342
11. Loops and Renormalization II: QED
Equation (11.56) shows that the effective strength α(q
2 ) tends to increase
at large |q
2
| (short distances). This is, after all, physically reasonable: the
reduction in the effective charge caused by the dielectric constant associated
with the polarization of the vacuum disappears (the charge increases) as we
pass inside some typical dipole length. In the present case, that length is m
−1
(in our standard units ħ = c = 1), the fermion Compton wavelength, a typical
distance over which the fluctuating pairs extend.
The foregoing is the reason why this whole phenomenon is called vacuum
[2]
polarization, and why the original diagram which gave Π γ is called a vacuum
polarization diagram.
Equation (11.56) is the lowest-order correction to α, in a form valid for
|q
2
| ≫ m
2 . It turns out that, in this limit, the dominant vacuum polarization
contributions (for a theory with one charged fermion) can be isolated in each
order of perturbation theory and summed explicitly. The result of summing
these ‘leading logarithms’ is
2
α(Q
2 ) =
α
for Q
2
≫ m
(11.57)
[1 − (α/3π) ln(Q 2 /Am 2 )]
where we now introduce Q
2 = −q
2 , a positive quantity when q is a momentum transfer. The justification for (11.57) – which of course amounts to the
very plausible return to (11.32) instead of (11.38) – is subtle, and depends
upon ideas grouped under the heading of the ‘renormalization group’. This
is beyond the scope of the present volume, but will be taken up again in
volume 2.
Equation (11.57) presents some interesting features. First, note that for
typical large Q
2
∼ (50 GeV)
2 , say, the change in the effective α predicted by
(11.57) is quite measurable. Let us write
α(Q
2 ) =
α
(11.58)
1 − Δα(Q 2 )
in general, where Δα(Q
2 ) includes the contributions from all charged fermions
with mass m such that m
2
≪ Q
2 . The contribution from the charged leptons
is then straightforward, being given by
∑
α
Δα leptons =
ln(Q
2 /Am l
2 )
(11.59)
3π
l
where m l is the lepton mass. Including the e, μ and τ one finds (problem 11.8)
Δα leptons (Q
2 = (50 GeV)
2 ) ≈ 0.03.
(11.60)
However, the corresponding quark loop contributions are subject to strong
interaction corrections, and are not straightforward to calculate. We shall not
pursue this in detail here, noting just that the total contribution from the five
quarks u, d, s, c and b has a value very similar to (11.60) for the leptons (see,
11. Loops and Renormalization II: QED
Equation (11.56) shows that the effective strength α(q
2 ) tends to increase
at large |q
2
| (short distances). This is, after all, physically reasonable: the
reduction in the effective charge caused by the dielectric constant associated
with the polarization of the vacuum disappears (the charge increases) as we
pass inside some typical dipole length. In the present case, that length is m
−1
(in our standard units ħ = c = 1), the fermion Compton wavelength, a typical
distance over which the fluctuating pairs extend.
The foregoing is the reason why this whole phenomenon is called vacuum
[2]
polarization, and why the original diagram which gave Π γ is called a vacuum
polarization diagram.
Equation (11.56) is the lowest-order correction to α, in a form valid for
|q
2
| ≫ m
2 . It turns out that, in this limit, the dominant vacuum polarization
contributions (for a theory with one charged fermion) can be isolated in each
order of perturbation theory and summed explicitly. The result of summing
these ‘leading logarithms’ is
2
α(Q
2 ) =
α
for Q
2
≫ m
(11.57)
[1 − (α/3π) ln(Q 2 /Am 2 )]
where we now introduce Q
2 = −q
2 , a positive quantity when q is a momentum transfer. The justification for (11.57) – which of course amounts to the
very plausible return to (11.32) instead of (11.38) – is subtle, and depends
upon ideas grouped under the heading of the ‘renormalization group’. This
is beyond the scope of the present volume, but will be taken up again in
volume 2.
Equation (11.57) presents some interesting features. First, note that for
typical large Q
2
∼ (50 GeV)
2 , say, the change in the effective α predicted by
(11.57) is quite measurable. Let us write
α(Q
2 ) =
α
(11.58)
1 − Δα(Q 2 )
in general, where Δα(Q
2 ) includes the contributions from all charged fermions
with mass m such that m
2
≪ Q
2 . The contribution from the charged leptons
is then straightforward, being given by
∑
α
Δα leptons =
ln(Q
2 /Am l
2 )
(11.59)
3π
l
where m l is the lepton mass. Including the e, μ and τ one finds (problem 11.8)
Δα leptons (Q
2 = (50 GeV)
2 ) ≈ 0.03.
(11.60)
However, the corresponding quark loop contributions are subject to strong
interaction corrections, and are not straightforward to calculate. We shall not
pursue this in detail here, noting just that the total contribution from the five
quarks u, d, s, c and b has a value very similar to (11.60) for the leptons (see,