Slides - indico in2p3

Transcription

Corrections non-perturbatives exponentielles et physique des
particules
S. Friot
I.P.N. Orsay
5 février 2010
Projet
PARTIE 1: Hyperasymptotique en théorie des champs – avec D. Greynat
1
Etape 1: des modèles simples entièrement sous contrôle
2
Etape 2: applications phénoménologiques
PARTIE 2: Le g − 2 du muon – avec D. Greynat et E. de Rafael
1
Contributions hadroniques
2
Contributions électrofaibles
⇒ Demande de CDD pour D. Greynat
S. Friot
Corrections non-perturbatives exponentielles et physique des particules
PARTIE 1: Hyperasymptotique en théorie des champs – avec D. Greynat
MOTIVATIONS
Diagrammes de Feynman et représentation de Mellin-Barnes:
Pôles de T. de Mellin ⇒ D. A. en puissances (et log.) d’un paramètre
Série perturbative non-Borel sommable:
Pôles de T. de Borel (renormalons, instantons) ⇒ corr. exp. (non-pert.)
Malgré noyau MB gère corrections exponentielles
⇒ donne à la fois série pert. et série non-pert. (dév. hyperasymptotique)
Dev. hyper. incluent automatiquement phénomène de Stokes
Evidemment rep. MB d’observables en QCD (par ex.) sont inconnues mais:
Théorie des terminants (Dingle):
Permet de ”terminer” une grande classe de séries divergentes et retombe sur
résultats hyperasymptotiques
S. Friot
Questions:
1
developpements hyperasymptotiques ⇒ objets que l’on manipule en
physique des particules?
2
Peut-on déduire de la théorie des terminants des moyens de calculer
corrections NP en théorie des champs?
3
Y a-t-il un lien entre phénomène de Stokes et ambiguı̈tés de Borel?
S. Friot
Etape 1: des modèles simples entièrement sous contrôle
But: Application théorie terminants sur modèle le plus simple possible: φ40 .
Vérification des résultats avec théorie de MB hyperasymptotique.
S. Friot and D. Greynat, arXiv:0907.5593 [hep-th].
résultats intéressants: résurgence, influence de la queue perturbative sur la
précision, prolongement analytique (ligne de Stokes incluse), partie
imaginaire non-perturbative
Extension de ces travaux:
1
Approche basée sur équation différentielle (Schwinger-Dyson)
2
Approche basée sur méthode des points selles
3
Traiter exemple avec ligne de Stokes sur axe réel positif
S. Friot
Etape 2: applications phénoménologiques
1
g − 2 des leptons chargés: une classe de corrections non-perturbatives
en QED
2
QCD à grand β0
3
Instantons en théorie effective de QED
S. Friot
g − 2 des leptons chargés: corrections non-perturbatives en QED
Insertions de polarisation du vide (type renormalons)
ℓ1
ℓ2
ℓN
B. Lautrup, Phys. Lett. B 69 (1977) 109.
1
2
3
Expression exacte des ak sous forme intégrale
⇒ obtention des expressions asymptotiques (pour k → ∞)
Calcul de Lautrup correct?
Lien avec QCD à grand β0
S. Friot
Instantons en théorie effective de QED
Phénomène de Stokes ⇒ instantons en théorie des cordes topologiques 2009
(Approche hyperasymptotique basée sur points selles)
Cas très similaire en théorie effective de QED (champ électrique constant)
1
Retrouver instantons à une boucle avec MB (partie imaginaire du
lagrangien effectif )
2
Ordres supérieurs? (conjecture à tous les ordres en QED scalaire)
S. Friot
PARTIE 2: g − 2 du muon – avec D. Greynat et E. de Rafael
Valeur actuelle:
−10
aexp
)
µ = 11659208.9 ± 5.4 ± 3.3 (×10
Déviation de 3σ
M. Davier arXiv:1001.2243 [hep-ph]
S. Friot
Corrections hadroniques
Contributions hadroniques (ordre α3 )
1
Le premier diagramme est le
moins bien connu
2
Diagrammes 3 et 4: calcul vérifié
pour électron dans polarisation et
fait pour tau
3
Diagrammes 5 à 16 : test pour
contributions électrofaibles
Calculs en cours
S. Friot
Corrections électrofaibles
Diagrammes testés
Plus généralement
1
2
A une boucle ⇒ résultat analytique.
Deux boucles ⇒ ordre dominant seulement (plusieurs échelles) .
Contributions avec boucles de quarks sont toujours calculées avec
masse des quarks légers nulle
MB multidimensionnelle permet de faire calculs à plusieurs échelles ⇒
vrai ordre de grandeur de ces contributions
J.-Ph.
Aguilar, D. Greynat and E. de Rafael, Phys.Rev.D 77 (2008).
S. Friot
Etat de l’art asymptotique
Berry (1989), Olver (1991)
Développement asymptotique
Obs.(α) =
N−1
X
k =0
ak αk +RN où RN = O(αN ).
x →0
On note:
O(α) ∼
α→0
∞
X
ak αk
k =0
Choix de N = N0 : développement superasymptotique
Obs.(α) =
N0 −1
X
k =0
k
„
ak α + RN0 où RN0 = O e
α→0
λ
− α0
«
Itération: développement hyperasymptotique
Obs.(α) =
N0 −1
X
k0 =0
(0)
A k0 +
N1 −1
X
Nj −1
(1)
Ak1 + ... +
X
(j)
Akj + RNj où
kj =0
k1 =0
λ1 < λ2 < ... < λj
„ λ «
j
R Nj = O e − α
x →0
Phénomène de Stokes
S. Friot
in collaboration with David GREYNAT
based on hep-th 0907.5593
Introductive remarks
• 4D-QFT perturbative series are expected to be (divergent) asymptotic
expansions in powers of the coupling constants
• Mellin-Barnes (MB) representation is a powerful tool in asymptotics which
has been used a lot for Feynman diagrams calculations in the perturbative
approach of 4D-QFT
Smirnov 2004, Friot-Greynat-de Rafael 2005, Aguilar-Greynat-de Rafael 2008
• MB representation also allows to deal with asymptotics beyond all orders
(hyperasymptotics)
Is MB representation able to tell us about non-perturbative effects?
S. Friot
Beyond asymptotic expansions
Berry (1989), Olver (1991)
Asymptotic expansion
f (x) =
N−1
X
k =0
ak x k +RN where RN = O(x N ).
x →0
f (x) ∼
We write
x →0
∞
X
ak x k
k =0
Superasymptotic expansion
f (x) =
N0 −1
X
k =0
k
„
ak x + RN0 where RN0 = O e
x →0
λ
− x0
«
Hyperasymptotic expansion
f (x) =
N0 −1
X
k0 =0
(0)
A k0
+
N1 −1
X
k1 =0
(1)
A k1
Nj −1
+ ... +
X
(j)
Akj + RNj where
kj =0
„ λ «
j
R Nj = O e − x
x →0
λ1 < λ2 < ... < λj
S. Friot
Mellin-Barnes representation
Mellin transform
M[f (x)](s) =
ˆ
∞
dx x s−1 f (x)
0
Mellin-Barnes representation (inverse Mellin transform)
f (x) =
1
2πi
c+i∞
ˆ
ds x −s M[f (x)](s)
c−i∞
Condition:
.
Re s = c ∈]α, β[
with
f (x) = O(x −α )
x →0+
and
f (x)
=
x →+∞
S. Friot
O(x −β )
0-dimensional euclidean φ4 theory
Let us consider the 4-dimensional euclidean φ4 action
–
»
ˆ
1
λ
1
∂µ φ(x)∂ µ φ(x) + m2 φ2 (x) + φ4 (x)
S = d 4x
2
2
4!
Its 0-dimensional version is
S=
1 2 2
λ
m φ + φ4
2
4!
and the associated generating functional, defined for Re λ > 0, is
ˆ +∞
Z (j) = N
dφ e −S+jφ
−∞
We will focus on the vacuum-to-vacuum transitions generated (for m = 1 and
N = √12π ) by
1
Z (0) = √
2π
ˆ
S. Friot
+∞
1
dφ e − 2 φ
2
λ φ4
− 4!
−∞
Z (0)
Formal approach
Rigorous approach
Perturbative
expansion
Mellin-Barnes
representation
Divergent tail
MB - remainder
Non-perturbative
expansion
Iterative procedure
Hyperasymptotic expansion
S. Friot
Perturbative expansion
1
Z (0) = √
2π
ˆ
+∞
1
dφ e − 2 φ
2
λ φ4
− 4!
−∞
Taking λ small, one has the perturbative expansion
`1
´ „ «k
∞
k
λ
35 2 385 3
1 X (−1) Γ 2 + 2k
1
Z (0) ∼ √
λ −
λ +O(λ4 )
= 1− λ+
λ→0
k!
3!
8
384
3072
π k =0
But ∀λ,
˛
˛
˛ 1 (2k + 3 )(2k + 1 ) ˛
˛
˛
2
2
λ
˛
˛ −−−−→ +∞
˛k + 1
˛ k →+∞
3!
therefore the perturbative series is a divergent series for all values of the
coupling.
S. Friot
Counting Feynman diagrams
Feynman diagrams contributions to Z (0) in 4 dimensions:
Order λ
(−1) λ
3
1! 4!
=−
1
λ
8
Order λ2
24
384
λ
2
1
0
A +
@
3
384
λ2
+
8
384
λ2
Since, in dimension 0,
Z (0) = 1 −
λ→0
1
35 2
385 3
λ+
λ −
λ + O(λ4 )
8
384
3072
the 0-dimensional theory performs a counting of the diagrams in 4
dimensions
S. Friot
Numerical analysis
1
3
5
7
9
11
1
For λ =
1
3
one has the exact result
13
1
0.98
0.98
0.96556048
Pert
Z (0)|λ= 1 ≈ 0.96556048...
Sn-1 0.96
0.96
0.94
0.94
3
0.92
0.92
1
3
5
7
9
11
13
n
An arbitrary resummation procedure for the perturbative series leads to the
property that |Rn | < |un | and |Rn | < |un−1 |. One then deduce
η−1
Z (0) =
X
uk +
k =0
1
1
|uη | ± |uη | ,
2
2
with
∀n, |uη | 6 |un |
This gives
Z (0)|λ= 1 = 0.96555187 ± 0.00140990
3
S. Friot
Inverse factorial expansion
Barnes’ lemma
Γ(s + a)Γ(s + b)
=
Γ(s + c)
ˆ
+i∞
−i∞
dt Γ(t + c − a)Γ(t + c − b)Γ(s + ϑ − t)Γ(−t)
2iπ
Γ(c − a)Γ(c − b)
with ϑ = a + b − c.
Inverse factorial expansion
M−1
X (−1)j
Γ(s + a)Γ(s + b)
=
(c − a)j (c − b)j Γ(s + ϑ − j)
Γ(s + c)
j!
j=0
ˆ
dt Γ(t + c − a)Γ(t + c − b)Γ(s + ϑ − t)Γ(−t)
+
2iπ
Γ(c − a)Γ(c − b)
M−δ+i R
M > Re (a − c) + δ, M > Re (b − c) + δ, M < Re (s + ϑ)
for 0 < δ < 1 and | arg s| < π2 .
Olver 1995
S. Friot
The starting point is now the expression of Z (0) as a divergent series
`1
`1
´ „ «k
´ „ «k
n−1
∞
k
k
λ
λ
1 X (−1) Γ 2 + 2k
1 X (−1) Γ 2 + 2k
Z (0) ∼ √
+√
λ→0
k!
6
k!
6
π k =0
π k =n
{z
}
|
{z
} |
.
.
=Rn
=S Pert.
n−1
Using the duplication formula,
Rn =
`
´ `
´
«k
∞
1
3 „
1 XΓ k+4 Γ k+4
2λ
√
−
Γ (k + 1)
3
π 2 k =n
We can apply the Inverse factorial expansion
`
´ `
´ m−1
X
Γ k + 41 Γ k + 34
(−1)j Aj Γ(k − j) +
=
Γ(k + 1)
j=0
1
where Aj =
j!
„ « „ «
1
3
4 j 4 j
and
S. Friot
ˆ
ds
f (s) Γ(k − s)
2iπ
c+m+i R
`
´ `
´
Γ(−s)Γ s + 14 Γ s + 43
` ´ ` ´
f (s) =
Γ 14 Γ 34
Resummation and terminant function
The tail then becomes
Rn =
„
«k
∞
m−1
X
−2λ
1 X
√
Γ(k − j)
(−1)j Aj
3
π 2 j=0
k =n
„
«k
ˆ
∞
X
ds
−2λ
1
Γ(k − s)
+ √
f (s)
2iπ
3
π 2
k =n
c+m+i R
One can now perform a Borel resummation on the two infinite sums in Rn ,
"∞
«k #
«n
„
„ «
„
X
2λ
3
−2λ
= Γ(n − j) −
Λn−j−1
Γ(k − j)
B
3
3
2λ
k =n
The function Λℓ is known as a terminant function
Λℓ (x) = x
ℓ+1
ˆ
∞
where
Γ(a, x) =
Dingle 1973
x
e Γ(−ℓ, x)
dy y a−1 e −y
x
is the incomplete Gamma function.
S. Friot
One then obtains (for m 6 n) the formal expression
`1
´ „ «k
n−1
k
1 X (−1) Γ 2 + 2k
λ
Z (0) = √
k!
6
π k =0
+
(−1)n
√ e
π 2
3
2λ
m−1
X
j=0
n
(−1)
+ √ e
π 2
3
2λ
(−1)j Aj Γ(n − j)
ˆ
c+m+i R
ds
2iπ
„
2
3λ
„
«
„
3
Γ −n + j + 1,
2λ
«
„
3
f (s) Γ(n − s) Γ −n + s + 1,
2λ
2λ
3
«−s
«j
From the MB hyperasymptotic theory, one may prove that it is the
hyperasymptotic expansion of Z (0) at first hyperasymptotic level. Paris
and
Kaminsky 2001
S. Friot
Resurgence phenomenon

„
«
3
3
λ
35 2
385 3 25025 4
1
+
λ −
λ +
λ − √ e 2λ Γ(5)Γ −4,
8
384
3072
98304
2λ
2π
„
«
„
«
„
«
3
35 2
3
385 3
3
1
+
λ Γ(3)Γ −2,
−
λ Γ(2)Γ −1,
− λ Γ(4)Γ −3,
8
2λ
384
2λ
3072
2λ
„
«ﬀ
ˆ c+5+i∞
3
3
1
1
25025 4
λ Γ(1)Γ 0,
− √ e 2λ
ds ...
+
98304
2λ
2iπ
2π
c+5−i∞
Z (0) = 1 −
This resurgence phenomenon also appears in higher order hyperasymptotic
levels.
S. Friot
Iterations...
We showed that it is possible to write the tail as
Rn =
(−1)n
√ e
π 2
3
2λ
j=0
n
+
m−1
X
(−1)
√ e
π 2
3
2λ
(−1)j Aj Γ(n − j)
ds
2iπ
ˆ
c+m+i R
„
2
3λ
„
«
„
3
Γ −n + j + 1,
2λ
«
„
3
f (s) Γ(n − s) Γ −n + s + 1,
2λ
2λ
3
«−s
«j
where
`
´ `
´
´ `
´
`
1
3
Γ(−s)Γ s + 41 Γ s + 34
1 Γ s+ 4 Γ s+ 4
π
` ´ ` ´
=− √
f (s) =
Γ(s
+
1)
sin
πs
Γ 14 Γ 43
π 2
Applying the Inverse factorial expansion on Rn,m one has
Functional equation
2
π 4
1
f (s) = − √
π 2 sin πs
′
m
−1
X
l=0
(−1)l Al Γ(s − l) +
S. Friot
ˆ
c+m′ −i R
3
dt
f (t) Γ(s − t)5
2iπ
MB hyperasymptotic theory
One may prove that the remainder of the asymptotic expansion of Z (0) in the
MB approach is
1
√
π
ˆ
ds
2iπ
c−n+iR
„ «−s
«
„
λ
1
− 2s = Rn
Γ (s) Γ
6
2
and, from this, that
˛ ˛n−c
˛ 2λ ˛
`
´
|Rn | = ˛˛ ˛˛
O e −n nn−c
n→∞
3
Superasymptotic Theorem
When |λ| is small, there exist a0 > 0 and |b0 | < ∞ so that if
n=
a0
+ b0
|λ|
then Rn is exponentially small in λ (non perturbative).
S. Friot
Optimal truncation schemes
This gives
a
|Rn | = O
|λ|→0
e
0
− |λ|
„
2a0
3
« a0 !
|λ|
and for a0 = 32 , we have the Optimal Truncation Scheme i.e. the smallest
remainder
“
”
− 3
|Rn | = O e 2|λ| .
|λ|→0
The remainder at next hyperasymptotic level is !Paris
˛ ˛−n−c
˛ 3 ˛
1
e −n (n − m)n−m mm+c− 2
|Rn,m | = O ˛˛ ˛˛
n→∞
2λ
For n =
a0
|λ|
+ b0 and m =
a1
|λ|
|Rn,m | = O
„
|λ|→0
and Kaminsky 2001
+ b1 , we have
p
a
|λ|e
1
a +ln 3−ln 2
− 0 |λ|
a1|λ| (a0
S. Friot
− a1 )
a0 −a1
|λ|
«
„
a
|Rn | = O
e
|λ|→0
|Rn,m | = O
|λ|→0
OTS 1
3
a0 =
2
OTS 2
a0 = 3
„
0
− |λ|
p
2a0
3
« a0 !
|λ|
a
|λ|e
1
a +ln 3−ln 2
− 0 |λ|
a1|λ| (a0
3
a1 =
4
3
a1 =
2
S. Friot
Rn = O(·)
e
3
− 2|λ|
−
− a1 )
p
a0 −a1
|λ|
«
Rn,m = O(·)
|λ|e
3 (1+ln 2)
− 2|λ|
p
|λ|e
3
− |λ|
OTS1 at first hyperasymptotic level
h
i
3
−1
2|λ|
X
Z (0) =
„
(−1)k Ak
k =0
1
− √
π 2
+O
„
h
3
2λ
i
3
−1
4|λ|
X
|λ|e
−
ˆ
(−1)j Aj
j=0
p
«k
dt
2iπ
„
dt
2iπ
„
i
h
3
+iR
−c+ 2|λ|
3(1+ln 2)
2|λ|
«−t
π
Γ (t − j)
sin πt
«−t
π
Γ (t − j)
sin πt
3
2λ
«
OTS2 at first hyperasymptotic level
h
i
3
−1
|λ|
Z (0) =
X
k
(−1) Ak
k =0
1
− √
π 2
+O
h
„
3
2λ
«k
i
3
−1
2|λ|
X
ˆ
(−1)j Aj
j=0
”
“p
− 3
|λ|e |λ|
h
i
3
−c+ |λ|
+iR
S. Friot
3
2λ
Numerical analysis
OTS 1
OTS 2
Pertur. expa.
OTS 2
m
m′
m′′
1
2
3
3
]=4
[ 2|λ|
3
[ 2|λ|
]=4
3
[ 2|λ|
]=4
3
]=2
[ 4|λ|
3
[ 4|λ|
]=2
3
[ 4|λ|
]=2
3
[ 8|λ|
]=1
3
[ 8|λ|
]=1
3
[ 8|λ|
]=0
1
2
3
3
[ |λ|
]=9
9
[ 2|λ| ] = 13
6
[ |λ|
] = 18
3
[ 2|λ|
]=4
3
[ |λ| ] = 9
9
[ 2|λ|
] = 13
Z (0)
0.965560481..
0.96555187
Mathematica
OTS 1
n
3
[ 2|λ|
]=4
3
[ |λ|
]=9
Sn
Sm
0.9638
0.9638
0.9638
0.9696
1.0573
-27.696
0.0017
0.0017
0.0017
-0.0040
-0.0917
28.662
3
[ 2|λ|
]=4
Sm ′
Sm′′
0.000041
0.000041
0
±0.001410990
1
2
3
1
2
3
0.96552297
0.96556492
0.96556492
0.965562911
0.965560477
0.965560486
S. Friot
0.0000061
-0.0001292
9 × 10−9
• Numerical stability
Olver et al.
’95
It has been proven that it is possible to obtain numerical stability by choosing
another OTS, taking
n+1−j
aj =
n+1
For the third hyperasymptotic level we have
n=8
m′ = 4
m=6
m′′ = 2
and
Z (0) = 0.965560480
S. Friot
C ONCLUSIONS
We started with the formal expression of a divergent (asymptotic) power
perturbative series and obtained a non-perturbative asymptotic
improvement, from the birth of hyperasymptotic expansions out of the
perturbative tail
We saw that the best OTS allows to reach much more precise results
than perturbation theory
The (exponentially suppressed) non-perturbative corrections are not
necessarily numerically small
Using MB representation extends the validity of the asymptotic
expansions beyond the complex λ-plane.
Aim: apply the method on less trivial examples
S. Friot
Perturbative expansion from Mellin-Barnes representation
The first step is to obtain an inverse Mellin-Barnes representation of Z (0):
„
«−s
ˆ
λ 4
λ 4
ds
φ
Γ(s) ,
e − 4! φ =
2iπ
4!
c+iR
with h0, +∞i and | arg λ| <
1
Z (0) = √
π
π
.
2
Then we obtain the following results
ˆ
ds
2iπ
c+iR
¸
˙
with 0, 41 and | arg λ| <
3π
2
«
„ «−s
„
λ
1
− 2s
Γ (s) Γ
6
2
(Analytic continuation).
S. Friot
Paris et al.
90’s
Im s
Re s
Applying the Cauchy’s theorem on the integral
over s inside a rectangle horizontally sized n,
one can prove that
`1
´ „ «k
n−1
k
1 X (−1) Γ 2 + 2k
λ
1
Z (0) = √
+√
k!
6
π k =0
π
ˆ
c−n+iR
ds
2iπ
«
„ «−s
„
λ
1
− 2s
Γ (s) Γ
6
2
therefore
1
Rn = √
π
ˆ
c−n+iR
ds
2iπ
„ «−s
«
„
λ
1
− 2s
Γ (s) Γ
6
2
S. Friot

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