University of California at San Diego – Department of Physics – Prof. John McGreevy
Quantum Field Theory C (215C) Spring 2015
Assignment 4 – Partial Solutions
Posted April 30, 2015
Due 11am, Tuesday, May 12, 2015
Problem Set 4
1. Spectral representation at finite temperature.
In lecture we (will) have derived a spectral representation for the two-point function
of a scalar operator in the vacuum state
iD(x) = h0|T O(x)O† (0)|0i
Derive a spectral representation for the two-point function of a scalar operator at a
nonzero temperature:
iDβ (x) ≡ tr
e−βH
1 X −βEn
T O(x)O† (0) =
e
hn|T O(x)O† (0)|ni.
Zβ
Zβ n
Here Zβ ≡ tre−βH is the thermal partition function. Check that the zero temperature
(β → ∞) limit reproduces our previous result.
The idea is again to insert a resolution of the identity in between the operators. All the
steps as for the vacuum correlators go through, the only difference being that instead
of arriving at a sum of squares of matrix elements of the operator between the vacuum
and an arbitrary state, we get matrix elements between pairs of states:
X
X
iD(x) = Zβ−1
e−βEn
|| Onm ||2 eix(pn −pm ) θ(t) + eix(pm −pn ) θ(−t) .
n
m
2. Cutting rules.
Work through the discussion of unitarity cuts for φ2 → φ2 scattering in φ4 theory at
one loop on pages 233-235 of Peskin and Schroeder.
3. Consequences of unitarity. [From Aneesh Manohar]
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.
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(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.
Although this problem is relevant right now as an exercise in studying singularities
and imaginary bits of amplitudes from real intermediate states, the remaining parts of
this problem will be easier in a couple of weeks (i.e. some of the ingredients will be
discussed explicitly in Chapter 4 of the lecture notes). I will re-post this problem on a
later problem set and you can hand it in then.
(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− .
(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.
4. Practice with cutting rules. [This problem is optional and not easy.]
[roughly, this is Zee problem III.8.2] Consider the massive scalar field theory in four
dimensions with 3!g φ3 interaction. Show that to order g 4 there is a ‘box diagram’
contributing to φφ → φφ scattering with amplitude
Z
d¯4 k
4
I=g
(k 2 − m2 )((k + p2 )2 − m2 )((k − p1 )2 − m2 )((k + p2 − p3 )2 − m2 )
(where m2 means m2 − i).
Here the momenta of the φ particles are specified to be p1 + p2 → p3 + p4 . This integral
is a function of s = (p1 + p2 )2 and t = (p3 − p2 )2 . Study the analyticity properties of
I(s) at fixed t – there should be a cut starting at s = (2m)2 . Evaluate the discontinuity
of I across the cut (recall from complex analysis that for a cut of a real function on
the real axis, this discontinuity is directly proportional to the imaginary part of the
function) and verify Cutkosky’s cutting rule. Check that optical theorem works.
Hint: The problem simplifies a bit if we consider instead a theory of three scalars φ, η, ξ
with only a gφηξ coupling. Consider the case where mη mξ (e.g. set mξ = 0) and
study φφ → φφ scattering. This reduces the number of cuts which can contribute an
imaginary part.
Bonus material: Think about the dependence of I(t) on t at fixed s.
In assigning this problem, I regret not making it clearer that these box integrals cannot
be written in terms of elementary functions. Let’s just put it behind us.
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In the next problems, we will study free massless bosons in two dimensions. This system
has many physical applications – e.g. in string theory, and at the edge of quantum Hall
systems. It is an example of a conformal field theory. It is a field theory of the kind I have
advertised several times, where the excitations are not particles.
5. There are no Goldstone bosons in two dimensions.
[Perhaps this is more of a diatribe than a problem.]
(a) Consider a massless scalar X in 2d, with action
Z
1
d2 σ∂a X∂ a X.
S[X] = −
4π
(1)
Show that the euclidean Green function G2 satisfies
∇2 G2 (z, z 0 ) = −2πδ 2 (z − z 0 )
(2)
(z = σ1E + iσ2E )1 and is given by
1
ln |z − z 0 |,
2π
G2 (z, z 0 ) =
for example by Fourier transform.
Translation invariance says
0
0
Z
G2 (σ, σ ) = G2 (σ − σ ) =
0
d¯D k eik(σ−σ ) G̃(k)
and (2) gives
−k 2 G̃(k) = −
1
.
2π
This means the massless Green’s function is
(
Z
cD
, D 6= 2
1 ikσ
D
|σ|D−2
GD (σ) = d¯ k
e
=
2πk 2
− log |σ|, D = 2
.
[I wasnt careful about the factors of −1 and π in doing the Fourier transform, but the
final coefficients can be checked by taking box of both sides and comparing to (2).]
Note that the bad IR (large σ) behavior of the Greens function gets even worse in
d < 2.
(b) The long-distance behavior of G2 has important implications for the physics of
massless scalars in two dimensions. Thinking of G2 as the two point function of the
massless scalar
G2 (z, z 0 ) = hX(z)X(z 0 )i
let’s ask the following question:
1
Recall Schwinger-Dyson equations from problem set 2.
3
There is no potential energy for the field X in (1). Someone used to physics in 3+1
dimensions might think that this means that there is a vacuum for every value of X.
Let’s try to fix the expectation value of the scalar hXi = x and see what happens.
Perturb the putative vacuum |xi a little bit at the position z by inserting the operator
X there. To measure what happens, insert the operator X at z 0 . The correlator G2 can
thus be interpreted as a measurement of how the effects of our perturbation fall off with
distance. What happens? Contrast this with the behavior you would see for a scalar
field with a flat potential in more than two dimensions (recall problem 1 of problem
set 2). Note that the case of (0+1) dimensional QFT (i.e. quantum mechanics) is even
more problematic in the infrared.
One way to arrive at an action like (1) is if the field X arises as a Goldstone boson
associated with a symmetry X → X + a, which would be broken by fixing the vacuum
|xi. Then it is guaranteed by Goldstone’s theorem that the action can only depend on
derivatives of X.
But note that the Goldstone-ness of the massless bosons (i.e. whether they are massless
because of a broken symmetry) is not crucial for this discussion. In non-supersymmetric
sigma models, one expects massless bosons whose masslessness is not protected by a
symmetry to be lifted quantumly, but in supersymmetric theories, this need not be the
case, and different points on the space of minima of the potential need not be related
by a symmetry.
This result is called the Coleman-Mermin-Wagner (sometimes Hohenberg, too) Theorem. Coleman’s paper on the subject is S. Coleman, “There are no Goldstone bosons
in two-dimensions,” Commun. Math. Phys. 31:259-264 (1973).
A familiar fact is that non-relativistic QM of a particle in a potential with degenerate
minima does not have degenerate vacua. Tunneling between the minima leads to
mixing; the true vacuum is some superposition. If the potential minima are related by
a symmetry, the vacuum does not spontaneously break the symmetry, it is a symmetric
superposition. A less extreme version of this in QM, which also wont surprise you, is
that for a particle with no potential, a localized wavefunction will spread out, and the
ground state is uniform. Though these effects arent usually described in these terms,
the long-distance growth of the d ≤ 2 propagator we saw in part (a) is a direct signal
of them. And the fact that ln |σ − σ 0 | grows at large separation means that a similar
thing happens in two dimensions: if we try to localize the vev of X, it will be washed
out by fluctuations, and the ground state wavefunctional for the free boson is really
a sum over values of X. This is exactly what we accomplish by integrating over the
target space of the boson in the path integral.
If the boson field space is not swept out by the action of some symmetry (for example
if the kinetic terms do not respect the symmetry, but nevertheless the boson remains
flat quantum mechanically, e.g. because of supersymmetry), it just means that this
sum is not weighted with the uniform distribution. In conclusion, 2d massless bosons
exist and can be quite sensible and interesting, they just aren’t Goldstones ever.
6. Correlators of composite operators made of free bosons in 1+1 dimensions.
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Consider a collection of n two-dimensional free bosons X µ governed by the action
Z
1
S=−
d2 σ∂a Xµ ∂ a X µ .
4πg
[The coupling g can be absorbed into the definition of X if we prefer, but it is useful to
leave this coupling constant arbitrary since different physicists use different conventions
for the normalization and as you will see this affects the appearance of the final answer.]
Until further notice, we will assume that X takes values on the real line.
(a) Rotate eiS to Euclidean space (d2 σ = −i(d2 σ)E ) and compute the Euclidean generating functional
Z
R 2
R 2
µ
(d σ)E J µ Xµ
−1
Z[J] = he
i ≡ Z0
[dX]eiS e (d σ)E J Xµ
(where Z0−1 ≡ Z[J = 0] but please don’t worry too much about the normalization of
the path integral).
[Hint: use the Green function from the previous problem, and Wick’s theorem. Or use
our general formula for Gaussian integrals with sources.]
[Warning: In the problem at hand, even the euclidean kinetic operator has a kernel,
namely the zero-momentum mode. You will need to do this integral separately.]
[Cultural remark 1: this field theory describes the propagation of featureless strings in
n-dimensional flat space IRn – think of X µ (σ) as the parametrizing the position in IRn
to which the point σ is mapped.
Cultural remark 2: this is an example of a conformal field theory. In particular recall
that massless scalars in D = 2 have engineering dimension zero.]
In euclidean space,
1
S=−
4πg
Z
(d2 σ)E (−i) (−∂a Xµ ∂ a X µ ) .
So the Boltzmann factor in the euclidean space path integral is
Z
1
iS
−SE
e =e
, SE =
(d2 σ)E ∂a X · ∂ a X .
4πg
The generating functional is
R
ZE [J] = he
1
= e2
(d2 σ)E Jµ X µ
R
i=
Z0−1
Z
[DX]e
R
(d2 σ1 )E (d2 σ2 )E J µ (σ1 )( 2πg
2 )
∂
R
σ1 ,σ2
1
(d2 σ)E ( 4πg
∂X·∂X )+J·X
Jµ (σ2 )
The Green function from the previous problem (with the extra factor of g) is
2πg
G(σ1 , σ2 ) ≡
= −g log |σ1 − σ2 | = −g log |z1 − z2 |.
∂ 2 σ1 ,σ2
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This is the inverse of ∂ 2 on the complement of its kernel. Ignoring the zeromode of X
we find
R 2 R 2
1
µ
?
Z[J] = e 2 g d z1 d z2 J (z1 ,z̄1 ) log |z1 −z2 |Jµ (z2 ,z̄2 ) .
The zero-momentum mode of X
µ
Z
d2 σ X µ (σ)
x ≡
does not appear in the action, but we still have to do the integral over it. The correct
answer is then:
Z
R 2 R 2
1
µ
Z[J] = dxµ e 2 g d z1 d z2 J (z1 ,z̄1 ) log |z1 −z2 |Jµ (z2 ,z̄2 ) .
(b) Show that
h
N
Y
: e−i
√
!
2α0 ki ·X(σ (i) )
X
:i = δ n
i=1
kiµ
i
N
Y
0
|zi − zj |−α gki ·kj
(3)
i,j=1
(i)
(i)
where σ (i) label points in 2d Euclidean space, zi ≡ σ1 + iσ2 , α0 is a parameter with
dimensions of [X 2 /g] (called the ‘Regge slope’), and kiµ are a set of arbitrary n-vectors
in the target space. The : ... : indicate the following prescription for defining composite
operators. The prescription is simply to leave out Wick contractions of objects within
a pair of : ... :. Give a symmetry explanation of the delta function in k.
[Cultural remark: this calculation is the central ingredient in the Veneziano amplitude
for scattering of bosonic strings at tree level.]
Simply choose the source from the previous problem to be
√
µ
J = −i
2α0
N
X
kiµ δ 2 (σ − σi ) .
i=1
The zeromode xµ only appears in the J · X term. The nonzero-mode integrals give
e
1
g
2
R
d2 z1
R
d2 z2 J µ (z1 ,z̄1 ) log |z1 −z2 |Jµ (z2 ,z̄2 )
α0 g
=e
PN
i6=j=1
ki ·kj log |zi −zj |
=
N
Y
0
|zi − zj |α gki ·kj .
i6=j
The zeromode integrals are
Z Y
n
X µ
µ
dxµ eiki xµ = δ n (
ki ).
µ=1
i
Why did this happen? The field theory has a symmetry under the shifts X µ 7→
X µ + constµ , a translation invariance in the target space. The resulting conserved
charge is a target-space momentum, k units of which is injected into the correlator by
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the operator eikX . The fact that the charge is conserved (and not eaten by the vacuum)
means that there must be a law saying that amplitudes predicting non-conservation
of
P µ
n
the charge should vanish – the amplitudes must be proportional to δ ( i ki ).
What is the interpretation of this charge? The model we are describing is analogous
to the 0 + 1-dimensional model with action
Z
1
i i
mẊ Ẋ
S = dt
2
– the worldline description of a free non-relativistic particle in n dimensions. Instead
of describing the propagation of a zero-dimensional object through spacetime, the
model in this problem describes the propagation of a one-dimensional object, a string.
Instead of a worldline, its image is a worldsheet. The analog of the (NR) mass m
1
is the coefficient of the kinetic term, which we called 4πα
0 above – this is the string
tension. In the 0 + 1d particle model, the symmetry X i → X i + i is just translation
(i)
invariance, and the conserved quantity is j0 = m∂Rt X i , the momentum. In the 1 + 1d
1
∂α φj α so the conserved current is
string model, the Noether method gives δS = 2πα
0
1
α
j α = 2πα
0 ∂ φ, and the conserved charge is
Z
Z
Z
1
1
(i)
(i)
1
i
1 i
jt =
Q =
dσ ∂σ0 X = ∂t
dσ X .
2πα0
2πα0
space
R
The quantity dσ 1 X i is just the location of the center-of-mass of the string. The
1
(i)
is just the
mass of the string is 2πα
0 times the length. So the conserved quantity Q
center-of-mass momentum.
(c) Conclude that the composite operator Oa ≡: eiaX : has scaling dimension ∆a =
in the sense that
1
hOa (z)Ob† (0)i = δ(a − b) 2∆a .
|z|
ga2
,
2
Notice that the correlation functions of these operators do not describe the propagation
of particles in any sense. The operator O produces some power-law excitation of the
CFT soup.
Just set N = 2 in the previous result.
(d) Suppose we have one field (n = 1) X which takes values on the circle, that is, we
identify
X ' X + 2πR .
What values of a label single-valued operators : eiaX : ? How should we modify (3)?
Just like in QM on a circle, the momentum is quantized:
!
eia(X+2πR) = eiaX =⇒ a =
n
, n ∈ ZZ.
R
The only change in (3) is that the Dirac delta becomes a Kronecker delta.
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