Recursion and growth estimates in quantum field theory

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1 Recurson and growth estmates n quantum feld theory Karen Yeats Boston Unversty Aprl 9, 2007 Johns Hopns Unversty Some recursve equatons Start n the mddle γ (x) = γ (x)( rx x )γ (x) γ,n = p(n) ( rj )γ,j γ,n j j= How do we get these? How do we analyze them? What does t mean for quantum feld theory? A Dyson-Schwnger equaton Example In the context of renormalzaton Hopf algebras consder X(x) = I x p()b (X(x)Q(x) ) where Q(x) = X(x) r wth r < 0 an nteger. Ths carres the combnatoral nformaton. Consder the ntegral ernels for each B, namely the Melln transforms F (ρ,...,ρ s ). Ths adds the analytc nformaton. Wrte the combnaton (X G, B F ) as G(x,L) = γ (x)l wth γ (x) = j γ,jx j. Worng wth systems of equatons only ncreases techncal messness. From Broadhurst and Kremer []. ( ) X(x) = I xb. X(x) So Q(x) = /X(x) 2 Combnatorally counts rooted trees. F(ρ) = q 2 Combne to get G(x,L) = x q 2 d 4 q ( 2 ) ρ ( q) 2 q 2 =µ 2 d 4 q 2 G(x,log 2 )( q) 2 q 2 =µ 2 where L = log(q 2 /µ 2 ). The (analytc) Dyson- Schwnger equaton for a bt of massless Yuawa theory. 2 3

2 The lnearzed coproduct Extractng γ (x) wth S Y Defne or n general ln = (P ln P ln ) ln = (P ln P ln ) }{{} n where P ln projects onto the lnear part of the Hopf algebra, that s, lls dsjont unons of graphs. By the Hochschld closedness of B we get ln X = P ln X P ln X P ln Q x x X where P ln Q = rp ln X as Wrtng the (analytc) Dyson-Schwnger equaton G(x,L) = γ (x)l, we now from Connes and Kremer [2] that f and then σ = L φ(s Y ) L=0 σ n = m (σ σ ) }{{} n γ (x) = σ (X(x)) where φ s the renormalzed Feynman rules, m s multplcaton, S s the antpode, and Y s the gradng operator. 4 5 Extractng γ (x) wth ln The power of prmtves But σ only sees the lnear part of the Hopf algebra so we can use ln n place of. Gvng γ (x) = γ (x)( rx x )γ (x), We need not restrct ourselves to connected prmtves. We can choose a bass for the prmtves whch nvolves only one nserton place. the frst of the recursve equatons we began wth. In the Yuawa example γ (x) = γ (x)( 2x x )γ (x). Melln transforms become unvarate: F (ρ). 6 7

3 More power of prmtves Fndng the messy γ recurson We can also expand ρ( ρ)f (ρ) as a seres mang new prmtves out of the hgher order terms: so that p = p p 2 = B p (B p (I)) 2 Bp (I)B p (I) F (ρ) = ρ n F p n (ρ) = r n ρ n ρ( ρ) Melln transforms become geometrc seres: F (ρ) = r ρ( ρ). Rewrte the (analytc) Dyson-Schwnger equaton γ L = p()x (γ ρ ) r ( e Lρ )F (ρ) where γ U = γ U. Tae an L dervatve and set L = 0 to get γ = p()x ( γ ρ ) r ρf (ρ) Ths determnes γ recursvely, but messly. 8 9 Usng the geometrc seres Fndng the nce γ recurson We had γ L = p()x ( γ ρ ) r ( e Lρ )F (ρ) γ = p()x ( γ ρ ) r ρf (ρ) Now use ρf (ρ) = r ( ρ ρ 2 ). Tae two L dervatves of the DSE and set L = 0 to get 2γ 2 = p()x ( γ ρ ) r r ( ρ ρ 2 ) = γ x p()r Suc r nto the defnton of p() gvng γ = p()x 2γ 2 ρ We had and the other recurson Together γ = p()x 2γ 2 γ (x) = γ (x)( rx x )γ (x). γ = p()x γ ( rx x )γ, or at the level of coeffcents γ,n = p(n) ( rj )γ,j γ,n j. j= the second of the recursve equatons we began wth. 0

4 Growth of γ Lower bound on a n How bad s the growth of γ? Assume γ, 0 and f(x) = p(n) x n has nonzero radus of convergence ρ. Let a(n) = γ,n. The recurson becomes Idea: = p(n) ( r )a a n = ) ( rn ) 2 a a n = a(n) s approxmately p(n) ra a { } gvng a radus of mn ρ, ra for a n x n. ) Implement the dea by boundng on each sde. Recall ( rn ) 2 a a n = so a n p(n) r n 2 n a a Let b = a, B(x) = b n x n and b n = p(n) r n 2 n b b ) Then B (x) = f (x) rb xb { (x) whch } can be solved for B (x) to gve radus mn ρ, ra for B(x). 2 3 Upper bound on a n The radus of C(x) Recall ( rn ) 2 a a n = ) so for any ǫ > 0 there s an N > 0 such that for n > N a n p(n) Let c = a, C(x) = c n x n, c n = p(n) ra a ǫ a j a n j j= rc c ǫ c j c n j j= f ths s greater than a n and c n = a n otherwse. Then C(x) = f(x) ra xc(x) ǫc(x) 2 P ǫ (x) where P ǫ s a polynomal to deal wth ntal terms. We have C(x) = f(x) ra xc(x) ǫc(x) 2 P ǫ (x) The radus comes from the dscrmnant Clear poles ( ra x) 2 4ǫ(f(x) P ǫ (x)) ( ra x) 2 f(x) ( 4ǫ P ) ǫ(x) f(x) Techncal computaton gves that P ǫ (x)/f(x) s bounded as ǫ 0 so conclude that the radus of C(x) s { mn ρ, ra }. 4 5

5 Why? Understandng the growth of γ s understandng the growth of the whole theory. Expect a Lpatov bound γ,n c n. Does the frst sngularty of γ,n xn come from renormalon chans or from nstantons? We ve shown that a Lpatov bound for the prmtves leads to a Lpatov bound on the whole theory. The radus s ether the radus from the prmtves or rγ,, the frst coeffcent of the beta functon. The moral s that the prmtves control matters. References [] D.J. Broadhurst and D. Kremer, Exact solutons of Dyson-Schwnger equatons for terated oneloop ntegrals and propagator-couplng dualty. Nucl.Phys. B 600, (200), (Also arxv:hep-th/00246). [2] Alan Connes and Dr Kremer, Renormalzaton n quantum feld theory and the Remann-Hlbert problem II, Commun.Math.Phys. 26 (200) (Also arxv:hep-th/000388) [3] Dr Kremer and Karen Yeats, An Étude n nonlnear Dyson-Schwnger Equatons. Nucl. Phys. B Proc. Suppl., 60, (2006), 6-2. (Also arxv:hep-th/ ) [4] Dr Kremer and Karen Yeats, Recurson and Growth Estmates n Renormalzable Quantum Feld Theory. arxv:hep-th/