Complex Langevin Simulations and Zeroes of the Measure. I.-O. Stamatescu (Heidelberg) Results in the frame of common work with: G. Aarts (Swansea), E. Seiler (Munich) and D. Sexty (Wuppertal) and further collaboration with F. Attanasio, L. Bongiovanni, B. J¨ ager (Swansea) and J. Pawlowski (Heidelberg). SIGN 2015, Debrecen 1
Items of the discussion 1. CLE: setup and problems. 2. Discussion of the formal proof of equivalence. 3. Dealing with the non-holomorphy in an effective model. 4. Taking over some conclusions to HD-QCD. 5. Discussion. Discussion and results are part of the analysis in an upcoming paper by the mentioned authors. 2
1. CLE set up and problems 3
Complex, holomorphic action − → complex drift K ( z ) = −∇ z S , − → imaginary parts for the variables − → Process on the complex extension of the original manifold: z ( t ) = x ( t ) + i y ( t ) , x ∈ M r , z ∈ M c The process realises a positive probability distribution P ( x, y ) . Formal equivalence theorem: for analytic observables O ( x, y ) the averages over the process reproduce the ensemble averages with the (complex!) distribution ρ ( x ) = exp( − S ( x )) : what we get: �O� P ( t ) = �O� ρ ( t ) : what we want ( t → ∞ ) Cf. G. Aarts, E. Seiler and I. -O. Stamatescu, Phys. Rev. D 81 (2010) 054508 [arXiv:0912.3360 [hep-lat]]. G. Aarts, F. A. James, E. Seiler and I. -O. Stamatescu, Eur. Phys. J. C 71 (2011) 1756 [arXiv:1101.3270], etc. 4
As with every numerical method the the proof of equivalence relies on certain conditions to be fulfilled, for CLE in particular: 1 holomorphy of the drift and of the observables, 2 sufficient fall off of P ( x, y ) in the y − direction. There are also ”Practical problems”: 3. Accumulation of numerical errors. Typical effect: run-aways, divergence of some quantities. K ( z ) becomes unbounded 4. Unprecise sampling - in the presence of trajectories going far in the y direction, a further effect of 2 above.. Require adaptive step size, controlling the distribution, etc. Notice: there are many processes K ( z ) ( P ( x, y ) ) leading formally to the desired EV’s. This can be used in controlling the method. 5
2. Discussion of the Formal Proof of Equivalence 6
Formal proof Consider first holomorphic drift. P ( x, y ) and the observables evolve with the Fokker-Planck equation and its adjoint: P ( x, y ; t ) = L T P ( x, y ; t ) , ˙ (1) O ( x + iy ; t ) = L O ( x + iy ; t ) = ˜ ˙ L O ( z ; t ) , (2) L = ( N R ∇ x + K x ) ∇ x + ( N I ∇ y + K y ) ∇ y (3) L T = ∇ x [ N R ∇ x − K x ] + ∇ y [ N I ∇ y − K y ] , (4) ˜ L = [ ∇ z − ( ∇ z S ( z ))] ∇ z , ( z = x + iy, N R − N I = 1) (5) K ( z ) = −∇ z S ( z ) = ρ ( z ) − 1 ∇ z ρ ( z ) , ρ ( z ) = e − S ( z ) (6) Here we used the Cauchy-Riemann eqs. We shall set N R = 1 , N I = 0 . 7
Consider the interpolation function: � F ( t, τ ) ≡ P ( x, y ; t − τ ) O ( x + iy ; τ ) dxdy , (7) � F ( t, 0) = P ( x, y ; t ) O ( x + iy ) dx dy = �O� P ( t ) , (8) � F ( t, t ) = O ( x ; 0) ρ ( x ; t ) dx = �O� ρ ( t ) (9) Equivalence is ensured if ∂ � � L T P ( x, y ; t − τ ) � ∂τ F ( t, τ ) = − O ( x + iy ; τ ) dxdy (10) � + P ( x, y ; t − τ ) L O ( x + iy ; τ ) dxdy = 0 (11) which follows with integration by parts if there are no boundary terms. 8
Boundary terms from large | z | (cf. Condition 1!) Typically x is compact, thus for holomorphic drift K boundary terms may only come from the non-compact direction y . For gauge-theories our method of Gauge Cooling efficiently concentrates the distribution around the compact domain! (which also helps with the practical problems) Cf. E. Seiler, D. Sexty and I. -O. Stamatescu, Phys. Lett. B 723 , 213 (2013) [arXiv:1211.3709 [hep-lat]]. 9
Meromorphy (cf. Condition 2!) Zeroes of ρ ( x ) (e.g. from fermionic determinants, first stressed by Molgaard and Splittorff) will lead to a meromorphic drift. The following discussion is due to E. Seiler. To define a holomorphy domain we can introduce further boundaries cutting small circles of radius ǫ p around the poles z p . In all examples we find P ( x p , y p ) = 0 . Then we can take the limit ǫ p → 0 if the evolution of the (originally analytic) observables ∞ t k O ( z ; t ) = exp(˜ � ˜ L k O Lt ) O ( z ) = (12) k ! n =0 does not introduce essential singularities. This can be shown to hold in special cases of second order zeroes. 10
Example: − S ( z ) = − ωz 2 K ( z ) = n + ln( z n ) , z − ωz, (13) 2 � d d 2 � n ˜ L = dz 2 + z − ωz (14) dz then it easy to see that Lz = n L 2 z = n (2 − n ) L 1 z = 2 − n + ω ˜ ˜ ˜ + ω 2 z, z − ω, z , · · · (15) z 3 z 3 and for n = 2 no higher singularities are produced! (This example can be solved, e.g. for O ( z ) = z 2 , n = 2 we obtain � z 2 − 3 � + 3 → 3 L z 2 = e − 2 ωt e t ˜ ω − ω = �O� ρ ( t → ∞ ) (16) ω which is the exact result.) 11
3. The one link SU(3) effective model One link in the field of its neighbors: Seen as one Polyakov line of a 4-dim lattice model in temporal gauge 12
Effective model for HD-QCD 3 e α i e i w i + e − α i e − i w i � � � − S = β + ln Det + ln H (17) i =1 = sin 2 w 2 − w 3 sin 2 w 3 − w 1 sin 2 w 1 − w 2 H (18) 2 2 2 O n = tr ( U n ) = e i nw 1 + e i nw 2 + e i nw 3 , w 1 + w 2 + w 3 = 0 , (19) K = −∇ S (20) The matrices A = { e α i } ∈ GL (3 , C ) simulate the staples (after diagonalisation) The process runs in three or (equivalently) two angles, with correspondingly three (two) noise terms. 13
The determinant is Det = ( D ˜ D ) 2 (21) D = 1 + C tr U + C 2 tr U − 1 + C 3 = (1 + C 3 )(1 + a P + b P ′ ) , (22) C tr U − 1 + ˜ C 3 = (1 + ˜ a P ′ + ˜ D = 1 + ˜ ˜ C 2 tr U + ˜ C 3 )(1 + ˜ b P ) , (23) 3 ˜ 3 C C C 3 , ˜ b = ˜ a = 1 + C 3 , b = C a, a = ˜ C ˜ a, (24) 1 + ˜ P = 1 3 tr U, P ′ = 1 C = 2 κ e µ , ˜ C = 2 κ e − µ , 3 tr U − 1 (25) The square corresponds to dimension 4 of the simulated lattice HD-QCD model, a, b have maxima at C = 2 − 1 / 3 and 2 1 / 3 , respectively with the same value 2 2 / 3 independently on C . 14
2.5 1.6 SU3, 2 angles, kappa=5.5E-6, mu=0, vs beta: O 1 ex 3*x/(1+x**3) O - 1 ex 3*x**2/(1+x**3) O 2 ex O - 2 ex 1.4 2 1.2 1.5 1 1 0.8 0.6 0.5 0.4 0 0.2 -0.5 0 0 0.5 1 1.5 2 0 0.5 1 1.5 2 2.5 3 Figure 1: Observables vs β at µ = 0 (left plot). The coefficients a = 3 C/ (1 + C 3 ) , b = 3 C 2 / (1 + C 3 ) vs C (right plot). We shall use β = 0 . 25 , 0 . 75 , 1 . 25 . a, ˜ b ) solely depend on C ( ˜ a, b (˜ C ) . 15
Setup as effective model to simulate an N 3 σ N τ lattice model. The relevant parameters are C, ˜ C and thus we should choose: 2 κ = (2 κ N τ ) N τ , µ = N τ µ N τ (26) For reference we take N τ = 8 , 16 . We shall work at κ N τ = 0 . 12 . 1.6 1.6 a, n=8, k=0.12 a, n=16, k=0.12 b, n=8, k=0.12 b, n=16, k=0.12 1.4 1.4 1.2 1.2 1 1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 0 1 1.2 1.4 1.6 1.8 2 1 1.2 1.4 1.6 1.8 2 Figure 2: a, b at κ N τ = 0 . 12 vs µ N τ for N τ = 8 and 16 . 16
Typical Behaviour 1.2 1.2 \beta=0.5, \kappa L =0.12, vs \mu L : a \beta=0.5, \kappa L =0.12, no cut, vs \mu L : O 1 ex O - 1 ex b O 1 ex O 2 ex 1 1 O - 1 ex O - 2 ex O 2 ex O 3 ex O - 2 ex O - 3 ex 0.8 0.8 O 3 ex O - 3 ex O 1 0.6 O - 1 0.6 O 2 O - 2 O 3 0.4 0.4 O - 3 0.2804 -0.2412 0.2 0.2 0.9311 0 0 -0.2 -0.2 -0.4 -0.4 -0.6 -0.6 -0.8 -0.8 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 1.41 1.415 1.42 1.425 1.43 1.435 1.44 1.445 Figure 3: κ N τ = 0 . 12 , β = 0 . 25 , ordered lattice α i = 0 : observables and 1 3 a, 1 3 b vs µ N τ (left), and zoom of the ”dangerous” region (right), t Langevin ∼ 200 . 17
200 250 250 O - 3 U - 3 O - 3 O 2 U 2 O 2 -log(min|D|) -log(min|D|) log(min|D|) 200 200 150 ReD<0 ReD<0 ReD<0 O - 3 av U - 3 av O - 3 av O 2 av U 2 av O 2 av 150 150 O - 3 ex U - 3 ex O - 3 ex 100 O 2 ex U 2 ex O 2 ex 100 100 50 50 50 0 0 0 -50 -50 -50 -100 -100 -100 -150 -150 -150 -200 -200 -200 -250 -250 -250 0 20 40 60 80 100 120 0 20 40 60 80 100 120 0 20 40 60 80 100 120 1 0.1 10 \beta=0.5, \kappa L =0.12, Det, \mu L =1.375, |D|>1E-6 (red) \beta=0.5, \kappa L =0.12, Det, \mu L =1.425, |D|>1E-=6 (red) \beta=0.5, \kappa L =0.12, Det , \mu L =1.425,|D|>1E-6 (red) |D|<1E-6 (blue) |D|<1e-6 (blue) 0.5 0.05 5 0 0 0 -0.5 -0.05 -5 -1 -0.1 -10 0 5 10 15 20 0 5 10 15 20 25 30 35 40 0 50 100 150 200 Figure 4: β = 0 . 25 , κ Nτ = 0 . 12 , α i = 0 , Trajectory analysis and Det scatter plots: µ Nτ = 1 . 375 , d min > 10 − 4 ; µ Nτ = 1 . 425 , d min ∼ 10 − 9 ; µ Nτ = 1 . 475 , d min > 10 − 4 , d min ≡ min | Det | . Red points from traj. with d min > d c = 10 − 6 , Blue from d min < d c . Observe clear correlation!: - trajectories with d min > d c = 10 − 6 allways produce correct results, - wrong results only come from trajectories with d min < d c . 18
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