What becomes massless at strong coupling?

Before this lesson: Electric–magnetic duality as a change of variables; Monopoles, dyons, and a physical BPS bound.

A singular coefficient in an effective action can mean that a field was removed just where it should have been kept. For example, expanding a massive propagator in powers of momentum produces coefficients proportional to inverse powers of its mass. Taking that mass to zero after the expansion is not a low-energy approximation.

Here the missing field will be a magnetic monopole. Read electric–magnetic dualization and the physical monopole and BPS bound first. We will construct a local theory in which this formerly solitonic particle is an ordinary charged field, then calculate the singularity obtained by integrating it out again.

The microscopic theory remains pure four-dimensional N=2 $SU(2)$ Yang–Mills: eight real supercharges and gauge rank one. We use flat Lorentzian signature $(+---)$ , with an N=1 presentation making four supercharges manifest. There is no adjoint mass deformation or Fayet–Iliopoulos term. A Coulomb-branch vacuum preserves all eight supercharges.

Start with a physical hypothesis, not a charge label

Keep the period column $(a_D,a)^T$ and charge row $(n_m,n_e)$ . The central charge and bound are

Z_\gamma=n_ma_D+n_ea,\qquad M_\gamma\ge\sqrt2\,|Z_\gamma|. \tag{SW07.1}

Equality applies to a BPS particle that exists. Assigning an integer pair does not create that particle. In particular, a zero of $a_D$ is a candidate massless-monopole location, not an existence proof by itself.

The local input is that near an isolated vacuum $u=u_0$ a single unit-monopole hypermultiplet becomes light, with $a_D(u_0)=0$ , while other relevant charged thresholds stay separated. The hypermultiplet includes the opposite-charge CPT partner. Its mass is $\sqrt2|a_D|$ on the nearby Coulomb branch. This is the physical interpretation proposed and tested in Seiberg–Witten, sections 5.2–5.4. The global solution supplies further tests; the calculation below determines the consequences of this local input.

The chosen duality frame has

b=a_D,\qquad b_D=-a,\qquad \tau_D=\frac{db_D}{db}=-\frac1\tau,\qquad (n_m,n_e)_D=(-n_e,n_m). \tag{SW07.2}

The monopole $(1,0)$ therefore has dual electric charge $+1$ . The opposite charge is $-1$ . They can couple locally to one dual photon. This is a change of effective variables; it does not turn the microscopic $SU(2)$ Lagrangian into a weakly coupled magnetic copy of itself.

Put the charged fields back into the action

Let $B$ be a neutral chiral superfield with lowest component $b$ , and $V_D$ the dual vector superfield. The full hypermultiplet consists of independent chiral superfields $M,\widetilde M$ of charges $+1,-1$ , with scalar components $m,\widetilde m$ . A tilde does not mean complex conjugation. Reserve $m_h$ for their positive physical mass.

At a fixed Wilsonian scale $\mu$ below the other charged thresholds, retain these fields. In locally normalized hypermultiplet variables their leading action is the dual vector action plus

\begin{aligned} K_h&=M^\dagger e^{2V_D}M+ \widetilde M e^{-2V_D}\widetilde M^\dagger,\\ W_h&=\sqrt2\,B M\widetilde M. \end{aligned} \tag{SW07.3}

All three chiral scalars have engineering dimension one. $K_h$ has dimension two and $W_h$ dimension three. The coupling is absorbed into $V_D$ and $b$ ; the vector kinetic coefficient is $G_D(\mu)=\operatorname{Im}\tau_{D,W}(\mu)/(4\pi)=1/g_D^2(\mu)$ . Thus $\mathscr D_\mu m=(\partial_\mu-i v_{D\mu})m$ . Rescaling to canonical vector and neutral scalar fields at this fixed scale restores the familiar cubic coefficient $\sqrt2 g_D$ . The interaction follows the N=2 hypermultiplet construction; the original paper’s (5.17) gives precisely (SW07.3) when its adjoint mass is zero.

The auxiliary terms make the coefficients testable. Put $W_i=\partial_i W_h$ . The chiral auxiliary density is $K_{i\bar j}F^i\bar F^{\bar j}+W_iF^i+\overline{W_iF^i}$ . Here $K_{b\bar b}=1/g_D^2$ and the two matter metrics equal one, so eliminating $F^i$ gives $K^{i\bar j}W_i\bar W_{\bar j}$ . The vector auxiliary density is $D_D^2/(2g_D^2)+D_D(|m|^2-|\widetilde m|^2)$ . Consequently

\begin{aligned} U={}&2|b|^2(|m|^2+|\widetilde m|^2) +2g_D^2|m\widetilde m|^2 +\frac{g_D^2}{2}(|m|^2-|\widetilde m|^2)^2\\ ={}&2|b|^2(|m|^2+|\widetilde m|^2) +\frac{g_D^2}{2}(|m|^2+|\widetilde m|^2)^2. \end{aligned} \tag{SW07.4}

In the Coulomb vacuum $m=\widetilde m=0$ , the two complex scalars have mass squared $2|b|^2$ . The fermion Hessian has $W_{M\widetilde M}=\sqrt2 b$ , producing one charged Dirac fermion with the same mass:

m_h=\sqrt2|b|. \tag{SW07.5}

This checks the central-charge normalization without inserting an extra trace factor into the Abelian action. At $b=0$ the hypermultiplet is massless. Equation (SW07.4) still forbids a nonzero condensate at finite positive $g_D(\mu)$ : masslessness alone is not monopole condensation or confinement.

Calculate the logarithmic threshold

The Abelian vector multiplet is neutral. The charged loop contains two Weyl fermions and two complex scalars of unit charge. Using the one-loop field-counting rule established in running couplings, its screening coefficient is

c_h=2\left(\frac23\right)+2\left(\frac13\right)=2,\qquad \mu\frac{dg_D}{d\mu}=\frac{g_D^3}{8\pi^2},\qquad \frac{d}{d\log\mu}\frac1{g_D^2}=-\frac1{4\pi^2}. \tag{SW07.6}

This sign is the opposite of the microscopic non-Abelian running. Lowering the energy weakens this magnetic QED. Its holomorphic Wilsonian logarithm is perturbatively one-loop exact in the N=2 two-derivative action; this does not assert an exact arbitrary-momentum scattering amplitude.

For $m_h>0$ , integrate down to its threshold and then remove the hypermultiplet only for external energies $E\ll m_h$ . If $\mu_0$ is a reference mass below the other thresholds, integration gives

\frac1{g_{D,\mathrm{below}}^2} =\frac1{g_{D,W}^2(\mu_0)} +\frac1{4\pi^2}\log\frac{\mu_0}{m_h} +\text{finite matching terms}. \tag{SW07.7}

The subscript “below” labels the photon-only coefficient after matching. Holomorphy completes the magnitude logarithm to a complex logarithm on a chosen local branch:

\boxed{\displaystyle \tau_D(b)=h(b)-\frac{i}{\pi}\log\frac b{\mu_0},\qquad \operatorname{Im}\tau_D= \operatorname{Im}h(b)+\frac1\pi\log\frac{\mu_0}{|b|}.} \tag{SW07.8}

The dimensionless function $h$ is analytic near $b=0$ and contains the regular matching data. In particular, the $\sqrt2$ in $m_h$ shifts its constant term. The singular coefficient is fixed; $h$ is not fixed by this loop. This reproduces the original paper’s (5.8), now with a dimensionless logarithm and the omitted regular part explicit.

Do not use (SW07.8) as the Wilsonian coupling of a theory that still contains all the monopole’s low-energy modes: that would count their threshold twice. At $b=0$ keep the hypermultiplet and a nonzero cutoff $\mu$ . The coupling runs with $\mu$ , and tends to zero only as $\mu\to0$ . The photon-only action instead has a vacuum-dependent logarithm and ceases to apply at any fixed nonzero $E$ sufficiently near $b=0$ .

Constrain the other period

Suppose $b$ is a good local coordinate: $b=c_0(u-u_0)+O((u-u_0)^2)$ with $c_0\ne0$ . Since $[u]=2$ , $[c_0]=-1$ . Integrating $da/db=-\tau_D$ on the same logarithm branch yields

\boxed{\displaystyle a(b)=a_0-\int_0^b h(w)\,dw +\frac{i}{\pi}\left[b\log\frac b{\mu_0}-b\right].} \tag{SW07.9}

Differentiating the bracket gives $\log(b/\mu_0)$ : the apparently harmless linear term is needed for this precise definition of $h$ . Thus one period vanishes analytically and the other has a $b\log b$ correction while tending to $a_0$ . Equivalently its leading nonanalytic dependence is $(u-u_0)\log(u-u_0)$ after introducing a scale inside that logarithm. This is the content of the original paper’s (5.9)–(5.10). We assume $a_0\ne0$ in this one-light-charge description; a simultaneously light mutually nonlocal sector would require a different analysis.

The loop has not determined $u_0$ , $a_0$ , $c_0$ or the analytic part. Nor does it show that a weak-coupling $W$ boson remains a stable particle throughout the intervening region. Continuing the logarithm around a loop will constrain how period patches fit together; that is the next calculation, rather than an additional local spectrum assumption.

Worked example: the two low-energy limits

As a local logarithmic model set $h=3i$ and take positive real $b$ . At $b=e^{-4\pi}\mu_0$ ,

\tau_D=7i,\qquad \tau=\frac i7,\qquad \frac{g_D^2}{4\pi}=\frac17,\qquad m_h=\sqrt2 e^{-4\pi}\mu_0\simeq4.93185\times10^{-6}\mu_0. \tag{SW07.10}

The dual loop parameter is $g_D^2/(16\pi^2)=1/(28\pi)$ . A probe with $E=10^{-8}\mu_0$ lies below this mass. But at $b=10^{-12}\mu_0$ , the same probe is above the charged mass and must include the hypermultiplet. The smaller threshold coupling does not rescue a theory with missing accessible particles. Unknown analytic terms would change the numerical model; the hierarchy test and the logarithmic slope do not depend on taking $h$ constant.

An independent loop check from a magnetic field

One can see both the logarithm and the failure of the truncation without assuming the beta function. For this check only, continue to flat Euclidean space with orientation $\epsilon_{1234}=+1$ and put the dual connection in a constant real magnetic field $f_{D12}=\mathcal B>0$ . This is a diagnostic external field, not a supersymmetric vacuum. Take the hypermultiplet mass positive by a phase convention. We compute the real magnetic response; its mass-phase anomaly belongs to the holomorphic coupling above.

The charged scalar Landau levels in the $(1,2)$ plane are $(2n+1)\mathcal B$ , with degeneracy $\mathcal B/(2\pi)$ per area. Integrating the two remaining momenta gives $1/(4\pi s)$ , where proper time $s$ has dimension minus two. Summing the geometric series gives the heat trace per four-volume

K_s(s)=\frac1{(4\pi s)^2}\frac{x}{\sinh x},\qquad K_f(s)=4\cosh x\,K_s(s),\qquad x=\mathcal B s. \tag{SW07.11}

The second formula follows from the two spin shifts $\pm\mathcal B$ , each with multiplicity two, in the squared Dirac operator. A complex scalar contributes $\operatorname{Tr}\log$ to the Euclidean action; a Dirac fermion contributes minus one half the logarithm of its squared operator. Using $\log A=-\int_0^\infty ds\,e^{-sA}/s$ up to subtraction constants, the two scalars and one Dirac fermion give

\frac{\Gamma_h}{\mathrm{Vol}_4} =\frac1{16\pi^2}\int_0^\infty\frac{ds}{s^3}e^{-m_h^2s} \underbrace{2x\tanh(x/2)}_{x^2-x^4/12+O(x^6)}. \tag{SW07.12}

The zero-field terms cancel. The quadratic term needs a gauge-kinetic subtraction. Subtracting its value at a positive reference mass $m_0$ gives

\Delta\mathcal L_{\mathcal B^2} =\frac{\mathcal B^2}{16\pi^2} \int_0^\infty\frac{ds}{s} \left(e^{-m_h^2s}-e^{-m_0^2s}\right) =\frac{\mathcal B^2}{8\pi^2}\log\frac{m_0}{m_h}. \tag{SW07.13}

Since the Maxwell density is $\mathcal B^2/(2g_D^2)$ , this is exactly (SW07.7)‘s logarithmic coefficient. After subtracting the quadratic term, the next finite term is

\mathcal L_{\mathcal B^4} =-\frac{\mathcal B^4}{192\pi^2m_h^4}, \qquad |\mathcal B|\ll m_h^2. \tag{SW07.14}

The proper-time integral here uses $\int_0^\infty ds\,s e^{-m_h^2s}=m_h^{-4}$ . This is an asymptotic weak-field expansion, not the full determinant. Its divergent coefficient as $m_h\to0$ diagnoses the loss of the photon-only truncation. Small momenta and small field strengths are separate tests. Direct Landau sums and convergent subtracted proper-time integrals provide numerical checks of these statements.

Exercises

1. Massless does not mean condensed

At a fixed nonzero Wilsonian cutoff with $g_D>0$ , find all zero-potential scalar configurations of (SW07.4), including $b=0$ . Check the scalar and fermion quadratic spectra about the Coulomb vacuum. Explain why substituting the photon-only limit $g_{D,\mathrm{below}}\to0$ into this matter theory gives an invalid argument for a new flat branch.

Hint 1

Combine the two quartic terms before imposing $b=0$ . Keep both independent complex scalars.

Hint 2

Use real canonical components $m=(x_1+ix_2)/\sqrt2$ and $\widetilde m=(x_3+ix_4)/\sqrt2$ . The fermion mass matrix is the Hessian of $W_h$ in the two charged chiral fields.

Solution

Both terms in the second line of (SW07.4) are nonnegative. The quartic vanishes only when $|m|^2+|\widetilde m|^2=0$ , so $m=\widetilde m=0$ for every $b$ . Gauge quotienting changes nothing at this point. The vacuum family is the Coulomb coordinate $b$ ; at $b=0$ there is no new nonzero Higgs vacuum for this one-hyper theory without an FI term.

The quadratic potential is $|b|^2\sum_{j=1}^4x_j^2$ , hence the real scalar Hessian is $2|b|^2\mathbf1_4$ . The charged fermion matrix is $\begin{pmatrix}0&\sqrt2 b\\\sqrt2 b&0\end{pmatrix}$ . Multiplication by its Hermitian conjugate gives $2|b|^2\mathbf1_2$ . Thus the two Weyl fields combine into one Dirac fermion with the same mass as the scalars; at $b=0$ all these quadratic masses vanish, but the quartic potential remains.

The zero of the limiting infrared coupling does not justify setting the finite-cutoff quartic identically to zero. More basically, $g_{D,\mathrm{below}}$ was obtained by removing the very matter fields whose potential is under discussion. Keeping the fields requires the Wilsonian coefficient at the chosen scale. The strictly free infrared limit forgets interactions that constrain the full theory’s vacua.

2. What information is contained in a logarithmic coefficient?

Replace the single hypermultiplet by full hypers of nonzero integer charges $q_j$ with superpotential $\sqrt2\sum_j q_j B M_j\widetilde M_j$ , and assume they all become light at $b=0$ . Derive the singular coefficient of $\tau_D$ and the coefficient of $\mathcal B^4$ . Can the logarithmic coefficient alone distinguish one charge-two hyper from four charge-one hypers?

Hint 1

The beta function contains charge squared. The physical masses are $\sqrt2|q_j b|$ , so the constant $\log|q_j|$ does not change the coefficient of $\log|b|$ .

Hint 2

In each heat trace replace $\mathcal B$ by $|q_j|\mathcal B$ and use that hypermultiplet’s mass. Compare which information survives in the quadratic and quartic terms.

Solution

Define $\kappa=\sum_jq_j^2$ . The charged contribution is $c_h=2\kappa$ , so

\tau_D=h(b)-\frac{i\kappa}{\pi}\log\frac b{\mu_0}.

Finite mass-dependent constants are absorbed into $h$ . A negative $q_j$ can also change a constant holomorphic phase; it does not change the singular coefficient. Each hyper contributes

-\frac{q_j^4\mathcal B^4}{192\pi^2m_j^4} =-\frac{\mathcal B^4}{768\pi^2|b|^4}.

Thus the quartic response counts hypers in this particular mass family, while the logarithm weighs charge squared. One charge-two hyper and four charge-one hypers both have $\kappa=4$ , but their quartic coefficients differ by a factor four. Each expression requires $|q_j\mathcal B|\ll m_j^2$ . Their elementary charged spectra and allowed screenings also differ; the logarithm does not specify global line probes. In particular, the local one-unit-hyper interpretation uses physical input beyond a generic logarithmic singularity.

3. Can the singular point be at finite scalar distance?

In a punctured local model take $h=iT$ with $T>0$ , choose $0<\rho\le R=\mu_0$ , and approach $b=0$ along $b=\rho>0$ . Use the scalar metric $ds_{\rm scalar}^2=G_D|db|^2$ , $G_D=\operatorname{Im}\tau_D/(4\pi)$ . Calculate the radial length from $0$ to $R$ in terms of the upper incomplete gamma function, or prove convergence directly. Does a finite answer allow the charged fields to remain integrated out at the endpoint?

Hint 1

Set $y=\log(R/\rho)$ . Then $d\rho=-R e^{-y}dy$ ; the logarithmic growth of the metric competes with an exponential.

Hint 2

Define $\Gamma(s,x)=\int_x^\infty t^{s-1}e^{-t}dt$ and shift $t=y+\pi T$ .

Solution

Since $\operatorname{Im}\tau_D=T+\log(R/\rho)/\pi$ , the radial length is

L=\frac R{2\pi}\int_0^\infty e^{-y}\sqrt{\pi T+y}\,dy =\frac R{2\pi}e^{\pi T}\Gamma\!\left(\frac32,\pi T\right).

It is finite: the integrand is bounded near $y=0$ and behaves as $e^{-y}\sqrt y$ at infinity. As a dimension check, $G_D$ is dimensionless and $L$ has the dimension of the scalar coordinate $b$ . A finite-length path suffices to show that the endpoint is not at infinite metric distance in this model; we have not claimed a globally shortest path in the full quantum moduli space.

The kinetic coefficient itself diverges, and $m_h=\sqrt2\rho$ vanishes. At any fixed nonzero probe energy the threshold condition eventually fails. Finite scalar distance therefore coexists with the need for extra light fields. It is not permission to extend the photon-only effective action through its own cutoff.

The local monopole theory predicts a definite logarithm, a definite charged threshold and a finite limiting value of the other period. It does not determine the entire Coulomb branch or its stable spectrum. Continue by following the fields around a singularity to calculate how the logarithmic periods change and how those changes fit together globally.