Download Part (a): Matrix Elements

Physics 217
Quantum Field Theory I
Fall 2015
Solution to Homework Set #6
Some Particle Dark Matter Model Building
Author: Michael C. Shamma
In this problem, we consider the process of scalar electron (φ− ) scalar
positron (φ+ ) pair annihilation to two photons in scalar QED. The scalar
QED Lagrangian is
L = −Fµν F µν − φ(∂µ ∂ µ + m2e )φ∗ − ieAµ (φ∂µ φ∗ − φ∗ ∂µ φ) + e2 A2µ |φ|2
(1)
Where me is the mass of the electron and of the positron. We are asked
to calculate the tree-level matrix elements (Part (a)) and the cross section
in the center of mass frame (Parts (b) and (c)) for the e− e− → γγ in the
case of a massless photon and for a massive photon. In part (d) we are asked
to determine suitable combinations of the mass of said photon and a ”dark
matter” fine structure constant αdark given specific electron/positron masses
and cross sections.
Part (a): Matrix Elements
There are three tree-level diagrams for e− e− → γγ in scalar QED for this
process, where the fields are created and annihilated according to the scalar
field equations for a particle and anti particle
Z
d3 p 1
(ap e−ipx + b†p eipx )
(2)
φ(x) =
3
(2π) ωp
Z
d3 p 1 † ipx
∗
φ (x) =
(ap e + bp e−ipx )
(3)
3
(2π) ωp
where b†p and a†p create positrons and electrons respectively and their Hermitian conjugates annihilate them.The three tree-level diagrams are the uchannel, the t-channel, and a 4-point diagram (arising from the e2 A2µ |φ|2 term
in the Lagrangian density). These are shown, respectively, below:
Physics 217
Quantum Field Theory I
e−
p3µ
Fall 2015
γ, ∗µ
3
p1µ
p4ν
e+
p2ν
γ, ∗ν
4
γ, ∗µ
3 , p3µ
e−
p1µ
p2ν
e+
γ, ∗ν
4 , p4ν
e−
γ, ∗µ
3 , p3µ
p1µ
p2ν
γ, ∗ν
4 , p4ν
e+
where the momentum of the positron in each diagram points opposite
the direction of the arrow. And each vertex gets a factor of (ie)2 , except
the 4-point diagram which gets and 2e2 coming once again from the e2 A2µ
and a symmetry factor for the two fields term. Using the momentum-space
Feynman rules, we can find the matrix element for this process. For the
t-channel, making sure to conserve momentum at each vertex, we obtain
iMt = (ie)2
i(2p1µ − p3µ )(p4ν − 2p2ν ) ∗µ ∗ν
3 4
t − m2e
(4)
Where p1µ and p2ν are the momenta of the electron and positron, respectively, and p3µ and p4ν are the momenta of the outgoing photons. However,
the inner product of the polarization vectors with the corresponding photon’s
momentum is null by definition:
∗ν
p3µ ∗µ
3 = p4ν 4 = 0
(5)
Physics 217
Quantum Field Theory I
Fall 2015
and equation (4) simplifies to
iMt = 4ie2
p1µ p2ν ∗µ ∗ν
t − m2e 3 4
Similarly for the u-channel, we can take p3µ → p4ν simplify according to
equation (5) and find
iMu = 4ie2
p1µ p2ν ∗µ ∗ν
u − m2e 3 4
Lastly, we write the matrix element of the 4-point diagram as
∗ν
iMseagull = 2ie2 gµν ∗µ
3 4
And the sum of the three diagrams is
1
1
1
∗ν
+
)]∗µ
iM = iMt + iMu + iMseagull = 4ie2 [ gµν + p1µ p2ν (
3 4
2
2
2
t − me u − me
(6)
However, since we are asked later to find the cross section in the center of
mass frame we can simplify the second term in the brackets and the matrix
element becomes
1
u + t − 2m2e
iM = 4ie2 [ gµν + p1µ p2ν (
)]∗µ
∗ν
2
2
2
(t − me )(u − me ) 3 4
and note that for this process u + t + s = 2m2e and we see that
1
−s
iM = 4ie2 [ gµν + p1µ p2ν (
)]∗µ
∗ν
2
2
2
(t − me )(u − me ) 3 4
(7)
Part (b): Cross Section
In order to calculate the cross section for this process we must find the square
of the matrix element and plug into the familiar equation for the differential
cross section for a 2 → 2 process (in center of mass frame)
dσ
1 |pf |
=
|M|2
dΩ
64π 2 s |pi |
To find |M|2 , it makes sense to first write
∗ν
M = ∗µ
3 4 Mµν
(8)
Physics 217
Quantum Field Theory I
Fall 2015
then square and sum over the polarizations of each outgoing photon:
2
|M| =
2 X
2
X
∗ν
2
|∗µ
3i 4j Mµν |
=
i=1 j=1
2 X
2
X
α ∗ν β
∗µ
3i 3i 4j 4j Mµν Mαβ
(9)
i=1 j=1
In order to go any further with this, the Ward Identity will be used. In
this process, take the initial e− and e+ to be traveling in opposite directions
along the z-axis. In the center of mass frame, the initial four-momenta of
the electron and positron are
p1µ = (E, 0, 0,
p
E 2 − m2e )
p2ν = (E, 0, 0, −
(10)
p
E 2 − m2e )
respectively, and the center of mass energy s = 4E 2 . There are two
distinguisable polarization vectors for each photon, namely
µ (1) = (0, 1, i, 0)
µ (2) = (0, 1, −i, 0)
. In order to compute the sum, I will complete the sum for one outgoing
photon and then generalize to N outgoing photons. For one outgoing photon,
we can transform to a frame in which the four-momentum is (E,0,0,E) (this
can be done since the total matrix element squared is Lorentz invariant), the
Ward identity (pµ Mµ = 0) is utilized and it is found that
E 2 M0 − E 2 M3 ⇒ M20 − M23 = 0
Now we compute
−1
µ (1)Mµ = √ (M1 + iM2 )
2
1
µ (2)Mµ = √ (iM2 − M1 )
2
Combining this with what was found with the Ward Identity
1
|M|2 = (2M1 + M2 ) = M1 + M2 + M3 − M0 = −gµν Mµν = −Mµ M∗ µ
2
Physics 217
Quantum Field Theory I
Fall 2015
This can be readily generalized to 2, 3,..., N outgoing photons as each
new photon gives an independent sum that yields another factor of −gµν . In
general, this can be written as
X ∗µ
2
N
µ1 ...µN
N
|i1 1 ...∗µ
M∗ µ1 ...µN
iN Mµ1 ...µN | = (−1) M
i1 ...iN
Thus, equation 9 can be rewritten as
|M|2 = Mµν M∗ µν
(11)
Putting all of the previous results together,
s2
1
]2
|M|2 = 16e4 [[ −
2
2
2 4(t − me )(u − me )
s2
m2e s
1
+
]2
−[ −
2
2
2
2
2 4(t − me )(u − me ) (t − me )(u − me )
s s
( − m2e )s2
2 4
−
(t − m2e )2 (u − m2e )2
1
− ]
2
The algebra on the road to simplification of this equation was not fun and neither will be entering all 2 pages (and 3 days of mistakes) of algebra required.
So, I will quote the result, after many cancellations and odd factorizations,
the matrix element is
|M|2 = 4e4 [1 + (1 −
2m2 s
)2 ]
(t − m2e )(u − m2e )
(12)
√
Plugging this back into equation 8, noting that |pf | = 2E 2 = s and |pi | =
p
p
2 E 2 − m2e = s − 4m2e we find the differential cross section to be
dσ − −
2m2 s
α2
(e e → γγ) = p
[1 + (1 −
)2 ]
2 )(u − m2 )
2
dΩ
(t
−
m
s(s − 4me )
e
e
where α =
e2
4π
Physics 217
Quantum Field Theory I
Fall 2015
Part (c): Cross Section with a Massive Photon
In the case of a massive photon, a few things change from of the sum over
polarization vectors to the sum of the Mandelstam variables. We note that
the sum over polarization vectors goes to 3 instead of 2 and there is now a
term that is porportional to the photon momenta over m2γ . However, when
taken with the total matrix element, the Ward identity implies that this term
exactly cancels and we are left with the same matrix element squared, with
one caveat. The sum of the Mandelstam variables is now
s + t + u = 2m2e + 2m2γ
p
and the p1µ p2ν term is now multiplied by 2m2γ − s and |pf | = s − 4m2γ and
a guess at the new differential cross section is
s
2
s − 4m2γ
2m2e (2m2γ − s) 2
dσ − −
α
(e e → γγ) =
)]
(13)
[1 + (1 +
dΩ
s
s − 4m2e
(t − m2e )(u − m2e )
Part (d): Dark Matter
Supposing that dark matter in the universe is made of scalar electron and
scalar positrons of mass m who interact with massive photons of mass mγ
where the strength is given by αdark , we can write the differential cross section
for the process e− e+ → γγ as the equation 13. To simplify the calculation
of the total cross section, we will go to a frame where the outgoing photon’s
√
have zero momentum and we will look at the case where s me > mγ .
We can therefore right their momentum four-vectors as
p3µ = (mγ , 0, 0, 0) = p4ν
and recalling the equations for p1µ and p2ν we can also rewrite the Mandelstam variables, t and s, as
t − m2e = (p1 − p2 )2 − m2e = (E − mγ )2 − (E 2 − m2e ) − m2e = m2γ − 2Emγ
u − m2e = (p1 − p4 )2 − m2e = m2γ − 2Emγ
Physics 217
Quantum Field Theory I
Fall 2015
√
Then, taking E =
s
2
the total cross section can be written as
2
4παdark
σ=
s
s
s
s − 4m2γ
2m2e (s − 2m2γ ) 2
√
[1
+
(1
−
)]
s − 4m2e
(m2γ − smγ )2
2m2
s − 4m2γ
2m2e s(1 − s γ )
=
[1 + (1 − m2
)]
s
s − 4m2e
s( √γs − 2mγ )2
√
And if approximation above ( s me > mγ ) is made we find
2
4παdark
σ≈
2
2
2m2γ
me 4
4παdark
me 4
4παdark
[2 + (
[2 + (
) (1 − 2 )] ≈
)]
s
2mγ
me
s
2mγ
After attempting to find suitable combinations of αdark and mγ with a
variety of reasonable center of mass energies for the given total cross sections,
my attempts came at no avail, probably because I over simplified the problem
and made an error (two?) on the way.