Dyons and KvBLL Instantons in QCD Tin Sulejmanpa si c Work done - PowerPoint PPT Presentation

Dyons and KvBLL Instantons in QCD Tin Sulejmanpaˇ si´ c Work done with E. Shuryak, F. Bruckmann and R. R¨ odle December 12, 2012 1 / 45

Outline Instantons from dyons The fermionic zero modes The fundamental zero modes The adjoint zero modes The caloron zero modes at finite chemical potential Interactions of dyons Classical interactions One loop effects (Debye screening) The Fermionic zeromode Interactions Chiral symmetry breaking via dyons Some lattice observations How topology breaks chiral symmetry The three models Shift of critical coupling as a function of N f Some comments on confinement Conclusion and outlook 2 / 45

KvBLL instantons ◮ Classical self-dual solutions of Yang-Mills eq. at finite temperature ◮ Generalizations of instantons with an additional parameter – � holonomy exp( i A 0 d τ ) ◮ When holonomy is “non-trivial”, they disassociate into static objects called dyons 3 / 45

KvBLL instantons ◮ Classical self-dual solutions of Yang-Mills eq. at finite temperature ◮ Generalizations of instantons with an additional parameter – � holonomy exp( i A 0 d τ ) ◮ When holonomy is “non-trivial”, they disassociate into static objects called dyons A 0 ( r → ∞ ) = v τ · ˆ ω – Acts like a an adjoint higgs field 2 µν = ( D i A 0 ) 2 + F 2 F 2 ij (Assuming time independent fields) 3 / 45

The dyon In the usual radial ansatz it looks One can see that functions as H = − 1 − vr coth( vr ) (3) A a r a 0 = H ( r )ˆ (1) r A = 1 v A a r j i = A ( r ) ǫ aij ˆ (2) r − (4) sinh( vr ) Imposing selfduality satisfy the self duality equations, F µν = 1 such that 2 ǫ µνρσ F ρσ � d 3 x F 2 = 4 π v with boundary condition which gives a fractional A ( r → ∞ ) = 0 , topological charge Q = v β 2 π . H ( r → ∞ ) = v 4 / 45

0.8 H(r) 0.6 0.4 A(r) 0.2 0.0 � 20 � 10 0 10 20 rv A a r a A a r j 0 = H ( r )ˆ i = A ( r ) ǫ aij ˆ 5 / 45

The dyon The fields look like r i A θ = o ( e − vr ) E i = ˆ r · τ ˆ A r = 0 , + o ( e − rv ) , (5) r 2 2 r ) τ 3 A 0 = ( v − 1 2 + o ( e − vr ) B i = E i (6) A ϕ = tan θ τ 3 Changing the gauge so that 2 2 + o ( e − vr ) r → τ 3 (which is not unique) r τ · ˆ reveals the abelian nature of the In this form the Dirac monopole solutions, for example field is evident, with the Dirac string along θ = π direction. 6 / 45

The second dyon Take v → 2 π T − v and using the time dependent gauge transform U ( t ) = e − it πτ 3 /β we get a new solution with same asymptotics A 0 ∼ ( v + 1 / r ) τ 3 2 but opposite charge of magnetic and electric fields, i.e. r i E i = − ˆ τ 3 2 , B i = E i . r 2 F 2 = 4 π (2 π − v β ) � This configuration, however, has and action topological charge ¯ Q = 1 − v β/ (2 π ), so that Q + ¯ Q = 1 7 / 45

The approximate higgs potential 6 5 H(r1,r2) 4 3 2 1 0 � 30 � 20 � 10 0 10 20 30 8 / 45

What can we conclude? 9 / 45

What can we conclude? ◮ There are two solutions which have A 0 → v τ 3 . 9 / 45

What can we conclude? ◮ There are two solutions which have A 0 → v τ 3 . ◮ Their individual topological charges are Q = v β/ (2 π ) and ¯ Q = 1 − v β/ (2 π ), where v ∈ [0 , π T ] is an angular variable. 9 / 45

What can we conclude? ◮ There are two solutions which have A 0 → v τ 3 . ◮ Their individual topological charges are Q = v β/ (2 π ) and ¯ Q = 1 − v β/ (2 π ), where v ∈ [0 , π T ] is an angular variable. ◮ The total topological charge of two objects is unity Q + ¯ Q = 1. 9 / 45

What can we conclude? ◮ There are two solutions which have A 0 → v τ 3 . ◮ Their individual topological charges are Q = v β/ (2 π ) and ¯ Q = 1 − v β/ (2 π ), where v ∈ [0 , π T ] is an angular variable. ◮ The total topological charge of two objects is unity Q + ¯ Q = 1. ◮ These are, in fact, two parts of the same object! 9 / 45

What can we conclude? ◮ There are two solutions which have A 0 → v τ 3 . ◮ Their individual topological charges are Q = v β/ (2 π ) and ¯ Q = 1 − v β/ (2 π ), where v ∈ [0 , π T ] is an angular variable. ◮ The total topological charge of two objects is unity Q + ¯ Q = 1. ◮ These are, in fact, two parts of the same object! ◮ At v = 0 the “heavy”’ dyon, with topological charge ¯ Q reduces to Harrington-Shepard caloron of infinite size (know to be a monopole) 9 / 45

What can we conclude? ◮ There are two solutions which have A 0 → v τ 3 . ◮ Their individual topological charges are Q = v β/ (2 π ) and ¯ Q = 1 − v β/ (2 π ), where v ∈ [0 , π T ] is an angular variable. ◮ The total topological charge of two objects is unity Q + ¯ Q = 1. ◮ These are, in fact, two parts of the same object! ◮ At v = 0 the “heavy”’ dyon, with topological charge ¯ Q reduces to Harrington-Shepard caloron of infinite size (know to be a monopole) ◮ Instanton has constituents! And their “masses” (action) depend on the holonomy parameter v . 9 / 45

What can we conclude? ◮ There are two solutions which have A 0 → v τ 3 . ◮ Their individual topological charges are Q = v β/ (2 π ) and ¯ Q = 1 − v β/ (2 π ), where v ∈ [0 , π T ] is an angular variable. ◮ The total topological charge of two objects is unity Q + ¯ Q = 1. ◮ These are, in fact, two parts of the same object! ◮ At v = 0 the “heavy”’ dyon, with topological charge ¯ Q reduces to Harrington-Shepard caloron of infinite size (know to be a monopole) ◮ Instanton has constituents! And their “masses” (action) depend on the holonomy parameter v . ◮ Indeed Kraan and van Baal found this solution! 9 / 45

The exact KvBLL Caloron 10 / 45

The exact KvBLL Caloron 10 / 45

The exact KvBLL Caloron 10 / 45

The exact KvBLL Caloron 10 / 45

The exact KvBLL Caloron 10 / 45

The exact KvBLL Caloron 10 / 45

The exact KvBLL Caloron 10 / 45

The fermionic zero modes 10 / 45

Fermionic zero modes Now we look for the solution of Dirac equation in the background of a single dyon. The result is well known from the 70s for 3D theory. Writing the equation for a chiral fermion r · τ ) ǫ ] A α e − i ϕ t /β , ψ α A ( r ) = [( α 1 ( r ) 1 + α 2 ( r ) ˆ we obtain for the Dirac equation d α 1 ( r ) + H + 2 A α 1 + ϕ T α 2 = 0 , (7) 2 dr � H − 2 A � d α 2 ( r ) + 2 + α 2 + ϕ T α 1 = 0 . (8) dr 2 r with φ = 0 (periodic fermions for now) we can get that / D ψ = 0 results in α 1 ( r ) = const tanh( vr / 2) ∼ e − vr / 2 , α 2 ( r ) = 0 . � vr sinh( vr ) But we can do better! In fact one can solve the Dirac equation with general φ . 11 / 45

The solution turns out to be (Shuryak, Sulejmanpasic - Phys. Rev. D86 036001) χ 1 , 2 ( vr ) α 1 , 2 ( r , ϕ ) = c (9a) � vr sinh( vr ) 2 ϕ � � χ 1 ( rv ) = v sinh( r ϕ/β ) − tanh( vr / 2) cosh( r ϕ/β ) , (9b) − 2 ϕ � � χ 2 ( rv ) = v cosh( r ϕ/β ) + coth( vr / 2) sinh( r ϕ/β ) , (9c) 12 / 45

Α 1 � r � � solid, Α 2 � r � � dashed 1.0 The answer lies in the second 0.8 dyon! Recall that the second 0.6 dyon differs from the first by a 0.4 0.2 time dependent gauge transform r 5 10 15 20 (in stringy gauge) � 0.2 � 0.4 exp( i π tT τ 3 ) Functions α 1 (solid) and α 2 (dashed) for ϕ = 0 (black) to ϕ = 0 . 55 v /β Therefore the solution has to be multiplied by an anti periodic When ϕ ≥ v β/ 2 the solution gauge transformation! This diverges. But since v ∈ [0 , π T ], means that the solution for anti what happens to antieriodic periodic fermions has the same fermions? profiles, but with v → ¯ v 13 / 45

The adjoint zero modes The Dirac equation: 1.0 1 + 2 = ϕ v α ′ r α 1 − 2 sinh( vr ) β 1 β α 2 , 0.5 v = ϕ α ′ 2 + 2 sinh( vr ) β 2 β α 1 , 0.0 = ϕ v β ′ 1 − sinh( vr ) α 1 + v coth( vr ) β 1 β β 2 , � 0.5 0 2 4 6 8 10 12 14 v = ϕ β 2 + sinh( vr ) α 2 + v coth( vr ) β 2 β β 1 . ϕ = (0 , 0 . 2 , 0 . 4 , 0 . 6 , 0 . 8 , 0 . 9 , 0 . 95) v β ψ m = ˆ r m α 2 ( r ) − i (ˆ r × � σ ) m β 2 ( r )+ α 1 ( r )ˆ r m ( � σ · ˆ r )+((ˆ r × � σ ) × ˆ r ) β 1 ( r ) and Ψ a = ψ a ε e i ϕ t /β ε - arbitrary 2-spinor 14 / 45

The adjoint zero modes The leading order Ψ M ∼ e − ( v − ϕ/β ) r Similarly we can construct a zero mode on top of the “heavy” dyon v − ϕ/β ) r . Ψ L ∼ e − (¯ v = 2 π − v ¯ However there are also solutions with ϕ → ϕ − 2 π Ψ L ∼ e − ( ϕ/β − v ) r . Ψ M ∼ e − ( ϕ/β − ¯ ˜ v ) r ˜ 15 / 45

The adjoint hopping v v 16 / 45

The adjoint hopping v Each dyon has a zeromode v 16 / 45

The adjoint hopping v L (heavy) dyon has a zeromode v 16 / 45

The adjoint hopping v v 16 / 45

Zeromodes at finite µ (with Bruckmann and R¨ odl) One needs to solve 0.8 ( / D + µγ 0 ) ψ = 0 Ψ 0 � �Μ � Ψ 0 � Μ ��� Tv 3 � 0.6 0.4 This is the same as solving the Dirac equation with Ψ = ψ e µ t , 0.2 so µ = i ϕ/β . However, since the 0.0 operator does not have any 1 2 3 4 rv Hermiticity property, it turns out The full solution for the caloron that one needs to redefine the is complicated, but will be bra vector and use ψ † ( − µ ) published soon! instead of ψ † ( µ ). 17 / 45

Interactions of dyons 17 / 45