Towards the gauge beta function at O (1 / N 2 f ) and O (1 / N 3 f ) Manuel Reichert Bridging perturbative and non-perturbative physics, Primosten, 07. October 2019 CP 3 -Origins, SDU Odense, Denmark Nicola Dondi, Gerald Dunne, MR, Francesco Saninno: arXiv:1903.02568 Nicola Dondi, MR, Francesco Saninno: in preparation
Which matter systems are asymptotically safe in d = 4 ? • Gauge-Yukawa theories at large N f & N c (perturbatively) [Litim, Sannino ’14] • How far does this extend to small N c ? • Test gauge theories at large N f non-perturbatively Standard QCD picture: 70 60 • Small N f : asymptotic freedom & Safe QCD 50 confinement in the IR 40 N f 30 • Medium N f : asymptotic freedom IR conformal 20 & IR Banks-Zaks fixed point 10 0 2 3 4 5 6 7 • Large N f : asymptotic freedom lost N c → asymptotic safety? [Antipin, Sannino ’17] 1
Beta functions of (S)QED and (S)QCD β ( K ) = β (0) ( K ) + β (1) ( K ) + . . . N f QED 10 QCD SQED SQCD β (1) 5 0 0 1 2 3 4 5 6 7 8 K UV fixed point for QED & QCD Landau pole for SQED & SQCD 2
How physical are these fixed points? • The fermion mass anomalous dimension goes to zero in QCD and to infinity in QED [Antipin, Sannino ’17] • Hints for FP in QCD at medium N f from resummations with Meijer G-functions [Antipin, Maiezza, Vasquez ’18] • Lattice studies inconclusive so far [Leino, Rindlisbacher, Rummukainen, Sannino, Tuominen ’19] • Poles might be resummable within the 1 / N f series [Alanne, Blasi, Dondi ’19] 3
How to go beyond 1 / N f • The next orders in the 1 / N f expansion would test the physical nature of the FP • No known resummation formula for two bubble-chains, needed for 1 / N 2 f and higher orders • Can we extract the location of the pole, the residuum, etc., with a finite amount of coefficients? 4
How to go beyond 1 / N f • The next orders in the 1 / N f expansion would test the physical nature of the FP • No known resummation formula for two bubble-chains, needed for 1 / N 2 f and higher orders • Can we extract the location of the pole, the residuum, etc., with a finite amount of coefficients? Two methods: • Large-order behaviour of expansion coefficients • Pad´ e approximants 4
Large-order behaviour: Darboux’s Theorem The nearby singularity determines the large-order growth of the expansion coefficients a n . E.g. for expansion around z = 0 • pole of order p at z 0 ( f ( z ) ∼ φ ( z )(1 − z / z 0 ) p + finite) � � a n ∼ 1 n + p − 1 φ ( z 0 ) + . . . z n n 0 • logarithmic branch cut at z 0 ( f ( z ) ∼ φ ( z ) ln(1 − z / z 0 ) + finite) a n ∼ 1 1 n φ ( z 0 ) + . . . z n 0 5
Large-order behaviour: Darboux’s Theorem The nearby singularity determines the large-order growth of the expansion coefficients a n . E.g. for expansion around z = 0 • pole of order p at z 0 ( f ( z ) ∼ φ ( z )(1 − z / z 0 ) p + finite) � � a n ∼ 1 n + p − 1 φ ( z 0 ) + . . . z n n 0 • logarithmic branch cut at z 0 ( f ( z ) ∼ φ ( z ) ln(1 − z / z 0 ) + finite) a n ∼ 1 1 n φ ( z 0 ) + . . . z n 0 Expectation for QED F QED = � n f n x n � 2 � 2 � 2 � n � n � n � � f n ∼ R 0 + R 1 + R 2 + . . . 15 21 27 5
Large-order behaviour of F QED 0 . 4 0 . 2 0 f n +1 − 0 . 2 f n 2 15 − 0 . 4 0 10 20 30 40 50 60 n Ratio test f n +1 reveals location of the first pole f n 6
Large-order behaviour of F QED 4 , 000 � 15 � n +1 f n 2 2 , 000 28 − 45 π 2 0 − 2 , 000 − 4 , 000 0 10 20 30 40 50 60 n 7
Large-order behaviour of F QED 1 . 5 � 15 � n +1 f n 2 1 28 − 45 π 2 0 . 5 0 − 0 . 5 25 30 35 40 45 50 55 60 n With the knowledge of the pole the residuum is computable 7
Large-order behaviour of F QED 0 . 4 0 . 2 0 ˜ f n +1 − 0 . 2 ˜ f n 2 21 − 0 . 4 0 10 20 30 40 50 60 n Subtracting the first pole reveals the second pole � − n − 1 28 � 15 ˜ f n = f n + 45 π 2 2 8
Large-order behaviour of F QED 0 log 10 | ( n + 1) c (1) n | log 10 | ( n + 1) c (1) n | large-order approximation − 20 − 40 0 10 20 30 40 50 60 n After ∼ 30 terms the large-order behaviour sets in (for subleading behaviour later) 9
How many coefficients are needed? 50 40 coefficients 30 20 QED QCD 10 SQED SQCD 0 1 2 3 4 n pole ”Closer” to the origin → less coefficients are needed 10
Pad´ e methods Analytic continuation of truncated Taylor series by ration of two polynomials M P [ R , S ] ( x ) = P R ( x ) � f n x n F QED ( x ) ≈ − → Q S ( x ) n =0 with R + S = M . 11
Pad´ e methods Analytic continuation of truncated Taylor series by ration of two polynomials M P [ R , S ] ( x ) = P R ( x ) � f n x n F QED ( x ) ≈ − → Q S ( x ) n =0 with R + S = M . Rewriting of resummed F QED ( x ) sin 2 � π x F QED ( x ) ∼ Γ(1 + x � 3 ) 3 � π x Γ( 1 2 + x � 3 ) cos 3 Pad´ e approximant with 2 R ≈ S should lead to best results. 11
Pad´ e approximants of F QED F QED 1 0 . 5 0 − 0 . 5 0 2 4 6 8 10 12 14 16 18 20 x 12
Pad´ e approximants of F QED F QED 1 M = 20 0 . 5 0 − 0 . 5 0 2 4 6 8 10 12 14 16 18 20 x 12
Pad´ e approximants of F QED F QED 1 M = 20 M = 30 0 . 5 0 − 0 . 5 0 2 4 6 8 10 12 14 16 18 20 x 12
Pad´ e approximants of F QED F QED 1 M = 20 M = 30 M = 40 0 . 5 0 − 0 . 5 0 2 4 6 8 10 12 14 16 18 20 x 12
Pad´ e approximants of F QED F QED 1 M = 20 M = 30 M = 40 0 . 5 M = 50 0 − 0 . 5 0 2 4 6 8 10 12 14 16 18 20 x 12
Pad´ e approximants of F QED F QED 1 M = 20 M = 30 M = 40 0 . 5 M = 50 M = 60 0 − 0 . 5 0 2 4 6 8 10 12 14 16 18 20 x 12
Pad´ e approximants of F QED F QED 1 M = 20 M = 30 M = 40 0 . 5 M = 50 M = 60 0 − 0 . 5 0 2 4 6 8 10 12 14 16 18 20 x • Need ∼ 30 coefficients to resolve first singularity (similar to large-order growth analysis) • Can resolve function beyond the first singularity 12
QED beta function 1 / N f : 13
QED beta function 1 / N f : 1 / N 2 f (subset): 13
QED beta function 1 / N f : 1 / N 2 f (subset): Master integral known / not know 13
Beyond 1 / N f : nested diagrams Nested sub-part of beta function: gauge & RG scale independent 1 /N f Computation up to K 44 14
Beyond 1 / N f : nested diagrams Nested sub-part of beta function: gauge & RG scale independent 1 /N f Computation up to K 44 At O (1 / N 3 f ) Computation up to K 32 14
Ratio test at O (1 / N 2 f ) β (2) � b n K n nested = n 1 0 . 5 0 b n +1 b n − 0 . 5 0 5 10 15 20 25 30 35 40 n New finite radius of convergence 15
Ratio test at O (1 / N 2 f ) β (2) � b n K n nested = n 0 . 34 0 . 33 0 . 32 0 . 31 b n +1 b n 0 . 3 28 30 32 34 36 38 40 42 n New finite radius of convergence but extreme slow convergence 15
Richardson extrapolation Enhance the convergence of the series a n = s + A n + B n 2 + C n 3 + . . . 16
Richardson extrapolation Enhance the convergence of the series a n = s + A n + B n 2 + C n 3 + . . . First Richardson ( B = C = . . . = 0) R (1) a n ≡ s = ( n + 1) a n +1 − na n 16
Richardson extrapolation Enhance the convergence of the series a n = s + A n + B n 2 + C n 3 + . . . First Richardson ( B = C = . . . = 0) R (1) a n ≡ s = ( n + 1) a n +1 − na n Second Richardson ( C = . . . = 0) R (2) a n ≡ s = 1 � ( n + 2) 2 a n +2 − 2( n + 1) 2 a n +1 + n 2 a n � 2 16
Richardson extrapolation Enhance the convergence of the series a n = s + A n + B n 2 + C n 3 + . . . First Richardson ( B = C = . . . = 0) R (1) a n ≡ s = ( n + 1) a n +1 − na n Second Richardson ( C = . . . = 0) R (2) a n ≡ s = 1 � ( n + 2) 2 a n +2 − 2( n + 1) 2 a n +1 + n 2 a n � 2 For oscillating series: Shanks transformation 16
Ratio test at O (1 / N 2 f ) 0 . 34 0 . 33 0 . 32 0 . 31 b n +1 b n 0 . 3 28 30 32 34 36 38 40 42 n Bare series: K ∗ = 3 . 14 17
Ratio test at O (1 / N 2 f ) 0 . 34 0 . 33 0 . 32 b n +1 0 . 31 b n R (1) b n +1 b n 1 3 0 . 3 28 30 32 34 36 38 40 42 n Bare series: K ∗ = 3 . 14 First Richardson: K ∗ = 3 . 003 17
Ratio test at O (1 / N 2 f ) 0 . 34 0 . 33 0 . 32 b n +1 b n R (1) b n +1 0 . 31 b n R (2) b n +1 b n 1 3 0 . 3 28 30 32 34 36 38 40 42 n Bare series: K ∗ = 3 . 14 First Richardson: K ∗ = 3 . 003 Second Richardson: K ∗ = 3 . 00008 17
Residue at O (1 / N 2 f ) − 0 . 49 − 0 . 5 − 0 . 51 3 n n 2 b n − 0 . 52 R (2) [3 n n 2 b n ] − 1 2 − 0 . 53 30 32 34 36 38 40 42 44 n Bare series: 3 n n 2 b n = − 0 . 512 Second Richardson: 3 n n 2 b n = − 0 . 500007 18
Subleading behaviour b n = b n + 1 1 1 ˜ 2 3 n n 2 19
Subleading behaviour b n = b n + 1 1 1 ˜ 2 3 n n 2 0 . 34 0 . 32 0 . 3 b n +1 b n R (2) b n +1 b n 1 3 0 . 28 32 34 36 38 40 42 n Bare series: K ∗ = 3 . 215 Second Richardson: K ∗ = 3 . 0003 19
Subleading behaviour b n = b n + 1 1 1 ˜ 3 n n 2 2 − 0 . 49 3 n n 3 ˜ b n R (2) [3 n n 3 ˜ b n ] − 1 2 − 0 . 5 − 0 . 51 − 0 . 52 32 34 36 38 40 42 44 n Bare series: 3 n n 3 ˜ b n = − 0 . 512 Second Richardson: 3 n n 3 ˜ b n = − 0 . 500007 20
Recommend
More recommend
