University of California at San Diego – Department of Physics – Prof. John McGreevy
Quantum Field Theory C (215C) Spring 2015
Assignment 6 – Solutions
Posted May 19, 2015
Due 11am, Thursday, May 28, 2015
Problem Set 6
1. Matching with massive electrons. [from Iain Stewart]
Consider QED in the regime where the photon momenta q µ are much smaller than
the electron mass me . In this regime, we can integrate out the electron and write an
effective field theory involving only the photon.
(a) Calculate the QED one-loop vacuum polarization diagram using dimensional
2
regularization, in the MS scheme. Expand Π(q 2 ) through first order in mq 2 .
e
We did this calculation in lecture and found
2 µν
δΠµν
− q µ q µ δΠ2 (q 2 )
2 (q) = q g
with
2 MS
δΠ2 (q )
2α
=
π
Z
1
dx x(1 − x) log
0
m2 − x(1 − x)q 2
µ2
.
To facilitate the expansion in q 2 , write
m2
q2
m2 − x(1 − x)q 2
= log 2 + log 1 − x(1 − x) 2 .
log
µ2
µ
µ
P
un
Taylor expanding the log log(1 − u) = − ∞
n=1 n gives
4
α
m2
α q2
q
2 MS
log 2 −
+
O
δΠ2 (q ) =
3π
µ
15π m2
m4
since
Z
0
1
1
dx x(1 − x) = ,
6
Z
0
1
dx x2 (1 − x)2 =
1
.
30
(b) Write down a Lagrangian involving only the photon field operator that reproduces the first two terms in the expansion. (Hint: it should be gauge invariant
and Lorentz invariant. There is essentially (up to integration by parts) only one
addition to the Maxwell term). Use the calculation above to match between
the photon-only EFT and QED at µ = me , at this order in the fine structure
constant α.
1
The leading term is the correction to the Maxwell term. The order-q 2 bit is
produced by an operator involving two gauge fields and two extra derivatives.
1
c2 1
δL = c1 Fµν F µν + 2 Fµν ∂ρ ∂ ρ F µν + ...
4
m 4
Other gauge-invariant dimension six operators made of two As either are related
by integration by parts ( ∂ρ Fµν ∂ ρ F µν ) or break parity ( (?F )µν ∂ρ ∂ ρ F µν ) and
are therefore not produced by QED, which is parity invariant. Matching to our
expression above gives:
c1 = δΠ2 (0), c2 = m2 ∂q2 δΠ2 (0).
(c) What symmetry of QED forbids dimension-6 operators involving three field
strengths?
Charge conjugation invariance acts by Fµν → −Fµν , and j µ → −j µ , where j is
the charge current. In the absence of any external charges, it acts like (−1)n
where n is the number of external photons in the amplitude. This is called
‘Furry’s theorem’.
(d) At dimension 8, there are operators in the photons-only EFT which describe
light-by-light scattering. Write them down (there are two). Draw the QED
Feynman diagram which matches to these terms, and determine the number
of factors of α in their coefficients. (Don’t do the integrals unless you find it
enjoyable.)
They are going to involve four F s and therefore go like 1/m4 . A single loop
of an electron with four photon lines coming off (a box diagram) has four vertices, which therefore goes like e4 ∼ α2 . We must decide what to do with the
indices on the F s. Parity demands that an even number of the F smust be
2 ?F s
µν 2
~ ·B
~ and
(can be zero). This leaves two terms, which are (Fµν (?F ) ) ∼ E
2
(Fµν F µν )2 ∼ (E 2 − B 2 ) . For the actual matching coefficients, see ItzyksonZuber.
(e) [bonus] Use dimensional analysis in the low-energy EFT to estimate the size of
the γγ → γγ cross section.
The amplitude goes like
p4
A ∼ α2 4
m
with one factor of p from each F in the interaction. So the cross section is
σ(p) ∼
It’s pretty small.
2
1
α 4 p6
2
|A|
∼
.
p2
m8
2. Consequences of unitarity, the second coming. [From Aneesh Manohar]
As foretold on problem set 4, this problem is returning from the past.
A general statement of the optical theorem is:
XZ
−i (M(a → b) − M(b → a)) =
dΦf M? (b → f )M(a → f ) .
f
Consider QED with electrons and muons.
(a) Consider scattering of an electron (e− ) and a positron (e+ ) into e− e+ (so a = b
in the notation above). We wish to consider the contribution to the imaginary
part of the amplitude for this process which is proportional to Q2e Q2µ where Qe
and Qµ are the electric charges of the electron and muon (which are in fact
numerically equal but never mind that). Draw the relevant Feynman diagram.
(a) Compute the imaginary part of this amplitude (just the Q2e Q2µ bit) as a function
of s ≡ (k1 + k2 )2 where k1,2 are the momenta of the incoming e+ and e− (so that
k1 + k2 is the 4-momentum of the intermediate photons).
We can focus on the part of the amplitude proportional to δΠ2 (s),
as defined in the lecture notes. The only difference from the discussion in the
lecture notes is that m there should be replaced by mµ , and e2 replaced by Q2µ :
2
Z
Q2µ 1
mµ − x(1 − x)s
(MS)
2
Π2 (s = (k1 + k2 ) ) = 2
dxx(1 − x) log
.
2π 0
µ2
A first satisfying observation is that the argument of the logarithm is
m2µ − x(1 − x)s
which (since x(1 − x) ∈ [0, 41 ] for x ∈ [0, 1]) can become negative only if s =
(k1 + k2 )2 ≥ (2mµ )2 , which is the threshold for creating a muon-antimuon pair
(In the COM frame of the initial e+ e− , we must have E > 2mµ .).
(b) Use the optical theorem and the fact that the total cross section for e+ e− →
µ+ µ− must be positive
σ(e+ e− → µ+ µ− ) ≥ 0
to show that a Feynman diagram with a fermion loop must come with a minus
sign.
If we hadn’t included the minus sign in the fermion loop, it would switch the sign
of ImM and hence the sign of σ inferred from the optical theorem. According
3
to Prof. Manohar, this was the method by which Feynman discovered that this
minus sign was required by his rules. Yet another lesson in the value of indirect
checks.
3. Right-handed neutrinos. [from Iain Stewart, and hep-ph/0210271]
Consider adding a right-handed singlet (under all gauge groups) neutrino NR to the
Standard Model. It may have a majorana mass M ; and it may have a coupling gν to
leptons, so that all the dimension ≤ 4 operators are
M
M
LN = N̄R i/
∂ NR − N̄Rc NR − N̄R NRc + gν N̄R HiT Lj ij + h.c.
2
2
T
where NRc = C N̄R is the the charge conjugate field, C = iγ2 γ0 (in the Dirac representation), H is the Higgs doublet, L is the left-handed lepton doublet, containing
νL and eL . Take the mass M to be large compared to the electroweak scale. Integrate
out the right-handed neutrinos at tree level. [Hint: you may find it useful to work in
terms of the Majorana field
N ≡ NR + NRc
which satisfies N = N c .]
Show that the leading term in the expansion in 1/M is a dimension-5 operator made
of Standard Model fields. Explain the consequences of this operator for neutrino
physics, assuming a vacuum expectation value for the Higgs field.
In terms of N , the lagrangian is
1
∂ − M ) N + gν N̄ Hi Lj ij + gν N̄ Hi? Lcj ij .
LN = N̄ (i/
2
The equation of motion for N (from varying N̄ ) is
(i/
∂ − M ) N = −gν Hi Lj + Hi? Lcj ij
which gives
1
1
gν (Hk L` + Hk? Lc` ) k` .
LN | = − gν L̄cj Hi + L̄j Hi? ij
2
i/
∂−M
As for our discussion of W -bosons, we expand this in powers of 1/M to get a local
effective field theory. The leading term is
O(5) =
gν2 c
L̄ Hi ij L` Hk k` + h.c.
M j
Plugging in hHi =
6 0, this is a neutrino mass.
Place a bound on M assuming that the observed neutrinos have masses mν < 0.5 eV.
2
c5 v
In terms of the parameterization from lecture, mν = 2Λ
. This gives Λnew ≥
new
14
14
10 GeV for c5 ∼ 1. We find Λnew /c5 ∼ M , so M ≥ 10 GeV.
4