228
8. Elementary Processes in Scalar and Spinor Electrodynamics
The Dirac equation may then be written (problem 4.3) as
(i∂ / − m)ψ = 0
(8.38)
where the ‘slash’ notation introduced in (7.59) has been used (i∂ / = iγ
μ ∂ μ ).
Defining ψ ¯ = ψ
† γ
0 , (8.35) becomes
∫
A e − = −i d
4 x (−eψ ¯′ (x)γ
μ ψ(x))A μ (x)
(8.39)
∫
μ
≡ −i d
4 x j
− (x)A μ (x)
(8.40)
em,e
where we have defined an electromagnetic transition current for a negatively
charged fermion:
μ
¯
j
− (x) = −eψ
′ (x)γ
μ ψ(x),
(8.41)
em,e
exactly analogous to the one for a positively charged boson introduced in
section 8.1.1. We know from section 4.1.2 that ψ ¯′ γ
μ ψ is a 4-vector, showing
that A e − of (8.40) is Lorentz invariant.
Inserting free-particle solutions for ψ and ψ
′ † in (8.41), we obtain
)·x
j
μ
(x) = −eu ¯(k
′ , s
′ )γ
μ u(k, s)e
−i(k−k
'
(8.42)
em,e −
so that (8.39) becomes
∫
¯′ γ
μ
−i(k−k
'
A e − = −i d
4 x (−eu ue
)·x )A μ (x)
(8.43)
where u = u(k, s) and similarly for u
′ . Note that the u’s do not depend on x.
For the case of the Coulomb potential in equation (8.13), A e − becomes
Ze
2 ′ †
A e − = i2πδ(E − E
′ )
u u
(8.44)
q 2
just as in (8.15), where q = k − k
′ and we have used ¯
u
′ γ
0 = u
′ † . Comparing
′
(8.44) with (8.15), we see that (using the covariant normalization N = N = 1)
the amplitude in the spinor case is obtained from that for the scalar case by
′ †
the replacement ‘2E → u u’ and the sign of the amplitude is reversed as
expected for e
− rather than s
+ scattering.
We now have to understand how to define the cross section for particles
with spin and then how to calculate it. Clearly the cross section is proportional
to |A e − |
2 , which involves |u
† (k
′ , s
′ )u(k, s)|
2 here. Usually the incident beam
is unpolarized, which means that it is a random mixture of both spin states
s (‘up’ or ‘down’). It is important to note that this is an incoherent average,
in the sense that we average the cross section rather than the amplitude.
Furthermore, most experiments usually measure only the direction and energy
8. Elementary Processes in Scalar and Spinor Electrodynamics
The Dirac equation may then be written (problem 4.3) as
(i∂ / − m)ψ = 0
(8.38)
where the ‘slash’ notation introduced in (7.59) has been used (i∂ / = iγ
μ ∂ μ ).
Defining ψ ¯ = ψ
† γ
0 , (8.35) becomes
∫
A e − = −i d
4 x (−eψ ¯′ (x)γ
μ ψ(x))A μ (x)
(8.39)
∫
μ
≡ −i d
4 x j
− (x)A μ (x)
(8.40)
em,e
where we have defined an electromagnetic transition current for a negatively
charged fermion:
μ
¯
j
− (x) = −eψ
′ (x)γ
μ ψ(x),
(8.41)
em,e
exactly analogous to the one for a positively charged boson introduced in
section 8.1.1. We know from section 4.1.2 that ψ ¯′ γ
μ ψ is a 4-vector, showing
that A e − of (8.40) is Lorentz invariant.
Inserting free-particle solutions for ψ and ψ
′ † in (8.41), we obtain
)·x
j
μ
(x) = −eu ¯(k
′ , s
′ )γ
μ u(k, s)e
−i(k−k
'
(8.42)
em,e −
so that (8.39) becomes
∫
¯′ γ
μ
−i(k−k
'
A e − = −i d
4 x (−eu ue
)·x )A μ (x)
(8.43)
where u = u(k, s) and similarly for u
′ . Note that the u’s do not depend on x.
For the case of the Coulomb potential in equation (8.13), A e − becomes
Ze
2 ′ †
A e − = i2πδ(E − E
′ )
u u
(8.44)
q 2
just as in (8.15), where q = k − k
′ and we have used ¯
u
′ γ
0 = u
′ † . Comparing
′
(8.44) with (8.15), we see that (using the covariant normalization N = N = 1)
the amplitude in the spinor case is obtained from that for the scalar case by
′ †
the replacement ‘2E → u u’ and the sign of the amplitude is reversed as
expected for e
− rather than s
+ scattering.
We now have to understand how to define the cross section for particles
with spin and then how to calculate it. Clearly the cross section is proportional
to |A e − |
2 , which involves |u
† (k
′ , s
′ )u(k, s)|
2 here. Usually the incident beam
is unpolarized, which means that it is a random mixture of both spin states
s (‘up’ or ‘down’). It is important to note that this is an incoherent average,
in the sense that we average the cross section rather than the amplitude.
Furthermore, most experiments usually measure only the direction and energy