231
8.2. Coulomb scattering of charged spin1 particles
2
for u and u
′ † were used. Fortunately we can avoid this by using a powerful
labour-saving device due to Feynman, in which the γ’s come into their own.
We need to calculate the quantity S given in (8.46). This will turn out to
be just the first in a series of such objects. With later needs in mind, we shall
here calculate a more general quantity than (8.46), namely the lepton tensor
∑
L
μν (k
′ ′ )γ
μ
′ ′ )γ
ν
′ , k) =
1
u ¯(k , s
u(k, s)[¯ u(k , s
u(k, s)]
∗
(8.56)
2
s ' ,s
∑
1
′
′
∗
=
′
| ˆ j
μ
(0)|e
− , k, s>
′
| ˆ j
ν
(0)|e
− , k, s> .
(8.57)
em,e
em,e
2e 2
s ' ,s
Clearly this will be relevant to the more general case in which A
μ contains
non-zero spatial components, for example. For our present application, we
shall need only L
00 .
We first note that L
μν is correctly called a tensor (a contravariant secondrank one, in fact – see appendix D), because the two uγ
μ
uγ
ν u’ factors are
‘¯ u, ¯
each 4-vectors, as we have seen. (We might worry a little over the complex
conjugation of the second factor, but this will disappear after the next step.)
′
Consider therefore the factor [¯ u(k , s
′ )γ
ν u(k, s)]
∗ . For each value of the index
ν, this is just a number (the corresponding component of the 4-vector), and
so it can make no difference if we take its transpose, in a matrix sense (the
transpose of a 1 × 1 matrix is certainly equal to itself!). In that case the
complex conjugate becomes the Hermitian conjugate, which is:
′
′
[¯ u(k , s
′ )γ
ν u(k, s)]
†
= u
† (k, s)γ
ν† γ
0† u(k , s
′ )
(8.58)
′
= ¯
u(k, s)γ
ν u(k , s
′ )
(8.59)
since (problem 8.6)
γ
0 γ
ν† γ
0
γ
ν
=
(8.60)
and γ
0 = γ
0† . Thus L
μν may be written in the more streamlined form
∑
L
μν
1
′ ′ )γ
μ
′ ′ )
=
u ¯(k , s
u(k, s)¯ u(k, s)γ
ν u(k , s
(8.61)
2
'
s ,s
which is, moreover, evidently the (tensor) product of two 4-vectors. However,
there is more to this than saving a few symbols. We have seen the expression
∑
u(k, s)¯ u(k, s)
(8.62)
s
before! (See (7.64) and problem 7.8.) Thus we can replace the sum (8.62)
over spin states ‘s’ by the corresponding matrix (k / + m):
∑
L
μν
1
′
′ ′ )
=
u ¯ α (k , s
′ )(γ
μ ) αβ (k / + m) βγ (γ
ν ) γδ u δ (k , s
(8.63)
2
s '
8.2. Coulomb scattering of charged spin1 particles
2
for u and u
′ † were used. Fortunately we can avoid this by using a powerful
labour-saving device due to Feynman, in which the γ’s come into their own.
We need to calculate the quantity S given in (8.46). This will turn out to
be just the first in a series of such objects. With later needs in mind, we shall
here calculate a more general quantity than (8.46), namely the lepton tensor
∑
L
μν (k
′ ′ )γ
μ
′ ′ )γ
ν
′ , k) =
1
u ¯(k , s
u(k, s)[¯ u(k , s
u(k, s)]
∗
(8.56)
2
s ' ,s
∑
1
′
′
∗
=
′
| ˆ j
μ
(0)|e
− , k, s>
′
| ˆ j
ν
(0)|e
− , k, s> .
(8.57)
em,e
em,e
2e 2
s ' ,s
Clearly this will be relevant to the more general case in which A
μ contains
non-zero spatial components, for example. For our present application, we
shall need only L
00 .
We first note that L
μν is correctly called a tensor (a contravariant secondrank one, in fact – see appendix D), because the two uγ
μ
uγ
ν u’ factors are
‘¯ u, ¯
each 4-vectors, as we have seen. (We might worry a little over the complex
conjugation of the second factor, but this will disappear after the next step.)
′
Consider therefore the factor [¯ u(k , s
′ )γ
ν u(k, s)]
∗ . For each value of the index
ν, this is just a number (the corresponding component of the 4-vector), and
so it can make no difference if we take its transpose, in a matrix sense (the
transpose of a 1 × 1 matrix is certainly equal to itself!). In that case the
complex conjugate becomes the Hermitian conjugate, which is:
′
′
[¯ u(k , s
′ )γ
ν u(k, s)]
†
= u
† (k, s)γ
ν† γ
0† u(k , s
′ )
(8.58)
′
= ¯
u(k, s)γ
ν u(k , s
′ )
(8.59)
since (problem 8.6)
γ
0 γ
ν† γ
0
γ
ν
=
(8.60)
and γ
0 = γ
0† . Thus L
μν may be written in the more streamlined form
∑
L
μν
1
′ ′ )γ
μ
′ ′ )
=
u ¯(k , s
u(k, s)¯ u(k, s)γ
ν u(k , s
(8.61)
2
'
s ,s
which is, moreover, evidently the (tensor) product of two 4-vectors. However,
there is more to this than saving a few symbols. We have seen the expression
∑
u(k, s)¯ u(k, s)
(8.62)
s
before! (See (7.64) and problem 7.8.) Thus we can replace the sum (8.62)
over spin states ‘s’ by the corresponding matrix (k / + m):
∑
L
μν
1
′
′ ′ )
=
u ¯ α (k , s
′ )(γ
μ ) αβ (k / + m) βγ (γ
ν ) γδ u δ (k , s
(8.63)
2
s '