Transcript
Indiana’s school lecture (black board); prepared by I.V.Danilkin (JLab, June, 2015) Literature: [1] M. E. Peskin and D. V. Schroeder. An Introduction to Quantum Field Theory. Westview Press, 1995. [2] Bastian Kubis lectures https://www.jlab.org/conferences/asi2012/Kubis/Kubis_lectures123.pdf [3] F. J. Yndurain. Low-energy pion physics. arXiv: hep-ph/0212282, 2002.
Dispersion relation (10 min)
Cauchy theorem 1
d s ' FHs 'L
FHsL = 2i
à
(1)
Π s' - s
G
for holomorphic function FHsL and G is a rectifiable path. We can always deform the contour until the closest singularity. If FHs ® ¥L ® 0 on the large semi-circle then 1
d s ' FHs 'L
FHsL = 2i 1
1 ®
à
¥
d s ' FHs ' + i ΕL
à
1 -
¥
d s ' FHs ' - i ΕL
à
=
Π s' - s 2 i 4 m2 Π s' - s 2 i 4 m2 Π s' - s ¥ d s ' FHs ' + i ΕL - FHs ' - i ΕL ¥ d s ' Disc FHs 'L 1 = à à 2 i 4 m2 Π s' - s 2 i 4 m2 Π s' - s Disc FHsL = FHs + i ΕL - FHs - i ΕL G
(2)
Analyticity: 1 FHsL = 2i
¥
à
4 m2
d s ' Disc FHs 'L Π
(3)
s' - s
Reflection principle: F* Hs + i ΕL = FHs - i ΕL
(4)
allows to relate imaginary part of the amplitude (unitarity) to analytical continuation of the amplitude (i.e. dispersion relation) Exercise: Disc FHsL = FHs + i ΕL - FHs - i ΕL = FHs + i ΕL - F* Hs + i ΕL = 2 i Im FHsL Subtractions: In general, one can always introduce subtractions
(5)
2
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FHsL = ¥ d s ' s ' - s + s - s0 Im FHs 'L ¥ d s ' Im FHs 'L d s ' Im FHs 'L Hs - s0 L ¥ d s ' Im FHs 'L =à =à + à 2 à 2 4m Π 4 m2 Π 4 m2 Π 4 m s ' - s0 s' - s s ' - s0 s' - s s ' - s0 Π s' - s ¥
(6)
As a result we got an improved convergence at ¥ (due to additional power of s ' in the denominator); sum rule (when the integral converges) d s ' Im FHs 'L
¥
FHs0 L = à
4 m2
(7)
s ' - s0
Π
Exercise: If the integral does not converge sufficiently fast on the large semi-circle, derive dispersive relation for (assuming FHs0 L is Real) FHsL - FHs0 L
(8)
s - s0
General formula. Note that by introducing subtractions you improve the convergence at s ® ¥, however there are additional parameters to determine. n-1
FHsL = â i=0
1
FHiL Hs0 L Hs - s0 Li +
i!
Hs - s0 Ln Π
d s'
¥
à
2
4m
Im FHs 'L n
Hs ' - s0 L
s' - s
(9)
Calculation of the Disc (10 min) Instead of making full loop calculations one can use Cutkosky (cutting) rule: 1 2
p - m2 + i Ε
® -2 Π i ∆Ip2 - m2 M
(10)
Example:
d4 l iM =à
1 4
H2 ΠL Il2 - m2 + i ΕM IHp - lL2 - m2 + i ΕM d4 l
Disc M = H-iL à
H2 ΠL4
H-2 Π iL ∆Il2 - m2 M H-2 Π iL ∆IHp - lL2 - m2 M l 2 + m2
∆ l0 4
2
2
2
2
2
d l = d l0 l d l dW, ∆Il - m M = ∆Il0 - l - m M = 2
2
2
2
2
∆IHp - lL - m M = ∆Ip + l - 2 p l - m M = ∆Is - 2 p l = p0 l0 - p l =
2 l0 s l0 M,
s l0 , cm : p = I s , 0M, l = Hl0 , lL (11)
Documentation_Indiana_school.nb
p l = p0 l0 - p l =
4 Π2
s l0 , cm : p = I s , 0M, l = Hl0 , lL
l2 d l
i Disc M =
à
d W ∆Is - 2 2 l0
s l0 M s 2
∆Kl0 2
l dl=
l02
3
2
- m l0 d l0 , ∆Is - 2
s l0 M = 2
s
4 m2
i Disc M =
O
1-
= 2 i Im M s
8Π Therefore
4 m2
1 Im M =
116 Π
1 =
s
ΡHsL
(12)
16 Π
Pion vector form factor (15 min)
We consider the process of a transition of a photon into a pair of pions. This is an important building block: resposible for a hadronic part of e+ e- ® Π+ Π- , Τ- ® Π- Π0 ΝΤ , ... The matrix element: YΠ+ HpL Π- HqL J Μ H0L 0] = Hp - qL Μ FΠV HsL
(13)
where J Μ is the EM current and FΠV HsL is the pion vector form factor (normalized FΠV H0L = 1). This process does not have any crossed channel exchanges and therefore no left-hand cuts. Note also, that the factor Hp - qL Μ insures gauge invariance when the two pions are on-shell (i.e. Hp - qL Μ Hp + qL Μ = p2 - q2 = m2 - m2 = 0). At very low energy the pion vector form factor can be calculated in Chiral Perturbation Theory (ΧPT). At NLO it reads
(11)
4
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L ~ i A Μ HΠ+ ¶ Μ Π- - Π- ¶ Μ Π+ + ...L FΠV HsL = 1 +
1
1
1 2
6 H4 Π fΠ L
HL6 - 1L s +
6
fΠ2
(14)
-
Is - 4 m2Π M J HsL + OIs2 M, L6 » 16
Improvement: dispersion relation ® need to calculate the Discontinuity (Cutkosky rules again)
Hp - qL Μ Disc FΠV HsL = d4 l
1 2
H-iL à
H2 ΠL4
H-2 Π iL ∆Il2 - m2Π M H-2 Π iL ∆IHps - lL2 - m2Π M TI* Hs, zL Hps - 2 lL Μ FΠV HsL (15)
ps = p + q, s = Hp + qL2 , s z = cos Θ is c.m. angle note also that Hp + qL = 9 s , 0=, p0 = q0 =
2
s = l0 ,
p = q = l =
ΡHsL 2
We obtain Hp - qL Μ Disc FΠV HsL = à Hp - qL ... Exercise: Show that
dW TI* Hs,
i 2
64 Π
ΡHsL FΠV HsL à dW TI* Hs, zL Hps - 2 lL Μ
zL Hps - 2 lL Μ = L1 Hp + qL Μ + L2 Hp - qL Μ ... then contract with Hp + qL,
(16)
Documentation_Indiana_school.nb
5
1 * * à dW TI Hs, zL Hps - 2 lL Μ = 2 Π Hp - qL Μ à â z z TI Hs, zL
(17)
-1
where TI Hs, zL is the ΠΠ amplitude ¥
TI Hs, zL = 32 Π âH2 l + 1L tlI HsL Pl HzL l=0
(18) 2 ∆l l'
1
à â z Pl HzL Pl' HzL = -1
2l+1
Since Disc FΠV HsL = 2 i Im FΠV HsL, we get Im FΠV HsL = ΡHsL FΠV HsL t11* HsL ΘIs > 4 m2Π M
(19)
If we consider only elastic scattering: 1
t11 HsL =
sin ∆11 HsL e i ∆1 HsL
1
ΡHsL t11* HsL = sin ∆11 HsL e -i ∆1 HsL
,
ΡHsL
(20)
Watson final state theorem: the phase of the form factor is determined by the two-particle scattering phase shift: 1 FΠV HsL = FΠV HsL e i ∆1 HsL 1
FΠV HsL = FΠV HsL e i ∆1 HsL
(21)
Arg IFΠV HsLM = ∆11 HsL
The full Omnes–Muskhelishvili problem (20 min) We want to find the most general representation for a function, FHsL, which is the analytic in the complex s plane with the cut from s= A4 m2 , ¥E, assuming that we know its phase on the cut, Arg HFHsLL = ∆HsL, s > 4 m2
(22)
Solution is not unique, if F0 HsL is a solution, then eΑ s F0 HsL is a solution too. Need to know the asymptotic information! We look for a solution in the from FHsL = PHsL WHsL 1 HWHs + i ΕL - WHs - i ΕLL = WHs + i ΕL sin ∆HsL e -i ∆HsL , 2i e i ∆HsL - e -i ∆HsL
1 WHs + i ΕL
2i
2i
e -i ∆HsL = WHs + i ΕL e -2 i ∆HsL
1
1 = WHs - i ΕL
2i
W Hs + i ΕL e-2 i ∆HsL = WHs - i ΕL, ln W Hs + i ΕL - 2 i ∆ = ln W Hs + i ΕL DiscHln WHsLL = 2 i ∆HsL Dispersion relation
2i
(23)
6
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1 ln WHsL =
d s ' DiscHln WHs 'LL
¥
¥
d s ' ∆Hs 'L
, 4 m2 Π s ' - s Π s' - s ¥ d s ' ∆Hs 'L s ¥ d s ' ∆Hs 'L WHsL = Exp à , WHsL = Exp a + à 4 m2 Π s ' - s Π 4 m2 s ' s ' - s 2i
=à
à
4 m2
s WHsL = Exp Π
¥
d s ' ∆Hs 'L
4 m2
s' s' - s
, WH0L = 1
à
(24)
(25)
One subtraction: normalization WH0L = 1. The function W(s) is known as the Omnes function. Many applications! Exercise: Show that Arg WHsL = ∆HsL 1 if ∆HsL ® Α Π, show WHs ® ¥L ®
(26)
Α
s Use f Hs 'L d s à
s' - s ¡ i Ε
f Hs 'L d s = p.v.à
± i Π f HsL
(27)
s' - s
If one assume asymptotic: FH¥L ® 1 s, then FΠV HsL = WHsL
Numerical implementation (25 min) s WHsL = Exp Π Tangent stretching Let’s now consider the integral
¥
d s ' ∆Hs 'L
4 m2
s' s' - s
à
(28)
Documentation_Indiana_school.nb
7
¥
f HyL â y
à a
In order to account the whole region, the replacement is needed y ® yHxL, which changes the integrating range (a,¥) into (0,1) f HyL â y = à f HyHxLL
a
dy
Π â x, yHxL = a + Cext tg
dx
0 ¥
à
dy
1
¥
à
Cext Π 2
x ;
=
2
dx
Π
cosI 2 xM
2
n
f HyL â y = â f Hyi L wi .
0
i=1
If the integrand is smooth, it is convenient to use Gaussian weights for integration. Few words about Cext : As you can see y Hx = 1 2L = a + Cext . It means that n/2 points will be accounted before a + Cext and n/2 after. It is very useful when you know the general behavior of your function. If you do not, just put it equal to unity, Cext = 1. ¥
2
Example 2: Ù2 e-y â y << NumericalDifferentialEquationAnalysis` n = 10; Cext = 1; wg = GaussianQuadratureWeights@n, 0, 1D; yn = 2 + Table@Cext * Tan@Π 2 * wg@@i, 1DDD, 8i, n
4 m2 M à Π 4 m2 s ' s ' - s - i Ε
(29)
The latter integral can be written as s Π
¥
âs'
f Hs 'L
s
¥
p.v.à
=
à
s' s' - s -i Ε
2
4m
Π
â s ' f Hs 'L
f HsL + iΠ
2
4m
(30)
s
s' s' - s
For the p.v. integral we use the following trick: ¥
p.v.à
2
4m
â s ' f Hs 'L s' s' - s
¥
= p.v.à
â s ' f Hs 'L - f HsL + f HsL
2
4m
¥
=à
4 m2
s'
s' - s
â s ' f Hs 'L - f HsL s'
s' - s
4 m2
f HsL ln
+ s
s - 4 m2
(31)
8
Documentation_Indiana_school.nb
1
¥
p.v.à
4 m2
4 m2
1
âs'
ln
= s ' Hs ' - sL
s
s - 4 m2
All together (Omnes function) s WHsL = Exp Π
¥
d s ' ∆Hs 'L
4 m2
s' s' - s
à
(32)
Exit@D; << NumericalDifferentialEquationAnalysis` SetDirectory@NotebookDirectory@DD; StyleList = [email protected], AbsolutePointSize@5D, ð< & 8Blue, Red, Green, Purple<; SetOptions@Plot, Frame -> True, Axes -> 8True, False<, PlotStyle -> StyleList, AspectRatio -> 0.8, FrameStyle -> Directive@13DD; << PiPi_Madrid.m; mpi = 0.138; mK = 0.4956745; Mrho = 0.770; Ε = 0.00001; LamPhShift = 1.3; OmnesNInt@sig_D@s_D := Exp@s Pi * NIntegrate@deltaFinal@1D@sbD Hsb Hsb - s - I sig ΕLL, 8sb, 4 mpi ^ 2, 4 mK ^ 2, Infinity<, AccuracyGoal ® 5, MaxRecursion ® 200DD; nOmn = 120; Cx = 1.0; wg = GaussianQuadratureWeights@nOmn, 0, 1D; s0 = 4 mpi ^ 2; sn = s0 + Table@Cx * Tan@Pi 2 * wg@@i, 1DDD ^ 2, 8i, nOmn 4 mpi ^ 2, Exp@I * sig * delta@1D@sDD * Exp@deltas Π Log@s0 Hs - s0LDD * OmnesTemp@sDDD;
(31)
Documentation_Indiana_school.nb
ShowBPlotBdeltaFinal@1D@par ^ 2D * 180 Pi, 8par, 2 mpi, 2.0<, PlotPoints ® 50, MaxRecursion ® 0, PlotStyle ® 8Black, Thick<, FrameLabel ® :"
s
@GeVD", "∆HsL">F, DeltaExp@1D, ImageSize ® 8250, 250F,
PlotB8Im@Omnes@1D@par ^ 2DD<, 8par, 2 mpi, 1.1<, PlotStyle ® 8Black<, MaxRecursion ® 0, PlotPoints ® 100, PlotRange ® All, ImageSize ® 8250, 250<, FrameLabel ® :"
s
@GeVD", "ImHWHsLL">F>F
6 3
5 ImHWHsLL
ReHWHsLL
2 1 0
4 3 2
-1 1 -2 0.4
0.6
0.8
s @GeVD
Omnes@1D@1D OmnesNInt@1D@1D -1.43762 + 0.680652 ä -1.43777 + 0.680767 ä
1.0
0 0.4
0.6
0.8
s @GeVD
1.0
9
