Recurrence relations in the large space-time dimension limit - PowerPoint PPT Presentation

Recurrence relations in the large space-time dimension limit P.A.Baikov (Moscow St.Uni.) 1

Recurrence relations (RR): R ( I − , I + , d ) F ( n 1 , ..., n k , d ) = 0 Result of the reduction procedure: F ( n, d ) = C 1 ( n, d ) F 1 + ... + C k ( n, d ) F k C i ( n, d ) are rational in d and obey the RR: R ( I − , I + ) C i ( n, d ) = 0 C i ( n, d ) = 0 if some ”hard” n l ≤ 0 (for lines in F i ) C i are much simpler then F Can we calculate C i directly (without reduction)? 2

Expand C i in 1 /d → 0 Calculate sufficiently many coefficients Reconstruct exact rational d dependence 3

Expansion R ( I − , I + ) = R (0) + 1 dR (1) + 1 + 1 C i = C (0) dC (1) d 2 C (2) + ... i i i 0 = R ( I − , I + ) C i ⇒ 0 = R (0) C (0) i 0 = R (0) C (1) + R (1) C (0) i i ... 0 = R (0) C ( k ) + R (1) C ( k − 1) i i ... 4

1-dimentional example −∞ ( x 2 + 2 x + 2) d /x n dx ∞ � f n = nf n +1 = ( d + 1 − n ) f n + ( d + 1 − n/ 2) f n − 1 Reduction is trivial, but let’s try 1 /d 0 = ( f n + f n − 1 ) + 1 /d ((1 − n/ 2) f n − 1 + (1 − n ) f n − nf n +1 ) � �� � � �� � R (0) f R (1) f + f (0) 0 = R (0) f (0) = f (0) ⇒ f (0) = ( − 1) n n − 1 n n R (0) f (1) + R (1) f (0) 0 = + f (1) = 1 / 4( n 2 + n )( − 1) n ⇒ f (1) = f (1) n − 1 + n/ 2( − 1) n n n 5

R ( I − ) C i = 0 can be solved in multi-dimentional case C i ( n ) = Π a r − n a , where R ( r a ) = 0 a R ( I − a ) C i = R ( r a ) C i = 0 R ( I − ) vs. R ( I − , I + ) like algebraic vs. differential equations It is convinient to choose R (0) = R ( I − ) It can be done in case of Feynman integrals 6

Feynman integrals � d d p 1 ..d d p L / ( E n 1 · · · E n a F ( n, d ) = a ) 1 E a = A ik a ( p i p k ) + m 2 a IBP: � d d p 1 ..d d p L ∂ p i ( p k · · · ) 0 = ∂ p i ( p k · ) = d δ i k + p k ( ∂ p i · ) = d δ i k + ( AA ) a b E a ( ∂ E b · ) RR: 0 = d δ i k F + ( AA ) a b E a ∂ E b F 7

0 = d δ i k F + ( AA i k ) a b E a ∂ E b F R (0) ? R (1) ? 0 = R (0) C (0) = δ i k C (0) C (0) ⇒ = 0 ??? n n 1/d does not work ? 8

No solutions like C i = C (0) d C (1) + 1 + ... i i Indeed, C k ≈ d − S ( n ) , S ( n ) = Σ (”hard” n i ) We apply 1 /d expansion to the subcase ”hard” n i = 1 n i > 1 can be reduced to n i = 1 by direct recursion 9

Modified IBP � d d p 1 ..d d p L ∂ p i ( p k Π ik ( E a ) · · · ) 0 = With some polinomials Π ik we come to diagonalized RR 0 = ∂ E a ( P ( E ) F ) − ( d − L − 1) / 2 ( ∂ E a P ( E )) F R (1) R (0) large Equations with ”hard” ∂ E a : n a → 1 Equations with ”soft” ∂ E a (”hard” E a = 0) 0 = R (0) F (0) = ( ∂ E a P ( E )) F (0) F (0) ( n ) = Π r n a ⇒ a r a : ∂ E a P ( E ) | E a = r a = 0 10

What is the optimal way to calculate C ( k ) ? i One can to obtain C ( k ) ( n ) as polinomials in n i One can construct recurrent procedure for C ( k ) ( n ) i More efficient is to expand in 1 /d → 0 auxiliary integrals � dx 1 .. dx a / ( x n 1 C ( k ) 1 .. x n a a ) P ( x ) ( d − L − 1) / 2 ( n ) = i In 1 /d → 0 they expand to Gaussian type integrals � dx 1 .. dx a x k 1 1 .. x k a a exp( − A ik x i x k ) 11

Possible applications Pro and Contra Massless 0-scale problems 1 /d coefficients are pure numbers Relatively small set of master integrals Very convinient 4-loop propagators are fine Few calendar months for R ( s, N f = 3) 12

1-mass 0-scale problems (bubbles) 1 /d coefficients are pure numbers But many master integrals ⇒ Many contributions to calculate Calculational efforts (setup + CPU) comparable to other approaches (Laporta, Smirnov’s, direct recursion) 13

Multi-scale problems 1 /d coefficients are multi-scale rationals ⇒ Reduction is possible, but difficult Many master integrals, difficult to calculate 14

Summary When d is large RR for FI become ”algebraic” ⇒ systematic reduction is possible 1 /d coefficients demands big amount of CPU But massless 4-loop propagators are reachable 15