QED Vertex Correction
Consider the dressed electron-electron-photon vertex in QED,
ieΓµ (p′ , p) =
(1)
1PI
We are interested in this vertex in the context of elastic Coulomb scattering,
e−
X
e−
X
(2)
so we take the incoming and the outgoing electrons to be on-shell, p2 = p′2 = m2 , but the photon
is off-shell, q 2 6= 0. Moreover, we put the vertex in the context of the complete electron line —
including the external line factors, thus ū(p′ ) × ieΓµ × u(p). As discussed in class, this simplifies
the Lorentz and Dirac structure of the vertex and allows us to write it as
Γµ (p′ , p) = Fel (q 2 ) ×
(p′ + p)µ
iσ µν qν
iσ µν qν
+ Fmag (q 2 ) ×
= F1 (q 2 ) × γ µ + F2 (q 2 ) ×
. (3)
2m
2m
2m
At the tree level, the electron is a point-like spin =
1
2
particle obeying Dirac equations, hence
F1 (q 2 ) ≡ 1 and F2 (q 2 ) ≡ 0. But the quantum corrections in QED mix the elementary electron
state with the composite states like e− γ , e− γγ , or e− e− e+ , and this leads to the non-trivial
q 2 –dependent form-factors.
In these notes we shall calculate the F1 (q 2 ) and the F2 (q 2 ) form-factors to the one-loop order
in QED.
1
Working Through the Algebra
Fortunately, there is only one 1PI one-loop diagram contributing to the dressing-up of the
electron-electron-photon vertex, namely
(4)
Using the Feynman gauge for the internal photon’s propagator, this diagram evaluates to
ieΓµ1 loop (p′ , p)
=
Z
reg
i
i
d4 k −ig νλ
× ieγν × ′
× ieγ µ ×
× ieγλ
4
2
(2π) k + i0
6 p + 6 k − m + i0
6 p+ 6 k − m + i0
Z
d4 k
6 p′ + 6 k + m
6 p+ 6 k + m
1
ν
×γ × ′
× γµ ×
× γν
4
2
2
2
(2π) k + i0
(p + k) − m + i0
(p + k)2 − m2 + i0
Z
d4 k N µ
(2π)4 D
3
= e
reg
3
= e
reg
(5)
where
N µ = γ ν (6 k + 6 p′ + m)γ µ (6 k + 6 p + m)γν
(6)
and
D =
2
k + i0 × (p + k)2 − m2 + i0 × (p′ + k)2 − m2 + i0 .
(7)
The purpose of this section of the notes is to simplify these numerator and denominator. Using
the Feynman parameter trick, we may combine the 3 denominator factors as
1
=
D
1
ZZZ
0
2
dx dy dz δ(x + y + z − 1) h
i3 . (8)
2
2
2
2
2
′
x((p + k) − m ) + y((p + k) − m ) + z(k ) + i0
2
Inside the big square brackets here we have
[· · ·] = x × (p + k)2 − m2
+ y × (p′ + k)2 − m2
+ z × k2
= k 2 × (x + y + z = 1) + 2kµ(xp + yp′ )µ + x(p2 − m2 ) + y(p′2 − m2 )
(9)
= (k + xp + yp′ )2 − ∆
where
∆ = (xp + yp′ )2 − xp2 − yp′2 + (x + y)m2
= xy × 2p · p′ = p2 + p′2 − (p′ − p)2 + (x2 − x) × p2 + (y 2 − y) × p′2 + (x + y) × m2
= −xy × q 2 − x(1 − x − y) × p2 − y(1 − x − y) × p′2 + (x + y) × m2
= −xy × q 2 − xz × p2 − yz × p′2 + (1 − z) × m2
(10)
For the on-shell electron momenta, p2 = p′2 = m2 , we may further simplify
2
2
(1 − z) × m − xz × p − yz × p
′2
2
= m × (1 − z) − (x + y)z = (1 − z)
2
(11)
which gives
∆ = (1 − z)2 × m2 − xy × q 2 .
(12)
Let us also define the shifted loop momentum
ℓ = k + xp + yp′ ,
(13)
then we can rewrite the denominator as
1
=
D
1
ZZZ
dx dy dz δ(x + y + z − 1)
2
.
[ℓ2 − ∆ + i0]3
(14)
0
As usual, we plug this denominator into the loop integral (5), then change the order of inteR
R
gration — over the loop momentum before over the Feynman parameters, — and then shift
3
the momentum integration variable from k to ℓ, thus
Γµ1 loop (p′ , p)
2
= −2ie
1
ZZZ
dx dy dz δ(x + y + z − 1)
Z
reg
0
d4 ℓ
Nµ
.
(2π)4 ℓ2 − ∆ + i0 3
(15)
But to make full use of the momentum shift, we need to re-express the numerator N µ in terms of
the shifted momentum ℓ. It would also help to simplify the numerator (6) in the context of this
monstrous integral.
The first step towards simplifying the N µ is obvious: Let us get rid of the γ ν and γν factors
using the γ matrix algebra, eg., γ ν 6 aγν = −2 6 a, etc.. However, in order to allow for the dimensional
regularization, we need to re-work the algebra for an arbitrary spacetime dimension D where
γ ν γν = D 6= 4. Consequently,
γ ν 6 aγν = −2 6 a + (4 − D) 6 a,
γ ν 6 a 6 bγν = 4(ab) − (4 − D) 6 a 6 b,
(16)
γ ν 6 a 6 b 6 cγν = −2 6 c 6 b 6 a + (4 − D) 6 a 6 b 6 c,
and therefore
def
N µ = = γ ν (6 k + 6 p′ + m)γ µ (6 k + 6 p + m)γν
= −2m2 γ µ + 4m(p′ + p + 2k)µ − 2(6 p+ 6 k)γ µ (6 p′ + 6 k)
(17)
+ (4 − D)(6 p′ + 6 k − m)γ µ (6 p+ 6 k − m).
The second step is to re-express this numerator in terms of the loop momentum ℓ rather than
k using eq. (13). Expanding the result in powers of ℓ, we get quadratic, linear and ℓ–independent
R
terms, but the linear terms do not contribute to the dD ℓ integral because they are odd with
respect to ℓ → −ℓ while everything else in that integral is even. Consequently, in the context of
4
eq. (15) we may neglect the linear terms, thus
N µ = −2m2 γ µ + 4m(p′ + p + 2ℓ − 2xp − 2yp′ )µ
− 2(6 p+ 6 ℓ − x 6 p − y 6 p′ )γ µ (6 p′ + 6 ℓ − x 6 p − y 6 p′ )
+ (4 − D)(6 p′ + 6 ℓ − x 6 p − y 6 p′ − m)γ µ (6 p+ 6 ℓ − x 6 p − y 6 p′ − m)
hhskipping terms linear in ℓ ii
(18)
∼
= −2m2 γ µ + 4m(p + p′ − 2xp − 2yp′ )µ
− 2 6 ℓγ µ6 ℓ − 2(6 p − x 6 p − y 6 p′ ) γ µ (6 p′ − x 6 p − y 6 p′ )
+ (4 − D) 6 ℓγ µ6 ℓ + (4 − D)(6 p′ − y 6 p′ − x 6 p − m) γ µ (6 p − x 6 p − y 6 p′ − m).
Next, we make use of p′ − p = q and 1 − x − y = z to rewrite
2xp + 2yp′ = (x + y) × (p + p′ ) + (x − y) × (p − p′ ),
p + p′ − 2xp − 2yp′ = z × (p′ + p) + (x − y) × q,
p − xp − yp′ = z × p − y × q
= z × p′ − (1 − x) × q,
(19)
p′ − xp − yp′ = z × p′ + x × q
= z × p + (1 − y) × q,
and consequently
Nµ ∼
= −2m2 γ µ + 4mz(p′ + p)µ + 4m(x − y)q µ
+ (−2 + 4 − D)× 6 ℓγ µ6 ℓ
− 2(z 6 p′ + (x − 1) 6 q) γ µ (z 6 p + (1 − y) 6 q)
(20)
+ (4 − D)(z 6 p′ + x 6 q − m) γ µ (z 6 p − y 6 q − m).
The third step is to make use of the external fermions being on-shell. This means more than
just p2 = p′2 = m2 : We also sandwich the vertex ieΓµ between the Dirac spinors ū(p′ ) on the left
and u(p) on the right. The two spinors satisfy the appropriate Dirac equations 6 pu(p) = mu(p)
and ū(p′ ) 6 p′ = ū(p′ )m, so in the context of ū(p′ )Γµ u(p),
A× 6 p ∼
= A × m and
6 p′ × B ∼
= m×B
(21)
for any terms in Γµ that look like A× 6 p or 6 p′ × B for some A or B. Consequently, the terms on
5
the last two lines of eq. (20) are equivalent to
(z 6 p′ + (x − 1) 6 q) γ µ (z 6 p + (1 − y) 6 q) ∼
= (zm + (x − 1) 6 q) γ µ (zm + (1 − y) 6 q)
= z 2 m2 × γ µ − (1 − x)(1 − y)× 6 qγ µ 6 q
+ z(x − y)m × 21 {γ µ ,6 q} = q µ
µν
1 µ
+ z(2 − x − y)m × 2 [γ ,6 q] = −iσ qν
(z 6 p′ + x 6 q − m) γ µ (z 6 p − y 6 q − m) ∼
= ((z − 1)m + x 6 q) γ µ ((z − 1)m − y 6 q)
(22)
= (1 − z)2 m2 × γ µ − xy× 6 qγ µ 6 q
− (1 − z)(x − y)m × 12 {γ µ ,6 q} = q µ
+ (1 − z)(x + y)m × 12 [γ µ ,6 q] = −iσ µν qν .
Let’s plug these expressions back into eq. (20), collect similar terms together, and make use of
(1 − x)(1 − y) = 1 − x − y + xy = z + xy. This gives us
Nµ ∼
= −(D − 2) 6 ℓγ µ6 ℓ + 4mz(p′ + p)µ
+ m2 γ µ × −2 − 2z 2 + (4 − D)(1 − z)2
µ
+ 6 qγ 6 q × 2(z + xy) − (4 − D)xy
+ mq µ × (x − y) 4 − 2z − (4 − D)(1 − z)
µν
2
+ imσ qν × 2z(1 + z) − (4 − D)(1 − z) .
(23)
6 qγ µ6 q = 2q µ 6 q − q 2 γ µ ∼
= −q 2 γ µ
(24)
Furthermore, in the context of the Dirac sandwich ū(p′ )Γµ u(p) we have
because ū(p′ ) 6 qu(p) = 0, and also
(p′ + p)µ ∼
= 2mγ µ − iσ µν qµ
(the Gordon identity). Plugging these formulae into eq. (23), we arrive at
Nµ ∼
= −(D − 2) 6 ℓγ µ6 ℓ + m2 γ µ × 8z − 2(1 + z 2 ) + (4 − D)(1 − z)2
µν
2 µ
− q γ × 2(z + xy) − (4 − D)xy − imσ qν × (1 − z) 2z + (4 − D)(1 − z)
+ mq µ × (x − y) 4 − 2z − (4 − D)(1 − z) .
6
(25)
(26)
To further simplify this expression, let us go back to the symmetries of the integral (15). The
R
integral over the Feynman parameters, the integral dD ℓ, and the denominator [l2 − ∆]3 are all
invariant under the parameter exchange x ↔ y. In eq. (26) for the numerator, the first two lines
are invariant under this symmetry, but the last line changes sign. Consequently, only the first two
lines contribute to the integral (15) while the third line integrates to zero and may be disregarded,
thus
Nµ ∼
= −(D − 2) 6 ℓγ µ6 ℓ + m2 γ µ × 8z − 2(1 + z 2 ) + (4 − D)(1 − z)2
− q 2 γ µ × 2(z + xy) − (4 − D)xy − imσ µν qν × (1 − z) 2z + (4 − D)(1 − z) .
(27)
R
Finally, thanks to the Lorentz invariance of the dD ℓ integral,
ℓ2
ℓλ ℓν ∼
= gλν × ,
D
(28)
ℓ2
ℓ2
µ
λ µ ν
=
−(D
−
2)γ
×
.
γ
γ
γ
×
g
6 ℓγ µ6 ℓ = γ λ γ µ γ ν × ℓλ ℓν ∼
=
λν
D
D
(29)
and hence
Plugging this formula into eq. (26) and grouping terms according to their γ–matrix structure, we
arrive at
N µ = N1 × γ µ − N2 ×
iσ µν qν
2m
(30)
where
N1
(D − 2)2
2
2
2
∼
× ℓ + 8z − 2(1 + z ) + (4 − D)(1 − z) × m2
=
D
− 2(z + xy) − (4 − D)xy × q 2
(D − 2)2
× ℓ2 − (D − 2) × ∆ + 2z × (2m2 − q 2 ),
D
∼
= (1 − z) 4z + 2(4 − D)(1 − z) × m2 .
=
N2
(31)
(32)
Note that splitting the numerator according to eq. (30) is particularly convenient for calculating
the electron’s form factors:
Γµ1 loop = F11 loop (q 2 ) × γ µ + F21 loop (q 2 ) ×
7
iσ µν qν
,
2m
(33)
F11 loop (q 2 )
F21 loop (q 2 )
2
= −2ie
2
= +2ie
1
ZZZ
0
1
ZZZ
dx dy dz δ(x + y + z − 1)
Z
dx dy dz δ(x + y + z − 1)
Z
0
N1
dD ℓ
,
D
(2π) ℓ2 − ∆ + i0 3
N2
dD ℓ
.
D
(2π) ℓ2 − ∆ + i0 3
(34)
(35)
Electron’s Gyromagnetic Moment
As explained earlier in class, electron’s spin couples to the static magnetic field as
−eg
S·B
Ĥ ⊃
2me
where g = 2 Fmag
= F1 + F2 q 2 =0
.
(36)
The electric form factor F1 ≡ Fel for q 2 = 1 is constrained by the Ward identity,
F1tot = F1tree + F1loops + F1counter−terms −−
−→ 1.
2
q →0
(37)
Therefore, the gyromagnetic moment is
g = 2 + 2F2 (q 2 = 0)
(38)
where F2 = F2loops because the there are no tree-level or counter-term contributions to the F2 ,
only to the F1 . Thus, to calculate the g − 2 at the one-loop level, all we need is to evaluate the
integral (35) for q 2 = 0.
Let’s start with the momentum integral
Z
dD ℓ
N2
D
(2π) ℓ2 − ∆ + i0 3
(39)
where ∆ = (1 − z)2 m2 for q 2 = 0 and N2 is as in eq. (32). Because the numerator here does not
depend on the loop momentum ℓ, this integral converges in D = 4 dimensions and there is no
8
need for dimensional regularization. All we need is to rotate the momentum into Euclidean space,
Z
d4 ℓ
N2
= N2 ×
(2π)4 ℓ2 − ∆ + i0 3
Z
−iN2
=
×
16π 2
i d 4 ℓE
1
2
(2π)4 −(ℓE + ∆)3
Z∞
dℓ2E
0
ℓ2E
(ℓ2E + ∆)3
−iN2
1
=
×
2
16π
2∆
−i
N2 = 4z(1 − z)m2 hhfor D = 4ii
=
×
32π 2
∆ = (1 − z)2 m2 hhfor q 2 = 0ii
4z
−i
×
.
=
2
32π
1−z
(40)
Substituting this formula into eq. (35), we have
F21 loop (q 2
e2
= 0) =
16π 2
1
ZZZ
dx dy dz δ(x + y + z − 1) ×
4z
.
1−z
(41)
0
The integrand here depends on z but not on the other two Feynman parameters, so we can
immediately integrate over x and y and obtain
ZZ1
0
1−z
Z
dx dy δ(x + y + z − 1) = dx = 1 − z.
(42)
0
Consequently,
F21 loop (q 2
e2
×
= 0) =
16π 2
Z1
4z
e2
α
dz (1 − z) ×
=
×2 =
2
1−z
16π
2π
(43)
0
and the gyromagnetic moment is
g = 2 +
α
+ O(α2 ).
π
9
(44)
The higher-loop corrections to this gyromagnetic moment are harder to calculate because the
number of diagrams grows very rapidly with the number of loops; at the 4-loop order there are
thousands of diagrams, and one needs a computer just to count them! Also, at higher orders one
has to include the effects of strong and weak interactions because the photons interact with hadrons
and W ± particles, which in turn interact with other hadrons, Z 0 , Higgs, etc., etc. Nevertheless,
people have calculated the electron’s and the muon’s anomalous magnetic moments
ae =
gµ − 2
ge − 2
= F2electron (0) and aµ =
= F2muon (0)
2
2
(45)
to the order α4 back in the 1970s, and more recent calculations are good up to the order α5 .
Meanwhile, the experimentalists have measured ae to a comparable accuracy of 12 significant
digits and aµ to 9 significant digits
aexp
= 0.001 159 652 180 28 (24),
e
aexp
= 0.001 165 920 89 (63).
µ
(46)
The theoretical value of the electron’s anomalous magnetic moment is in good agreement with the
theory
experimental value, while for the muon there is a small discrepancy aexp
≈ (29±8)·10−10 .
µ −aµ
This discrepancy might stem from some physics beyond the Standard Model, maybe supersymmetry, maybe something else. Note that the effect of heavy particles on the aµ is proportional to
(mµ /Mheavy )2 , that’s why the muon’s anomalous magnetic moment is much more sensitive to the
new physics than the electron’s.
However, the discrepancy between the aexp
and the atheory
might also stem from a small
µ
µ
theoretical error in modeling the photon-hadron interactions, which affects the atheory
via 2+ loop
µ
diagrams like
(47)
hadrons
For a recent review of the muon’s high-precision anomalous magnetic moment — both the experiments and the theory — see http://arxiv.org/abs/1311.2198 and the references cited therein.
10
I would like to complete this section of the notes by calculating the F21 loop (q 2 ) form factor for
q 2 6= 0. Proceeding as in eq. (40) but letting ∆ = (1 − z)2 m2 − xyq 2 , we have
Z
and hence
F21 loop (q 2 )
4z(1 − z)m2
N2
d4 ℓ
−i
×
=
(2π)4 ℓ2 − ∆ + i0 3
32π 2 (1 − z)2 m2 − xyq 2
e2
=
16π 2
1
ZZZ
4z(1 − z)m2
dx dy dz δ(x + y + z − 1) ×
.
(1 − z)2 m2 − xyq 2
(48)
(49)
0
To evaluate this integral over Feynman parameters, we change variables from x, y, z to w = 1 − z
and ξ = x/(x + y),
x = wξ,
y = w(1 − ξ),
z = 1 − w,
dx dy dz δ(x + y + z − 1) = w dw dξ.
(50)
Consequently,
F21 loop (q 2 )
e2
=
16π 2
Z1 Z1
dξ dw w ×
0
e2
=
16π 2
4(1 − w)w × m2
w 2 × m2 − w 2 ξ(1 − ξ) × q 2
0
Z1
m2
dξ 2
×
m − ξ(1 − ξ)q 2
0
e2
=
×
8π 2
Z1
dξ
Z1
4w(1 − w)
dw w ×
w2
0
(51)
m2
m2 − ξ(1 − ξ)q 2
0
4m2
s
q2
4m2 − q 2
p
p
α
4m2 − q 2 + −q 2
4m2
=
×p
× log
.
2π
2m
(−q 2 ) × (4m2 − q 2 )
=
α
×p
× arctan
2π
q 2 × (4m2 − q 2 )
For q 2 < 0 and −q 2 ≫ m2 ,
F21 loop (q 2 ) ≈
2m2
−q 2
α
×
×
log
.
2π
−q 2
m2
11
(52)
The Electric Form Factor
Now consider the electric form factor F1 (q 2 ). In the first section we have obtained
F11 loop (q 2 )
2
= −2ie
1
ZZZ
dx dy dz δ(x + y + z − 1)
0
for
Z
dD ℓ
N1
,
D
(2π) ℓ2 − ∆ + i0 3
(34)
(D − 2)2
N1 ∼
× ℓ2 − (D − 2) × ∆ + 2z × (2m2 − q 2 )
=
D
(31)
and ∆ = (1 − z)2 m2 − xyq 2.
Let’s start by calculating the momentum integral in eq. (34). The numerator N1 depends on
ℓ as aℓ2 + b, so there is a logarithmic UV divergence for ℓ → ∞; to regularize this divergence, we
work in D = 4 − 2ǫ dimensions. Thus,
aℓ2 + b
d4 ℓ
≡ −iµ4−D
−i
(2π)4 [ℓ2 − ∆ + i0]3
Z
reg
Z
dD ℓ
aℓ2 + b
=
(2π)D [ℓ2 − ∆ + i0]3
idD ℓE −aℓ2E + b
(2π)D −[ℓ2E + ∆]3
Z D
d ℓE
aℓ2E − b
a
a∆ + b
4−D
=
= µ
×
− 2
(2π)D
(ℓE + ∆)2
(ℓ2E + ∆)3
(ℓE + ∆)3
Z D
Z∞ d ℓE
4−D
−t(∆+ℓ2E )
1 2
dt
a
×
t
−
(a∆
+
b)
×
= µ
×
e
t
2
(2π)D
= −iµ
4−D
Z
0
Z∞ Z D
d ℓE −tℓ2e
4−D
1 2 −t∆
= dt a × t − (a∆ + b) × 2 t e
e
×µ
(2π)D
(53)
0
Z∞ = dt a × t − (a∆ + b) × 21 t2 e−t∆ ×
0
µ4−D
=
(4π)D/2
µ4−D
µ4−D
(4πt)D/2
Z∞
dt e−t∆ × a × t1−(D/2) − 12 (a∆ + b) × t2−(D/2)
0
D
2
D
∆ 2 −2
1
2 (a∆ + b)
a×Γ 2−
×
−
(4π)D/2
a
a∆ + b
(4πµ2)ǫ Γ(1 + ǫ)
.
×
×
−
→
16π 2
∆ǫ
ǫ
2∆
=
12
×Γ 3−
D
2
D
× ∆ 2 −3
Going back to eq. (31), we identify a and b in eq. (52) as
(D − 2)2
2(1 − ǫ)2
=
,
D
2−ǫ
b = 2z × (2m2 − q 2 ) − (D − 2) × ∆ = b̂ − 2(1 − ǫ)∆ ,
a =
(54)
where b̂ = 2z(2m2 − q 2 ).
Consequntly, on the last line of eq. (53) we have
a
a∆ + b
−
= a×
ǫ
2∆
1
1
−
ǫ
2
+
2(1 − ǫ)∆
b̂
−
2∆
2∆
2(1 − ǫ)2 2 − ǫ
b̂
×
+ (1 − ǫ) −
2−ǫ
2ǫ
2∆
b̂
1−ǫ
× ((1 − ǫ) + ǫ = 1) −
=
ǫ
2∆
2
2
1−ǫ
z(2m − q )
=
−
,
ǫ
∆
=
(55)
so the momentum integral for the electric form factors evaluates to
2 4−D
−2ie µ
Z
N1
dD ℓ
=
2
D
(2π) [ℓ − ∆ + i0]3
ǫ 4πµ2
z × (2m2 − q 2 )
α
Γ(ǫ) × (1 − ǫ) − Γ(1 + ǫ) ×
.
=
2π
∆
∆
(56)
The next step in our claculation is to integrate the result in eq. (56) over the Feynman parameters. Changing the integration variables from x, y, z to w and ξ according to eq. (50), we
have
F11 loop (q 2 ) =
α
(4πµ2)ǫ
2π
0
where
or equivalently,

1



[∆(w, ξ)]ǫ
dξ dw w ×

(1 − w)(2m2 − q 2 ) 


 − Γ(1 + ǫ) ×

0
[∆(w, ξ)]1+ǫ
Z1



 (1 − ǫ)Γ(ǫ) ×
Z1
∆(w, ξ) = (1 − z)2 m2 − xyq 2 = w 2 × m2 − ξ(1 − ξ)q 2 ,
def
∆(w, ξ) = w 2 × H(ξ) where H(ξ) = m2 − ξ(1 − ξ)q 2.
13
(57)
(58)
(59)
R
The form (59) is particularly convenient for evaluating the dw integral in eq. (57), which becomes
Z1 2(1 − ǫ)Γ(ǫ)
w
2m2 − q 2 w(1 − w)
dw
× 2ǫ − 2Γ(1 + ǫ) ×
×
.
Hǫ
w
H 1+ǫ
w 2+2ǫ
(60)
0
Near the lower limit w → 0, the integrand is dominated by the second term, which is proportional
to w −1−2ǫ . But for any ǫ ≥ 0 — i.e., for any dimension D ≤ 4 — the integral
positive
Z
dw
w 1+2ǫ
(61)
0
diverges: For D = 4 the divergence is logarithmic while for D < 4 it becomes power-like.
The Infrared Divergence
Physically, the divergence (61) is infrared rather than ultraviolet, that’s why it gets worse as
we lower the dimension D. Indeed, let’s go back to the diagram (4) and look at the denominator
D in eqs. (5) and (7). Taking the electron’s momenta p and p′ on-shell before introducing the
Feynman parameters, we have
(p + k)2 − m2 = k 2 + 2kp + p2\ − m2\ = k 2 + 2kp = O(|k|) when k → 0,
(62)
and likewise
(p′ + k)2 − m2 = k 2 + 2kp′ = O(|k|) when k → 0.
(63)
Combining these two electron propagators with the O(1/k 2) photon propagators, we see that the
net denominator behaves as D ∝ |k|4 for k → 0 the numerator N µ remains finite, which makes
the integral
Z
Nµ
d k
∝
D
D
Z
dD k
|k|4
(64)
diverge for k → 0. In D = 4 dimensions, the infrared divergence here is logarithmic, while in
lower dimensions D < 4 it becomes power-like, i.e. O (1/kmin )4−D — precisely as in eqs. (61)
and (60).
14
We can regularize the infrared divergence (64) — and also (61) — by analytically continuing
the spacetime dimension to D > 4. Such dimensional regularization of the IR divergences is used in
many situations in both QFT and condensed matter. However, taking D > 4 makes the ultraviolet
divergences worse, so if some amplitude has both UV and IR divergences, we cannot cure both
of them at the same time by analytically continuing to D 6= 4. In particular, when calculating
the electric form factor F1 (q 2 ) of the electron, we need D < 4 to regulate the momentum integral
R D
d ℓ, but then we need D > 4 to regulate the integral over the Feynman parameters.
A common dirty trick is to first continue to D < 4, shift the loop momentum from k µ to
R
ℓµ = k µ + shift, evaluate the dD ℓ momentum integral in D < 4 dimension, then analytically
continue the result to D > 4 and integrate over the Feynman parameters, and ultimately continue
the final result to D = 4. However, in this kind of dimensional regularization it is hard to
disentangle the 1/ǫ poles coming from the UV divergence log(Λ2 /µ2 ) from the 1/ǫ poles coming
2 ), so we are not going to use it here.
from the IR divergence log(µ2 /kmin
Instead, we are going to use DR for the UV divergence only, while the IR divergence is regulated
by a tiny but not-quite-zero photon mass m2γ ≪ m2e . Strictly speaking, a massive vector particle
has three polarization states and its propagator is
−i
kµkν
µν
= 2
.
× g −
k − m2γ + i0
m2γ
(65)
However, the longitudinal polarization of the massive but ultra-relativistic photon does not couple
to a conserved current, so we are going to disregard the k µ k ν terms in the propagator (65) and
use
=
−ig µν
.
k 2 − m2γ + i0
(66)
In other words, we use the Feynman gauge in spite of the photon’s mass; this is not completely
consistent, but the inconsistencies go away in the mγ → 0 limit.
Using this infrared regulator for the internal photon line in the one-loop diagram (4), we get
the vertex amplitude that looks exactly like eq. (5) except for one factor in the denominator,
k2
1
+ i0
becomes
k2
1
.
− m2γ + i0
(67)
In terms of the integral (15), this change has no effect on the numerator N µ or the loop momentum
15
ℓ (which remains exactly as in eq. (13)), but the ∆ in the denominator becomes
∆′ (x, y, z) = ∆(x, y, z) + z × m2γ .
(68)
Consequently, the electric form factor is
F11 loop (q 2 )
=
Z
Z
d(F P ) µ4−D
dD ℓ −2ie2 × N1
,
(2π)D [ℓ2 − ∆′ + i0]3
(69)
exactly as in eq. (34), except for the ∆′ instead of the ∆ in the denominator. The momentum
integral here converges for any D < 4 and it evaluates exactly as in eq. (53). The only subtlety
here is that in the numerator, the ℓ–independent term b involves the un-modified ∆ instead of ∆′
(cf. eq. (54)), but we can fix that by writing
b = 2z × 2m2e − q 2 + (1 − ǫ)m2γ
− 2(1 − ǫ) × ∆′ .
(70)
Hence, instead of eq. (57) we get


1




Z1 Z1
 (1 − ǫ)Γ(ǫ) × [∆′ (w, ξ)]ǫ

α
1 loop 2
2 ǫ
F1
(q ) =
(4πµ ) dξ dw w ×
(71)
(1 − w)(2m2e − q 2 + (1 − ǫ)m2γ ) 

2π


 − Γ(1 + ǫ) ×

0
0
[∆′ (w, ξ)]1+ǫ
where
∆′ (w, ξ) = (1 − z)2 m2e − xyq 2 + zm2γ = w 2 × H(ξ) + (1 − w) × m2γ .
(72)
Note that the photon’s mass is tiny, m2γ ≪ m2e , q 2 ; were it not for the IR divergences, we
would have used m2γ = 0. This allows us to neglect various O(m2γ ) terms in eq. (71) except when
it would cause a divergence for w → 0; in particular, we may neglect the (1 − ǫ)m2γ term in the
numerator of the second term in the integrand. As to the denominators, in eq. (72) the second
term containing the photon’s mass becomes important only in the w → 0 limit, and in that limit
16
(1 − w)m2γ → m2γ . Thus, we approximate
∆′ (w, ξ) ≈ w 2 × H(ξ) + m2γ
(73)
R
and the dw integral in eq. (71) becomes
Z1
dw w × (1 − ǫ)Γ(ǫ) ×
0
(1 − ǫ)Γ(ǫ)
=
×
Hǫ
1
(1 − w)(2m2e − q 2 )
−
Γ(1
+
ǫ)
×
[w 2 H(ξ) + m2γ ]ǫ
[w 2 H(ξ) + m2γ ]1+ǫ
Z1
0
+ Γ(1 + ǫ)
− Γ(1 + ǫ)
[w 2
dw w
+ (m2γ /H)]ǫ
2m2e − q 2
H 1+ǫ
×
2m2e − q 2
×
H 1+ǫ
Z1
0
Z1
(74)
dw w 2
[w 2 + (m2γ /H)]1+ǫ
[w 2
0
For 0 < ǫ <
1
2
dw w
.
+ (m2γ /H)]1+ǫ
— i.e., for 3 < D < 4 — the integrals on the second and third lines here converge
even for m2γ = 0,
Z1
dw w
1
=
2
ǫ
[w ]
2 − 2ǫ
for ǫ < 1,
0
Z1
dw w 2
[w 2 ]1+ǫ
(75)
=
1
1 − 2ǫ
for ǫ < 12 ,
0
so we may just as well evaluate them without the photon’s mass. Only on the last line of eq. (74)
we do need m2γ 6= 0 to make the integral converge for some D ≤ 4:
Z1
0
1
ǫ
dw w
1
H
−1
1
=
=
−1 .
[w 2 + (m2γ /H)]1+ǫ
2ǫ [w 2 + (m2γ /H)]ǫ 0
2ǫ
m2γ
17
(76)
Combining all these
R
dw integrals together, we get
ǫ
Z1 n o
Γ(ǫ)
Γ(1 + ǫ) 2m2e − q 2
Γ(1 + ǫ) 2m2e − q 2
H
dw · · · =
+
×
−
×
×
−1
2H ǫ
1 − 2ǫ
H 1+ǫ
2ǫ
H 1+ǫ
m2γ
0
ǫ Γ(ǫ)
2m2e − q 2
2m2e − q 2
2ǫ
1
H(ξ)
=
−
× 1 +
×
+1 =
×
2H ǫ
H(ξ)
1 − 2ǫ
1 − 2ǫ
H(ξ)
m2γ
(77)
and hence
F11 loop (q 2 )
α
=
4π
ǫ ǫ Z1
2m2e − q 2
4πµ2
1
H
dξ Γ(ǫ)
× 1 +
×
−
H(ξ)
H
1 − 2ǫ
m2γ
(78)
0
where
H(ξ) = m2e − ξ(1 − ξ)q 2 .
(59)
Before we even try to perform this last integral, let’s remember that
µ
+ Γµloops + δ1 × γ µ
Γµnet = γtree
(79)
F1net (q 2 ) = 1tree + F1loops (q 2 ) + δ1 .
(80)
and hence
Also, the net electric charge does not renormalize, so we must have
F1net (q 2 ) → 1 for q 2 → 0
(81)
δ1 = −F1loops (q 2 = 0).
(82)
and hence
In particular, the δ1 counterterm to the order α follows from eq. (78) for q 2 = 0, in which case
R
H(ξ) ≡ m2e and the dξ integral becomes trivial (the integrand does not depend on ξ at all).
18
Thus,
δ1
α
= − Γ(ǫ)
4π
4πµ2
m2e
ǫ
2 ǫ 2
me
× 1+
+ O(α2 ).
− 2
1 − 2ǫ
m2γ
(83)
This formula holds for any dimension D between 3 and 4 (i.e., 0 < ǫ < 12 ). In the D → 4 limit, it
becomes
α
= −
×
4π
δ1
m2e
1
4πµ2
+
4
−
2
log
− γE + log
ǫ
m2e
m2γ
+ O(α2).
(84)
Now let’s go back to the electric form factor F1net (q 2 ) for q 2 6= 0. According to eqs. (80)
and (82), at the one-loop level
F1net (q 2 ) − 1 = F11 loop (q 2 ) − F11 loop (0) + O(α2)
(85)
where F11 loop (q 2 ) is given by eq. (78). Taking the ǫ → 0 limit of that formula, we arrive at
F11 loop (q 2 )
α
=
4π
Z1 1
4πµ2
2m2e − q 2
H(ξ)
dξ
,
− γE + log
+
× 2 − log
ǫ
H(ξ)
H(ξ)
m2γ
(86)
0
and now we should subtract a similar a similar expression for q 2 = 0. This subtraction cancels
the UV divergence and the associated 1/ǫ pole but not the IR divergence. Moreover, not only the
subtracted one-loop amplitude depends on the IR regulators, but the coefficient of the log m2γ has
a non-trivial momentum dependence. Indeed,
F11 loop (q 2 ) − F11 loop (0) =
Z1 m2e
m2e
2m2e − q 2
H(ξ)
α
− 2 2 − log 2
dξ log
+
× 2 − log
=
4π
H(ξ)
H(ξ)
m2γ
mγ
0
α
=
4π
Z1 m2e
H(ξ)
m2e
2m2e − q 2
m2e
dξ log
− 2 2 − log 2
+
× 2 − log 2 − log
H(ξ)
H(ξ)
mγ
m2e
mγ
0
2
Z1 2me − q 2
m2e
m2e
2m2e − q 2
dξ log
+ 2 − log 2 ×
× 1+
−2
H(ξ)
H(ξ)
mγ
H(ξ)
0
m2e
α
2
2
2
2
× fIR (q /me ) × log 2 + h(q /me )
= −
2π
mγ
(87)
α
=
4π
19
where
Z1 2
2me − q 2
2
2
fIR (q /me ) = dξ
−1 ,
2H(ξ)
(88)
0
h(q
2
/m2e )
= −2fIR (q
2
/m2e )
1
−
2
Z1 2
2me − q 2
m2e
dξ
+ 1 × log
.
H(ξ)
H(ξ)
(89)
0
Note that both of these integrals are finite in the mγ → 0 limit. Also, both fIF and h vanish for
q = 0, while for non-zero but small q 2 ≪ m2e both fIF and h diminish as O(q 2/m2e ):
For q 2 ≪ m2e ,
fIR (q 2 /m2e ) ≈ −
q2
,
3m2e
h(q 2 /m2e ) ≈ −
5q 2
.
12m2e
(90)
In the opposite limit of very large q 2 , the integrals (88) and (89) grow as polynomials of log(q 2 ):


−q 2

 fIR (q 2 /m2e ) ≈ log 2 − 1,

m
(91)
For q 2 ≫ m2e
e 2
2

π
−q
1

2
2
2
2

log 2 − 2 × fIR (q /me ) −
.
 h(q /me ) ≈
2
me
6
Finite Cross-sections for IR–Divergent Amplitudes
Remarkably, the function fIR (q 2 /m2e ) from eq. (88) which governs the infrared divergence of
the electron’s electric form factor F1 (q 2 ) also govern the IR divergence of the the soft-photon
bremsstrahlung. In terms of §6.1 of the Peskin & Schroeder textbook, I(v, v′ ) = 2fIR (q 2 /m2e ).
Let me postpone the proof of this identity towards the end of these notes (or you can find it in
the textbook $6.4, eqs. (6.69–70).) Meanwhile, let me explore the physical consequences of the
related IR divergences of the F1 form factor and of the bremsstrahlung.
Consider elastic Coulomb scattering of an electron off a very heavy point-like particle X:
e−
X
e−
X
20
(92)
2 , we may approximate the X particle as a static (non-recoiling) source of electric
For q 2 ≪ MX
field, and in its rest frame (the lab frame), the scattering amplitude evaluates to
M(eX → eX)
4παeff (q 2 )
=
× ū(p′ )Γ0 (p′ , p)u(p).
2MX
q2
(93)
At the tree level Γµ (p′ , p) ≡ γ µ , while at the one-loop level we get non-trivial form-factors. Focusing
at their infrared divergences, we have
F21 loop (q 2 ) =
2α
× finite
2π
(94)
while
F1tree+1 loop (q 2 )
α
= 1 −
2π
m2e
fIR (q ) × log 2 + finite ,
mγ
2
(95)
and therefore
Γµtree+1 loop (p′ , p)
α
m2e
α
2
× fIR (q ) × log 2 × Γµtree +
× finite.
= 1−
2π
mγ
2π
(96)
Plugging this one-loop vertex into the Coulomb scattering amplitude (93), we obtain
M
tree+1 loop
(eX → eX) = M
tree
α
α
m2e
2
(eX → eX) × 1 −
× fIR (q ) × log 2 +
× finite
2π
mγ
2π
(97)
and hence the partial cross-section
dσ tree+1 loop (eX → eX)
dσ tree (eX → eX)
=
×
dΩ
dΩ
α
m2e
α
2
2
× finite + O(α ) .
× 1 − × fIR (q ) × log 2 +
π
mγ
π
(98)
Note the IR divergence of the one-loop term in this cross-section. For future reference, I would
21
like to rephrase it in terms of log(Ee /mγ ) rather than log(me /mγ ), thus
dσ tree (eX → eX)
dσ tree+1 loop (eX → eX)
=
×
dΩ
dΩ
α
Ee
α
2
2
× 1 − × 2fIR (q ) log
+
× finite + O(α ) .
π
mγ
π
(99)
Now consider the inelastic scattering in which a photon is emitted, eX → eXγ,
γ
X
e−
e−
(100)
X
In general, an extra incoming or outgoing photon costs a factor e in the amplitude and hence a
factor α in the cross-section, thus at similar loop levels σ(eX → eXγ) = O(α) × σ(eX → eX).
Specifically, as explained in detail in §6.1 of the Peskin & Schroeder textbook, for a soft photon
whose energy is much smaller than the electron’s, ωγ ≪ Ee ,
M
tree
(eX → eXγ) = M
tree
(eX → eX) × e
p′ · ǫ∗γ
p · ǫ∗γ
−
p′ · k γ
p · kγ
(101) (6.22)
and therefore
dσ tree (eX → eX) α
I
dσ tree (eX → eXγ)
=
× ×
dΩe dωγ
dΩ
π ωγ
(102) (6.25)
" 2 #n0 =|n|=1
′µ
µ
p
p
−
−
.
′
4π
(np )
(np)
(103) (6.13)
where
def
I(p′ , p) =
Z
d2 Ωn
Integrating the partial cross-section (102) over the photon’s frequencies, we immediately run into
the infrared divergence:
Z
dσ
dωγ
∝
dωγ
ωmax
Z ∼Ee
0
dωγ
= ∞.
ωγ
(104)
To regulate this divergence, we need to impose a minimal energy requirement on the emitted
photon, and the simplest way to do this is to assume a tiny but non-zero photon mass mγ .
22
Consequently
Z
reg
dωγ
ωmax
Ee
= log
+ finite = log
+ finite,
ωγ
mγ
mγ
(105)
and therefore
dσ tree (eX → eXγ)
dσ tree (eX → eX) α
=
×
dΩ
dΩ
π
Ee
I(p , p) × log
+ finite .
mγ
′
(106)
We shall see momentarily that I(p′ , p) = 2fIR (q 2 ). Consequently, the one-loop-level crosssection (99) and the tree-level cross-section (106) have exactly the same infrared divergence, except
for opposite signs,
dσ tree (eX → eX)
dσ tree+1 loop (Xe → Xe)
=
×
(99)
dΩ
dΩ
α
α
Ee
+
× 1 − × 2fIR (q 2 ) log
× finite + O(α2 ) ,
π
mγ
π
dσ tree (Xe → Xeγ)
dσ tree (eX → eX)
=
×
dΩ
dΩ
α
α
Ee
2
+
× 0 + × 2fIR (q ) log
× finite ,
π
mγ
π
(106)
and therefore the combined cross-section has no infrared divergence:
dσ tree+1 loop (Xe → Xe) + dσ tree (Xe → Xeγ)
α
dσ tree (Xe → Xe) =
× 1 + × finite + O(α)2 .
dΩe
dΩe
π
(107)
But what do we do with the IR divergences of the partial cross-sections (99) and (106)? While
it is OK to UV-regulate or IR-regulate the intermediate stages of a calculation, but a finite result
for a measurable quantity like a partial cross-section must be finite and it cannot depend on an IR
regulator like mγ . Nevertheless, eqs. (106) and (99) seem to contradict this rule, so what gives?
To resolve this paradox, consider a real-life scattering experiment. No photon detector can
detect a photon with an arbitrarily low energy ωγ , there is always a threshold ωthr > 0 below
which the detector is blind. Thus, a final state |Xeγi where the photon’s energy is below the
threshold — ωγ < ωthr — will be seen by the detector as simply |Xei since the photon would
23
not be detected. In other words, observationally we should not classify the final states by the net
number of photons with any energy at all but rather by the number pf detectable photons with
ωγ > ωthr. As to the soft photons with ωγ < ωthr , the detector would not tell us if they are
there or not, so the physically measureable cross-sections should include all possible numbers of
undetectably low-energy photons. In particular,
dσ(X + e → observed X + e) = dσ(X + e → X + e) + dσ(X + e → X + e + γ(ω < ωthr))
+ dσ(X + e → X + e + γ(ω < ωthr ) + γ(ω < ωthr )) + · · · ,
dσ(X + e → observed X + e + γ) = dσ(X + e → X + e + γ(ω > ωthr))
+ dσ(X + e → X + e + γ(ω > ωthr ) + γ(ω < ωthr )) + · · · ,
etc., etc.
(108)
To the order O(α × σ tree (Xe → Xe) = O(α3), we should stop at one final-state photon,
detectable or not, and calculate the dσ(Xe → Xe) to the one-loop level while the dσ(Xe → Xeγ)
just to the tree level. Thus,
dσ(X + e → observed X + e + γ)
dσ tree (X + e → X + e + γ(ω > ωthr))
≈
dΩe
dΩe
ωmaxZ=O(Ee )
dσ tree (X + e → X + e + γ)
dωγ
=
dΩe dωγ
ωthr
dσ tree (X + e → X + e) α
Ee
2
=
2fIR (q ) log
×
+ finite ,
dΩe
π
ωthr
(109)
where the last equality comes from
dσ tree (eX → eX) α 2fIR (q 2 )
dσ tree (eX → eXγ)
=
× ×
dΩe dωγ
dΩ
π
ωγ
hh for ωγ ≪ Ee ii
(102) (6.25)
and
ωmaxZ=O(Ee )
ωthr
Ee
dω
= log
+ finite.
ω
ωthr
(110)
Note that observed the cross-section (109) is infrared finite and does not depend on the mγ (as
long as mγ ≪ ωthr). Instead, it depends on the photon detector’s low-energy threshold ωthr .
24
Similarly, to the same order O(α × σ tree (Xe → Xe) = O(α3),
dσ(X + e → observed X + e)
≈
dΩe
dσ tree+1 loop (X + e → X + e)
dσ tree (X + e → X + e + γ(ω < ωthr ))
≈
+
dΩe
dΩe
tree
α
dσ (X + e → X + e)
α
Ee
2
≈
× 1 − × 2fIR (q ) log
+
× finite
dΩe
π
6mγ
π
α
ωthr
α
dσ tree (X + e → X + e)
2
× 0 + × 2fIR (q ) log
+
× finite
+
dΩe
π
6mγ
π
α
Ee
dσ tree (X + e → X + e)
α
2
× 1−
× 2fIR (q ) log
× finite .
=
+
dΩe
π
ωthr
π
(111)
Note how the IR regulator mγ cancels between the two contributions to the net observed crosssection. Again, the observed cross-section is IR-finite, but it depends on the photon detector’s
threshold ωthr.
Similar cancellations of IR divergences from the observed cross-sections happen at the higher
loop orders. In general, go get a finite cross-section to the order O(αL × σ tree ) we should combine
an L-loop cross-section with no soft photons, an L − 1 loop cross-section with one soft photon, an
L − 2 cross-section with 2 soft photons, etc., etc., ending with a tree-level cross-section with L soft
photons. Individually, all these formal cross-sections are infrared-divergent, but once we combine
them together into a complete observed cross-section, the IR divergence should cancel out. Please
read (or at least skim) textbook §6.5 to see how this works.
⋆
⋆
⋆
Let me conclude these notes by proving that I(p′ , p) = 2fIR (q 2 /m2 ). By definition of the soft
bremsstrahlung factor I(p′ , p) — or as the textbook calls it I(v, v′ ), since it only depends on the
velocities and not the electron’s mass, —
I(p′ , p) =
Z
" 2 #n0 =|n|=1
′µ
µ
p
p
−
.
−
′
4π
(np )
(np)
d2 Ωn
(103) (6.13)
Inside the integral here, we have
−
pµ
p′µ
−
(np′ )
(np)
2
= −
m2
m2
2(pp′ ) = 2m2 − q 2
−
+
.
(np)2
(np′ )2
(np)(np′ )
(112)
Now let’s integrate each of these tree terms over the directions of the unit 3-vector n. For the first
25
term, we have
Z
d2 Ωn 1
2π
=
2
4π (np)
4π
Z+1
d cos θ
1
1
1
=
×
2
(E − |p| × cos θ)
2 |p|
−1
=
1
2|p|
E+|p|
Z
d(E − |p| × cos θ)
(E − |p| × cos θ)2
E−|p|
1
1
−
E − |p|
E + |p|
=
E2
(113)
1
1
=
.
2
− |p|
m2
Likewise, for the second term
Z
d2 Ωn 1
1
=
.
2
′
4π (np )
m2
(114)
Finally, for the third term we use the Feynman parameter trick:
1
=
(np) × (np′ )
Z1
dξ
1
=
[(1 − ξ) × (np) + ξ × (np′ )]2
0
Z1
dξ
0
1
(npξ )2
(115)
where
pµξ = (1 − ξ) × pµ + ξ × p′µ .
(116)
Consequently,
Z
1
d2 Ωn
=
4π (np) × (np′ )
Z1 Z 2
Z1
1
d Ωn
dξ
,
dξ
=
2
4π (npξ )
p2ξ
0
(117)
0
where
p2ξ =
2
(1 − ξ)p + ξp′
= (1 − ξ)2 × (p2 = m2 ) + ξ 2 × (p′2 = m2 ) + ξ(1 − ξ) × (2pp′ = 2m2 − q 2 )
(118)
= m2 − ξ(1 − ξ)q 2 .
Plugging all these formulae back into eqs. (103) and (112), we arrive at
m2
m2
+ (2m2 − q 2 ) ×
I(p′ , p) = − 2 −
m
m2
Z1
dξ
=
2
m − ξ(1 − ξ)q 2
0
Z1
0
dξ
2m2 − q 2
−2 .
m2 − ξ(1 − ξ)q 2
(119)
26
Finally comparing the last formula with the definition of the fIR ,
fIR (q
2
/m2e )
Z1 = dξ
0
2m2e − q 2
−1 ,
2[m2 − ξ(1 − ξ)q 2 ]
we immediately see that I(p′ , p) = 2fIR (q 2 /m2e ), quod erat demonstrandum.
27
(88)