Yang-Mills Theory in (2+1) Dimensions

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1 1/30 Yang-Mills Theory in (2+1) Dimensions Vacuum wave function, string tension, etc. V. P. NAIR City College of the CUNY Seminar at Rutgers University, April 19, 2005

2 Why is YM(2+1) interesting? Interesting in its own right YM(1+1) YM(2+1) YM(3+1) 2/30 No propagating Propagating degrees Highly nontrivial degrees of freedom of freedom, Exactly solvable Nontrivial Too difficult Dimensional coupling Super-renormalizable A real physical context for YM(2+1) Mass gap of YM(2+1) Magnetic screening mass of YM(3+1) at high temperatures

3 Magnetic screening necessary to define thermal perturbation theory due to magnetic-type infrared divergences YM(3+1) YM(3) as T 3/30 YM(2+1) by Wick rotation We can get an estimate of magnetic mass Work in collaboration with D. Karabali and Chanju Kim We use a Hamiltonian approach, some exact results are possible.

4 Hamiltonian Analysis Choose A 0 = 0 A i, i = 1, 2. A g i = g A i g 1 i g g 1 Wave functions are gauge-invariant (Equivalent to imposing Gauss law) 4/30 Requires 3 basic ingredients 1. Inner product Matrix variables Gauge-invariant measure Identification of proper gauge-invariant variables (CFT argument) 2. Hamiltonian H in the new variables (Propagator mass, comparison with resummation techniques,...)

5 3. Solve the Schrödinger equation HΨ = EΨ Ψ 0, the vacuum wave function 5/30 string tension, comparison with lattice estimates

6 Matrix variables, volume element Complex coordinates, z = x 1 ix 2, z = x 1 + ix 2 A A z = 1 2 (A 1 + ia 2 ), Ā = 1 2 (A 1 ia 2 ) 6/30 1. Parametrize A as A = M M 1 Ā = M 1 M G = SU(N) = M SL(N, C) = SU(N) C More generally G G C Parametrization well-known in 2 dim. YM context, gauged WZW models, etc. M and MV ( z) = same A (Come back to this later)

7 2. Gauge transformation M M g = g M, A g i = ga ig 1 i gg 1 H = M M is gauge-invariant 7/30 3. Calculation of volume δa = (δmm 1 ) + [ MM 1, δmm 1 ] = D(δMM 1 δā = D(M 1 δm ) ds 2 A = = ds 2 SL(N,C) = d 2 x Tr(δAδĀ) Tr [ (M 1 δm )( DD)(δMM 1 ) ] Tr(M 1 δm δmm 1 )

8 dµ A = det( DD) dµ(m, M ) }{{} Haar measure for SL(N, C) dµ(m, M ) = dµ(u) }{{} Haar for SU(N) dµ(h) }{{} Haar for SL(N,C) SU(N) 8/30 H = M M, H = e ta ϕ a, H 1 δh = δϕ a R ab (ϕ)t b dµ(h) = [dϕ] det R dµ A = det( DD) dµ(h) dµ(u) dµ(a/g) = det( DD) dµ(h) = dµ(h) exp [2C A S wzw (H)]

9 C A δ ab = f amn f bmn = Nδ ab for SU(N). S wzw (H) = 1 Tr( H 2π H 1 ) (WZW action for H) 4. Inner product 1 2 = i 12π Tr(H 1 dh) 3 dµ(h) exp [2C A S wzw (H)] Ψ 1Ψ 2 9/30 T wo remarks 1. YM (2+1) has Gribov problem. But inner product formula has no difficulty due to this, it is exact 2. It shows Matrix elements in YM(2+1) = Correlators of a hermitian WZW model

10 An intuitive argument for mass gap H = 1 ] [e 2 E 2 + B2 2 e 2 [E, B] p (in momentum space) E B p, E p B E 1 [e 2 p 2 ] 2 ( B) + ( B)2 2 e 2 Minimize with respect to B ( B) 2 p E p. This is the photon. 10/30

11 For us, H = dµ(h) exp [2C A S wzw (H)] 1 ] [e 2 E 2 + B2 2 e [ 2 dµ(h) exp C A B 1 ] ] 1 [e 2π p 2B E 2 + B2 2 e 2 11/30 Gaussian ( B) 2 p 2 More detailed analysis mass gap. m mag = m = g2 T C A 2π

12 Calculation of det( DD) Γ = log det( DD) = Tr log( DD) 12/30 δγ δa a (x) = Tr [ D 1 Reg (x, y)( ita ) ] y x Re-integrate to get the determinant. = Tr [ G(x, x)( it a ) ] = 1 π Tr [ MM 1 ( it a )]

13 The hermitian WZW model Unitary model Level k exp[ks wzw (U)] Hermitian model Level k + 2C A exp[(k + 2C A )S wzw (H)] 13/30 κ = k + C A κ = (k + C A ) Compare using (κ κ) Integrable rep s spin k Nonintegrable Correlators =0 Nonintegrable Correlators = Finite norm = Integrable rep s

14 k = 0 for us, = Ψ s are functions of the current J = C A H H 1 π ( W (C) = Tr Pe π A = Tr P exp C A All gauge-invariant quantities can be made from J. ) J 14/30

15 Construction of H H = e2 E a E a + 1 B a B a } 2 {{}} 2e 2 {{} T + V T Ψ = e2 2 = e2 2 = δ 2 x δa(x)δā(x) Ψ δj(u) δj(v) δa(x) }{{ δā(x) } Ω δ 2 Ψ Ω ab (u, v) δj a (u)δj b (v) + δ 2 Ψ δj(u)δj(v) + ω a δψ (u) δj a (u) δ 2 J(u) } δa(x)δā(x) {{} ω δψ δj(u) 15/30

16 ω a = e2 δ 2 J a (u) 2 x δa b (x)δāb (x) ( e 2 ) [ ] C A = M 2π am(x)tr t m D 1 (y, x) = m J a m = e2 C A 2π y x cf. dµ(a/g) 16/30 [ T = m Ω ab (u, v) = C A π 2 Recheck J δ δj J a δ δj + a δ ab (u v) if abcj c (v) 2 u v by self-adjointness. δ 2 ] Ω ab (u, v) δj a (u)δj b (v) + O(ɛ)

17 V = 1 2e 2 = π mc A B a B a = 2π2 e 2 C A 2 : J J : : J a J a : 17/30 Regularization G ma (x, y) = [ 1 δ ma π(x y) All results checked using regularization Useful result e (x y)2 ɛ T V = 2m V ] [H(x, ȳ)h 1 (y, ȳ)]

18 Vacuum wave function Ignore V for the moment Vacuum: Ψ 0 = 1, this is okay since T Ψ 0 = 0 18/30 Normalizable dµ(h)e 2C A S wzw (H) Ψ 0Ψ 0 < Include V perturbatively, Ψ 0 = e P P = π : m 2 C J J : ( A ) 2 π : m 2 C JD J : + 1 : A 3 J[J, 2 J] : + + } {{ } } {{ } sum sum

19 [ P = 2 π 2 ( ) J a 1 e 2 CA 2 (x) m + J a (y) m 2 2 x,y ] + f abc J a (x)j b (y)j c (z)f(x, y, z) + O(J 4 ) 19/30 f(x, y, z) = e ikx+ipy+iqz (2π) 2 δ(k + p + q) ( ) 3 π ( k 2 + m 2 m)( p 2 + m 2 m) 2C A k2 + m 2 + p 2 + m 2 + q 2 + m 2 [ 1 B 1 ] B k 2 Ψ 0 = e P exp exp 2e [ 2 1 4e 2 m B 2 ] O(J 3, J 4 ) terms are small at k e 2 and at k e 2 m 1 k m 1 k p kp

20 String tension W R (C) = Tr R P e π C A J 20/30 (constant) e σa C σ = e 2 CA C R 4π C R = Casimir for representation R C A = Casimir for the adjoint representation Consistent with large N expectations, even though we did not use large N analysis

21 Comparison with lattice calculations Compare σ/e 2 with numerical (lattice)estimates by Teper et al. Our predictions are in black, lattice values are in red. (difference 3%) Group k=1 k=2 k=3 k=2 k=3 k=3 Fund. antisym antisym sym sym mixed SU(2) SU(3) SU(4) SU(5) SU(6) /30

22 Comment on higher corrections Contribution from O(J 3 ), etc.? Two types of corrections possible 22/30 Corrections to coupling, purely numerical. ratios σ R /σ F are unaffected Corrections via new diagrams to Wilson line expectation value (under investigation)

23 Magnetic mass, resummed perturbation theory [ T = m J δ δj + Ω δ ] δ δj δj = T J a = m J a 23/30 (T + V ) J a = p 2 + m 2 J a + Ψ e C AS wzw (H) Ψ, H e C AS wzw (H) H e C AS wzw (H) H = 1 ] [ δ2 2 δφ + 2 φ( 2 + m 2 )φ + Propagator mass for gauge particles (magnetic mass) m = e2 C A 2π 0.32 e2 for SU(2)

24 Comparison with other methods m/e 2 = 0.35 Common factor for glueball masses (lattice, Philipsen) 24/ Max. Abelian gauge (lattice, Karsch et al) 0.52 Landau gauge ( ) 0.44 λ 3 = 2 gauge ( ) 0.38 Resummation of P.T. (Alexanian & Nair ) 0.28 Resummation of P.T. (Buchmuller & Philipsen, Jackiw & Pi) 0.37 Gauge-invariant lattice lattice definition (Philipsen)

25 Glueballs T J a = m J a J a is not a good state, need holomorphic invariance. A = MM 1, M, MV ( z) same A 25/30 J V JV 1 V V 1 J V J V 1 A 2J state (0 ++ ) Ψ 2 = f(x, y) : J a (x) [H(x, ȳ)h 1 (y, ȳ)] ab J b (y) Take the same x, y f(x, y) = σ(x, y, ɛ) T Ψ 2 = 2m Ψ 2

26 Higher number of J s Ψ n : J a 1 a2 J J a n : ω a1 a }{{ 2 a } n invariant tensor of SU(N) 26/30 T Ψ n = mn Ψ n Can include center-of-mass motion from B 2 /2e 2. Relative motion higher radial excitations Regge trajectory Back to Ψ 2 : { ( ) 2 2m + 1 2m x y + m log 2m 2ɛ } + f(x, y) = E f(x, y)

27 Yang-Mills +Chern-Simons [ Ψ = exp ks wzw (M ) k ] A a Ā a Φ(H) 4π 1 2 = dµ(h)e (k+2c A)S wzw (H) Φ 1Φ 2 T = e2 4π (k + 2C A) J δ δj + e2 C A Ω δ δ [ 2π δj δj Φ 0 exp π ] 1 J mc A m + J m 2 2 m = e2 4π (k + 2C A) New integrable operators from CFT screening of W F (C) Large k standard perturbation theory A number of eigenstates of T can be constructed 27/30

28 Comment on Gribov problem G A A/G Nontrivial bundle 28/30 Π 2 (A/G ) = Z Π n (A) = 0 Noncontractible S 2 in A/G H = cosh 2f + J sinh 2f f = 1 ( ) z z + w w + µ 2 2 log z z + w w

29 J = ( ) z z w w 2 wz 2w z w w z z w w = 0, z = 0 singular point. Move singularity by H V H V, V = exp [ σ 3 (log S wzw (H) unchanged, finite )] z z ā 29/30

30 What is next? Many things are still not understood Is a proof of mass gap possible? Better handle on glueballs Higher order corrections (*) Screening of adjoint and string breaking (*) YM(3+1), do we dare? 30/30