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1 S-matrix for scattering of electron in external field

Remember the following formula for the time evolution of an electron wave function (positive energy) in an external electromagnetic fieldA^µ (see No. 8):

i Z

d³x⁰ SF(x−x⁰)γ⁰ Ψn(x⁰) = θ(t−t⁰) Ψn(x) (1.1) i

Z

d³x Ψ_n(x)γ⁰S_F(x−x⁰) = θ(t−t⁰) Ψ_n(x⁰) (1.2) HereS_F is the exact Feynman propagator of the electron in an external field, and Ψ(x) is the exact wave function of the electron.

Consider the process of electron scattering by an external electromagnetic field, which may be created by another particle (for example, a nucleus):

time (t)

0 detector

Suppose we have an exact wave function for the electron (Ψi(x)), which satisfies the following initial condition fort → −∞:

Ψ_i(x) ^(t→−∞)−→ ψ_i(x) (1.3)

Here ψ_i(x) is a free (positive energy) solution of the Dirac equation. Take the limit t⁰ → −∞ on both sides of Eq.(1.1) forn =i:

Ψ_i(x) = i lim

t⁰→−∞

Z

d³x⁰S_F(x−x⁰)γ⁰ψ_i(x⁰) Insert here the Dyson equation SF =SF0+SF(e6A)SF0 (see No. 8):

Ψ_i(x) =i lim

t⁰→−∞

Z d³x⁰

S_F0(x−x⁰) + Z

d⁴y S_F(x−y)(e6A(y))S_F0(y−x⁰)

γ⁰ψ_i(x⁰) (1.4)

Eq.(1.1) for t⁰ → −∞ and A^µ= 0 becomes i lim

t⁰→−∞

Z

d³x⁰S_F0(x−x⁰)γ⁰ψ_i(x⁰) =ψ_i(x) Using this in Eq.(1.4) we get

Ψ_i(x) =ψ_i(x) + Z

d⁴y S_F(x−y) (e6A(y))ψ_i(y) (1.5) At time t → +∞, place a detector which can filter out any free state ψ_f(x) from the exact wave function (1.5). The probability amplitude to detect a particular state ψ_f(x), which is contained in (1.5), is given by

S_{f i}≡ lim

t→∞

Z

d³x ψ^†_f(x) Ψ_i(x) (1.6)

For all possible states (f, i) this is a matrix, which is called the S-matrix. Inserting here the formula (1.5) we obtain

S_{f i} = lim

t→∞

Z

d³x ψ_f^†(x)

ψ_i(x) + Z

d⁴y S_F(x−y) (e6A(y))ψ_i(y)

= δ_{f i}+ lim

t→∞

Z d³x

Z

d⁴y ψ_f^†(x)S_F(x−y) (e6A(y))ψ_i(y) (1.7) Insert here the Dyson equation

S_F(x−y) =S_F0(x−y) + Z

d⁴z S_F0(x−z) (e6A(z))S_F(z−y) and use Eq.(1.2) for t→ ∞ and A^µ = 0:

t→∞lim Z

d³x ψ^†_f(x)S_F0(x−y) =−iψ_f(y)

Then, from (1.7), we obtain the following convenient form of the S-matrix:

S_{f i}=δ_{f i}−i Z

d⁴y ψ_f(y)(e6A(y))ψ_i(y)−i Z

d⁴y Z

d⁴z ψ_f(z) (e6A(z))S_F(z−y) (e6A(y))ψ_i(y) (1.8) Note that in (1.8) all wave functions are free wave functions, but S_F(z −y) is the exact Feynman propagator, which can be expanded in perturbation theory according to the Dyson equation SF = SF0 +SF0(e6A)SF0+. . .. Because the Feynman propagator has two time orderings (see No. 8), the interaction terms in Eq.(1.8) can be graphically expressed as follows (up to second order perturbation

+ +

y

z

y z

y

Example: Scattering by a Coulomb potential (produced by a heavy nucleus).

ψ_i(x) =

rm E_p

√1

V u(~p, s)e⁻ⁱ(^{Epy0−~p·~y}) ψ_f(x) =

r m E_p⁰

√1

V u(~p⁰, s⁰)e⁻ⁱ(^Ep0y⁰−~p⁰·~y) A^µ(y) =

A⁰ =− Ze

4π|~y|, ~A= 0

(1.9) Then the S-matrix (1.8), forf 6=i, to order e² becomes:

S_{f i} =iZ e² 4π

1 V

s m²

E_pE_p⁰ u(~p⁰, s⁰)γ⁰u(~p, s)

Z d⁴y

|~y| e^{i(Ep0−Ep)y0}e^{−i(~p0−~p)·~y}

Use here the relations

Z +∞

−∞

dy⁰e^{i(Ep0−Ep)y0} = (2π)δ(E_p⁰−E_p) (1.10) Z d³y

|~y| e^−i~q·~y = 4π

~

q² (~q=~p⁰ −~p) Then we obtain

Sf i = iZe² V

m E_p

u(~p⁰, s⁰)γ⁰u(~p, s)

~

q² (2π)δ(Ep⁰ −Ep) (1.11)

p,s p’,s’

q -Ze

The probability for electron scattering i = (~p, s) →f = (~p⁰, s⁰) is then given by |S_{f i}|². However, in a finite volumeV, a state with sharp value of~p⁰ cannot be observed. To calculate the probability for electron scattering, we need the number of final states in the volume V and in the momentum interval d³p⁰. This number is called the phase space factor.

Consider first only 1 space dimension:

In classical physics, each point of the phase space (x, p) corresponds to one state. In quantum mechanics, because of the uncertainty principle dxdp > h= 2π~, each cell of size h corresponds to one state:

x p

dx dp

cell dx dp = h

corresponds to 1 state with spin up (or down)

.

In 3 space dimensions:

The number of states, with spin up or spin down, in the volume V and in the momentum interval d³p⁰, is equal to the number of cells in the phase space volume (V ·d³p⁰), and is given by

V d³p⁰

(2π~)³ (1.12)

Therefore, the probability for scattering into any of these ^Vd3p0

(2π~)³ states is given by

|S_{f i}|² V d³p⁰

(2π~)³ (1.13)

Definition of differential cross section:

d³σ ≡ number of particles scattered (per unit time) into d³p⁰ number of incoming particles (per, time, per area)

= N_in

|S_{f i}|²· V d³p⁰ (2π~)³

1

∆T

/|~j_in| (1.14)

Here

• ∆T is the observation time ' time it takes the electron to go from the accelerator to the

• N_in is the number is incoming particles per unit time ;

• ~j_in is the incoming flux of particles:

~j_in = ψ_i(x)~γ ψ_i(x)

·N_in = m E_p

1

V (u(~p, s)~γu(~p, s))·N_in

= ~p E_p

N_in

V =~v N_in V

where~v is the velocity of the incoming electrons.

Then we get the differential cross section as follows:

d³σ = V

v |S_{f i}|²· V d³p⁰ (2π~)³

1

∆T

= d³p⁰ v(2π)³

1

∆T Ze²2 m E_p

2

|u(~p⁰, s⁰)γ⁰u(~p, s)|²

~

q⁴ (2πδ(E_p⁰ −E_p))² (1.15) What is the meaning of the square of the delta function?

If the observation time is ∆T ' 10⁻⁸ s, then the integration over time (see Eq.(1.10)) should be replaced byR+∆T /2

−∆T /2 dy⁰, and therefore

(2πδ(E_p⁰ −E_p))² −→

Z +∆T /2

−∆T /2

dt e^i(Ep0−Ep)t

!2

= 4sin^{2 Ep0}₂^−Ep∆T

(E_p⁰ −E_p)² = (2π∆T) sin^{2 ∆E}₂ ∆T π^(∆E)₂ ²∆T

!

(1.16) Here ∆E ≡E_p⁰ −E_p. Consider now

sin^{2 ∆E}₂ ∆T

π^(∆E)₂ ²∆T = ∆T 2π

sin^∆E₂ ∆T

∆E 2 ∆T

!2

as a function of ∆E:

(In the figure below, t ≡∆T, and Ω ≡∆E.) Because ∆T ' 10⁻⁸ s is a “macroscopic time”, we have ^{1 2π}_∆T^~ ' 2π×10⁻⁷eV, which is very small compared to a typical resolution energy of ' 1 eV.

Therefore, ^2π_∆T^~ is practically zero, which means that the macroscopic time ∆T can be replaced by

∆T → ∞. Then we can use the relation sin^{2 ∆E}₂ ∆T

π^(∆E)₂ ²∆T

∆T→∞

−→ δ(∆E)

1Note that~'10⁻¹⁵ eV·s.

Therefore, finally, we get for (1.16),

(2πδ(E_p⁰−E_p))^{2 ∆}−→^T→∞ (2π∆T) δ(E_p⁰ −E_p) (1.17) Note: A shorthand “derivation” of this result is simply as follows:

(2πδ(E_p⁰ −E_p))² = (2πδ(E_p⁰ −E_p)) (2πδ(E_p⁰−E_p))

= (2πδ(E_p⁰ −E_p)) (2πδ(0)) ≡(2πδ(E_p⁰−E_p)) ∆T

Physically, it means that if we observe the particle for a macroscopic time ∆T, there is no uncertainty of the energy. (Note that the static Coulomb potential cannot transfer energy, so we must get energy conservation: E_p =E_p⁰.)

Using the result (1.17) in the differential cross section (1.15), and d³p⁰ =p⁰²dp⁰dΩ⁰ withp=p⁰ from the energy conserving delta function, we get (with α≡e²/(4π)'1/137)

d³σ

dp⁰dΩ⁰ = 4Z²α²p² v

m Ep

2

δ(E_p⁰ −E_p)|u(~p⁰, s⁰)γ⁰u(~p, s)|²

~ q⁴

We can integrate this over p⁰, using δ(Ep⁰ −Ep) = ¹_vδ(p⁰ − p) to get the usual “differential cross section”:

dσ

dΩ⁰ ≡ d²σ

dΩ⁰ = 4Z²α²m²

~

q⁴ |u(~p⁰, s⁰)γ⁰u(~p, s)|² (1.18) Here the momentum transfer is given in terms of the scattering angle θ by

2 0− ^{2 2} − ^{2 2} θ

In the nonrelativistic limit we haveu(~p⁰, s⁰)γ⁰u(~p, s)'1, and Eq.(1.18) becomes the Rutherford cross section.

Relativistic effects from electron spin, Mott cross section:

Here we calculate the “unpolarized cross section”:

• The initial electrons have equal probability for spin up (s = 1/2) and spin down (s=−1/2)

⇒average ¹₂P

s;

• Both spin directions of the final electron are observed, i.e., the detector does not differentiate between the spin directions

⇒sum P

s.

Then the unpolarized differential cross section becomes dσ

dΩ⁰ = 2Z²α²m²

~ q⁴

X

ss⁰

|u(~p⁰, s⁰)γ⁰u(~p, s)|² (1.20) We now use the following identity for the square of spinor matrix elements: If Γ is a Diracγ-matrix, then

|u(f) Γu(i)|² = (u(f) Γu(i)) u^t(f)γ⁰Γ^∗u^∗(i)

= (u(f) Γu(i)) u^†(i) Γ^†γ⁰u(f)

= (u(f) Γu(i)) u(i) γ⁰Γ^†γ⁰ u(f)

If we define Γ≡γ⁰Γ^†γ⁰, and indicate the Dirac indices α, β, α⁰, β⁰ explicitly, we obtain X

ss⁰

|u(f) Γu(i)|² = X

ss⁰

u_α(~p⁰, s⁰)Γ_αβu_β(~p, s)u_α⁰(~p, s)Γ_α⁰_β⁰u_β⁰(~p⁰, s⁰)

= Tr Λ₊(~p⁰) Γ Λ₊(~p) Γ

where the positive energy projection operator is given by (see No. 6) Λ₊(~p) =6p+m

2m (here 6p=E_pγ⁰ −~p·~γ)

In our case (Eq.(1.20)) we have Γ = Γ =γ⁰, and the unpolarized differential cross section becomes dσ

dΩ = Z²α² 2~q⁴ Tr

(6p⁰ +m)γ⁰ (6p+m)γ⁰

(1.21) There are many theorems about traces of products of γ-matrices Γ = (γ⁰, γⁱ). Here we just need the following two theorems:

• The trace of a product of an odd number of Γ - matrices is zero:

Tr (Γ₁. . .Γ_n) = 0 if n= odd (1.22) Proof: Using the matrix γ₅ = iγ⁰γ¹γ²γ³, which satisfies γ₅² = 1 and {γ₅,Γ} = 0, and the property of the trace Tr(AB) = Tr(BA), we have for n=odd

Tr (Γ1. . .Γn) = Tr (Γ1. . .Γnγ5γ5)

= Tr (γ₅Γ₁. . .Γ_nγ₅) = (−1)ⁿ Tr (Γ₁. . .Γ_n) =−Tr (Γ₁. . .Γ_n) = 0

• For the product of two and fourγ-matrices we have the following formulas:

Tr (γ^µγ^ν) = 4g^µν (1.23)

Tr γ^µγ^νγ^λγ^σ

= 4 g^µνg^λσ−g^µλg^νσ+g^µσg^νλ

(1.24) Proof of (1.23): Using Tr(AB) = Tr(BA) and the anticommutation relations ofγ-matrices, we have

Tr (γ^µγ^ν) = 1

2Tr{γ^µ, γ^ν}=g^µνTr 1 = 4g^µν Eq.(1.24) can be derived by using similar methods.

Note that the formula (1.23) gives the following results for any 4-vectors a, b:

Tr (6a6b) = a_µb_νTr (γ^µγ^ν) = 4a_µb_νg^µν = 4a·b

In the same way, the formula (1.24) gives the following results for any 4-vectors a, b, c, d:

Tr (6a6b6c6d) = 4 (a·b c·d−a·c b·d+a·d b·c)

Now we can continue with the calculation of the unpolarized cross section (1.21): The trace factor becomes

Tr

(6p⁰+m)γ⁰ (6p+m)γ⁰

= Tr m²+6p⁰γ⁰6pγ⁰

= 4 m²+ 2E_pE_p⁰ −p·p⁰

(1.25) Using now (withE ≡E_p =E_p⁰)

· ^{0 2}− ^{2 2 2} − ^{2 2 2}θ

we finally get for the trace factor (1.25) Tr

(6p⁰ +m)γ⁰ (6p+m)γ⁰

= 8E²

1−~v²sin² θ 2

Inserting this into the cross section (1.21), and using also the relation (1.19), we finally obtain dσ

dΩ = Z²α² 4p²v²sin^{4 θ}₂

1−v²sin² θ 2

(1.26) Here we denote the magnitude squared of 3-vectors by p² ≡ ~p² and v² ≡~v². In our natural units, α=e²/(4π)'1/137 is the “fine structure constant”.

The formula (1.26) is called the Mott cross section, and describes elastic scattering of electrons on a Coulomb potential (for example, produced by a heavy spinless nucleus).