PHYSICS 218 PROBLEM SET #4
BEN LEHMANN
1.
We denote the lepton momentum by p` and the neutrino momentum by pn to avoid confusion with
Lorentz index. The squared matrix element is
1
(1)
|M|2 = Fπ2 G2F m2π pµ pν ū(pn )γ µ (1 − γ5 )v(p` )v̄(p` )(1 − γ5 )γ ν u(pn )
2
Let us work in the rest frame, where pµ γ µ = mπ γ 0 . Now we can sum over the spins of the products,
finding
|M|2 = 4Fπ2 G2F m2π 2p0` p0n − p` · pn
(2)
From the kinematics in the rest frame, assuming a massless neutrino, we have
Thus p` · pn =
1
2
m2 − m2`
m2 + m2`
,
p0n = |pn | = |p` | = π
p0` = π
2mπ
2mπ
2
2
mπ − m` , and our squared matrix element becomes
|M|2 = 2Fπ2 G2F m2` m2π − m2`
(3)
(4)
Now we can find the decay width using
λ1/2 m2π , m2n , m2`
Γ=
|M|2
16πm3π
(5)
where λ(M, m1 , m2 ) = (s − m21 − m22 )2 − 4m21 m22 is the triangle function. With our matrix element,
we have
Fπ2 G2F m3` − m` m2π 2
+
+
(6)
Γ π → ` ν` =
8πm3π
The branching ratio for ` = e versus ` = µ is determined by the respective lepton masses. Plugging
in numbers, we find that
3
2
Γ (π + → e+ ν` )
me − me m2π
=
≈ 1.3 × 10−4
(7)
Γ (π + → µ+ ν` )
m3µ − mµ m2π
2. We have just one diagram contributing to this process at tree level (fig. 1). We evaluate the
p1
p3
e
e
H
Z; k µ
p2
Z
p4
Figure 1. The only tree-level diagram contributing to e+ e− → HZ.
1
PHYSICS 218 PROBLEM SET #4
2
matrix element as1
−i g αν −
kα kν
m2Z
2
mZ
ig
ig m2Z νµ µ∗
γα (cV − cA γ5 ) u(p1 )
g 4
cos θw
sin θw mW
k2 −
αµ − kα kµ µ∗
−i
g
2
2
2
4
mZ g
mZ
v̄(p2 )γα (cV − cA γ5 ) u(p1 )
2
2
cos θw sin θw
k − mZ
iM = −v̄(p2 )
=
mW
(8)
(9)
We take the adjoint as
†
(iM) =
i g αµ −
m2Z g 2
kα kµ
m2Z
2
k − m2Z
mW cos θw sin θw
µ4
ū(p1 ) (cV − cA γ5 ) γα v(p2 )
(10)

β ν
ν
g βν − kmk2 µ∗
4 4
Z

2
2
2
k − mZ
(11)
and thus find the squared matrix element
|M|2 =
2
m2Z g 2
mW cos θw sin θw

g αµ −
kα kµ
m2Z

× ū(p1 )γα (cV − cA γ5 ) v(p2 )v̄(p2 ) (cV − cA γ5 ) γβ u(p1 )
Since me this is
√
s, we take the electrons to be massless. Then, summing over spins and polarizations,
|M|2 =
1
4
m2Z ge
mW cos θw sin θw
2
 −
g αµ −
kα kµ
m2Z
g βν −
k2 −
g µν −
kβ kν
m2Z
2
m2Z
pµ
4 2
pν4 p4


(12)
× tr [p1 γα (cV − cA γ5 ) p2 (cV − cA γ5 ) γβ ]
We evaluate the trace as
i
h
tr p1 γα (cV − cA γ5 ) p2 (cV − cA γ5 ) γβ
= p1ρ p2σ tr [γρ γα (cV − cA γ5 ) γσ (cV − cA γ5 ) γβ ]
i
h
= 2 (cA − cV ) (cA + cV ) sg αβ − 2 pβ1 pα2 + pα1 pβ2
where we have used p1 · p2 = 21 s. Putting it all together, we have
 αµ − kα kµ
βν − kβ kν
2
g
g
g µν −
2
2
2
mZ ge
1
mZ
mZ
−
|M|2 =
2
4 mW cos θw sin θw
k 2 − m2Z
h
i
× 2 (cA − cV ) (cA + cV ) sg αβ − 2 pβ1 pα2 + pα1 pβ2
(13)
(14)
ν
pµ
4 p4
2
mZ


(15)
Now we contract the remaining indices and simplify. Since k 2 = s and p1 · p2 = 12 s, we have
|M|2 =
2m2Z g 4
(cA − cV ) (cA + cV ) csc2 (2θW ) 4(p1 · p4 )(p2 · p4 ) + m2Z s
2
2
2
mW (mZ − s)
We can further simplify this result with a bit of kinematics:
1
1
p2 · p4 =
m2z − t ,
p1 · p4 =
m2z − u ,
s − t − u = 2s − (m2z + m2H )
2
2
(16)
(17)
1I started working this out using a reference that wrote out G in terms of m . I decided not to change this
F
W
back, but I used GF in problem 3.
PHYSICS 218 PROBLEM SET #4
3
Making these substitutions, we have
|M|2 = −
2m2Z g 4 csc2 (2θw )
(cA − cV )(cA + cV ) m2H (m2Z − t) + t(s + t) − m2Z (2s + t)
2
2
2
mW (mZ − s)
(18)
Before we find the differential cross section, we rewrite this result in terms of the scattering angle.
From kinematics, we have
√ m2Z − m2H + s
1√
√
−
s |p4 | cos θ
(19)
p2 · p4 = E2 E4 − p2 · p4 = 12 s
2
2 s
We can thus write t in terms of θ, as
q
1
2
2
2
2
2
2
2
t=
mH + mZ − s − mH + (mZ − s) − 2mH (mZ + s) cos θ
(20)
2
Making this substitution in our matrix element and simplifying, we have
|M|2 = −
g 4 m2Z csc2 (2θw )
(cA − cV ) (cA + cV )
(21)
2m2W (m2Z − s)2
× −(m2H − m2Z )2 + 2(m2H − 3m2Z )s − s2 + (mH − mZ )2 − s (mH + mZ )2 − s cos2 θ
Now the differential cross section can be written in the form
dσ
1
|p4 |
√ |M|2
=
dΩ
(2E1 )(2E2 ) |vrel,i | (2π)2 4 s
q
|M|2
=
m4H + (m2Z − s)2 − 2m2H (m2Z + s)
64π 2 s2
Finally, we can integrate to find the total cross section:
Z
dσ
σ = dΩ
dΩ
m2 g 4 (cA − cV )(cA + cV ) csc2 (2θw ) 4
mH + m4Z + 10m2Z s + s2 − 2m2H (m2Z + s)
= Z
2
2
2
2
48πmW (mZ − s) s
(22)
(23)
(24)
(25)
Inserting numerical values gives σ ≈ 6.5 pb, so we expect about 452 Higgs bosons2 to be produced
at LEP.
2This result is different from what I heard some other folks mention. I think I might be using the wrong values
for cA and cV .
PHYSICS 218 PROBLEM SET #4
q1 ; µ; a
4
q1 ; µ; a
k − q1 ; i
p
p
k; j
k + q2 ; `
q2 ; ν; b
q2 ; ν; b
Figure 2. Diagrams contributing to H → gg at 1-loop.
3.
(a) We have two diagrams related under the interchange of q1 , q2 and µ, ν (fig. 2). We evaluate
the first diagram as
Z
√
k + mt ) ν b i(
k + q2 + mt ) i(
k − q1 + mt ) 1∗ 2∗
d4 k
1/2
µ a i(
tr iγ Tij 2
iM1 = −(−1)4πiαs ( 2GF ) mt
iγ Tj`
δi`
(2π)4
k − m2t
(k + q2 )2 − m2t (k − q1 )2 − m2t µ ν
(26)
b = tr(T a T b ) = 1 δ ab . Pulling everything out, we can write this
The color factor is just δi` Tija Tj`
2
2∗ M µν for
as iM1 = 1∗
µ ν
1
Z
√
d4 k
k + mt ) ν i(
k + q2 + mt ) i(
k − q1 + mt )
µν
1/2
ab
µ i(
M1 = −2πiαs ( 2GF ) mt δ
tr γ
γ
(27)
(2π)4
k 2 − m2t
(k + q2 )2 − m2t (k − q1 )2 − m2t
The trace is straightforward, if cumbersome, to compute; we simplify with q1 · q2 = 12 m2H . But
now, by Lorentz invariance and symmetry of q1µ and q2µ , we know that M µν must have the form
M µν = Ag µν + B(q1µ q2ν + q1ν q2µ ). Eliminating other terms, we must have
Z
Z
−m2H g µν + 2q1ν q2µ + 2q1µ q2ν
d4 k
d4 k
tr(·
·
·
)
=
−2im
(28)
t
(2π)4
(2π)4 k 2 − m2t (k + q2 )2 − m2t (k − q1 )2 − m2t
Putting it all together, we have
M1µν
=
−m2H g µν
+
2q1ν q2µ
+
2q1µ q2ν
Z
√
d4 k
4παs ( 2GF )1/2 m2t δ ab
(2π)4 k 2 − m2t (k + q2 )2 − m2t (k − q1 )2 − m2t
(29)
The second diagram contributes M2µν identical to M1µν up to the interchange of µ and ν and
of q1 and q2 , which gives us the same amplitude. Thus,
√
Z
8παs ( 2GF )1/2 m2t δ ab
d4 k
µ ν
µν
2 µν
ν µ
(30)
M = −mH g + 2q1 q2 + 2q1 q2
(2π)4 k 2 − m2t (k + q2 )2 − m2t (k − q1 )2 − m2t
(b) All of the Lorentz indices on M µν lie in the −m2H g µν + 2q1ν q2µ + 2q1µ q2ν term, so we have
q1µ M µν ∝ −m2H q1ν + 2q1ν (q1 · q2 ) + 2q12 q2µ
(31)
But q1 · q2 = 12 m2H and q12 = 0, so this is
q1µ M µν ∝ −m2H q1ν + m2H q1ν = 0
(32)
Since this term is symmetric in q1 and q2 , it is clear that q2µ M µν = 0 for the same reason.
(c) We use Feynman parameters to evaluate the integral. Recall that
Z 1
1
2δ(x + y + z − 1)
=
dx dy dz
(33)
ABC
(Ax + By + Cz)3
0
PHYSICS 218 PROBLEM SET #4
Thus, our integrand can be written in the form
1
=
2
2
2
k − mt (k + q2 ) − m2t (k − q1 )2 − m2t
Z 1
2δ(x + y + z − 1)
dx dy dz
3
0
k 2 − m2t x + (k + q2 )2 − m2t y + (k − q1 )2 − m2t z
Expanding the denominator and dropping q12 = q22 = 0, this is
Z 1
2δ(x + y + z − 1)
dx dy dz
3
0
−2(k · q1 )z + 2(k · q2 )y + k 2 (x + y + z) − m2t (x + y + z)
Taking advantage of the delta function, we set x + y + z = 1, which leaves
Z 1
2δ(x + y + z − 1)
dx dy dz
3
2
0
k − 2(k · q1 )z + 2(k · q2 )y − m2t
5
(34)
(35)
(36)
We can complete the square in this expression, since
(k + q2 y + q1 z)2 = k 2 − 2(k · q1 )z + 2(k · q2 )y − 2(q1 · q2 )yz
(37)
Define ` = k + q2 y + q1 z and ∆ = −2(q1 · q2 )yz + m2t , so that our integral can be written in
the form
Z 1
Z
2δ(x + y + z − 1)
d4 k
dx dy dz
(38)
4
(2π) 0
(`2 − ∆)3
But ` is just a shift of k, so we can equivalently write
Z 1
Z
d4 ` 2δ(x + y + z − 1)
dx dy dz
(39)
(2π)4
(`2 − ∆)3
0
The integral with respect to ` can be performed directly:
Z
δ(x + y + z − 1)
d4 ` 2δ(x + y + z − 1)
√
=
3
4
2
(2π)
16π 2 −∆2
(` − ∆)
=
δ(x + y + z − 1)
q
2
16π 2
m2H yz − m2t
(40)
(41)
where we have again used q1 · q2 = 12 m2H . In evaluating the remaining integrals, we expend the
delta function on the integral with respect to z, replacing z → 1 − x − y. Then
Z 1
δ(x + y + z − 1)
q
dx dy dz
(42)
2 =
2
2
0
2
16π
mH yz − mt
We can do the integrals in x and y by hand, finding
2 Z 1
Z 1
mH
δ(x + y + z − 1)
1
dz
q
dx dy dz
log 1 −
z(1 − z)
2 = − 16π 2 m2
m2t
0
H 0 z
16π 2
m2H yz − m2t
We are given the value of this integral:
2 Z 1
mH
1
dz
1
2 1 mH
−
log 1 −
z(1 − z) = 2 2 arcsin
2 mt
16π 2 m2H 0 z
m2t
8π mH
(43)
(44)
Since the rest of M µν is real, let us find the imaginary part of this expression. We know the
integral will be purely real for mH /mt < 2, which we note is exactly the condition required for
the top quarks to be produced on-shell.
PHYSICS 218 PROBLEM SET #4
To find the imaginary part, observe that
2
p
p
arcsin2 z = − log iz + 1 − z 2 = − log 2iz + 2 1 − z 2
6
(45)
In general, log(r eiθ ) = log r + iθ, so log(r eiθ )2 = (log r)2 − θ2 + 2iθ log r, so we have
Im (log z)2 = 2(log |z|) arg z
(46)
√
√
For real masses with z > 1, we have that 1 − z 2 = i z 2 − 1 is pure imaginary, so the
argument is π/2, and the modulus is the imaginary part. If instead z < 1, the modulus is 1,
so log |z| vanishes, and the imaginary part with it. Thus, minding factors of 2, we have


s
2
m
m
m
1
m
H
H
H
H
2
= −π log 
+
− 1Θ
−1
(47)
Im arcsin
2 mt
2mt
2mt
2mt
Now we can write M µν and its imaginary part:
√
m2t αs ( 2GF )1/2 δ ab
µ ν
µν
2 µν
ν µ
2 1 mH
M =
−mH g + 2q1 q2 + 2q1 q2 arcsin
,
(48)
2 mt
πm2H


s
√
2
2 α ( 2G )1/2 δ ab
m
m
mH
m
F
H
H
µ ν
t s
2 µν
ν µ
µν


−mH g + 2q1 q2 + 2q1 q2 log
+
−1
−1 Θ
Im M = −
2mt
2mt
2mt
m2H
(49)
M µν
(d) As mt → ∞, we can expand
in a power series in mH /mt . We find
√
αs ( 2GF )1/2 δ ab
−m2H g µν + 2q1ν q2µ + 2q1µ q2ν + O(mH /mt )
M µν =
2π(mH /mt )
(50)
Note that M µν is real in this limit.
(e) We reattach the polarization vectors and square the matrix element:
!2
√
1/2 δ ab
α
(
2G
)
s
F
2∗
2 µν
|M|2 =
δ ab δab 1∗
+ 2q1ν q2µ + 2q1µ q2ν 1ρ 2σ −m2H g ρσ + 2q1σ q2ρ + 2q1ρ q2σ
µ ν −mH g
2π(mH /mt )
(51)
Summing over color gives a factor of the number N of generators. Summing over polarizations,
we have, for nµ = (1, 0, 0, 0),
!2 √
1/2
q1ρ nµ + q1µ nρ
q1µ q1ρ
α
(
2G
)
s
F
2
|M| = N
−gρµ +
−
−m2H g µν + 2q1ν q2µ + 2q1µ q2ν
2
2π(mH /mt )
q1 · n
(q1 · n)
q2σ nν + q2ν nσ
q2σ q2ν
ρ σ
2 ρσ
σ ρ
× −gσν +
−
−m
g
+
2q
q
+
2q
q2
(52)
H
1
2
1
q2 · n
(q2 · n)2
α2 N GF m2 m2
= s √ H t
(53)
2π 2
Using eq. (5), this corresponds to a decay width of
Γ(H → gg) =
αs2 N GF mH m2t
√
16 2π 3
(54)
Now consider the much simpler decay H → bb̄ at tree level, shown in fig. 3. We evaluate the
diagram as
1/2
√
iM = i
2GF
ū(q2 )v(q1 )
(55)
PHYSICS 218 PROBLEM SET #4
7
q1 ; a
p
q2 ; b
Figure 3. H → bb̄ at tree level.
We will get a factor of N from color, as before. Taking the adjoint, squaring, and summing
over fermion spins, we have
√
|M|2 = 4 2N GF m2b (q1 · q2 )
(56)
1
2
2
From kinematics, q1 · q2 = 2 mH − 2mb , so we have
√
(57)
|M|2 = 2 2N GF m2b m2H − 2m2b
and this gives us a decay width of
q 2
N GF m2b m2H − 2m2b
mH − 4m2b
√
Γ(H → bb̄) =
4 2πm2H
(58)
The ratio is then
Γ(H → gg)
=
Γ(H → bb̄)
4π 2 m2b m2H − 2m2b
q 2
mH − 4m2b
α2 m3H m2t
≈ 0.026
(59)