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 

Download Presentation
Download Policy: The content available on the website is offered to you 'AS IS' for your personal information and use only. It cannot be commercialized, licensed, or distributed on other websites without prior consent from the author. To download a presentation, simply click this link. If you encounter any difficulties during the download process, it's possible that the publisher has removed the file from their server.

Recommend


More recommend