PHY 396 T: SUSY Solutions or problem set #. Problem a: In priniple the non-perturbative superpotential o the theory may depend on the dual quark and antiquark ields q and q as well as the singlets Φ but or simpliity let s ous on the terms whih depend only on the Φ s. In the m 0 limit the theory has anomaly-ree SUN SUN symmetry and the only holomorphi ombination o the Φ ields whih is invariant under this symmetry is the determinant o the Φ matrix. Thereore the non-perturbative superpotential depends on the Φ ields only via the determinant detφ W n.p. Φ λ Λ W n.p. detφ λ Λ. S. To work out the dependene o this superpotential on the Yukawa oupling λ onsider a ield redeinition Φ e t Φ aompanied by the oupling redeinition λ e t λ. In light o equation µ λ 6π N Λ 3N N Λ 3N N /N.5 rom the previous homework set the redeinition o λ leads to µ e t µ and hene µφ µφ whih preserves the duality relation µφ M. Consequently the entire superpotential o the theory inluding both the tree-level and the non-perturbative terms should be invariant under this redeinition. Thereore the Φ ields enter the superpotential only in ombinations λφ Yukawa oupling λ only in ombination λφ hene or µφ. In partiular the W n.p. depends on the Φ and the W n.p. detφ λ Λ W n.p. detλφ Λ. S. Next onsider the axial U symmetry o the theory q e+ia q q e+ia q Φ e ia Φ
Θ Θ + N a Λ 3N N e+in a Λ 3N N. S.3 where the seond line ompensates or the axial anomaly. The non-perturbative superpotential should be invariant under this symmetry whih restrit its dependene on the detλφ and the Λ to the invariant ombination detλφ Λ 3N N thus W n.p. Φ λ Λ F detλφ Λ 3N N S.4 or some holomorphi untion F. To ind this untion onsider the anomaly ree R symmetry whose harges you should have ound in problem b o the previous homework: R µ 0 Rgaugino + Rsquark Rantisquark N N Rquark Rantiquark N N S.5 RΦ + N N Rψ φ + N N Rsuperpotential + RΛ 0 where the last line ollows rom the olor anomaly anellation. In partiular detλφ Λ 3N N has R N N N + 0 +N S.6 while F detλφ Λ 3N N should have R +. S.7
These R harges immediately imply F detλφ Λ 3N N detλφ Λ 3N N /N a numeri onstant S.8 and hene W n.p. Φ λ Λ detλφ Λ 3N N /N a numeri onstant. For uture reerene I am going to denote this onstant C N. Note: or the problem at hand 3N < N the theory is IR-ree and Λ is the UV Landau pole rather than an IR sale. The irrelevant operators originating at the Λ sale should arry negative powers o Λ and indeed the non-perturbative superpotential arries a negative power o Λ. Problem b: The IR-ree theory does not have a gaugino ondensate and hene does give quantum orretions to the squark-antisquark bilinears M q q. Consequently the bilinear matrix has rank N < N so it annot anel all the O Raieartaigh terms µm. Without the non-perturbative superpotential terms the un-aneled O Raieartaigh terms would lead to spontaneous supersymmetry breaking see arxiv:/hep-th/06039 by Intriligator Seiberg and Shih. On the other hand the dual theory with rankm N has N vaua with S Λ 3N N /N detm M m S 6π. stable supersymmetri S.9 To get similar stable supersymmetri vaua in the theory we need the non-perturbative superpotential. In general hiral VEVs in supersymmetri vaua o any theory ollow rom the onshell hiral ring equations but sine the theory is IR-ree we may use the lassial ield 3
equations W net q 0 W net q 0 W net Φ 0. S.0 In partiular the squark obey q Φ 0 Φ q 0 S. so or rank Φ N as we shall see momentarily all squarks and antisquarks have zero VEVs q q 0. s to the Φ ields we have log detλφ Φ Φ S. and hene W n.p. Φ Φ C detλφ Λ 3N N /N. S.3 where C or rather C N o the theory we must have is the numeri onstant in eq.. Thereore in a SUSY vauum Φ CdetλΦ Λ 3N N /N W tree +µm Φ S.4 or equivalently the matrix produt m Φ N N C µ detλφ Λ 3N N /N. S.5 In partiular we must have rankφ N and hene zero squark VEVs. 4
To solve the equation S.5 let s take the determinants o both sides o the equation thus hene detm detφ N C detλφ Λ 3N N N /N µ S.6 N N detm] detφ] N C N ] λ N N detφ µ Λ 3N NF ] n S.7 and thereore ] µ N N N N detφ detm µ N Λ N 3N λ N C N N detm] N. S.8 On the right hand side here eq..5 or the µ ator gives us µ N Λ N 3N λ N C N P Λ 3N N S.9 where P N 6π N C N S.0 is a numeri onstant. On the let hand side o S.8 µ N detφ detm det µφ m S. and sine aording to eq. S.5 µφ m is proportional to a unit N N matrix µφ m X N N det µφ m X N. S. Plugging all these ormulae bak into eq. S.8gives us X N N P Λ 3N N detm] N S.3 and hene µ Φ m P Λ 3N N detm] /N. S.4 Comparing this ormula to the eq. S.9 or the meson VEVs o the dual theory we 5
immediately see peret agreement between µφ and M up to an overall numeri onstant. Quod erat demonstrandum. PS: To get µφ M without any extra numeri onstants we need P 6π N. S.5 In light o eq. S.0 this alls or C N /N 6π S.6 In other words the non-perturbative superpotential o the theory should be W n.p. N 6π /N det λφ S.7 Λ N 3N with this partiular numeri ator inluding the signs. Problem the exerise part: Consider a dyon in eetive QED whih obtains rom Higgsing o an SU gauge theory by VEV o a salar triplet. I the salars are omplex as in supersymmetri theories their VEVs should have orm Φ a 0 0 eiα v modulo gauge symmetry S.8 or some phase α. In my notations below I assume omplex Φ a although he real salars would an be analyzed in a similar manner. The mass o the dyon omes rom the net energy o the magneti eletri and salar ields thus M dyon d 3 x E a + a + DΦ a + V Φ Φ S.9 where DΦ a are the ovariant spae derivatives o the salars and V Φ Φ V salar Φ Φ V salar vauum. S.30 Note: i V vauum 0 we must subtrat it rom the salar potential so that eq. S.9would 6
not mix the energy due to dyons existene with the vauum energy o the theory. Sine salar potential has its minimum in the vauum state the subtrated V Φ Φ is non-negative whih gives us a lower bound or the dyon s mass: M dyon d 3 x E a + a + DΦ a. S.3 In the integrand here E a + a E a + i a S.3 and hene E a + a + DΦ a E a + i a e iβ DΦ a + R e iβ DΦ a E a + i a S.33 or any omplex phase e iβ. The irst term in the RHS is non-negative so the seond term provides us with a lower bound or the LHS. Plugging this bound into the integral in eq. S.3 we arrive at M dyon R e iβ d 3 x DΦ a E a + i a. S.34 Moreover this bound must hold or any phase e iβ whih requires M dyon d 3 x DΦ a E a + i a. S.35. Now let s alulate the integral in this ormula. First let s extrat a total derivative rom the integrand DΦ a E a + i a Φ a E a + i a Φ a D E a + i D a. S.36 In the seond term here we may use the Yang Mills equations or the ovariant divergenes 7
o the SU eletri and magneti ields D a 0 D E a J a 0 S.37 where J a µ are the ovariant Yang Mills urrents. ssuming the only soure o these urrents are the salar ields we have J a µ igɛ ab Φ b D µ Φ Φ b D µ Φ. S.38 For a stati oniguration o the salar ields the time omponent µ 0 o this urrent vanishes whih leads to D E a D a 0. S.39 Thus the seond term in eq. S.36 vanishes whih allows us to onvert the spae integral in S.35 to a surae integral at ininite radius d 3 x DΦ a E a + i a d 3 x Φ a E a + i a R drea Φ a E a + i a radial. S.40 Far away rom the monopole the SU eletri and magneti ields are restrited to the unbroken U subgroup o the SU while the salar ields take their vauum values. In the unitary gauge Φ a ve iα 0 0 S.4 while E a + i a E + i EM 0 0. S.4 In other gauges the salar and gauge ields have dierent orms but the produt Φ a E a + i a ve iα E + i EM S.43 8
is gauge invariant and have the same orm as in the unitary gauge. Thereore R drea Φ a E a + i a radial ve iα R drea E + i radial EM. S.44 Finally or a dyon with eletri harge Q and magneti harge µ E EM Q n 4πR EM µ n 4πR S.45 and hene drea E + i radial R EM Q + iµ. S.46 Consequently d 3 x DΦ a E a + i a ve iα Q + iµ. S.47 Plugging this integral into the lower bound S.35 or the dyon mass we arrive at the ogomol nyĭ Prasad Sommerield PS bound or the dyon mass: M dyon v Q + iµ. S.48 9