PDF 1 Relativistic treatment of electron-proton scattering - Tokai University

1 Relativistic treatment of electron-proton scattering

Remember: The S-matrix (in second order perturbation theory) for the scattering of an electron in an external electromagnetic field A^µ(x) is given by (see No. 12, Eq.(1.8)):

S_{f i} =−ie Z

d⁴x j_{f i}^µ(x)A_µ(x) (1.1)

wherej_{f i}^µ(x) is the electron “transition current” given by

j_{f i}^µ(x) =ψ_f(x)γ^µψi(x) =e^i(p0−p)xj_{f i}^µ(x= 0) (1.2) with

j_{f i}^µ(x= 0) = r m

E_p⁰ r m

E_p⁰ 1

V u(~p⁰, s⁰)γ^µu(~p, s) (1.3)

i=(p,s) f=(p’,s’)

I=(P,S) F q

electron proton

final hadrons (F), total momentum P’

x x’

The vector potential A^µ(x), which is produced by the hadron in the transition I → F, is obtained from the Maxwell equation

A^µ(x) = e_pJ_{F I}^µ (x) (1.4)

whereep =−eis the proton charge, and=∂µ∂^µ is the d’Alembert operator. The solution to (1.4) is obtained as

A^µ(x) =−e_p Z

d⁴x⁰D(x−x⁰)J_{F I}^µ (x⁰) (1.5) HereD(x−x⁰) is the Green function of the d’Alembert operator defined by

xD(x−x⁰) = −δ⁽⁴⁾(x−x⁰) (1.6)

Using (1.5), the S-matrix (1.1) becomes Sf i =ieep

Z d⁴x

Z

d⁴x⁰j_{f i}^µ(x)D(x−x⁰)Jµ,F I(x⁰) (1.7) Because the proton initial state is described by a plane wave e^−iP·x0 (independent of its internal structure), and the final hadron state (whatever it is) is also described by a plane wavee^iP0·x0, thex⁰ - dependence of the hadron transition current is of the form

J_{F I}^µ (x⁰) = e^{i(P0−P)·x0}J_{F I}^µ (x⁰ = 0) (1.8) Then we can perform the integrations over x and x⁰ in (1.7) by

Z d⁴x

Z

d⁴x⁰e^i(p0−p)·xe^{i(P0−P)·x0}D(x−x⁰) ^(x,x

0)→(z=x−x⁰,x)

=

Z d⁴x

Z

d⁴z e^{i(p0−p+P0−P)·x}e^{−i(P0−P)·z}D(z)

= (2π)⁴ δ⁽⁴⁾(P⁰ +p⁰−P −p)D(q) (1.9) whereq =p−p⁰, and the Fourier transformed Green function D(q) is obtained from (1.6) as

D(q) = 1

q² (1.10)

Then the S-matrix (1.7) becomes S_{f i} = iee_p

q² (2π)⁴ δ⁽⁴⁾(P⁰+p⁰−P −p) j_{f i}^µ J_{µ,F I} (1.11) where the transition currents are now defined atx= 0.

Remember from No.12: In order to calculate the differential cross section from (1.11), we have to do the following:

• Calculate |S_{f i}|², by using the following rule:

(2π)⁴ δ⁽⁴⁾(P⁰+p⁰−P −p)2

≡ (2π)⁴ δ⁽⁴⁾(P⁰+p⁰ −P −p) ·(2π)⁴ δ⁽⁴⁾(k = 0)

≡ (2π)⁴ δ⁽⁴⁾(P⁰+p⁰ −P −p) V∆T whereV is the volume of the system, and ∆T is the observation time.

• Divide by the flux of incident particles. If we consider the laboratory frame (proton target at rest), this is the flux of incoming electrons (see No. 12):

|~j_in|=|~v|N_in V

where~v is the velocity of the incoming electrons, and N_in is the number of incoming electrons per second.

• Multiply a factor

N_in

∆T

Vd³p⁰ (2π)³

Vd³P⁰ (2π)³ After these steps, we get for the differential cross section:

dσ = e⁴

q⁴ (2π)⁴ δ⁽⁴⁾(P⁰+p⁰−P −p)1 v

d³p⁰ (2π)³

d³P⁰

(2π)³ |j_{f i}^µ J_{µ,F I}|² (1.12) To obtain this result, we have taken out the factors _V¹, which are contained in the transition currents (see (1.3) for the electron):

j_{f i}^µ → 1

V j_{f i}^µ , J_{F I}^µ → 1 V J_{F I}^µ where the transition currents j_{f i}^µ and J_{F I}^µ now have no factors 1/V.

We will always assume that the spin of the electron in the final state (s⁰) is not observed, and that the momentum and spins of the final hadrons (P~⁰ and S⁰, where S⁰ symbolically denotes the final spin of the hadrons) are also not observed. So, only the momentum of the electron in the final state (~p⁰) is observed. In this case, we have to sum overs⁰ andS⁰, and integrate over P~⁰ in Eq.(1.12). The factor |j_{f i}^µ J_{µ,F I}|² in (1.12) can be expressed as follows:

|j_{f i}^µ J_{µ,F I}|² = j_{f i}^µ j_{f i}^ν∗

(J_{F I}^µ J_{F I}^ν∗) (1.13)

We now define the leptonic tensor `^µν and the hadronic tensor W^µν by the following relations ¹:

`^µν ≡ 2m²X

s⁰

(u(~p⁰, s⁰)γ^µu(~p, s)) (u(~p⁰, s⁰)γ^νu(~p, s))^∗ (1.15) W^µν ≡ E_P

Z

d³P⁰δ⁽⁴⁾(P⁰+p⁰−P −p)X

S⁰

J_{F I}^µ J_{F I}^ν∗ (1.16) Using also d³p⁰ =p⁰²dp⁰dΩ⁰ =p⁰E_p⁰dE_p⁰dΩ⁰ we obtain the following important result:

dσ

dE_p⁰dΩ⁰ = e⁴ 8π²

p⁰ p

1

E_P q⁴ `_µνW^µν (1.17)

Remember that q² = q²₀ −~q² is the square of the 4-momentum transfer in the scattering process, and in the laboratory system the initial proton 4-momentum is P^µ =

E_P =M, ~0

. Therefore the Bjorken variablex=−q²/(2P ·q) can be measured by observing only the electron in the final state.

In the following, we will consider the following 2 processes:

• Elastic scattering: The final state (F) is a proton (x= 1).

• Inelastic inclusive scattering: The final state is not a proton (x < 1), but is not observed in the experiment. In this case, we have to sum over all possible final hadronic statesF.

(1) Elastic scattering (x= 1):

What is the form of the transition current J_{F I}^µ ? If the proton were a point particle, it must be of the same form as for the electron, i.e., J_{F I}^µ =q

M E_P0

qM

EP u(P~⁰, S⁰)γ^µu(P , S). The effects of proton~

1The hadronic tensor (1.16) is defined in a general frame (not only the laboratory frame). It is often written in the form

W^µν = 2E_p 2π

Z d³P⁰ (2π)³

X

S⁰

(2π)⁴δ⁽⁴⁾(P⁰+p⁰−P−p) J_{F I}^µ J_{F I}^ν∗ ≡ 2E_P 2π

ˆ X

F(2π)⁴δ⁽⁴⁾(P⁰+p⁰−P−p)J_{F I}^µ J_{F I}^ν∗

≡ 2EP

2π ˆ X

F(2π)⁴δ⁽⁴⁾(P⁰+p⁰−P−p)hF|Jˆ^µ|Ii hF|Jˆ^ν|Ii^∗ (1.14) where the sum over the final proton states is defined as ˆP

F ≡R d³P⁰ (2π)³

P

S⁰, and ˆJ^µ is the current field operator. The factor 2Ep in the above expression of W^µν reflects our non-covariant normalization of state vectors throughout this course. The normalization and completeness relations in this convention are

h~p⁰|~pi = (2π)³δ⁽³⁾(~p⁰−~p)⇔ h~p|~pi=V Z d³p

(2π)³|~pi h~p| = 1

In many texts, covariant normalization (c) is used, with|~pic=p

2E_p|~pi. Therefore our formulas for matrix elements h~p|. . .|~pi have a factor of 2E_p, which does not appear in covariant normalization. Note that in our non-covariant normalization, the matrix elementsJ_{F I}^µ are dimensionless.

i=(p,s) f=(p’,s’)

I=(P,S) F=(P’,S’) q

electron proton proton

γµ Γµ

structure can be described by using a modified vertex:

γ^µ−→Γ^µ(P⁰, P) (1.18)

This vertex must contain the electric and magnetic form factors, and will be specified later. So, we write for the transition current

J^µ(P⁰, P) = r M

E_P⁰ rM

E_P u(P~⁰, S⁰) Γ^µ(P⁰, P)u(P , S~ ) (1.19) Consider now the hadronic tensor (1.16):

W^µν = 2M²

Z d³P⁰

2E_P⁰ δ⁽⁴⁾(P⁰+p⁰−P −p) X

S⁰

u(P~⁰, S⁰) Γ^µ(P⁰, P)u(P , S~ )

×

u(P~⁰, S⁰) Γ^ν(P⁰, P)u(P , S)~ ∗

(1.20) We can rewrite this as a 4-dimensional momentum integral by using the identity ²

Z d³P⁰ 2E_P⁰ =

Z

d⁴P⁰δ

P⁰²−M²

θ(P₀⁰) Then the hadronic tensor (1.25) becomes

W^µν = 2M²θ(P₀+p₀−p⁰₀)δ

(P +p−p⁰)²−M²

× X

S⁰

u(P~⁰, S⁰) Γ^µ(P⁰, P)u(P , S)~ u(P~⁰, S⁰) Γ^ν(P⁰, P)u(P , S)~ ∗

whereP⁰ =P +p−p⁰.

In order to calculate the differential cross section dσ

d Ω⁰ = Z ∞

m

dσ

dE_p⁰dΩ⁰ dEp⁰

2This is an identity, because on the r.h.s. the integration over P₀⁰ gives: R∞

−∞dP₀⁰δ

P⁰²−M²

θ(P₀⁰) =

1 2E_P0

R∞

−∞dP₀⁰δ(P₀⁰−E_P⁰) = _2E¹

P0.

from Eq.(1.17), we also have to integrate over E_p⁰. This can be done by using the following relation (in the laboratory frame):

Z ∞

m

dE_p⁰θ(M +E_p−E_p⁰)δ

(P +p−p⁰)²−M²

F(E_p⁰) = 1 2M

1

1 + ^2E_M^psin^{2 θ}₂ F(E_p⁰ =E⁰) (1.21) where the final electron energyE⁰ is given by the initial electron energy (E =E_p) and the scattering angle by (see also No. 14, Eq.(1.8)):

E⁰ = E

1 + ^2E_M sin^{2 θ}₂ (1.22)

Home work: Derive Eq.(1.21) by using the following identity in the laboratory frame:

(P +p−p⁰)²−M² = 2M(E_p−E_p⁰)−4E_pE_p⁰sin² θ 2 (Remember that we always neglect the electron mass: E_p ' |~p|and E_p⁰ ' |~p⁰|.)

Then we get the following relation for the differential unpolarized cross section (withα=e²/(4π))³: dσ

dΩ⁰ = α² 4

E⁰ E

1 M²q⁴

1 1 + ^2E_M sin^{2 θ}₂

X

sS

`µνw^µν (1.23)

where the leptonic and hadronic tensors (for the elastic case) are given by

`^µν = 2m²X

s⁰

(u(~p⁰, s⁰)γ^µu(~p, s)) (u(~p⁰, s⁰)γ^νu(~p, s))^∗ (1.24) w^µν = 2M²X

S⁰

u(P~⁰, S⁰)Γ^µu(P , S~ ) u(P~⁰, S⁰)Γ^νu(P , S~ )∗

(1.25)

What is the form of the effective vertex Γ^µ? Remember: For a point particle (electron) we have Γ^µ=γ^µ. What is the difference between electron and proton (besides opposite charge)?

• Spin g-factor (magnetic moment): For electron g_s = 2, but for protong_s= 5.58.

• Proton has an extended charge distribution. Its Fourier transform is the electric form factor ⁴ G_E(Q²).

• Proton has an extended magnetic moment distribution. Its Fourier transform is the magnetic form factorG_M(Q²).

3The cross section for unpolarized scattering is obtained from (1.17) by an average over initial spins and sum over final spins: ¹₄P

4In the relativistic case, the form factors must be Lorentz invariant, and therefore depend on the 4-momentumsS

transfer squared: G_E(Q²), where Q²=−q²>0.

To get some hint about the form of the effective vertex Γ^µ, we first proof the following identity for the electron transition current (Gordon identity):

u(~k⁰, s⁰)γ^µu(~k, s) =u(~k⁰, s⁰)

(k⁰+k)^µ

2m +iσ^µνqν

2m

u(~k, s) (1.26)

Hereq=k⁰−k, and the Dirac matrixσ^µν is defined by the following commutator betweenγ-matrices:

σ^µν = i

2[γ^µ, γ^ν] (1.27)

Proof of (1.26): Use the following identity for any 4-vectors a^µ and b^µ: 6a6b = aµbν

1

2[(γ^µγ^ν+γ^νγ^µ) + (γ^µγ^ν −γ^νγ^µ)]

= a_µb^µ−iσ^µνa_µb_ν (1.28)

By using the Dirac equation (6k−m)u(~k, s) = u(~k⁰, s⁰) (6k⁰−m) = 0, we then have for any 4-vector a^µ

0 = u(~k⁰, s⁰) [(6k⁰−m)6a+6a(6k−m)]u(~k, s)

= u(~k⁰, s⁰)

k⁰·a−iσ^µνk_µ⁰a_ν+k·a−iσ^µνa_µk_ν−2m6au(~k, s)

= 2m u(~k⁰, s⁰)

(k⁰+k)^µ

2m a_µ+iσ^µνaµqν

2m −γ^µa_µ

u(~k, s)

Because a^µ is arbitrary, the Gordon formula (1.26) follows.

What is the physical meaning of the 2 terms on the r.h.s. of (1.26)? Consider the nonrelativistic limit, where the Dirac spinor is simply u(~k, s) = (χ(s),0) with χ(s) the 2-component Pauli spinor.

Then we get from (1.26)

u(~k⁰, s⁰)γ⁰u(~k, s)'χ^†(s⁰)χ(s) = δ_s⁰_s (1.29) u(~k⁰, s⁰)~γu(~k, s)'χ^†(s⁰)

"

(~k+~k⁰)

2m +i~σ×~q 2m

#

χ(s) (1.30)

Homework: Proof Eqs.(1.29) and (1.30), usingσ^ij =^ijkΣ^k, whereΣ is the relativistic spin matrix.~

• The first terms in (1.29) and (1.30) are the (transition) charge density and convection current in momentum space for a point particle =⇒We multiply them by the charge form factorGE(Q²) (=charge density in momentum space, normalized by G_E(Q²) = 1) to take into account the extended charge distribution.

• The second term in (1.30) has the form of the spin current in momentum space for a point particle with g_s = 2 (see No. 13, Eq.(2.2)): ~j_spin = i(m~ ×~q) where the magnetic moment is

~

m = _2m¹ χ^†(s⁰)~σχ(s) _gs

2 with g_s = 2 =⇒ we multiply this term by the magnetic form factor G_M(Q²) (=magnetic moment density in momentum space, normalized by G_M(Q² = 0) = ^g₂^s).

Result: For the proton (mass M), we modify the two terms in the transition current (1.26) as follows:

u(~k⁰, s⁰)Γ^µu(~k, s) = u(~k⁰, s⁰)

(k⁰ +k)^µ

2M G_E(Q²) +iσ^µνq_ν

2M G_M(Q²)

u(~k, s)

If we use here again the Gordon formula (1.26) to eliminate the (k⁰+k)^µ term, we get u(~k⁰, s⁰)Γ^µu(~k, s) = u(~k⁰, s⁰)

γ^µGE(Q²) +iσ^µνq_ν

2M GM(Q²)−GE(Q²)

= u(~k⁰, s⁰)

γ^µF1(Q²) +iσ^µνq_ν

2M F2(Q²)

(1.31) where we defined the Dirac form factor F₁(Q²) = G_E(Q²) and the Pauli form factor F₂(Q²) = G_M(Q²)−G_E(Q²). The normalizations are F₁(Q² = 0) = 1, F₂(Q² = 0) = ^g₂^s −1 ≡ κ, where κ is called the “anomalous part” of ^g₂^s, and is κ_p = 1.79 for proton andκ_n=−1.91 for neutron.

Note: The formula (1.31) is the final result for the elastic transition current for a spin 1/2 hadron.

For the calculation of the cross section, however, it is more convenient to use the Gordon formula (1.26) to eliminate thei^σ_2M^µνqν term:

u(~k⁰, s⁰)Γ^µu(~k, s) = u(~k⁰, s⁰)

γ^µ F₁(Q²) +F₂(Q²)

− (k+k⁰)^µ

2M F₂(Q²)

(1.32) Calculation of elastic cross section:

We go back to (1.23) to calculate the relativistic elastic cross section. Using the leptonic tensor (1.24), and the method to perform spin sums by traces of Dirac matrices (see No. 12, p.7), we get

X

s

`^µν = 2m²X

ss⁰

(u(~p⁰, s⁰)γ^µu(~p, s)) (u(~p⁰, s⁰)γ^νu(~p, s))^∗

= 1

2Tr [(6p⁰ +m)γ^µ(6p+m)γ^ν]

Using the two theorems about traces of Dirac matrices (see No. 12, p. 8) we obtain X

s

`^µν = 2h

p^0µp^ν +p^0νp^µ−g^µν p⁰·p−m²i

(1.33)

Using the hadronic tensor (1.25) with the form (1.32), we get X

S

w^µν = 2M²X

SS⁰

u(P~⁰, P⁰)Γ^µu(P , S)~ u(P~⁰, S⁰)Γ^νu(P , S)~ ∗

= 1

2Tr [(6P⁰+M) Γ^µ(6P +M) Γ^ν]

≡ w^µν₁ +w^µν₂ (1.34)

wherew^µν₁ is similar to the electron part and given by w^µν₁ = 2 F₁(Q²) +F₂(Q²)2h

P^0µP^ν +P^0νP^µ−g^µν P⁰·P −M²i

(1.35) The part w₂^µν contains all the other pieces, and is given by

w^µν₂ = − 1

4MF₂(Q²) F₁(Q²) +F₂(Q²)

× [(P⁰+P)^µTr [(6P⁰+M) (6P +M)γ^ν] + (P⁰ +P)^νTr [(6P⁰+M)γ^µ(6P +M)]]

= (P⁰+P)^µ(P⁰+P)^ν

−2F₂(Q²) F₁(Q²) +F₂(Q²)

+ (F₂(Q²))²P⁰·P +M² 2M²

(1.36) Finally, one has to calculate the contraction P

sS`_µνw^µν by using (1.33) for the electron part, and the sum of (1.35) and (1.36) for the proton part. Inserting the result into (1.23) gives the Rosenbluth formula of No. 13, Eq.(2.11):

dσ dΩ =

dσ dΩ

Mott

"

(G_E(Q²))²+b (G_M(Q²))²

1 +b + 2b G_M(Q²)2

tan² θ 2

#

(1.37) whereQ² =−q² =~q²−q₀² >0 is the 4-momentum transfer squared, and b =Q²/(4M²).

Home work (a bit long . . .): Calculate the contraction P

sS`_µνw^µν, and insert it into (1.23) to derive (1.37).