non abelian strings and monopoles
play

Non-Abelian strings and monopoles in supersymmetric gauge theories - PowerPoint PPT Presentation

Non-Abelian strings and monopoles in supersymmetric gauge theories Mikhail Shifman and Alexei Yung 1 Introduction Seiberg and Witten 1994 : Abelian confinement in N = 2 QCD Cascade gauge symmetry breaking: SU(N) U(1) N 1 VEVs of


  1. Non-Abelian strings and monopoles in supersymmetric gauge theories Mikhail Shifman and Alexei Yung

  2. 1 Introduction Seiberg and Witten 1994 : Abelian confinement in N = 2 QCD Cascade gauge symmetry breaking: • SU(N) → U(1) N − 1 VEV’s of adjoint scalars • U(1) N − 1 → 0 (or discrete subgroup) VEV’s of quarks/monopoles At the last stage Abelian Abrikosov-Nielsen-Olesen flux tubes are formed. π 1 ( U (1) N − 1 ) = Z N − 1 ( N − 1) infinite towers of strings. In particular ( N − 1) elementary strings → Too many degenerative hadron states

  3. In search for non-Abelian confinement non-Abelian strings were suggested in N = 2 U(N) QCD Hanany, Tong 2003 Auzzi, Bolognesi, Evslin, Konishi, Yung 2003 Shifman Yung 2004 Hanany Tong 2004 Z N Abelian string: Flux directed in the Cartan subalgebra, say for SO (3) = SU (2) /Z 2 flux ∼ τ 3 Non-Abelian string : Orientational zero modes Rotation of color flux inside SU(N).

  4. 2 Bulk theory N = 2 QCD with gauge group U ( N ) = SU ( N ) × U (1) and N f = N flavors of fundamental matter – quarks + Fayet-Iliopoulos term of U(1) factor The bosonic part of the action � 1 � 2 + 1 ( F µν ) 2 + 1 | D µ a a | 2 + 1 � � | ∂ µ a | 2 d 4 x F a S = µν 4 g 2 4 g 2 g 2 g 2 2 1 2 1 2 + 2 + V ( q A , ˜ � � � ∇ µ q A � � A � � ∇ µ ¯ q A , a a , a ) + q ˜ . � � � � � � Here ∇ µ = ∂ µ − i µ T a . 2 A µ − iA a

  5. The potential is � i � 2 g 2 a b a c + ¯ q A T a q A − ˜ q A T a ¯ A V ( q A , ˜ q A , a a , a ) 2 f abc ¯ = q ˜ g 2 2 2 g 2 A − Nξ � 2 q A q A − ˜ � q A ¯ 1 + ¯ q ˜ 8 2 + g 2 2 � q A T a q A � � q A q A � 2 g 2 1 + � ˜ � ˜ � � � � 2 � 2 � √ N 1 �� 2 2 m A + 2 T a a a ) q A � � + � ( a + � � 2 � A =1 √ 2 � � A � 2 m A + 2 T a a a )¯ + � ( a + q ˜ . � � �

  6. Vacuum   m 1 . . . 0 � Φ � = � 1 2 a + T a a a � = − 1   √    , . . . . . . . . .   2    0 . . . m N For special choice m 1 = m 2 = ... = m N U(N) gauge group is classically unbroken.   1 . . . 0   � kA � = 0 , � ¯ � q kA �   = ξ  , q ˜ . . . . . . . . .      0 . . . 1 k = 1 , ..., N A = 1 , ..., N ,

  7. Note • Color-flavor locking Both gauge U ( N ) and flavor SU ( N ) are broken, however diagonal SU ( N ) C + F is unbroken � q � → U � q � U − 1 � a � → U � a � U − 1 • Two ways to make it valid in quantum regime: • N f > 2 N The theory is not assymptotically free and stays at weak coupling ( Argyres, Plesser and Seiberg,1996 ). • Another way to stay at weak coupling: � ξ ≫ Λ √ ξ 8 π 2 Λ ≫ 1 2 ( ξ ) = N log g 2

  8. 3 Non-Abelian strings Z N string solution   φ 2 ( r ) 0 ... 0       ... ... ... ...     q = ,     0 ... φ 2 ( r ) 0         e iα φ 1 ( r ) 0 0 ...   1 ... 0 0       ... ... ... ... = 1   A SU( N )   ( ∂ i α ) [ − 1 + f NA ( r )] , i   N   0 ... 1 0         0 0 ... − ( N − 1)

  9. ≡ 1 2 A i = 1 A U(1) N ( ∂ i α ) [1 − f ( r )] i Magnetic U(1) flux of this Z N string is d 2 xF 12 = 4 π � N

  10. BPS string. First order equations r d dr φ 1 ( r ) − 1 N ( f ( r ) + ( N − 1) f NA ( r )) φ 1 ( r ) = 0 , r d dr φ 2 ( r ) − 1 N ( f ( r ) − f NA ( r )) φ 2 ( r ) = 0 , drf ( r ) + g 2 − 1 d 1 N ( N − 1) φ 2 ( r ) 2 + φ 1 ( r ) 2 − Nξ � � = 0 , r 4 drf NA ( r ) + g 2 − 1 d φ 1 ( r ) 2 − φ 2 ( r ) 2 � � 2 = 0 . r 2 Tension of the elementary Z N string T = 2 π ξ

  11. Profile functions of the string (for N = 2) 1 φ 2 φ 1 f 3 f r

  12. Non-Abelian string l     1 ... 0 0                       ... ... ... ...   1 p + 1       U − 1 = − n l n ∗ N δ l   U p ,   N   0 ... 1 0                           0 0 ... − ( N − 1)     p with l n l = 1 n ∗ Then 1 n · n ∗ − 1 � � N [( N − 1) φ 2 + φ 1 ] + ( φ 1 − φ 2 ) q = , N n · n ∗ − 1 x j � � A SU( N ) = ε ij r 2 f NA ( r ) , i N 1 x j A U(1) = N ε ij r 2 f ( r ) , i

  13. CP ( N ) model on the string 4 String moduli: x 0 i , i = 1 , 2 and n l , l = 1 , ..., N Make them t, z -dependent Z N solution breaks SU ( N ) C + F down to SU ( N − 1) × U (1) Thus the orientational moduli space is SU( N ) SU( N − 1) × U(1) ∼ CP( N − 1) � S (1+1) = 2 β � ( ∂ k n ∗ ∂ k n ) + ( n ∗ ∂ k n ) 2 � dt dz where the coupling constant β is given by a normalizing integral � φ 2 � ∞ � � 2 � β = 2 π − d NA + d = 2 π r f 2 1 dr drf NA + drf NA g 2 φ 2 g 2 0 2 2 2

  14. Gauge theory formulation of CP ( N − 1) model � d 2 x |∇ k n l | 2 , S CP ( N − 1) = 2 β where ∇ k = ∂ k − iA k n l , N complex fields l = 1 , ..., N Constraint: | n l | 2 = 1 . Gauge field can be eliminated: A k = − i ↔ ∂ k n l 2 ¯ n l Number of degrees of freedom = 2 N − 1 − 1 = 2( N − 1)

  15. 5 Confined monopoles Classical picture Z N Abelian strings ⇐ ⇒ N classical vacua of CP ( N − 1) model n · n ∗ − 1 x j � � A SU( N ) = ε ij r 2 f NA ( r ) i N n l = δ ll 0 CP ( N − 1) classical vacua: ⇒ confinement of monopoles Higgs phase for quarks = Elementary monopoles – junctions of two Z N strings monopole flux = 4 π × diag 1 2 { ... 0 , 1 , − 1 , 0 , ... }

  16. In 2D CP ( N − 1) model on the string we have N vacua = N Z N strings and kinks interpolating between these vacua Kinks = confined monopoles monopole string 1 string 2 4D vacuum 1 vacuum 2 2D kink

  17. Quantum picture Non-Abelian limit m 1 = m 2 = ... = m N   m 1 . . . 0 � Φ � = � 1 2 a + T a a a � = − 1   √    , . . . . . . . . .   2    0 . . . m N t’Hooft-Polyakov unconfined monopoles: M monopole = 4 π | m l 0 +1 − m l 0 | → 0 g 2 2 monopole size ∼ 1 /m W ∼ 1 / | m l 0 +1 − m l 0 | → ∞ Classically monopole disappear

  18. Confined monopoles = kinks are stabilized by quantum (non-perturbative) effects in CP ( N − 1) model on the string worldsheet Classically n l develop VEV ( | n | 2 = 1) There are 2( N − 1) massless Goldstone states. In quantum theory this does not happen SU ( N ) C + F global symmetry is unbroken Mass gap ∼ Λ CP no massless states ( �| n | 2 � = 0) M monopole = M kink ∼ Λ CP monopole size ∼ Λ − 1 CP Λ -1 - ξ 1/2

  19. 6 Less supersymmetry • bulk N = 2 SUSY = ⇒ N = (2 , 2) CP ( N − 1) on the string Hanany, Tong 2003 Auzzi, Bolognesi, Evslin, Konishi, Yung 2003 Shifman, Yung 2004 Hanany, Tong 2004 • bulk N = 1 SUSY = ⇒ N = (0 , 2) hetrotic CP ( N − 1) on the string Edalati, Tong 2007 Tong 2007 Shifman, Yung 2008 Shifman, Yung 2008 • bulk non-SUSY = ⇒ non-SUSY CP ( N − 1) on the string Gorsky, Shifman, Yung 2004 Gorsky, Shifman, Yung 2005

  20. 7 Large N solutions Witten 1979 : solved N = (2 , 2) and non-SUSY CP ( N − 1) in large N appreoximation Shifman, Yung 2008 : generalized Witten’s solution to N = (0 , 2) CP ( N − 1) N = (2 , 2) CP ( N − 1) N = 2 Bulk = ⇒ Consider limit e 2 → ∞ in |∇ k n l | 2 + 1 kl + 1 e 2 | ∂ k σ | 2 + 1 � � d 2 x 4 e 2 F 2 2 e 2 D 2 S 1+1 = 2 | σ | 2 | n l | 2 + iD ( | n l | 2 − 2 β ) + fermions � + Complex scalar σ , A k and D form gauge multiplet.

  21. The model has U(1) axial symmetry which is broken by the chiral anomaly down to discrete subgroup Z 2 N ( Witten 1979 ). The field σ which is related to the fermion bilinear operator transforms under this symmetry as 2 πk N i σ, σ → e k = 1 , ..., N − 1 . Z 2 N symmetry is spontaneously broken by the condensation of σ down to Z 2 , Witten’s solution: ⇒ N = (2 , 2) SUSY is unbroken E vac = D = 0 =

  22. √ 2 πk N i 2 � σ � = Λ CP e k = 0 , ..., N − 1 . There are N strictly degenerate vacua σ σ ∼ ¯ ψ L ψ R

  23. N = (0 , 2) CP ( N − 1) N = 1 Bulk ⇒ = Bulk: W 3+1 = µ A 2 + ( A a ) 2 � � , 2 String: Consider limit e 2 → ∞ in |∇ k n l | 2 + 1 kl + 1 e 2 | ∂ k σ | 2 + 1 � � d 2 x 4 e 2 F 2 2 e 2 D 2 S 1+1 = 2 | σ | 2 | n l | 2 + iD ( | n l | 2 − 2 β ) + N 2 π u | σ | 2 + fermions � + u is the deformation parameter N = (2 , 2) → N = (0 , 2) ,  const β g 4 2 | µ | 2 W , small µ m 2 u =  const β log g 2  2 | µ | m W , large µ

  24. Solution: 2 | σ | 2 = Λ 2 e − u , iD = Λ 2 � 1 − e − u � . Vacuum energy E vac = N 4 π iD = N 4 π Λ 2 � 1 − e − u � . 2 Λ iD 2 2|σ| u SUSY is broken spontaneously

  25. √ N i e − u/ 2 2 πk 2 � σ � = Λ e k = 0 , ..., N − 1 . There are N strictly degenerate vacua σ

  26. non-SUSY CP ( N − 1) σ = 0 N vacua split  � 2  � 2 πk   E vac = const N Λ 2  1 + const CP N 

Recommend


More recommend