# Quantum Field Theory C (215C) Spring 2015 Assignment 3 – Solutions ```University of California at San Diego – Department of Physics – Prof. John McGreevy
Quantum Field Theory C (215C) Spring 2015
Assignment 3 – Solutions
Posted April 21, 2015
Due 11am, Thursday, April 30, 2015
Relevant reading: Zee, Chapter IV.3. Zee, Chapter III.3 also overlaps with recent lecture
material.
Problem Set 3
1. Propagator corrections in a solvable field theory.
Consider a theory of a scalar field in D dimensions with action
S = S0 + S1
where
Z
S0 =
dD x
and
1
∂&micro; φ∂ &micro; φ − m20 φ2
2
Z
1
dD x δm2 φ2 .
2
We have artificially decomposed the mass term into two parts. We will do perturbation
theory in small δm2 , treating S1 as an ‘interaction’ term. We wish to show that the
organization of perturbation theory that we’ve seen lecture will correctly reassemble
the mass term.
S1 = −
(a) Write down all the Feynman rules for this perturbation theory.
The propagator is the usual one, with amplitude p2 −mi 2 +i . The only ‘vertex’ is a
2-point vertex, which comes with amplitude −iδm2 .
(b) Determine the 1PI two-point function in this model, defined by
X
iΣ ≡
(all 1PI diagrams with two nubbins) .
The only feynman diagrams are concatenations of free propagators and the mass
insertion. The only 1PI diagram is the single mass insertion (with two nubbins
sticking off).
iΣ = −iδm2 .
1
(c) Show that the (geometric) summation of the propagator corrections correctly
produces the propagator that you would have used had we not split up m20 + δm2 .
The geometric series for the two-point function results in
1
G(p) = G0 (p) + G0 (p)iΣG0 (p) + G0 (p)iΣG0 (p)iΣG0 (p) + ... = G0 (p)
1 − iΣG0 (p)
!
1
i
= 2
2
p − m0 + i 1 − iΣ p2 −mi 2 +i
0
i
i
= 2
.
= 2
2
2
p − m0 + Σ + i
p − m0 − δm2 + i
2. Coleman-Weinberg potential.
(a) [Zee problem IV.3.4] What set of Feynman diagrams are summed by the ColemanWeinberg calculation?
[Hint: expand the logarithm as a series in V 00 /k 2 and associate a Feynman diagram
with each term.]
Recall that this calculation was the order-~ contribution to the saddle-point expansion of
Z
2
[Dφ]ei((∂φ) −V (φ)+Jφ)
about φ = φc + ϕ. They are diagrams with a single loop of the fluctuation ϕ, no
external ϕ legs, and arbitrary numbers of φc insertions. Then one perspective is to
use the Feynman rules for the expansion φ = φc + ϕ, thinking of the potential as
a perturbation. For simplicity consider the case where V (φ) = φp is a monomial.
The CW potential comes from just one-loop diagrams, so vertices for ϕ in the
expansion of V (φ = φc + ϕ) of order higher than two do not contribute – they
would turn the one ϕ running in the loop into more than one, and then we would
need an extra loop to get rid of it.
In the figure, we show the example with V (φ) = φ4 , so that the ‘interaction
vertex’ for the fluctuations ϕ, V 00 (φc ) is quadratic in φc . This is an interaction
vertex in the same sense as in the previous problem – it’s just an insertion in
the propagator. The one-loop diagram with n insertions of the propagator has a
2
cyclic symmetry which rotates the diagram like a clock; this group has order n
and so the answer is of the form
n
Z
1 V 00 (φc )
D
.
d&macr; k
n k 2 + i
(Actually the diagram also has a reflection symmetry, since ϕ is real; this produces
an extra overall factor of 12 .) A little bit more explicitly, the propagators for the
fluctuations are
−i
hϕ(k)ϕ(−k)i = 2
k
00
and the 2-point vertex is −iV (φc ). The solid lines are ϕ propagators, and the
dotted lines represent insertions of φc .
(b) [Zee problem IV.3.3] Consider a massless fermion field coupled to a scalar field φ
by a coupling gφψ̄ψ in D = 1 + 1. Show that the one loop effective potential that
results from integrating out the fluctuations of the fermion has the form
2
φ
1
2
(gφ) log
VF =
2π
M2
In this problem, the terms in the boson action which do not involve the fermions
just go along for the ride. The fermion path integral is explained in Zee &sect;IV.
Notice that the fermion mass is m(φ) ≡ m + gφ. The two important differences
from the boson case are:
• the determinant of the kinetic operator ends up in the numerator
Z
dψdψ̄ eiψ̄Dψ = det D = e+tr log D
R
• the trace over single particle states, which previously was tr... = d&macr;D khk|...|ki
now also involves a trace over the spin space:
Z
X
tr log D = d&macr;D k
hk|Dαα (k)|ki.
α
It is still diagonal in momentum space. Zee explains a trick for getting rid of
the gamma matrices, which I will not repeat. One reason we know they must go
away is that they have a vector index and we are computing a rotation-invariant
quantity which doesn’t depend on any vectors. This gives
2
Z
k + m(φ)2
D
.
VF = d&macr; k log
k2
As in lecture, I added a (cutoff dependent) constant to eliminate the contribution to the vacuum energy (it also makes Mathematica happier about doing the
integral). We can regulate this integral by euclidean cutoff. That gives
2
Z Λ
kE + m(φ)2
D
Λ
VF =
d&macr; kE log
kE2
3
Z Λ2
2π
u + m2
1
=
du
log
(2π)2 0 2
u
1
−m2 log m2 − Λ2 log Λ2 + (m2 + Λ2 ) log(m2 + Λ2
=
2π
(1)
Don’t forget that m = m(φ) = mψ + gφ here. Interesting effects occur if we set
mψ = 0 so that setting φ = 0 makes the fermion massless. We shouldn’t think
about values of φ that are close to the cutoff, so we may assume φ Λ in which
case the 2nd and 3rd terms in (??) may be expanded as
φΛ
−Λ2 log Λ2 + (m2 + Λ2 ) log(m2 + Λ2 ' m2 1 + log Λ2 .
Since m = gφ, to get a cutoff-independent effective action for φ, we must add a
counterterm for the boson mass, δL = B(Λ)φ2 . The problem would have been
better-posed had I specified what you can measure – i.e. if I had told you what
renormalization conditions to impose – rather than the answer, which determines
a family of renormalization conditions with hindsight. A good example of a reasonable RC which gives the stated answer is analogous to our quartic condition
in four dimensions:
!
00
Veff
(φc )|φc =M 2 = &micro;2M .
This demands that
&micro;2M = 2B(Λ) + VF00 (φ = M ) = 2B − 6 − 2g 2 log(g 2 M 2 ) + 2g 2 1 + log Λ2
so
1
B(Λ) =
2
g2M 2
2
2
&micro;M + 6 + 2g log
.
Λ2
The important leftover φ dependent terms are
Veff (φ) =
φ2
1 2 2
g φ log 2 .
2π
M
3. 2PI generating functional.
In this problem, we will consider 0 + 0-dimensional φ4 theory for definiteness, but the
discussion also works with labels. Parametrize the integral as
Z ∞
g 4 1
1 2 2
2
Z[K] ≡
dq e− 2 m q − 4! q − 2 Kq ≡ e−W [K] ;
−∞
we will regard K as a source for the two-point function.
The 2PI generating functional is defined by the Legendre transform
Γ[G] ≡ W [K] − K∂K W [K]
with K = K(G) determined by G = ∂K W [K]. Here G is the 2-point Green’s function
G ≡ hq 2 iK .
4
(a) Show from the definition that Γ satisfies the Dyson equation
1
∂G Γ[G] = − K.
2
(b) Show that the perturbative expansion of Γ can be written as
A
Γ[G] =
1
1
log G−1 + m2 G + γ2P I + const.
2
2
(The A over the equal sign means that this is a perturbative statement.) Here
γ2P I is defined as everything beyond one-loop order in the perturbation expansion.
(c) Show that −γ2P I is the sum of connected two-particle-irreducible (2PI) vacuum
graphs, where 2PI means that the diagram cannot be disconnected by cutting any
two propagators.
(d) Define the 1PI self-energy in this Euclidean setting as
X
A
−Σ =
(1PI diagrams with two nubbins) .
Show that
∂γ2P I
∂K
by examining the diagrams on the BHS.
Taking a 2PI diagram and cutting a single line produces a 1PI diagram. Differentiating with respect to K produces a sum of ways of cutting a single line.
A
Σ=
(e) Show that the expression for G that results from the Dyson equation
G−1 = K + 2∂G γ2P I
agrees with our summation of 1PI diagrams.
(f) I may add another part to this problem which shows why 2PI effective actions
are useful.
See this paper and this paper.
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