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

F(s)=(1/(2 i))(∫_Γ)((d s')/π)((F(s'))/(s'-s))

for holomorphic function F(s) and Γ is a rectifiable path. We can always deform the contour until the closest singularity. 
If F(s->∞)-> 0 on the large semi-circle then

F(s)=(1/(2 i))(∫_Γ)((d s')/π)((F(s'))/(s'-s))-> (1/(2 i))(∫_(4(m^2))^∞)((d s')/π)((F(s'+i ϵ))/(s'-s))-(1/(2 i))(∫_(4(m^2))^∞)((d s')/π)((F(s'-i ϵ))/(s'-s))=
		      (1/(2 i))(∫_(4(m^2))^∞)((d s')/π)((F(s'+i ϵ)-F(s'-i ϵ))/(s'-s))=(1/(2 i))(∫_(4(m^2))^∞)((d s')/π)((Disc F(s'))/(s'-s))
		Disc F(s)=F(s+i ϵ) - F(s-i ϵ)

Analyticity:

F(s)=(1/(2 i))(∫_(4(m^2))^∞)((d s')/π)((Disc F(s'))/(s'-s))

Reflection principle:

(SuperStar[F])(s+ i ϵ)=F(s-i ϵ)

allows to relate imaginary part of the amplitude (unitarity) to analytical continuation of the amplitude (i.e. dispersion relation)

Exercise:

Disc F(s)=F(s+i ϵ)-F(s-i ϵ)=F(s+i ϵ)-(SuperStar[F])(s+i ϵ)=2 i Im F(s)

Subtractions: In general, one can always introduce subtractions

F(s)=(∫_(4(m^2))^∞)((d s')/π)((Im F(s'))/(s'-s))=(∫_(4(m^2))^∞)((d s')/π)(((s'-s+s-(Subscript[s, 0]))/(s'-(Subscript[s, 0]))))((Im F(s'))/(s'-s))=(∫_(4(m^2))^∞)((d s')/π)((Im F(s'))/(s'-(Subscript[s, 0])))+(((s-(Subscript[s, 0])))/π)(∫_(4(m^2))^∞)((d s')/(s'-(Subscript[s, 0])))((Im F(s'))/(s'-s))

As a result we got an improved convergence at ∞ (due to additional power of s' in the denominator); sum rule (when the integral converges)

F((Subscript[s, 0]))=(∫_(4(m^2))^∞)((d s')/π)((Im F(s'))/(s'-(Subscript[s, 0])))

Exercise: If the integral does not converge sufficiently fast on the large semi-circle, derive dispersive relation for (assuming F((Subscript[s, 0])) is Real)

((F(s)-F((Subscript[s, 0])))/(s-(Subscript[s, 0])))

General formula. Note that by introducing subtractions you improve the convergence at s->∞, however there are additional parameters to determine.

F(s)=(∑_(i=0)^(n-1))(1/(i!))(F^((i)))((Subscript[s, 0])) (((s-(Subscript[s, 0])))^i)+((((s-(Subscript[s, 0])))^n)/π)(∫_(4(m^2))^∞)((d s')/(((s'-(Subscript[s, 0])))^n))((Im F(s'))/(s'-s))

Calculation of the Disc (10 min)

Instead of making full loop calculations one can use Cutkosky (cutting) rule:

(1/((p^2)-(m^2)+i ϵ))-> -2π i δ((p^2)-(m^2))

Example:

i M=∫(((d^4) l)/(((2π))^4))(1/(((l^2)-(m^2)+i ϵ)((((p-l))^2)-(m^2)+i ϵ)))
		Disc M=(-i) ∫(((d^4) l)/(((2π))^4))(-2π i)δ((l^2)-(m^2))(-2π i)δ((((p-l))^2)-(m^2))
		(d^4)l=d (Subscript[l, 0]) (l^2)d l dΩ, δ((l^2)-(m^2))=δ((Subscript[l, 0])-(l^2)-(m^2))=((δ((Subscript[l, 0])-(Sqrt[(l^2)+(m^2)])))/(2(Subscript[l, 0])))
		δ((((p-l))^2)-(m^2))=δ((p^2)+(l^2)-2 p l -(m^2))=δ(s-2(Sqrt[s])(Subscript[l, 0])),
		 p l=(Subscript[p, 0])(Subscript[l, 0])-p l=(Sqrt[s])(Subscript[l, 0]), cm: p=((Sqrt[s]),0), l=((Subscript[l, 0]),l)
		
		
		Disc M=(i/(4(π^2)))∫(((l^2)d l)/(2(Subscript[l, 0])))d Ω δ(s-2(Sqrt[s])(Subscript[l, 0]))
		(l^2)d l=(Sqrt[(Subscript[l, 0]^2)-(m^2)])(Subscript[l, 0])d (Subscript[l, 0]), δ(s-2(Sqrt[s])(Subscript[l, 0]))=((δ((Subscript[l, 0])-((Sqrt[s])/2)))/(2(Sqrt[s])))

		Disc M=(i/(8π))(Sqrt[1-((4(m^2))/s)])=2i Im M

Therefore

Im M=(1/(16π))(Sqrt[1-((4(m^2))/s)])=(1/(16π))ρ(s)

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 (SuperPlus[e])(SuperMinus[e])-> (SuperPlus[π])(SuperMinus[π]), (SuperMinus[τ])-> (SuperMinus[π])(π^0)(Subscript[ν, τ]), ... The matrix element:

⟨(SuperPlus[π])(p)(SuperMinus[π])(q)|(Subscript[J, μ])(0)|0⟩=(Subscript[(p-q), μ])(Subscript[F, π]^V)(s)

where (Subscript[J, μ]) is the EM current and (Subscript[F, π]^V)(s) is the pion vector form factor (normalized (Subscript[F, π]^V)(0)=1). This process does not have any crossed channel exchanges and therefore no left-hand cuts. Note also, that the factor (Subscript[(p-q), μ]) insures gauge invariance when the two pions are on-shell 
(i.e. (Subscript[(p-q), μ])(((p+q))^μ)=(p^2)-(q^2)=(m^2)-(m^2)=0). At very low energy the pion vector form factor can be calculated in Chiral Perturbation Theory (χPT). At NLO it reads

L~ i (Subscript[A, μ])((SuperPlus[π])(∂^μ)(SuperMinus[π])-(SuperMinus[π])(∂^μ)(SuperPlus[π])+...)		
		(Subscript[F, π]^V)(s)=1+(1/6)(1/(((4π (Subscript[f, π])))^2))((Subscript[L, 6])-1)s+(1/(6 (Subscript[f, π]^2)))(s-4(Subscript[m, π]^2))(Overscript[J, -])(s)+O((s^2)), (Subscript[L, 6])≈16

Improvement: dispersion relation

-> need to calculate the Discontinuity (Cutkosky rules again)

(Subscript[(p-q), μ])Disc (Subscript[F, π]^V)(s)=(1/2)(-i) ∫(((d^4) l)/(((2π))^4))(-2π i)δ((l^2)-(Subscript[m, π]^2))(-2π i)δ(((((Subscript[p, s])-l))^2)-(Subscript[m, π]^2))(Subscript[T, I]^( * ))(s,z)(Subscript[((Subscript[p, s])-2 l), μ])(Subscript[F, π]^V)(s)
		(Subscript[p, s])=p+q, s=(((p+q))^2), z=cos θ  is c.m. angle   (note also that (p+q)={(Sqrt[s]),0}, (Subscript[p, 0])=(Subscript[q, 0])=((Sqrt[s])/2)=(Subscript[l, 0]), |p|=|q|=|l|=((Sqrt[s])/2)ρ(s))

We obtain

(Subscript[(p-q), μ])Disc (Subscript[F, π]^V)(s)=(i/(64(π^2)))ρ(s)(Subscript[F, π]^V)(s)∫dΩ (Subscript[T, I]^( * ))(s,z)(Subscript[((Subscript[p, s])-2 l), μ])
		∫dΩ (Subscript[T, I]^( * ))(s,z)(Subscript[((Subscript[p, s])-2 l), μ])= (Subscript[(Subscript[L, 1])(p+q), μ])+(Subscript[(Subscript[L, 2])(p-q), μ])   ... then contract with (p+q), (p-q) ...

Exercise: Show that

∫dΩ (Subscript[T, I]^( * ))(s,z)(Subscript[((Subscript[p, s])-2 l), μ])=2π (Subscript[(p-q), μ])(∫_(-1)^1)dz z (Subscript[T, I]^( * ))(s,z)

where (Subscript[T, I])(s,z) is the ππ amplitude

(Subscript[T, I])(s,z)=32 π(∑_(l=0)^∞)(2l+1)(Subscript[t, l]^I)(s) (Subscript[P, l])(z)		
		(∫_(-1)^1)dz (Subscript[P, l])(z) (Subscript[P, l'])(z)=(( 2(Subscript[δ, l l']))/(2l+1))

Since  Disc (Subscript[F, π]^V)(s)=2 i Im(Subscript[F, π]^V)(s), we get

Im (Subscript[F, π]^V)(s)=ρ(s) (Subscript[F, π]^V)(s) (Subscript[t, 1]^(1* ))(s) θ(s>4(Subscript[m, π]^2))

If we consider only elastic scattering:

(Subscript[t, 1]^1)(s)=((sin (Subscript[δ, 1]^1)(s) (e^( i (Subscript[δ, 1]^1)(s))))/(ρ(s))),     ρ(s) (Subscript[t, 1]^(1* ))(s) =sin (Subscript[δ, 1]^1)(s) (e^( -i (Subscript[δ, 1]^1)(s)))

Watson final state theorem: the phase of the form factor is determined by the two-particle scattering phase shift: (Subscript[F, π]^V)(s)=|(Subscript[F, π]^V)(s)| (e^( i (Subscript[δ, 1]^1)(s)))

(Subscript[F, π]^V)(s)=|(Subscript[F, π]^V)(s)| (e^( i (Subscript[δ, 1]^1)(s)))		
		Arg ((Subscript[F, π]^V)(s))= (Subscript[δ, 1]^1)(s)

The full Omnes–Muskhelishvili problem (20 min)

We want to find the most general representation for a function, F(s), which is the analytic in the complex s plane
with the cut from s= [4(m^2),∞], assuming that we know its phase on the cut,

Arg (F(s))=δ(s), s>4(m^2)

Solution is not unique, if (Subscript[F, 0])(s) is a solution, then (e^(α s))(Subscript[F, 0])(s) is a solution too. Need to know the asymptotic information! We look for a solution in the from

F(s)=P(s) Ω(s)
		(1/(2i))(Ω(s+i ϵ)-Ω(s-i ϵ))=Ω(s+i ϵ) sin δ(s) (e^( -i δ(s))), 
		Ω(s+i ϵ) ((1/(2i))-((((e^( i δ(s)))-(e^( -i δ(s))))/(2i))) (e^( -i δ(s))))=Ω(s+i ϵ)(e^( -2i δ(s)))(1/(2i))=Ω(s-i ϵ)(1/(2i))
		Ω (s+i ϵ) (e^(-2 i δ(s)))=Ω(s- i ϵ), ln Ω (s+i ϵ)-2i δ=ln Ω (s+i ϵ)
		
		Disc(ln Ω(s))=2i δ(s)

Dispersion relation

ln Ω(s)=(1/(2 i))(∫_(4(m^2))^∞)((d s')/π)((Disc(ln Ω(s')))/(s'-s))=(∫_(4(m^2))^∞)((d s')/π)((δ(s'))/(s'-s)),    
		Ω(s)=Exp((∫_(4(m^2))^∞)((d s')/π)((δ(s'))/(s'-s))),   Ω(s)=Exp(a+(s/π)(∫_(4(m^2))^∞)((d s')/(s'))((δ(s'))/(s'-s)))

Ω(s)=Exp((s/π)(∫_(4(m^2))^∞)((d s')/(s'))((δ(s'))/(s'-s))),   Ω(0)=1

One subtraction: normalization Ω(0)=1. The function Ω(s) is known as the Omnes function. Many applications!

Exercise: Show that

Arg Ω(s)=δ(s)
		if δ(s)-> α π, show Ω(s->∞)-> (1/(s^α))

Use

∫((f(s') d s)/(s'-s∓ i ϵ))=p.v.∫((f(s')d s)/(s'-s))±i π f(s)

If one assume asymptotic: F(∞)->1/s, then (Subscript[F, π]^V)(s)=Ω(s)

Numerical implementation (25 min)

Ω(s)=Exp((s/π)(∫_(4(m^2))^∞)((d s')/(s'))((δ(s'))/(s'-s)))

Tangent stretching 
Let’s now consider the integral

(∫_a^∞)f(y)dy

In order to account the whole region, the replacement is needed y->y(x), which changes the integrating range (a,∞) into (0,1)

(∫_a^∞)f(y)dy=(∫_0^1)f(y(x))((d y)/(d x))dx,  y(x)=a+(Subscript[C, ext])  tg((π/2)x) ; ((d y)/(d x))=(((Subscript[C, ext]) π/2)/((cos((π/2)x))^2))
			(∫_0^∞)f(y)dy=(∑_(i=1)^n) f((Subscript[y, i]))  (OverTilde[(Subscript[w, i])]) .

If the integrand is smooth, it is convenient to use Gaussian weights for integration. Few words about (Subscript[C, ext]): As you can see  y(x=1/2)=a+(Subscript[C, ext]). It means that n/2 points will be accounted before a+(Subscript[C, ext]) 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, (Subscript[C, ext])=1.

Example 2: (∫_2^∞)(e^(-(y^2)))dy

<<NumericalDifferentialEquationAnalysis`

n=10; Cext=1;
wg=GaussianQuadratureWeights[n,0,1];
yn=2+Table[Cext*Tan[π/2*wg[[i,1]]],{i,n}];
wgn=Table[wg[[i,2]]((Cext π/2)/(Cos[π/2 wg[[i,1]]]^2)),{i,n}];
Sum[Exp[-yn[[i]]^2] wgn[[i]],{i,1,n}]
Integrate[Exp[-y^2],{y,2.,Infinity}]

If the integral contains a square root singularity (something like(∫_a^∞)f(y, (Sqrt[y-a]))dy) it is useful to introduce another replacement

y(x)=a+(Subscript[C, ext])  ((tg((π/2)x))^2) ; ((d y)/(d x))=(((Subscript[C, ext]) π/2 )/((cos((π/2)x))^2))2 tg((π/2)x)

Principle values integrals
Very often one has to take integral of the following form

F(s)=(s/π)(∫_(4(m^2))^∞)((ds')/(s'))((f(s'))/(s'-s-i ϵ))=θ(s<4(m^2))(s/π)(∫_(4(m^2))^∞)((ds')/(s'))((f(s'))/(s'-s))
						+θ(s>4(m^2))(s/π)(∫_(4(m^2))^∞)((ds')/(s'))((f(s'))/(s'-s-i ϵ))

The latter integral can be written as

(s/π)(∫_(4(m^2))^∞)((ds')/(s'))((f(s'))/(s'-s-i ϵ))=(s/π)(p.v.(∫_(4(m^2))^∞)((ds')/(s'))((f(s'))/(s'-s))+ i π ((f(s))/s))

For the p.v. integral we use the following trick:

p.v.(∫_(4(m^2))^∞)((ds')/(s'))((f(s'))/(s'-s))=p.v.(∫_(4(m^2))^∞)((ds')/(s'))((f(s')-f(s)+f(s))/(s'-s))
					=(∫_(4(m^2))^∞)((ds')/(s'))((f(s')-f(s))/(s'-s))+((f(s))/s)ln(((4(m^2))/(s-4(m^2))))
		p.v.(∫_(4(m^2))^∞)ds'(1/(s'(s'-s)))=(1/s)ln(((4(m^2))/(s-4(m^2))))

All together (Omnes function)

Ω(s)=Exp((s/π)(∫_(4(m^2))^∞)((d s')/(s'))((δ(s'))/(s'-s)))

Exit[];

<<NumericalDifferentialEquationAnalysis`
SetDirectory[NotebookDirectory[]];
StyleList = {AbsoluteThickness[1.0], AbsolutePointSize[5], #} & /@ {Blue, Red, Green, Purple};
SetOptions[Plot, Frame -> True, Axes -> {True, False}, PlotStyle -> StyleList, AspectRatio -> 0.8, FrameStyle -> Directive[13]];
<<PiPi_Madrid.m;

mpi=0.138;
mK=0.4956745;
Mrho=0.770;
ϵ=0.00001;
LamPhShift=1.3;

OmnesNInt[sig_][s_]:=Exp[s/Pi*NIntegrate[deltaFinal[1][sb]/(sb(sb-s-I sig ϵ)),{sb,4mpi^2,4mK^2,Infinity},AccuracyGoal->5,MaxRecursion->200]];

nOmn=120;
Cx=1.0;
wg=GaussianQuadratureWeights[nOmn,0,1];
s0=4mpi^2;
sn=s0+Table[Cx*Tan[Pi/2*wg[[i,1]]]^2,{i,nOmn}];
wgn=Table[Cx*2*Tan[Pi/2*wg[[i,1]]]*wg[[i,2]]*Pi/2/(Cos[Pi/2*wg[[i,1]]]^2),{i,nOmn}];
Deltan=Table[deltaFinal[1][sn[[i]]],{i,nOmn}];

Clear[deltas];
OmnesTemp0[s_]=Exp[s/π*Sum[Deltan[[i]]/(sn[[i]](sn[[i]]-s))*wgn[[i]],{i,1,nOmn}]];
OmnesTemp[s_]=Exp[s/π*Sum[(Deltan[[i]]-deltas)/(sn[[i]](sn[[i]]-s))*wgn[[i]],{i,1,nOmn}]];

Omnes[sig_][s_]:=Which[sig==0, OmnesTemp0[s],
sig==1||sig==-1, deltas=delta[1][s]; Which[s≤4mpi^2, OmnesTemp0[s], 
s>4mpi^2,Exp[I*sig*delta[1][s]]*Exp[deltas/π Log[s0/(s-s0)]]*OmnesTemp[s]]];

Show[Plot[deltaFinal[1][par^2]*180/Pi,{par,2mpi,2.0},PlotPoints->50,MaxRecursion->0,PlotStyle->{Black,Thick},FrameLabel->{"(Sqrt[s]) [GeV]","δ(s)"}],DeltaExp[1],ImageSize->{250,250}]

GraphicsRow[{Plot[Re[Omnes[1][par^2]],{par,2mpi,1.1},PlotStyle->{Black},MaxRecursion->0,PlotPoints->100,PlotRange->All,ImageSize->{250,250},FrameLabel->{"(Sqrt[s]) [GeV]","Re(Ω(s))"}],Plot[{Im[Omnes[1][par^2]]},{par,2mpi,1.1},PlotStyle->{Black},MaxRecursion->0,PlotPoints->100,PlotRange->All,ImageSize->{250,250},FrameLabel->{"(Sqrt[s]) [GeV]","Im(Ω(s))"}]}]

Omnes[1][1]
OmnesNInt[1][1]