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i e 2
�
d 4 x e iqx 〈0|Tj µ em (x)jν em (0)|0〉 = −(q2 g µν − q µ q ν )Π ′ γ (q2 ) , (64)
which is purely transversal by virtue of electromagnetic current conservation ∂µj µ em(x) = 0.
In terms of the fine structure constant α = e2
4π Eq. (63) reads
α(q 2 α
) =
1 − ∆α(q2 ; ∆α(q
)
2 ) = −Re � Π ′ γ(q 2 ) − Π ′ γ(0) � . (65)
The various contributions to the shift in the fine structure constant come from the leptons (lep = e, µ and
τ), the 5 light quarks (u, b, s, c, and b) and/or the corresponding hadrons (had). The top quark is too heavy
to give a relevant contribution. The hadronic contributions will be considered later.
The renormalized photon self–energy is an analytic function and satisfies the dispersion relation (DR)
− Π′ γ ren(k 2 )
k 2
=
�∞
0
ds
s
1
π Im Π′ γ(s)
1
k2 . (66)
− s
Note that the only k dependence under the convolution integral shows up in the last factor. Thus, in a
generic VP contribution
µ
γ
γ
X
γ
where the “blob” is the full photon propagator, including all kinds of contributions as predicted by the SM
or beyond, the free photon propagator in the 1–loop vertex graph in the next higher order is replaced by
−igµν/k 2 → −igµν/(k 2 − s) ,
which is the exchange of a photon of mass square s. This result then has to be convoluted with the imaginary
part of the photon vacuum polarization. The calculation of the contribution from the massive photon
proceeds exactly as in the massless case. Again FM(0) most simply may be calculated using the projection
method which allows one to work at q 2 = 0. The result is [153,154]
(2) heavy γ
a µ ≡ α
π K(2) µ (s) = α
π
�1
0
dx
x2 (1 − x)
x2 + (s/m2 , (67)
µ)(1 − x)
which is the leading order contribution to aµ from an exchange of a photon with square mass s. For s = 0
we get the known Schwinger result. Utilizing this result and Eq. (66), the contribution from the “blob” to
g − 2 reads
a (X)
µ = α
π2 �∞
0
ds
s Im Π′ (X)
γ (s) K (2)
µ (s) . (68)
If we exchange integrations and evaluating the DR we arrive at [19]
a (X)
µ = α
�1
π
0
= α
π
�
0
1
dx (1 − x)
dx (1 − x)
�∞
0
ds
s
1
π Im Π′ (X) x
γ (s)
2
x2 + (s/m2 µ)(1 − x)
�
−Π ′ �
(X)
γ (sx) , with sx = − x2
1 − x m2 µ . (69)
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