs phase, not a gauge-invariant order parameter that violates gauge symmetry. Their product and the physical mass below are gauge invariant. The holomorphic product is fixed in the stated superfield normalization; a finite field renormalization must be accompanied by its kinetic and superpotential conversion.

Action and vacuum supersymmetry

For this section use the vacuum test with F and D fields. Before the perturbation the action has N=2 supersymmetry: eight real supercharges, four manifest in N=1 notation. A fixed nonzero $\mu$ explicitly leaves N=1, with four real supercharges. At (SW13.6), all its auxiliaries vanish and the homogeneous vacuum has zero potential. It preserves all four remaining charges. There is no Goldstino for spontaneous breaking of those charges.

There is a second vacuum near $u=-\Lambda^2$. In the previously chosen continuation the light charge there is the dyon $(n_m,n_e)=(1,-1)$, and its local electric coordinate is $a_D-a$. One must repeat the local construction in that frame; the monopole and dyon are mutually nonlocal and cannot both be elementary charged fields of this one Abelian Lagrangian. The two singularities give the two vacua exchanged by the broken discrete chiral symmetry, as explained after the source’s (5.21). They are not the two signs of the square root at one singularity.

Calculate the photon mass and the range of control

Expand the charged kinetic terms around (SW13.6). Their terms quadratic in $v_{D\mu}$ sum to $2v^2v_{D\mu}v_D^\mu$. Set $v_{D\mu}=g_D A_\mu^{\rm can}$ to obtain a canonical Maxwell term. Comparing with $m_\gamma^2 A_\mu^{\rm can}A_{\rm can}^\mu/2$ gives

$$\boxed{\ m_\gamma^2=4g_D^2v^2 =4\sqrt2\,g_D^2|\mu|\Lambda.\ } \tag{SW13.7}$$

The relative phase of the two condensates supplies the longitudinal polarization. The photon acquires a mass even though $b=0$. Indeed $\sqrt2|b|$ was the monopole particle mass on the undeformed Coulomb branch; it is not the mass formula for fluctuations in this new Higgs vacuum.

The magnetic theory is infrared free. With the light fields present its leading running is $g_D^{-2}(E)=g_D^{-2}(E_0)+(4\pi^2)^{-1}\log(E_0/E)$. For the massive vacuum this running stops at scales of order the Higgs masses. Use a matched coupling there in (SW13.7), rather than inserting $b=0$ into the singular photon-only logarithm. That logarithm had integrated out exactly the charged field that condenses. The cutoff by the new mass scale is the mechanism described in section 5.6, printed page 32.

The hierarchy is $v/\Lambda=2^{1/4}\sqrt{|\mu|/\Lambda}\ll1$ and $m_\gamma/\Lambda=2g_Dv/\Lambda\ll1$. For sufficiently small deformation the dual coupling is also weak. Higher-derivative and kinetic corrections can then be separated from this leading calculation. The protected superpotential alone does not make (SW13.7) an exact pole-mass formula at arbitrary $\mu$.

As a numerical example, choose $|\mu|/\Lambda=10^{-6}$ and a matched $g_D=0.4$. Then $v/\Lambda=1.1892071\times10^{-3}$ and $m_\gamma/\Lambda=9.5136569\times10^{-4}$. The specified coupling is input matching data, not a value deduced from the mass ratio alone.

See the electric flux tube in magnetic variables

A charged condensate expels magnetic flux of its own gauge field except in vortex cores. Here that field is the dual photon. Its magnetic flux is original electric flux, so separated original electric probes can be connected by a tube with energy proportional to its length.

We can exhibit the mechanism with its approximation explicit. Retain only the constant and linear terms of $u(b)$, hold $g_D$ fixed, and set

$$b=0,\qquad m=\frac{e^{i\vartheta/2}}{\sqrt2}\,q,\qquad \widetilde m=\frac{e^{i\vartheta/2}}{\sqrt2}\,\bar q,\qquad \xi=2v^2. \tag{SW13.8}$$

This is a consistent bosonic sector of that leading truncation; it is not a conjugacy constraint on the two independent chiral multiplets. Its kinetic term is $|\mathscr Dq|^2$, its D term vanishes, and its potential is $g_D^2(|q|^2-\xi)^2/2$.

For a straight static string in the third spatial direction, let $B_D=f_{D12}$. Its energy per unit length is

$$T=\int d^2x\left[ \frac{B_D^2}{2g_D^2}+|\mathscr D_1q|^2+|\mathscr D_2q|^2 +\frac{g_D^2}{2}(|q|^2-\xi)^2\right]. \tag{SW13.9}$$

Finite energy requires $|q|\to\sqrt\xi$ and $\mathscr D_iq\to0$ at large radius. If its phase winds $n$ times, Stokes’ theorem gives $\int B_D\,d^2x=\oint v_D=2\pi n$. For $n>0$, integrating the covariant-derivative cross term and completing the magnetic square gives

$$T=\int d^2x\left[ |(\mathscr D_1+i\mathscr D_2)q|^2+ \frac{\bigl[B_D-g_D^2(\xi-|q|^2)\bigr]^2}{2g_D^2} \right]+2\pi n\xi . \tag{SW13.10}$$

The omitted boundary derivative vanishes for the stated finite-energy asymptotics. Reverse both signs for negative winding. The leading model therefore has the bound $T\ge2\pi|n|\xi=4\pi|n|v^2$. Its unit-winding first-order equations can be checked using $q=\sqrt\xi f(\rho)e^{i\varphi}$, $v_D=a(\rho)d\varphi$ and $\rho=g_D\sqrt\xi\,r$:

$$f'=\frac{(1-a)f}{\rho},\qquad a'=\rho(1-f^2),\qquad f(0)=a(0)=0,\quad f(\infty)=a(\infty)=1. \tag{SW13.11}$$

These display a smooth core and localized flux. Substituting a numerical solution into the original energy (SW13.9), rather than just its completed square, checks the coefficient $2\pi\xi$.

This Bogomolny value belongs to the explicitly linearized local model. Restoring $u''(0)=-1/2$ sources $b$ inside a vortex, as Exercise 3 shows. Neither an exactly BPS string of the full mass-deformed theory nor an exact tension at finite $\mu/\Lambda$ follows from (SW13.10). The controlled leading mechanism is a mass gap and a positive flux-tube tension.

Which electric charge is confined? In the fixed first-paper units a W has $n_e=1$, while an allowed external fundamental $SU(2)$ probe has $n_e=\pm1/2$. This is the charge convention stated before the source’s (3.13). A fundamental probe has one unit of electromagnetic pairing with the unit monopole and terminates one elementary dual flux tube; an adjoint electric charge has an even pairing. Dynamical adjoint fields can screen integer electric charge, so an adjoint string need not remain unbreakable at arbitrarily long distance. With no fundamental matter, a probe with odd $SU(2)$ center charge cannot be screened this way. That is the asymptotic confinement statement here. Changing the global group or adding dynamical fundamentals changes the allowed probes and screening problem.

Exercises

1. Count the vacua without counting gauge copies

Let $\mu=|\mu|e^{i\beta}\ne0$. Solve (SW13.4) in the monopole patch including all scalar phases, quotient by the compact dual gauge rotation, and explain why a choice of square-root sign is not the second confining vacuum. What changes when $\mu=0$?

Hint 1

Write $m=v e^{i\alpha_1}$, $\widetilde m=v e^{i\alpha_2}$. Separate their sum and difference.

Hint 2

The charges are opposite. At zero deformation solve the product and D equations together before interpreting the photon.

Solution

The nonzero product and the two equations proportional to $b$ require $b=0$. Its phase is $\alpha_1+\alpha_2=\beta+\pi/2\pmod{2\pi}$, and $v^2=\sqrt2|\mu|\Lambda$. A gauge rotation shifts $(\alpha_1,\alpha_2)$ by $(\chi,-\chi)$, so every solution is equivalent to equal phases $(\beta+\pi/2)/2$. The remaining simultaneous sign change is $\chi=\pi$. There is one local gauge orbit.

The other vacuum belongs to the distinct dyon singularity and a different local frame. Counting a sign twice would incorrectly turn each of the two patches into two vacua.

At $\mu=0$, $m\widetilde m=0$ and $|m|=|\widetilde m|$ force both charged scalars to vanish. The remaining $b$ is arbitrary in the local Coulomb patch. At $b=0$ charged particles become massless, but they do not acquire a condensate. The undeformed photon remains massless. Thus the limit removes the Higgs phase and restores the Coulomb family; “a massless monopole” alone was never a confinement criterion.

2. Check that the leading vacuum is gapped

In the constant-coupling, linear-$u$ model take $\mu=-i|\mu|$, so the vacuum can be $m=\widetilde m=v>0$, and put $\phi=b/g_D$. Let $S=(\delta m+\delta\widetilde m)/\sqrt2$ and $D=(\delta m-\delta\widetilde m)/\sqrt2$; here $D$ names a scalar fluctuation, not the eliminated vector auxiliary. Find the quadratic superpotential, the real scalar left in $D$, and the eaten mode. Compare all physical masses with $m_\gamma$.

Hint 1

Expand $\sqrt2 g_D\phi(m\widetilde m-v^2)$ through second order in fluctuations.

Hint 2

Write $D=(d_R+i d_I)/\sqrt2$. The linear part of $|m|^2-|\widetilde m|^2$ is $2v d_R$.

Solution

The canonical kinetic terms are $|\partial\phi|^2+|\partial S|^2+|\partial D|^2$. The quadratic superpotential is

$$W^{(2)}=2g_Dv\,\phi S=m_\gamma\phi S.$$

It gives $m_\gamma^2(|\phi|^2+|S|^2)$, so these four real scalars have mass $m_\gamma$. Their two Weyl fermions have off-diagonal mass $m_\gamma$, hence the same physical mass.

The D-term potential is $g_D^2(2v d_R)^2/2=m_\gamma^2d_R^2/2$. Thus $d_R$ has mass $m_\gamma$. The relative phase $d_I$ is the gauge direction and is eaten; it is not a physical massless modulus. The gauge Yukawa term mixes the gaugino with the relative hypermultiplet fermion with mass $2g_Dv$.

There are three massive vector polarizations and five real physical scalars, matched by four massive Weyl fermions. All have mass $m_\gamma$ in this leading model. This degeneracy checks its coefficients; it is not a claim that the full N=1 deformation retains eight exact supercharges or this entire degeneracy after the omitted terms are restored.

3. Test the vortex after restoring the next term

Keep the full potential (SW13.3) through the quadratic term of $u(b)$. On the leading vortex ansatz (SW13.8), compute $\partial U/\partial\bar b$ at $b=0$. Is that ansatz a solution of the full $b$ equation inside the core? What does the answer allow you to conclude about the string bound just derived?

Hint 1

Only the first square can give a nonzero derivative at $b=0$. Differentiate its complex-conjugate factor.

Hint 2

Use $\sqrt2 m\widetilde m+\mu u'(0)=e^{i\vartheta}(|q|^2-\xi)/\sqrt2$. This expression vanishes in the vacuum, not throughout a vortex.

Solution

At constant coupling,

$$\left.\frac{\partial U}{\partial\bar b}\right|_{b=0} =g_D^2\bar\mu\,\overline{u''(0)} \left[\sqrt2m\widetilde m+\mu u'(0)\right] =-\frac{g_D^2\bar\mu e^{i\vartheta}}{2\sqrt2} \left(|q|^2-\xi\right).$$

It is nonzero where the condensate is depleted. A configuration with identically zero $b$ has no gradient term to balance that force, so the full solution must adjust its neutral scalar profile. In the homogeneous vacuum $|q|^2=\xi$, the force vanishes, consistently with (SW13.6).

The completed square was an exact identity for the leading reduced energy. It supplies its Bogomolny bound and leading tension. It does not prove an exact bound saturated by a string of the full theory once that reduced sector is no longer a solution. Small $|\mu|/\Lambda$, a weak matched magnetic coupling and momenta below the other thresholds make the leading analysis useful; they do not turn an omitted interaction into zero. An exact finite-deformation tension would require solving the enlarged effective problem and controlling its kinetic and higher-derivative corrections.

The mechanism is now concrete: a holomorphic perturbation selects isolated supersymmetric vacua, a charged condensate gives the dual photon a mass, and its vortices carry original electric flux. Continue with fundamental matter, masses and decoupling to see how additional matter changes the singularities and the screening charges.