Elastic ep scattering and higher radiative corrections Part II

Elastic ep scattering and higher radiative corrections

Part II

Krzysztof M. Graczyk

Institute of Theoretical Physics Neutrino Physics Division

University of Wrocław Poland

December 6, 2011

Motivation

Jeśli nie przestaniecie udowadniać tego, co już zrobili inni, nabierać pewności, komplikować rozwiązań – po prostu dla przyjemności – wtedy, pewnego dnia rozejrzycie się, i stwierdzicie, że tego jeszcze nikt nie zrobił! To jest sposób zostania uczonym!

R. P. Feynman.

PT and Rosenbluth

There is a systematic discrepancy between ratio µpGE/GM data extracted from PT and cross section measurements!

Figure:

Nucl. Phys. 59 (2007) 694.

PT and Rosenbluth

I The two-photon exchange (TPE) correction (Born-like) is responsible for that!

I The PT data is less affected by TPE correction than cross section

measurements! P. A. M. Guichon and M. Vanderhaeghen, Phys. Rev. Lett. 91 (2003) 142303, P. G. Blunden, W. Melnitchouk and J. A. Tjon, Phys. Rev. Lett.

91 (2003) 142304.

electron/positron scattering off proton

I At least two new experiments dedicated to the investigation of the TPE contribution!

σ(e⁺p → e⁺p)

σ(e⁻p → e⁻p) ≈ 1 −2∆C2γ

σ1γ

. (1)

I J. Arrington et al., Two-photon exchange and elastic scattering of electrons / positrons on the proton. (Proposal for an experiment at VEPP-3), arXiv:nucl-ex/0408020;

I Jefferson Lab experiment E04-116, Beyond the Born Approximation: A Precise Comparison of e⁺p and e⁻p Scattering in CLAS, W. K. Brooks, et al., spokespersons.

The Proton Radius

I The Proton Radius is extracted from CODATA, as it has been already explained,

p

Mod. Phys. 80, 633 (2008).) I Lamb shift in muonic atom,

p

hr²i = 0.84184 ± 0.00067 fm, R. Pohl, A.

Antognini, F. Nez et al., Nature 466, 213 (2010).

I The results are 5σ away of each other!

I The Lamb shift is a small difference in energy between two energy levels²S^1/2 and²P_1/2 of the hydrogen atom. According to Dirac, the²S_1/2and²P_1/2 orbitals should have the same energies. However, the interaction between the electron and the vacuum causes a tiny energy shift on²S_1/2. (see e.g. K.

Pachucki, Phys. Rev. A60 (1999) 3593.)

Lexp = 206.2949 ± 0.0032 meV (2)

L_th = 209.9779(49) − 5.2262

p

hr²i + 0.00913

p

hr³i₍₂₎ (3)

where hr³i₍₂₎ is the third Zemach moment defined as:

hr³i₍₂₎=

Z

d³rd³r⁰|r − r⁰|³ρ(r⁰)ρ(r) (4)

p

LAMB = 0.84184 ± 0.00067 fm (5)

p

CODATA = 0.8768 ± 0.0069 fm (6)

p

dipole = 0.81 fm (7)

p

Ei

NN = 0.85 fm (8)

p

NN = 0.82 fm (9)

(10) NN from K.M. Graczyk, Phys. Rev. C 84, 034314 (2011).

Electron Scattering off Coulomb Potential

Electron Scattering off Coulomb Potential

Our attention is concentrated on the first and second order Born diagrams.

|M|² ≈ |M⁽¹⁾|²

| {z }

α²

+ 2Re M^(1)∗M⁽²⁾

| {z }

α³

+ |M⁽²⁾|²

| {z }

α⁴

,(11)

d σcou.

d Ω = |p⁰| 16π²|p|·1

2

X

d Ω +d σ⁽²⁾cou.

d Ω +d σ⁽³⁾cou.

d Ω (12)

T-Matrix: 1st Born

hp⁰|iT⁽¹⁾

fi |pi = hp⁰| −i

Z

d⁴xeψ(x )γ^µψ(x )Aµ(x ) | pi (13)

= −ieu(p⁰)γ^µu(p)

Z

d⁴xe^{i (p0−p)x}Aµ(x ) (14)

= −ieu(p⁰)γ^µu(p)˜Aµ(p⁰− p) (15)

= −(2πi)δ(Ef − Ei)eu(p⁰)γ^µu(p)˜Aµ(p⁰− p) (16)

where we have assumed that Aµis time independent, and ˜Aµ(p⁰− p) is the Fourier transform:

A˜µ(p⁰− p) = (2π)δ(E_f− E_i)˜Aµ(p⁰− p), (17) and

A˜µ(p⁰− p) =

Z

d³re^{−i(p0−p)·r}Aµ(r). (18)

T-Matrix: 2st Born

hp⁰| iT⁽²⁾| pi = −1 2hp⁰|

Z

d⁴xd⁴ye²ψ(x )γ^µΨ(x )Aµ(x )ψ(y )γ^νΨ(y )Aν(y ) | pi (19)

= − ie²

Z

u(p⁰)γ^µ(ˆl + me)γ^νu(p) l²− m²_e+ i 

A˜µ(p⁰− l)˜Aν(l − p)

(20)

Coulomb Potential and M-matrix Point-like Coulomb potential

A^µ_poin(~r) = g^µ0 Ze

4π|r|. (21)

A˜⁰_poin(p⁰− p) =

Z

d³re^{−i(p0−p)·r} Ze 4π|r| = Ze

4πq²

Z

d³r 4 e^{−i(p0−p)·r}

|r|(22)

= Ze

4πq²

Z

d³re^{−i(p0−p)·r}4



|r|



= −Ze

q². (23)

Hence

hp⁰|iT⁽¹⁾|pi = (2πi)δ(E_f − E_i)Ze²

q² u(p⁰)γ⁰u(p). (24) hp⁰|iT⁽²⁾|pi = −(2πi)δ(p⁰₀− p0)Z²e⁴

Z

u(p⁰)(γ⁰E + ~γ · l + me)u(p) (l²− p²+ i )(p⁰− l)²(l − p)², (25) where we have integrated the energy component of dl0, namely

Z

2π(2π)δ(p⁰₀− l₀)(2π)δ(l0− p₀) = 2πδ(p⁰₀− p₀).

and we substituted, l0= E as well as l²= l₀²− l²− m²_e= p²− l². Hence

γ⁰(ˆl + me)γ⁰= γ⁰(γ⁰l⁰− ~γ · l + me)γ⁰= γ⁰l⁰+ ~γ · l + me→ γ⁰E + ~γ · l + me (26)

Coulomb Potential and M-matrix

In order to compute the cross section one has to compute the M matrix,

hp⁰| iT | pi = M(2πi)δ(p⁰₀− p₀). (27)

Similarly as in the case of the T matrix, the M matrix can be written in as the perturbative series:

M = M⁽¹⁾+ M⁽²⁾+ ... (28)

I First order:

M⁽¹⁾=Ze²

q² u(p⁰)γ⁰u(p). (29)

I Second order:

M⁽²⁾= −Z²e⁴

Z

u(p⁰)(γ⁰E + ~γ · l + me)u(p)

(p²− l²+ i )(p⁰− l)²(l − p)². (30)

Arbitrary Potential – Form Factor

Suppose that the Coulomb potential is not point-like but has its own spherical distribution.

A^µ(~r) = g^µ0φ(r), φ(r) = Z 4π

Z

d³r⁰ ρ(r⁰)

|r − r⁰|, Z =

Z

d³r ρ(r) (31)

The Fourier transformation of the potential reads

A⁰(q) = Ze 4π

Z

d³rd³r⁰e^−iq·r ρ(r⁰)

|r − r⁰|= Ze 4πq²

Z

d³rd³r⁰e^−iq·r4r

ρ(r⁰)

|r − r⁰|

= Ze q²

Z

d³rd³r⁰e^−iq·rρ(r⁰)δ⁽³⁾(r − r⁰)

= Ze q²

Z

d³re^−iq·rρ(r) ≡ Ze q² F (q)

|{z}

form factor

. (32)

F (q) =

Z

d³re^−iq·rρ(r) ρ(r) =

Z

d³re^{i q·r}F (q). (33)

Arbitrary Potential – Form Factor

d σ1γ

d Ω = Z²α² 4β²p²sin^{4 θ}₂



1 − β²sin²



2



, (34)

where

β²=|p²|

E² = v_electron² . (35)

Notice that for β → 0 we have well known Mott scattering formula,

d σ⁽¹⁾_{coul .}

d Ω ≈ Z²α²

4β²p²sin^{4 θ}₂

Notice that if instead of the point-like potential the one given by (31) is discussed the cross section in the first order Born approximation reads

d σ⁽¹⁾

d Ω = α²

4β²p²sin^{4 θ}₂



1 − β²sin²



2



F²(q). (37)

I Point-like static potential:

d σ

d Ω= α²E

4E³sin^{4 θ}₂

I Spatial-charge distribution but still static:

d σ

d Ω= α²E 4E³sin^{4 θ}₂

2(q²). (39)

I Proton with spin 1/2:

d σ

d Ω= α²E⁰ 4E³sin^{4 θ}₂ ·



cos²θ 2+ Q²

2M²sin²θ 2



, (40)

(recoil correction,spin 1/2 correction)

I Proton with spin 1/2, and magnetic anomalous moment

d σ d ΩLAB

= α²E⁰ 4E³sin^{4 θ}₂ ·



cos²θ 2



F12+ Q² 4M²F22



+ Q² 2M²sin²θ

2(F1+F2)²



(41) (γ^µ,σ^µν).

Notice that for Q²→ 0, sin^{2 θ}₂→ 0, cos^{2 θ}₂ → 1 the static potential F (q²) form factor has the same meaning as F1(Q²), and G_E because G_E= F1− τ F2.

Photon Mass

M⁽²⁾ = −Z²e⁴

Z

u(p⁰)(γ⁰E + ~γ · l + me)u(p)

(p²− l²+ i )(p⁰− l)²(l − p)² (42)

= Z²e⁴u(p⁰) (γ0E I1+ ~γ · I2) u(p) (43)

I Long-range character of the Coulomb forces → infrared singularities I One has to extract the divergent term from the amplitude

I R. H. Dalitz Proc. R. Soc. Lond. A206, 509 (1951).

1 q² → 1

q²+ µ². (44)

It corresponds to the screened Coulomb interaction

A^µ(r ) = g^µ0Zeexp(−µ|r|) 4π|r| = −

Z

δ(q0)

q²− µ²e^iq·r (45) The problem is seen already in QM, if the potential V (r ) does not converge faster then 1/r then the partial wave solution can not be obtained. For potential 1/r distorted wave functions are obtained!

Properties of the integrals I1and I2

Notice that I1and I2are symmetric under exchange p ↔ p,

I1=

Z

1

(l²− p²− i)(p⁰− l)²(l − p)², I2=

Z

l

(l²− p²− i)(p⁰− l)²(l − p)² (46) hence I2∼ p + p⁰. Notice that u(p⁰, s⁰)~γu(p, s) = χ^†_s0(p + p⁰+ i q × τ )χs.

Feynman’s Identity 1

(a + λ)(b + λ) = −∂

∂λ

Z

d α 1

αa + (1 − α)b + λ= −

Z

d α ∂

∂(αa)

1 αa + (1 − α)b + λ

(47) Then

I1 = −

Z

d α ∂

∂µ²

Z

1

(l²− p²− i)((l − P)²+ M₀²) (48)

= −

Z

d α ∂

∂M₀²

Z

1

(l²− p²− i)((l − P)²+ M₀²) (49)

I^k₂ =

Z

d α



2∂P^k − P^k ∂

∂M₀²

 Z

1

(l²− p²− i)((l − P)²+ M₀²)(50) where

Properties of the integrals I1and I2

I1 = −

Z

d α ∂

∂M₀²I (52)

I^k₂ =

Z

d α



2∂P^k − P^k ∂

∂M₀²



I (53)

I =

Z

1

(l²− p²− i)((l − P)²+ M₀²) (54) Notice that

I ∼

Z

|M|²

|M|²≈ |M⁽¹⁾|²+2 Re M^(1)∗M⁽²⁾

| {z }

∗

+ |M⁽²⁾|², (55)

X

spin

∗ = Z³e⁶

|q|²

X

spin

u(p⁰)γ⁰u(p)



I1E u(p⁰)γ⁰u(p) + I2· u(p⁰)~γu(p)

= Z³e⁶

|q|²

"

I1E Tr(ˆpγ⁰pˆ⁰γ⁰) +

3

X

k=1

I^k₂Tr(ˆpγ⁰pˆ⁰γ^k)

#

= Z³e⁶

|q|²





(56)

= 4Z³e⁶E³cos^{2 θ}₂ q²



cos^{2 θ}₂ I1+I2· (p + p⁰) E²cos^{2 θ}₂



(57)

|M|²: useful expressions

Trˆpγ⁰pˆ⁰γ⁰ = 4



2EiEf − p · p⁰



(58)

= 4(E²+ p · p⁰) = 4E²(1 + β²cos θ) (59) Trˆpγ⁰pˆ⁰γ^k = 4



p^kE + p^0kE



(60)

3

X

k=1

I^kTrˆpγ⁰pˆ⁰γ^k = 4E I2· (p + p⁰) (61)

(62)

(p + p⁰)²E = 2|p|(p²+ p · p⁰) = 4|p|³ β cos²θ

2 (63)

p²+ 2p · p⁰ = |p|²(4 cos²θ

2− 1) (64)

I

I(p², P, M₀²) =

Z

1

(l²− p²− i)((l − P)²+ M₀²(µ²)) (65)

=

Z

−1

dt

Z

dl (2π)²

l² (l²− p²− i)

1

l²− 2tPl + P²+ M₀² (66)

= 1

2

Z

−1

dt

Z

0

dl (2π)²

l² (l²− p²− i)

1

l²− 2tPl + P²+ M₀²

+1 2

Z

−1

dt

Z

dl (2π)²

l² (l²− p²− i)

1

l²+ 2tPl + P²+ M²₀(67)

t → −t,

Z

−1

dt →

Z

−1

dt (68)

(69)

I

I = 1

2

Z

−1

dt

Z

0

dl (2π)²

l² (l²− p²− i)

1

l²− 2tPl + P²+ M₀²

+1 2

Z

−1

dt

Z

−∞

dl (2π)²

l² (l²− p²− i)

1

l²− 2tPl + P²+ M₀²,(70)

l → −l ,

Z

0

dl →

Z

−∞

dl (71)

I = 1

2

Z

−1

dt

Z

−∞

dl (2π)²

l² (l²− p²− i)

1

l²− 2tPl + P²+ M₀² (72)

l²− p²− i = (l − p − i )(l + p + i ) (73)

= 1

2

Z

−1

dt

Z

−∞

dl (2π)²

l²

(l − p − i )(l + p + i )(l − l+)(l − l−)(74), where

l± = Pt ± i

p

(1 − t²)P²+ M₀² (75)

M₀² = p²+ µ²− P²= µ²+ 4α(1 − α)q² (76)

I

Z

−∞

(...)

| {z }

+

Z

R

(...)

| {z }

= 2πi

X

Res(top semicircle) (77)

I

I = ip

8π

Z

−1

dt 1

p²− 2tPp + P²+ M₀²+ 1 8π

Z

−1

dt l₊²

(l₊²− p²)

 p

(1 − t²)P²+ M₀²



(78)

= i

16πP



− ln p²− 2Pp + P²+ M₀²



+1 8π

Z

−1

dt l₊²

(l₊²− p²)

 p

(1 − t²)P²+ M₀²



In the second integral we do the change of the variables, t → l+, indeed

dl+= − iPl+dt

p

(80)

where 1 8π

Z

−1

dt l₊²

(l₊²− p²)

 p

(1 − t²)P²+ M₀²



i 8πP

Z

dl+

l+

(l₊²− p²) = i

16πPln(l₊²−p²) (81)

I

I = i

16πP



− ln p²− 2Pp + P²+ M₀²



+ i 16πP





, (82)

= i

8πPln



0

p − P + iM0



(83)

I1

I1= −

Z

d α ∂ 2M0∂M0

I (84)

Hence

I1 = 1 8π

Z

d α 2PM0

h

p + P + iM0

− 1

p − P + iM0

i

= − 1

8π

Z

d α 1 M0

1

(p + iM0)²− P² (85)

= − 1

8π

Z

d α 1 M0

1

−µ²+ i 2pM0

(86)

= 1

8πµ²

Z

d α M0

− 1

8πµ²

Z

d α 2pi

−µ²+ i 2pM0

(87)

I1

M₀² = q²α(1 − α) + µ²=



 

1 − q²

q² 4 + µ²



α −1 2





(88)

α⁰ = 2|q|

p



α −1 2



(89)

Now,

Z

= 1

|q|

Z

p

= 1

|q|arcsin(α⁰)= 1

|q|arcsin 2|q|

p



α −1 2



!

(90)

Z

d α M0

= 2

|q|arcsin |q|

p

!

= 2

|q|arcsin





1

q

1 +^4µ_q2²





π

|q| (91)

where

√ 1 ≈ 1 +x

, arcsin(1 + x ) ≈ π (1 + x )

I1

Z

d α 1

−µ²+ i 2pM0

= 1

2p

Z

d α 1

−^µ_2p² + iM0

(92)

= 1

2p

Z

0

d α 1

−^µ2

2p + i

p

µ²+ α(1 − α)q²

(93)

= 1

p|q|

Z

0

d α⁰ 1

−_2p|q|^µ2 + i

q

q² − α⁰² (94)

α⁰ = α −1

2 (95)

From Mathematica

Z

−B + i√

A²− x² = −i arctan



√A²− x²



+ (96)

−iB tanh⁻¹



√ Bx

−A²−B²

√

A²−x²



+ B arctan



−A²−B²



√−A²− B²

∗ =

Z

0

d α⁰ 1

−_2p|q|^µ2 + i

q

4+^µ_q2²− α⁰²

(97)

= 1

µ²







−i arctan



2µ



−

i_2p|q|^µ2

q

q² + ^µ4

4p²q²









i 2

q

q² + ^µ4

4p²q²





−i tanh⁻¹ −i µ 4p



1 +4µ² q² + µ⁴

p²q²



!#)

(98)

= −i π 2µ²− 1

p|q|tanh⁻¹



1 +4µ² q² + µ⁴

p²q²



!

=−i π 2µ²− 1

p|q|lnp sin^θ₂

µ (99)

tan(−iz) =1

i tanh(z), tanh(iz) = i tan(z)



1 +4µ² q² + µ⁴

p²q²



≈ 1 −1 2



p²q²



≈ 1 −2µ²

q² (100)





!

1

 

2µ²



1 q² 2p sin^θ

I1

I1 = 1 8π



|q|µ²− 2i

|q|



−i π 2µ² − 1

p|q|ln2p sin^θ₂ µ



(101)

= i

4πq²pln2p sin^θ₂

µ (102)

= i

16π sin^{2 θ}₂p³ln2p sin^θ₂

µ (103)

It is divergent when µ → 0, but does not contribute to the spin averaged interference term 2Re(M⁽¹⁾M^(2)∗)

I2

p · I2 =

Z

p · l

(l²− p²− i)((p⁰− l)²+ µ²)((l − p)²+ µ²) (104) p⁰· I₂ =

Z

p⁰· l

(l²− p²− i)((p⁰− l)²+ µ²)((l − p)²+ µ²). (105) Notice that

1

(l − p)²+ µ² − 1

(l²− p²− i) = 2l · p − 2p²− µ²

(l²− p²− i)((l − p)²+ µ²) (106) 1

(l − p⁰)²+ µ² − 1

(l²− p²− i) = 2l · p⁰− 2p²− µ²

(l²− p²− i)((l − p⁰)²+ µ²) (107) hence

l · p

(l²− p²− i)((l − p)²+ µ²) = 1 2

h

(l − p)²+ µ²− 1

(l²− p²− i)+ 2p²+ µ² (l²− p²− i)((l − p)²+ µ²)

i

l · p⁰

2− p²− i)((l − p^{02 2} = 1

h

02 2− 1

2− p²− i)+ 2p²+ µ² 2− p²− i)((l − p^{02 2}

i

I2

p · I₂ = 1

2

Z

1 ((p⁰− l)²+ µ²)



(l − p)²+ µ²)

− 1

(l²− p²− i)



+



p²+µ² 2



I₁

= 1

2(A − B_a) +



p²+µ² 2



I₁ (108)

p⁰· I₂ = 1 2

Z

1 ((p − l)²+ µ²)



(l − p⁰)²+ µ²− 1 (l²− p²− i)



+



p²+µ² 2



I₁

= 1

2(A − B_b) +



p²+µ² 2



I₁ (109)

where

A =

Z

1

((p⁰− l)²+ µ²)(l − p)²+ µ²) (110)

Ba =

Z

1

((p⁰− l)²+ µ²)(l²− p²− i) (111)

B_b =

Z

1

((p − l)²+ µ²)(l²− p²− i) (112)

It is easy to show that B_a= B_b≡ B.

A

We use the Feynman’s trick,

1 AB=

Z

dx [Ax + B(1 − x )]²

A =

Z

1

((p⁰− l)²+ µ²)(l − p)²+ µ²)=

Z

dx

Z

1





=

Z

dx

Z

1





0= l − (x p + (1 − x )p⁰) (113)

=

Z

dx

Z

1





2

0= µ²+ q²x (1 − x ) (114)

=

Z

dx

Z

−∞

dl 4π²

l²





1 8π

Z

dx M₀ = 1

8π

Z

dx

p

(115)

= 1

8|q| (116)

I2

Analogically w obtain

B = i 8πp



ln



µ



− iπ 2



= I(p², p⁰, µ²) (117)

hence

(p + p⁰) · I₂ = A − B + 2p²I₁= 1 8|q|

− i 8πp



ln



µ



− iπ 2



+ ip

4πq²ln2p sin^θ₂

µ (118)

= 1 − sin^θ₂ 8|q|

− i

16πp



2 ln



µ



− 1

sin^{2 θ}₂ ln2p sin^θ₂ µ



| {z }

divergent

(119)

σ⁽²⁾

d σ_cou⁽²⁾

d Ω = 1

16π²

X

spin

Re M^(1)∗M⁽²⁾

(120)

= 1

16π² Z³e⁶

|q|²





(121)

= 2Z³πα³E

|q|³



1 − sinθ 2



(122)

= Z³πα³ q²β



1 − sinθ 2



(123)

R_{coul .}⁽²⁾ =σ_{coul .}⁽²⁾ σ1γ

= πZαβ 1 − sin^θ₂

1 − β²sin^{2 θ}₂

θ

2 (124)

0 0.002 0.004 0.006 0.008 0.01 0.012

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 R(2) cou.

sin(θ/2) rel

non-rel

0 0.002 0.004 0.006 0.008 0.01 0.012

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 R(2) cou.

ε

rel, Q²=0.01 non-rel, Q²=0.01

Form-Factors: Time-like-Region

Interesting property:

ImF_1,2= p³_t

√

tΓ^∗_1,2(t) (125)

p_tpion momentum in the crossed (t−)channel, Γ_1,2– P-amplitudes for the ππ − NN.

Form-Factors: Time-like-Region

F (q²) = 1 2πi

I

C dt F (t)

t − q² = 1 2πi

Z

∞

dt F (t − i ) t − q²− i+

Z

dt F (t + i ) t − q²+ i 

!

(126)

= 1

π

Z

dtImF (t + i )

t − q² , F (s^∗) = F^∗(s) (127)

= F (0) +q² π

Z

dtImF (t + i )

t(t − q²) (128)

hr²i = 6 π

Z

dtImF₁^p(t) t²

| {z }

≈0.65 fm2

+ F₂^p(0) 4M²

|{z}

3κp 2M2≈0.12 fm2

(129)