Master Dissertation

Master Dissertation
The Method of Epstein and Glaser
Author: Asger Jacobsen Supervisor: J.P. Solovej
21st August 2005 Temp. version

Contents
1 Introductory Theory 6
1.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2 Formal Power Series . . . . . . . . . . . . . . . . . . . . . . . 8
1.3 Affine Transformations of Distributions . . . . . . . . . . . . 12
2 The Poincaré Group 14
2.1 The Lorentz Group . . . . . . . . . . . . . . . . . . . . . . . . 14
2.2 The Poincaré Group . . . . . . . . . . . . . . . . . . . . . . . 16
2.3 Spinor Representations of the Lorentz Group . . . . . . . . . 17
3 The Scattering Matrix 19
4 The Mathematical Setting of QFT 22
4.1 The Fock Space . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4.2 The Wightman Axioms . . . . . . . . . . . . . . . . . . . . . 25
5 The Method of Epstein and Glaser 27
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
5.2 Example - In the Hilbert Space Setting of Quantum Mechanics 36
5.3 Splitting of Distributions . . . . . . . . . . . . . . . . . . . . . 38
6 Regularly Varying Functions 41
7 Splitting of Numerical Distributions 49
7.1 The Singular Order of a Distribution . . . . . . . . . . . . . . 49
7.2 Case I: Negative Singular Order . . . . . . . . . . . . . . . . . 54
7.2.1 Existence . . . . . . . . . . . . . . . . . . . . . . . . . 54
7.2.2 Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . 59
7.3 Case II: Positive Singular Order . . . . . . . . . . . . . . . . . 64
8 Application to QED 66
8.1 Using the Game Plan . . . . . . . . . . . . . . . . . . . . . . 66
8.2 The Adiabatic Limit . . . . . . . . . . . . . . . . . . . . . . . 69
ii

9 The Microlocal Approach - A Condition on the Wave Front
Set. 71
iii

Abstract
The main purpose of this thesis is to prove that the causality axiom allows
a well-defined perturbation theory of quantum electrodynamics. This
will be done by using the method of Epstein and Glaser. The thesis introduces
the theory of formal power series and regularly varying functions, both
tools needed in the main proofs. Basic theory of the scattering matrix, the
Poincaré Group and its representations will be introduced. We will use the
concept of singular order to investigate the splitting of distributions into an
advanced and a retarded part and show when they exist and when they are
unique. Applications and the adiabatic limit will be discussed. The thesis
is concluded with a consideration of the microlocal approach to the method
of Epstein and Glaser and by showing that the translations invariance can
be substituted by a condition on the wave front set.
Sammenføjning
Hovedform˚alet med dette speciale er at vise at man ved hjælp af kausalitetsaxiomet
kan indføre en veldefineret pertubationsteori for kvanteelektrodynamik.
Dette gøres ved hjælp af Epstein og Glasers metode. Specialet introducerer
teorierne for formelle potensrækker og regulært varierende funktioner,
som begge skal bruges under hovedbeviserne. Basal teori for spredningsmatrice,
Poincaré gruppen og dens repræsentationer vil blive introducerede.
Vi vil bruge singulær orden til at undersøge opdelingen af distributioner
i en fremtids og fortids del og vise, hvorn˚ar de eksisterer, og hvorn˚ar
de er entydige. Anvendelser og den adiabatiske grænse vil blive diskuteret.
Specialet afsluttes med at overveje den mikrolokale tilgangsvinkel til Epstein
og Glasers metode og med at vise at translations varians kan afløses af en
betingelse p˚a bølgefront mængden. Specialet er skrevet p˚a engelsk.
1

Preface
Overview
This dissertation is about the causal approach to finite quantum electrodynamics
(QED). The word causal refers to the main assumption on which the
method we will introduce is based. In short it is the statement that what
happens at time t > s doesn’t influence what happens at earlier times t < s.
The word finite means that there do not arise any ultraviolet divergences.
The traditional formalism of quantum field theory (QFT) is known to
have problems with ultraviolet divergences. These occur because of misuse
of the mathematics. The problem lies in the fact that QFT deals with distributions
instead of functions and that the rules of calculus for these are
not the same. One cannot simply multiply a distribution by a discontinuous
step-function and other simple rules of calculus like multiplication of distributions
can be a very complicated matter. In fact, this is the main problem
of [4].
In practice the ultraviolet divergencies are dealt with by regularization
and renormalization FIXME: explain these. But instead of using these firstaid
features on an ill-defined theory one could introduce QFT using the
mathematics in the right way from the beginning. This is the basic aim of
this dissertation.
Structure
Chapter 1
In Chapter 1 I introduce the tools needed in our later work. Section 1.2
is about formal power series which is an abstraction of the usual power
series used in calculus. A generalization which lets us use most of the wellknown
machinery from calculus to settings which have no natural notion
of convergence. This will become useful later on when we introduce the
scattering matrix as a formal power series.
In section 1.3 we will introduce Affine Transformations of distributions.
2

Chapter 2
The Poincaré Group will be introduced in Sections 2.1 and 2.2 in Chapter 2.
Further, in Section 2.3 the spinor representation of the Lorentz Group will
be presented.
Chapter 3
In Chapter 3 we will take a look at the key player of this dissertation: The
Scattering Matrix. In this chapter we introduce it in the well-known setting
of quantum mechanics.
Chapter 4
In Chapter 4 we consider the setting of QFT. In Section 4.1 we introduce
the Fock space and in Section 4.2 we take a look at the Wightman axioms.
Chapter 5
In Chapter 5 we will finally introduce the method of Epstein and Glaser.
Section 5.1 is about the main ideas and a game plan is sketched. In Section
5.2 the method is applied to the of quantum mechanics which is based
on the well-known theory of Hilbert spaces. In Section 5.3 we will look at
the splitting of distributions into an advanced and a retarded part suggested
in the game plan. We focus on the support properties.
Chapter 6
In Chapter 6 we will leave the theory for a while introducing an important
tool for the splitting of numerical distributions in Chapter 7.
Chapter 7
In Section 7.1 of this chapter we develop some tools for studying the splitting
of numerical distributions. The notion of singular order of a distribution will
be a tool to categorize the type of splitting. Thus we will in Sections 7.2
and 7.3 look at the splitting of of distributions of positive and negative
singular order, separately. We will find out that the splitting may be done
by multiplication of a step-function (this will be done in a technical manner)
in the case of negative order. Further in this case the splitting is unique,
which will be the topic of Subsection 7.2.2. The splitting in the case of
positive singular order turns out to be a more complicated matter and only
works for a special class of distributions. Further the splitting is not unique.
3

Chapter 8
In Chapter 8 we will turn our attention to applications. In Section 8.1 we
find an expression for the difference distribution which is essential in the
splitting process. The expression can be considered a correspondent to the
Feynman Rules. We find the the first term describes photon exchange.
In Section 8.2 we will give an idea of how to take the adiabatic limit.
Chapter 9
In this chapter we consider the microlocal approach to the method of Epstein
and Glaser. The idea is to generalize the theory to manifolds. As
translations are substituted by parallel transport on manifolds the property
of translational invariance has to be substituted by a condition on the
smoothness. We find a condition on the wave front set which makes the
terms of the scattering matrix well-defined.
Thanks
I would like to thank my supervisor Jan Philip Solovej for his help and
support during the writing of this project. Further I would like the numerous
people who’s work and research this dissertation is based on; especially
Scharf, Fredenhagen, and of course Epstein and Glaser. Finally, I will also
thank my family who has had to bear over with my during the final work
under time pressure.
4

Prerequisites
Fixme: Reevaluate the needs as the theory develops.
This is a dissertation for the masters degree in mathematics at the University
of Copenhagen. It is thus the conclusion of 5 years of study. The
thesis is meant to rest on the mathematical fundament the student has obtained
during those years. Clearly, I will not use all the mathematical tools
I am equipped with from the wide variety of disciplines, I been acquainted
with through my studies. It would be more correct to say that it is a narrow
specialization of some of the knowledge I have gained. It is thus difficult
to point out what the prerequisites are, exactly. I will never the less try to
narrow down the fundament on which this thesis rests.
First of all I will emphasize that this thesis is concerned with quantum
electrodynamics. That is, it is concerned with problems arising from both
quantum mechanics and general relativity. Although I will only be concerned
with a few principles of the two, the whole idea and machinery of them are
luring in the background. I have through my studies read several books on
the subjects most important probably being the basic theory to be found in
[7], [8] and [1]. Note that Stone’s Theorem will be taken for granted and
will be normalized.
Also a good deal of functional analysis will be taken for granted. At least
the content of [5]. Especially the theory of distributions will be imperative
the thesis. I will, in some sense, consider the project [4], which I wrote with
my fellow student Morten Bakkedal, as a part of this thesis. The project
covers a good deal of the theory of distributions and wavefront sets resting
on the lecture notes on distributions by Gerd Grubb [2].
5

Chapter 1
Introductory Theory
1.1 Notation
I will mostly adopt the notation used in [4], but for convenience there will be
some differences in the notation of a distribution. Whereas we in [4] used to
write a distribution u acting on a test-function φ in the functional manner
u(φ) I will in this project use the “inner product” notation 〈u, φ〉.
Also note that I will use the definition of the fourier transform found in
[4] rather than the one found in [6]. This will only result in some differences
in the factors.
I will occasionally use Einstein notation along with the convention that
indices from the Greek alphabet take the values 0, 1, 2, 3 and indices from
the Latin alphabet take the values 1, 2, 3.
By a t-dependent operator O(t) : E → F we mean an indexed family of
operators {O(t)}t∈R all from E to F .
Given a t-dependent operator O(t) : E → F . We write O = s- limt→∞ O(t)
if it converges in the strong operator topology to an operator O. That is, if
Oφ = lim
t→∞ (O(t)φ) for all φ ∈ E.
Beware of the many meanings of the symbol δ. Usually it means the
Dirac delta-distribution, but it is also used as a small real, the Kronecker
symbol δi,j and the diagonal map. It should be transparent from the context
which meaning the symbol has.
For convenience I have listed some general notation on the next page:
6

General notation.
Let α = (α1, . . . , αn) ∈ N n 0 be a multi-index and x = (x1, . . . , xn) ∈ R n a
point, then
|α| = α1 + α2 + . . . + αn.
α! = (α1, α2, . . . , αn)! = α1!α2! · · · αn!.
x α = x α1
1
· · · xαn
n
∂j = ∂/∂xj for j = 1, . . . , n.
∂ α = ∂ α1
1
· · · ∂αn
n .
Dj = −i∂j for j = 1, . . . n.
f (α) (x) = ∂ α f(x).
7

1.2 Formal Power Series
The concept of formal power series will be of much importance to this
project. They make it possible to employ much of the analytical machinery
of usual power series but in settings which have no natural notion of
convergence.
Let R be a commutative ring with identity. We call the elements of R the
scalars. Let R N be the set of all infinite sequences in R. Now we define
addition and multiplication of two such sequences in the usual way, that is,
and the Cauchy product
(an) + (bn) = (an + bn),
n
(an) × (bn) =
k=0
akbn−k
=
h+k=n
ahbk,
which in a sense is discrete convolution. Note that addition and
multiplication are well-defined operations because each term only depends
on a finite number of terms from the original sequences.
Theorem 1.1. (R N , +, ×) is a commutative ring with the multiplicative
identity 1 := (1, 0, 0, . . .) and additive identity 0 := (0, 0, . . .).
Proof. The only non-trivial property is multiplicative associativity. We
need to prove that
((a × b) × cn) = (a × (b × c)n).
This follows from
((a × b) × cn) =
(a × b)hck =
and
h+k=n
(a × (b × c)n) =
r+h=n
r+s+k=n
ar(b × c)h =
r+s+k=n
arbsck
arbsck
It is easy to see from the definition of the Cauchy product that if we define
X = (0, 1, 0, 0, . . .) then every element of R N on the form
(a0, . . . , aN, 0, . . .) can be written on the form
N
anX n .
n≥0
8
.

Theorem 1.2. Let d : R N × R N → R be defined by
d((an), (bn)) = 2 −k , where k ∈ N0
is the smallest index such that ak = bk. If no such k exists we define
d((an), (bn)) = 0. Then (R N , d) is a metric space.
Proof. The only non-trivial part is the triangle inequality.
Let d(a, b) = 2−k1 −k2 , d(b, c) = 2
exist.
−k3 and d(a, c) = 2 assuming k1, k2 and k3
Say, k3 ≥ k1, then 2−k3 −k1 −k1 −k2 ≤ 2 ≤ 2 + 2 , hence we may assume
= bk3 and bk3 = ck3 thus ak3 = ck3 ,
k3 < ki for i = 1, 2. Then ak3
contradicting the definition of k3.
If k3 doesn’t exist the inequality is obvious. Say, k1 doesn’t exist, then
(an) = (bn) hence k2 = k3 if they exist. If not, then (an) = (cn).
Theorem 1.3. The operations of addition and multiplication are
continuous.
Proof. Say, fn → f and gn → g then we need to prove that
fn + gn → f + g, that is d(fn + gn, f + g) → 0 for n → ∞.
We need to prove that given ɛ > 0 there exists a δ such that
max{d(fn, f), d(gn, g)} < δ ⇒ d(fn + gn, f + g) < ɛ.
Given ɛ > 0 find k such that 2 −k < ɛ. Let δ = 2 −k /2 = 2 −(k+1) , then
|(f − fn)h| + |(g − gn)h| = 0, for all h ≤ k + 1. (1.1)
Hence since 2 −(k+1) + 2 −(k+1) = 2 −k
d(fn + gn, f + g) < 2 −k < ɛ.
Note that equation (1.1) also implies |(fngn)h| − |(fg)h|=0 for h ≤ k + 1,
hence multiplication is also continuous.
The cases where no k exists are obvious.
This leads to the following result.
Corollary 1.4. (RN , d) is a topological ring and
(an) =
anX n .
n≥0
Note that in this ring convergence is absolute, in fact any rearrangements
of the series converges to the same limit.
Definition 1.5. We denote the topological ring (R N , d) by R[[X]] and call
it the ring of formal power series.
9

Theorem 1.6. Let anX n be a formal power series. Then anX n has
a unique inverse if and only if a0 has a multiplicative inverse in R.
Proof. Say, bnX n is the inverse of anX n . Then by the definition of
multiplication we must have
(i) a0b0 = 1
(ii) a0b1 + a1b0 = 0
(iii) a0b2 + a1b1 + a2b0 = 0.
and so on. . .
Now it is clear from (i) that if the inverse exists then a0 must also be
invertible.
If on the other hand a0 ∈ R ∗ , the invertible elements of R, then the inverse
of anX n is given by bnX n , where b0 = a −1
0 and
bn = a −1
0 (−a1bn−1 − . . . − anb0), for n ≥ 1.
Say, both bnX n and cnX n are inverse to anX n . Then a0b0 = a0c0,
that is b0 = c0, since a0 = 0. Assume bk = ck, for all k < n, then
a0bn + a1bn−1 + · · · + anb0 = a0cn + a1cn−1 + · · · + anc0
By the induction assumption bk = ck for all k < n hence a0bn = a0cn and
thus bn = cn.
The geometric series formula is valid in R[[X]].
Theorem 1.7. Let anX n = 1 be a formal power series, then
anX n m
= (1 − anX n ) −1 .
Proof. Let g =
anX n
m . Then
1 − anx n
g = 1 − anx n
as wanted.
lim
M→∞
m=0
M
anx n ) m
= lim 1 −
M→∞
anx n M
m=0
= 1 − anx n
+ anx n −
= 1.
anx n ) m
Setting an = δ1,n we get the usual geometric series formula.
10
anx n 2
+ · · ·

Corollary 1.8.
∞
n=0
X n = (1 − X) −1 .
Note the following nice properties of R[[X]].
Lemma 1.9. The topology on R[X] is equal to the product topology on
R N , where R is equipped with the discrete topology.
Proof. Let B(a, ɛ) be an open ball in R[[X]], then
B(a, ɛ) = {b ∈ R[[X]]|d(a, b) < ɛ}
= {b ∈ R[[X]]|d(a, b) < 2 −k , where 2 −k ≤ ɛ < 2 −k+1 }
= {b ∈ R[[X]]|ah = bh for all 0 ≤ h < k}
=
k−1
h=0
π −1
h ({ah}),
where {ah} is open in the discrete topology.
If on the other hand
B = π −1
β1
(Uβ1 ) ∩ π−1(Uβ2
) ∩ · · · ∩ π−1(Uβn
),
β2
is a basis element of R N . Let a ∈ B. Define m = max{β1, . . . , βn}. Then
a ∈ B a, 2 −(m+1) which is clearly a subset of B.
Proposition 1.10. The metric space (R[[X]], d) is
1. Complete.
2. Compact if and only if R is finite.
Proof. Completeness: Let (an) be a cauchy sequence in R[[X]]. Then we
know
∀k∃N ∈ N∀n, m ≥ N : d(an, am) < 2 −k .
Hence for each h, πh(an) converges to, say, bh. Then (an) converges to the
formal power series b =
h bhX h .
Compactness: If R is finite R[[X]] is compact by Tychonoff’s theorem.
The topology of R is the discrete topology, hence if it is infinite,
a∈R a is
an open cover that doesn’t contain a finite subcollection covering R.
11
βn

1.3 Affine Transformations of Distributions
Let A = (aij) be an invertible n × n matrix and f a continuous function on
R n . We define a linear continuous composition ⋆ by
A ⋆ f(x) = f(Ax), x ∈ R n .
Let’s look at the distribution induced by f acting on a test-function φ.
Letting T (x) = A −1 x, where A −1 = (ãij), by change of variables theorem
〈A ⋆ f, φ〉 =
=
=
R n
since (A −1 x)i = n
j=1 ãijxj and
T ′ (x) =
⎛
⎜
⎝
∂(A −1 x)1
∂x1
.
∂(A −1 x)n
∂x1
f(Ax)φ(x)dx
f(T (Ax))φ(T (x))| det T ′ (x)|dx
Rn f(x)φ(A −1 x)| det A −1 |dx
= | det A| −1 〈f, A −1 ⋆ φ〉,
. . .
. ..
. . .
∂(A −1 x)1
∂xn
.
∂(A −1 x)n
∂xn
⎞
⎟
⎠ =
⎛
⎞
ã11 . . . ã1n
⎜
⎝
.
. ..
⎟
. ⎠ = A −1 .
ãn1 . . . ãnn
Definition 1.11. Let u ∈ D ′ (R n ) and let A be a real invertible n × n
matrix. Then the distribution A ⋆ u is defined by
for φ ∈ C ∞ 0 (Rn ).
〈A ⋆ u, φ〉 = | det A| −1 〈u, A −1 ⋆ φ〉, (1.2)
We should check that this actually defines a distribution.
Linearity:
〈A ⋆ u, rφ + sψ〉 = | det A| −1 〈u, A −1 ⋆ (rφ + sψ)〉
Continuity: Assume φn → φ, then
= | det A| −1 〈u, rA −1 ⋆ φ + sA −1 ⋆ ψ〉
= r| det A| −1 〈u, A −1 ⋆ φ〉 + s| det A| −1 〈u, A −1 ⋆ φ〉
〈A ⋆ u, φn〉 = | det A| −1 〈u, A −1 ⋆ φn〉 → | det A| −1 〈u, A −1 ⋆ φ〉,
since u and ⋆ are continuous.
12

Proposition 1.12. We state some often occurring affine transformations.
In the following u ∈ D ′ (R n ) and φ ∈ C(R n ).
1. Reflection: 〈u(−x), φ(x)〉 = 〈u(x), φ(−x)〉.
2. Scaling: 〈u(tx), φ(x)〉 = t −n 〈u(x), φ(x/t)〉 = t −n 〈u(x), λtφ(x)〉, letting
λtφ(x) := φ(x/t).
Proof. 1. and 2. follow directly from (1.2) by choosing A = −I and A = tI
respectively.
Note that λt : S(R n ) → S(R n ) is continuous. This follows since
λtφα,β = sup |x α D β (λt(φ(x)))| = sup |x α t −|β| (D β φ)(x/t)|
Further we define
= t |α|−|β| sup |(x/t) α (D β φ)(x/t)|
= t |α|−|β| φα,β.
Definition 1.13. Let u ∈ D ′ (R n ), h ∈ R n and φ ∈ C(R n ). We define the
translation τh of u by
〈u(x − h), φ(x)〉 = 〈τhu, φ〉 = 〈u, τ−hφ〉 = 〈u(x), φ(x + h)〉.
Clearly this defines a distribution.
Recall that a function f is said to be homogeneous of order λ ∈ C if
f(tx) = t λ f(x) for all t > 0 and x ∈ R n . We need to define the similar
property for distributions.
Definition 1.14. A distribution u is said to be homogeneous of degree
λ ∈ C if
〈u(tx), φ(x)〉 = 〈t λ u(x), φ(x)〉,
for all t > 0 and φ ∈ C ∞ 0 .
13

Chapter 2
The Poincaré Group
2.1 The Lorentz Group
We define the Lorentz Metric as the bilinear form on R4 seen as a vector
space:
〈x, y〉 = x 0 y 0 − x 1 y 1 − x 2 y 2 − x 3 y 3
for all x, y ∈ R 4 . (2.1)
Obviously it is symmetric and non-degenerate in the sense that there for
all x = 0 exists a y such that 〈x, y〉 = 0. But it is not positive definite.
Letting g be the Metric Tensor i.e.
−1 0T g = (gµν) =
, where 0
0 I3
T = (0, 0, 0).
Equation (2.1) becomes
〈x, y〉 = x T gy = gµνx µ y ν
The Minkowski Space M is the vector space R 4 endowed with the Lorentz
metric.
While y µ denotes the components of the vector y in M, yµ denotes the
components of the corresponding linear form y ′ given by y ′ (x) = 〈y, x〉, for
all x. In the dual space with the canonical basis the components of the
linear form y ′ are (y 0 , −y 1 , −y 2 , −y 3 ).
Definition 2.1. A Lorentz transformation of R 4 is a linear map
Λ : R 4 → R 4 satisfying
〈Λx, Λy〉 = 〈x, y〉 for all x, y ∈ R 4 . (2.2)
Since e T i Aej = Aij for a matrix A it is clear that (2.2) is equivalent to
Λ T gΛ = g. (2.3)
14

Clearly the composition of Lorentz transformations is itself a Lorentz
transformation. And since
1 = − det g = − det(Λ T gΛ) = − det Λ det g det Λ = (det Λ) 2 ,
we must have det Λ = ±1. Thus all Lorentz transformations are invertible
and the set of all Lorentz transformations form group called the Lorentz
Group L, also denoted O(3, 1).
Further we denote the group of Lorentz transformations Λ with det Λ = 1
by L+ = SO(1, 3). From equation (2.3) we get 10 independent quadratic
equations for the components of Λ. The first is
Which implies that either
When Λ 0 0
(Λ 0 0) 2 − (Λ 1 0) 2 − (Λ 2 0) 2 − (Λ 3 0) 2 = 1. (2.4)
Λ 0 0 ≥ 1 or Λ 0 0 ≤ −1. (2.5)
≥ 1 time is not reversed and we call the subgroup defined by
the Proper Lorentz Group.
L ↑
+ := {Λ ∈ L+|Λ 0 0 ≥ 1},
The Lorentz group is normally split into 4 disjoint classes. Namely, L ↑
+
and the 3 following
Name det Λ Λ 0 0
L ↑
− −1 ≥ +1
L ↓
− −1 ≤ −1
L ↓
+ +1 ≤ −1
Further we mention 3 important discrete Lorentz transformations, namely:
I, the identity, P = g, space inversion or the parity transform, and
T = −g, time inversion.
There are 3 types of vectors in M. A vector x is said to be time-like if
x 2 > 0, space-like if x 2 < 0 and light-like if x 2 = 0. Since x 2 is constant
under Lorentz transformation the categories stay the same under the
transformation.
Information cannot propagate faster than the speed of light. Equivalently,
two events at x µ and y ν cannot influence each other if they are separated
by a space-like distance,
(x − y) 2 < 0.
They have no causal relation. Causality plays a key role in this thesis.
15

Definition 2.2. Given some point x the forward cone of x is defined by
and the backwards cone of x is
V + (x) = {y|(y − x) 2 ≥ 0 and y 0 ≥ x 0 },
V − (x) = {y|(y − x) 2 ≥ 0 and y 0 ≤ x 0 }.
The n-dimensional generalizations are
Γ ± n (x) = {(x1, . . . , xn)|xj ∈ V ± (x) and ∀j = 1 = 1, . . . , n}.
That (y − x) 2 ≥ 0 just means that they have to be causally connected.
Later we will need a symbol to distinguish points in time.
Definition 2.3. Let A, B ⊂ M. If for all x ∈ A and y ∈ B we have
x 0 < y 0 , we write A < B.
2.2 The Poincaré Group
Definition 2.4. Let Λ ∈ L and a ∈ R 4 . By a Poincaré transformation
Π : R 4 → R 4 we mean Π(x) = Λx + a, and we write Π = (a, Λ).
The set P of all Poincaré transformations form a group under the
composition law
(a1, Λ1)(a2, Λ2) = (a1 + Λ1a2, Λ1Λ2).
We see that the Poincaré Group P is the semidirect product of L and the
group of space-time translations (R 4 , +), that is
P = R 4 ⊙ L
Further we define the Proper Poincaré Group as
P ↑
+ = R4 ⊙ L ↑
+
16

2.3 Spinor Representations of the Lorentz Group
We want to extend the spinor representation to four dimensions. 1 Doing
this we add the unit matrix
1 0
σ0 = , (2.6)
0 1
to the usual Pauli matrices:
σ1 =
0 1
1 0
, σ2 =
0 −i
i 0
1 0
, σ3 =
0 −1
. (2.7)
These form a basis of H(2), the vector space over R of all complex
Hermitian 2 × 2 matrices. Actually, given a 2 × 2 matrix H ∈ H(2) we get
by direct calculation that H can be represented by a vector x ∈ R4 in the
following way:
H := σ(x) = x µ σµ = 1
3
tr(Hσµ)σµ
(2.8)
2
The map x ↦→ σ(x) defined in this way is indeed an isomorphism of R 4
onto H(2).
Fixme: should this be written out?
Further we can define another isomorphism by
µ=0
x ↦→ σ ′ (x) = x 0 σ0 − x k σk =
3
xµσµ.
Now consider the transform of σ(x) with a complex 2 × 2 matrix A:
σ(x) ↦→ Aσ(y)A ∗ .
Clearly σ(y) := Aσ(x)A ∗ ∈ H(2). And from equation (2.8) we see that its
components in the basis of Pauli Matrices are given by
y ν = 1
2 tr(σ(y)σν) = 1
2 tr(Aσ(x)A∗σν) = 1
2
µ=0
3
µ=0
tr(AσµA ∗ σν)x µ .
In this way we get a linear map ΛA : x → y of R 4 into it self. Note that
Hence if det A = 1, then
σ(x) = x µ
x0 + x3 x1 − ix2 σµ =
x 1 + ix 2 x 0 − x 3
.
1 The two-component spinor formalism is treated in [8] chapter 3.2-3.3.
17

〈y, y〉 = det σ(y) = det A det σ(x) det A ∗ = det σ(x) = 〈x, x〉.
Clearly the parallelogram identity holds for the Lorentz metric 2 , thus
〈x, y〉 =
= det σ(x) + det σ(y) − det σ(x) − det σ(y)
〈x + y, x + y〉 − 〈x, x〉 − 〈y, y〉
2
.
2
From this we see that 〈ΛAx, ΛAy〉 = 〈x, y〉, thus it is a Lorentz
transformation.
Since
fixme: make sure the following is correct
Hence
ΛB(x) = y ⇔ σ(y) = Bσ(x)B ∗ and ΛA(y) = z ⇔ σ(z) = Aσ(y)A ∗ ,
ΛAB(x) = z ⇔ ABσ(x)B ∗ A ∗ = Aσ(y) = σ(z).
ΛAB = ΛAΛB.
In this way we can construct a representation Λ : A ↦→ ΛA of SL(2, C), the
group of all complex 2 × 2 matrices with determinant 1, onto L ↑
+ . We call
SL(2, C) the spinor representation. The vectors in the representation space
C2 are called spinors.
Note that the kernel of the representation Λ is
ker Λ = Λ −1 ({I4}) = {A ∈ SL(2, C)|AHA ∗ for all H ∈ H(2)}.
Letting H = I2 we see that A ∈ ker Λ must be unitary and A ∈ ker Λ if and
only if AH = HA for all H ∈ H(2). Hence A = −I2 or A = I2 and
L ↑
+ ∼ = SL(2)/{I2, −I2}.
Hence Λ(A) = Λ(−A) and SL(2, C) is a two-valued representation of L ↑
+ .
2 The proof is exactly the same as the usual one for inner product spaces.
18

Chapter 3
The Scattering Matrix
As the notion of the scattering matrix, or the S-matrix for short, plays the
central role in scattering theory I will give a brief introduction to the
development of it in the case of quantum mechanics. Later our main aim
will be to construct the S-matrix of quantum electrodynamics, QED, by
perturbation theory. First a definition:
Definition 3.1. Let H be a Hilbert space. Then a two-parameter family
U(s, t), (s, t) ∈ R 2 of bounded operators on H is called a unitary
propagator, if the following is satisfied:
(1) U(s, t) is unitary for all s, t ∈ R,
(2) U(t, t) = 1 for all t ∈ R,
(3) U(r, s)U(s, t) = U(r, t) for all r, s, t ∈ R,
(4) The map (s, t) ↦→ U(s, t)ψ is continuous for all ψ ∈ H.
Consider a quantum mechanical system described by the Hamiltonian
H = H0 + V (t), where H0 is the free Hamiltonian and V is the
time-dependent interaction. Then the time evolution is given by a unitary
propagator in the sense that
ψ(t) = U(t, s)ψ(s) 1 . (3.1)
Definition 3.2. Let U(s, t) be a unitary propagator then we define the
wave operators as follows
Win = s- lim
t→−∞ U(t, 0)∗ e −iH0t = s- lim
t→−∞ U(0, t)e−iH0t .
Wout = s- lim
t→∞ U(t, 0) ∗ e −iH0t = s- lim
t→∞ U(0, t)e −iH0t .
Provided the strong limits exist.
1 We remember that if H is time independent then then time evolution is given by the
unitary transform ψ(t) = e −iHt ψ(t0).
19

Note that
〈Woutφ, ψ〉 = 〈 lim
t→∞ U(0, t)e −iH0t φ, ψ〉
= lim
t→∞ 〈U(0, t)e −iH0t φ, ψ〉
= lim
t→∞ 〈φ, e iH0t U(0, t) ∗ ψ〉
= 〈φ, lim
t→∞ e iH0t U(t, 0)ψ〉.
Hence W ∗ out = s- limt→∞ e iH0t U(t, 0). Further
W ∗ outWinψ = lim
t→∞ lim
s→−∞ eiH0t U(t, 0)U(0, s)e −iH0s ψ
This leads us to the following definition.
= lim
t→∞ lim
s→−∞ eiH0t U(t, s)e −iH0s ψ.
Definition 3.3. The scattering matrix is defined as follows
S = W ∗ outWin = s- lim
t→∞ s- lim
s→−∞ eiH0t U(t, s)e −iH0s .
The physical meaning of the scattering matrix is now apparent. A
normalized initial asymptotic state ψ considered at time t = 0, say, is first
transformed to s = −∞ by free dynamics, then it is evolved from −∞ to
t = ∞ by full interacting dynamics and finally it is transformed back from
∞ to t = 0 again by free dynamics. Thus Sψ is in fact the outgoing
scattering state transformed to t = 0 by free dynamics. The probability for
a transition form ψ to φ is given by
P (ψ → φ) = |〈φ, Sψ〉| 2 .
Now ψ(t) as given by equation (3.1) is the solution to the Schrödinger
equation
i d
dt ψ(t) = (H0 + V (t))ψ(t).
If we go over to the interaction picture 2 by substituting φ = e iH0t ψ, we get
i d d
φ(t) = i
dt dt (eiH0tψ(t)) = −H0e iH0t iH0t
ψ(t) + e (H0 + V (t))ψ(t)
= e iH0t −iH0t
V e φ(t)
= ˜ V (t)φ(t).
FIXME: Make sure you can differentiate like this.
2 See [8] page 318.
20

Thus we note that the scattering matrix is in fact just the limit of the time
evolution in the interaction picture. That is
S = lim lim Ũ(t, s).
t→∞ s→−∞
If the interaction V (t) is a bounded operator we may write the unitary
operator in terms of the Dyson series 3
Ũ(t, s) = 1 +
∞
(−i) n
n=1
Claim. t
s
s
s
t
s
dt1
t1
s
tn−1
dt2 · · · dtn
s
˜ V (t1) · · · ˜ V (tn). (3.2)
tn−1
dt1 . . . dtn =
s
1
(t − s)n
n!
Proof. It is obvious for n = 1.
Assume it is right for n = k − 1 then
t tk−1
t
t1 1
dt1 · · · dtk = dt1
(k − 1)!
=
This proves the claim by induction.
s
1
(k − 1)!
t
1
=
(k − 1)!
= 1
(t − s)k
k!
s
t−s
0
s
k−1 dτ
(t1 − s) k−1 dt1
h k−1 dt1
It follows that the sum (3.2) converges in the operator-norm since each
term is bounded by
t
s
dt1 . . .
tn−1
s
dtn ˜ V (t1) . . . ˜ V (tn) = 1
n! (t − s)n ˜ V ) n
If the the interaction decreases for large times such that
then the S-matrix is given by
S =
∞
(−i) n
n=0
+∞
−∞
+∞
−∞
dt1
which is also norm-convergent.
3 See [8] page 326.
t1
dsV (s) < ∞,
dt2 · · ·
−∞
21
tn−1
−∞
dtn ˜ V (t1) · · · ˜ V (tn),

Chapter 4
The Mathematical Setting of
QFT
4.1 The Fock Space
Let H n = H ⊗ · · · ⊗ H be the tensor product space of n single-particle
Hilbert spaces H. Remembering that identical particles must obey either
Bose or Fermi statistics 1 , we symmetrize the states φn ∈ H n
S + n φn = 1
n!
π
φn(xπ1 , . . . , xπn)
in the case of bosons and antisymmetrize
(−1) π (xπ1 , . . . , xπn)
S − n φn = 1
n!
π
in the case of fermions. We then form a Hilbert space from the direct sum
of tensor products.
Definition 4.1. The Fock space F is defined as
F ± = ⊕ ∞ n=0S ± n H n
with H0 := {λ|0〉|λ ∈ C} where |0〉 is called the vacuum.
It is easy to prove that F is indeed a Hilbert space. It follows from the
continuity of the inner products. We simply state it here.
Proposition 4.2. The Fock space F is a Hilbert space with the inner
product
∞
〈Φ, Ψ〉 = 〈Φn, Ψn〉n,
n=0
where 〈·, ·〉n is the inner product on Hn.
1 See [8] chapter 6
22

We site Corollary 1.3 of [6] here 2 .
Corollary 4.3. Any (bounded) operator A ∈ F can be expressed as
A = A (−) + A (+)
where A (−) contain annihilation and A (+) creation operators, only.
Definition 4.4. We say that an operator A ∈ F is normally ordered if all
annihilation operators are placed to the right of the creation operators.
The normal ordering of A is expressed by : A :.
In the light of Corollary 4.3 we may define.
Definition 4.5. A contraction between two field operators A and B is
defined by
| AB | := [A (−) , B (+) ]±,
where [·, ·]± is the commutator in the case of Bose fields and
anti-commutator in the case of Fermi fields.
The following theorem tells us how to order field operators normally. We
state it without its proof. 3
Theorem 4.6 (Theorem of Wick). A product of n field operators can be
normally ordered as follows:
A1A2 · · · An =: A1A2 · · · An : + | A1A1 | · · · An + permutations
+ : | A1 | A2 · · · Aj | . . .An : + · · ·
+ : | A1A2 | | A3A4 | · · · : +permutations,
where the sum contains all normal products with all possible pairings of
contractions.
FIXME: think about proving it
We assume the contractions are complex numbers, which therefore can be
taken out of the normal products. Though we have to remember to take
the sign into account in the case of Fermi operators.
In QED there exists only three contractions.
| ψa(x)ψb(y) | := {ψ (−)
a (x), ψ (+)
b (y)} =: 1
i S(+)
ab
| ψa(x)ψb(y) | := {ψ (−)
a (x), ψ (+)
b (y)} =: 1
i S(−)
ba
(x − y) (4.1)
(y − x) (4.2)
| Aµ(x)Aν(y) | := [A (−)
µ (x), A (+)
ν (y)] = gµνiD (+)
0 (x − y), (4.3)
2 I have taken the liberty of reformulating the corollary such that its meaning will be
more apparent for my application. I have not altered the meaning of it, though.
3 See [6]
23

where ψ is the time-dependent Dirac field 4 and ψ = ψ (+) γ 0 is the Dirac
adjoint. The gamma matrices are
γ µ
0 σµ
=
σµ 0
with σµ given by (2.6) and (2.7). We here regard (4.1), (4.2) and (4.3) as
definitions of the operators S = S (−) + S (+) and D (+)
0 , with Aµ given by [6]
page 148.
4 Se [6] page 83
24
,

4.2 The Wightman Axioms
We will treat a quantum field theory as a tuple
(H, U, φ, D, |0〉),
where H is a separable Hilbert space, U a unitary representation of the
proper Poincaré group P ↑
+ , φ are field operators, D a dense subspace of H
and |0〉 the vacuum state.
We expect the tuple to satisfy the Wightman axioms listed on the next
page. 5
FIXME: make sure the syntax is right
We will not go in to the deeper meaning of the axioms. Instead we will
note the most important properties for our use. We should note that the
fields φ are assumed to be operator valued distributions, which are
well-defined on a dense domain D ⊂ H.
FIXME: note that D0 is contained in axiom 4
In the construction of the scattering matrix, however, it suffices to regard
the dense domain D0 ⊂ D ⊂ H given by,
D0 = span{φ(g1) · · · φ(gn)|0〉|g1, . . . , gn ∈ S, n ∈ N}.
5 We cite them from [9] page 103 − 104.
25

The Wightman Axioms.
Axiom 1 (quantum field) The operators φ1(f), . . . , φn(f) are given
for each C ∞ -function f with compact support on the Minkowski space
R 4 . Each φj(f) and its Hermitian conjugate operator φj(f) ∗ are defined
at least on a common dense linear subset D of the Hilbert space H and
D satisfies
φj(f)D ⊂ D, φj(f) ∗ D ⊂ D,
for any f and j = 1, . . . , n. For any Φ, Ψ ∈ D,
is a complex valued distribution.
f ↦→ (Φ, φj(f)Ψ),
Axiom 2 (relativistic symmetry) On H there exists a unitary repre-
sentation U(a, A) of ˜ P ↑
+ (a ∈ R4 , A ∈ SL(2, C)), satisfying U(a, A)D =
D (invariance of the common domain of the fields).
U(a, A)φj(f)U(a, A) ∗ = S(A −1 )jkφk(f (a,A))
f (a,A)(x) = f(Λ(A) −1 (x − a)),
where the matrix (S(A)j,k) is an n-dimensional representation of A ∈
SL(2C).
Axiom 3 (local commutativity) If the support of f and g is spacelike
separated, then for any vector Φ ∈ D
[φj(f) (∗) , φk(g) (∗) ]±Φ = 0,
where (∗) indicates that the equation holds for any choice.
Axiom 4 (vacuum state) There exists a vector |0〉 in D satisfying the
following conditions:
(i) (Ua, A)|0〉 = |0〉 (invariance).
(ii) The set of all vectors obtained by acting an arbitrary polynomial
P of the fields on |0〉 is dense in H (cyclicity).
(iii) The Spectrum of the translation group U(a, 1) on |0〉 ⊥ is contained
in
V m = {p|(p, p) ≥ m 2 , p 0 > 0} (m > 0)
(Spectrum condition).
26

Chapter 5
The Method of Epstein and
Glaser
5.1 Introduction
We want to express the S-matrix as a formal power series in R[[λ]], where
λ is the coupling constant λ = e, which is the unit of charge and R is the
ring
R = {Υ : D0 → D0|Υ is linear}.
We note that we are not concerned about the convergence of the series and
it might not have any convergence radius at all.
We begin from the expression.
S(g) = 1 +
∞
n=1
=: 1 + T
n 1
λ
n!
d 4 x1 · · · d 4 xnTn(x1, . . . , xn)g(x1) · · · g(xn)
We shall usually omit the λ in the notation. Further may assume that
Tn(x1, . . . , xn) is symmetric in x1, . . . xn. Otherwise we can always
symmetrize it. This is easy to see in 2 dimensions where we given
T2(x1, x2) can choose the symmetric map defined by
(T2(x1, x2) + T2(x2, x1))/2. The same way we can choose a symmetric map
in n-dimensions as
Tn(x1, x2, . . . , xn) + Tn(x2, x1, x3, . . . , xn) + · · · + Tn(xn, xn−1, . . . , x1)
.
n!
Since Tn(x1, . . . , xn) is symmetric we may occasionally use the short hand
notation Tn(x) where x = {xj ∈ M|j = 1, . . . , n} is disordered.
Along with the Wightman axioms we have to assume that each term of the
S-matrix is well-defined.
27

Axiom 0 (Well-definedness) Let g = g1, . . . , gn ∈ S(R n ), then
〈Tn, (g1 ⊗ · · · ⊗ gn)〉 : D0 → D0,
is a well-defined operator-valued distribution with
〈Tn, (g1 ⊗ · · · ⊗ gn)〉 = dx1 · · · dxnTn(x1, . . . , xn)g(x1) · · · g(xn).
Note that the S-matrix and Tn are operator-valued distributions, in the
sense that if φ ∈ D0, then
∞
1
S(g)φ = 1 +
n!
n=1
∞
1
= φ +
n!
n=1
d 4 x1 . . . d 4
xnTn(x1, . . . , xn)g(x1) . . . g(xn) φ
d 4 x1 · · · d 4 xnTn(x1, . . . , xn)(φ(g(x1)) · · · φ(g(xn))).
We can express the inverse of S(g) by a similar perturbation series as
follows
S(g) −1 ∞
1
= 1 + d
n!
n=1
4 x1 . . . d 4 xn ˜ Tn(x1, . . . , xn)g(x1) . . . g(xn)
= (1 + T ) −1 ∞
= 1 + (−T ) r ,
using Theorem 1.7. From
=
∞
n=1
∞
r=1
we see that
n 1
λ
n!
−
∞
n=1
r=1
d 4 x1 . . . d 4 xn ˜ Tn(x1, . . . , xn)g(x1) . . . g(xn) =
1
n! λn
˜Tn(X) =
∞
(−T ) r
r=1
d 4 x1 . . . d 4 r xnTn(x1, . . . , xn)g(x1) · · · g(xn) ,
n
(−1)
r=1
r
where Pr is the set of all partitions of X
Pr
Tn1 (X1) . . . Tnr(Xr), (5.1)
X = X1 ∪ . . . ∪ Xr , |X| = n , Xj = ∅ , |Xj| = nj.
28

Further,
1 = S(g)S(g) −1
=
1 +
= 1 +
∞
n1=1
× 1 +
∞
λ n
n=1
n1 1
λ
n1!
∞
n2=1
n2 1
λ
n1!
n1+n2=n
d 4 x1 · · · d 4 xn1Tn1 (x1, . . . , xn1 )g(x1) · · · g(xn1 )
1
n!
where of course |X| = n1 and T0(∅) = 1 = ˜ T0(∅).
By (5.2)
0 =
∞
λ n
n=1
n1+n2=n
1
n!
d 4 ˜x1 · · · d 4 ˜xn2 ˜ Tn2 (˜x1, . . . , ˜xn2 )g(˜x1) · · · g(˜xn2 )
d 4 x1 · · · d 4 xnTn1 (X) ˜ Tn2 ( ˜ X)g(x1) · · · g(xn),
d 4 x1 · · · d 4 xnTn1 (X) ˜ Tn2 ( ˜ X)g(x1) · · · g(xn)
Then by the definition of the metric on formal power series each power
must vanish. That is,
P 0 2
(5.2)
Tn1 (X) ˜ Tn−n1 ( ˜ X)g(x1) · · · g(xn) = 0, (5.3)
where P 0 2 is the set of all partitions {x1, . . . , xn} = X ∪ ˜ X.
In constructing the scattering matrix, we should be aware of the properties
we think it should have. We list such expected properties on the next page.
29

Expected properties of the S-matrix.
1 Unitarity i.e. S(g) −1 = S(g) ∗ . That is
˜Tn(x1, . . . , xn) = Tn(x1, . . . , xn) ∗ .
Fixme: this is not enough according to page 162 probably not important for
this project.
2 Translational invariance. Let U(a, 1) be the unitary translation operator
in the Fock space F, that is
Then we require
(U(a, 1)Φ)j(x) = Φj(x1 + a, . . . , xj + a).
U(a, 1)S(g)U(a, 1) −1 = S(ga), where ga(x) = g(x − a).
Hence for the Tn’s (and of course the ˜ Tn’s as well) we get:
U(a, 1)Tn(x1, . . . , xn)U(a, 1) −1 = Tn(x1 + a, . . . , xn + a).
3 Lorentz covariance Letting U(0, Λ) be the representation of L ↑
+ .
U(0, Λ)S(g)U(0, Λ) −1 = S(gΛ), where gΛ = g(Λ −1 x).
Note that 2. and 3. together form a condition of Poincaré invariance.
4 Causality. Suppose there exists a reference frame in which the
test-functions g1 and g2 have disjoint supports in time. Assuming
supp g1 < supp g2. That is, for some r ∈ R,
supp g1 ⊂ {x ∈ M|x 0 ∈ (−∞, r)} and supp g2 ⊂ {x ∈ M|x 0 ∈ (r, ∞)}.
We require that
S(g1 + g2) = S(g2)S(g1). (5.4)
This is a statement about the fact that what happens at time t < s is not
influenced by what happens at some later time t > s.
30

Of these properties, causality plays a key role in the method of Epstein
and Glaser. Equation (5.4) leads to the condition stated in the following
theorem.
Theorem 5.1. If {x1, . . . , xm} > {xm+1, . . . , xn}, then
Tn(x1, . . . , xn) = Tm(x1, . . . , xm)Tn−m(xm+1, . . . , xn),
Before proving this result we first note that if (5.4) is satisfied,
S(g1 + g2) =
=
∞
n=0
n 1
λ
n!
d 4 x1 · · · d 4 xnTn(x1, . . . , xn)
× (g1(x1) + g2(x1)) · · · (g1(xn) + g2(xn))
∞ n
λ n
1
m!(n − m)!
n=0 m=0
d 4 x1 · · · d 4 xnTn(x1, . . . , xn)
× (g2(x1) · · · g2(xm)) · · · (g1(xm+1) · · · g1(xn)),
as there are 2n terms from the product of the test-functions and
ways to pick g2 m times. On the other hand
S(g2)S(g1) =
=
=
∞
m=0
m 1
λ
m!
d 4 x1 · · · d 4 xmTm(x1, . . . , xm)
× g2(x1) · · · g2(xm)
∞
k 1
× λ
k!
k=0
∞
m=0 k=0
∞
n=0 m=0
d 4 ˜x1 · · · d 4 ˜xkTk(˜x1, . . . , ˜xk)
× g1(˜x1) · · · g1(˜xk)
∞
m+k 1
λ
m!k!
d 4 x1 · · · d 4 xmd 4 ˜x1 · · · d 4 ˜xk
× Tm(x1, . . . , xm)Tk(˜x1, . . . , ˜xk)
n!
m!(n−m)!
× g2(x1) · · · g2(x1) · · · g2(xm)g1(˜x1) · · · g1(˜xk)
n
λ n
1
dx1 · · · dxnTm(x1, . . . , xm)
m!(n − m)!
× Tn−m(xm+1, . . . , xn)g2(x1) · · · g2(xm)
× g1(xm+1) · · · g1(xn).
From the definition of the metric on formal power series we know that
31

equal powers must have equal coefficients. We conclude that for every m, n
=
d 4 x1 · · · d 4 xnTn(x1, . . . , xn)g2(x1) · · · g2(xm)g1(xm+1) · · · g1(xn)
d 4 x1 · · · d 4 xnTm(x1, . . . , xm)Tn−m(xm+1, . . . , xn)
× g2(x1) · · · g2(xm)g1(xm+1) · · · g1(xn) (5.5)
Further note that from the calculus of the tensor product
〈T2, (f1 + f2) ⊗ (f1 + f2)〉 = 〈T2, f1 ⊗ f1〉 + 〈T2, f1 ⊗ f2〉
+ 〈T2, f2 ⊗ f1〉 + 〈T2, f2 ⊗ f2〉,
FIXME: maybe better to write it out in n dimensions
and since Tm is symmetric
〈Tm, f1 ⊗ f2〉 = 1
2 (〈Tm, (f1 + f2) ⊗ (f1 + f2)〉 − 〈Tm, f1 ⊗ f1〉
− 〈Tm, f2 ⊗ f2〉). (5.6)
Now for transparency and to get a good idea of how to prove Theorem 5.1
let us prove it for n = 3 and m = 2. That is,
Claim. 〈T3, f1 ⊗ f2 ⊗ f3〉 = 〈T2, f1 ⊗ f2〉〈T1, f3〉.
Proof. By 5.6
Hence by 5.5
〈T2, f1 ⊗ f2〉 = 1
〈T2, (f1 + f2) ⊗ (f1 + f2)〉 − 〈T2, f1 ⊗ f1〉
2
− 〈T2, f2 ⊗ f2〉
〈T2, f1 ⊗ f2〉〈T1, f3〉
= 1
〈T3, (f1 + f2) ⊗ (f1 + f2) ⊗ f3〉 − 〈T3, f1 ⊗ f1 ⊗ f3〉
2
− 〈T3, f2 ⊗ f2 ⊗ f3〉
= 1
〈T3, f1 ⊗ f1 ⊗ f3〉 + 〈T3, f2 ⊗ f2 ⊗ f3〉 + 2〈T3, f1 ⊗ f2 ⊗ f3〉
2
− 〈T3, f1 ⊗ f1 ⊗ f3〉 − 〈T3, f2 ⊗ f2 ⊗ f3〉
= 〈T3, f1 ⊗ f2 ⊗ f3〉.
Which proves the claim.
Now before proving this for arbitrary n and m we need the following
lemma.
32

Lemma 5.2. The product (x1 + x2 + . . . + xn) m has
n−1
m m − k1 m − i
ki
· · · ·
,
k1
terms on the form x k1
1
· · · xkn
n .
k2
Proof. We note that the number of ways to pick x1, k1 times from m
possibilities
is given by the binomial coefficient. That is, there must be
m
k1
terms on the form x k1
1 (x2 + · · · + xn) m−k1 . Now there are m − k1
multiplications left and thus m−k1 ways to pick x2, k2 times. Hence there
k2
are m m−k1
k1
terms on the form x k1 k2
1 xk2 2 (x3 + · · · xn) m−k1−k2 . This way we
end up with
n−1
m m − k1 m − i
ki
· · · ·
,
k1
terms on the form x k1
1
· · · xkn
n .
Proof of Theorem 5.1. First note that by the lemma
k2
〈Tm, (f1 + f2 + · · · + fm) ⊗ · · · ⊗ (f1 + f2 + · · · + fm)〉
=
n−1
m m − k1 m − i
ki
· · · ·
k1+···+km=m
k1
since Tm is symmetric. Hence
〈Tm, f1 + f2 + · · · + fm〉
k2
kn
kn
× 〈Tm, f k1
1
kn
⊗ · · · ⊗ f km
m 〉,
= 〈Tm, (f1 + f2 + · · · + fm) ⊗ . . . ⊗ (f1 + f2 + · · · + fm)〉
−
n−1
m m − k1 m − i
ki
· · · ·
〈Tm, f k1
1
k1+···+km=m
(k1,...,km)=(1,...,1)
Hence by 5.5
k1
〈Tm, f1 ⊗ · · · ⊗ fm〉〈Tn−m, fm+1 ⊗ · · · ⊗ fn〉
k2
kn
⊗ · · · ⊗ f km
m 〉.
= 〈Tn, (f1 + f2 + · · · + fm) ⊗ . . . ⊗ (f1 + f2 + · · · + fm) ⊗ (fm+1 ⊗ · · · ⊗ fn)〉
−
n−1
m m − k1 m − i
ki
· · · ·
k1+···+km=m
(k1,...,km)=(1,...,1)
× 〈Tn, f k1
1
k1
k2
⊗ f km
m ⊗ (fm+1 ⊗ · · · ⊗ fn)〉
= 〈Tn, f1 ⊗ f2 ⊗ · · · ⊗ fm ⊗ fm+1 ⊗ · · · ⊗ fn〉
as wanted.
33
kn

We see that the Tn’s are time-ordered products (therefore the T ).
Similarly the causality condition for S −1 (g) is that if supp g1 < supp g2,
then
S(g1 + g2) −1 = S(g1) −1 S(g2) −1 .
This again implies that
˜Tn(x1, . . . , xn) = ˜ Tm(x1, . . . , xm) ˜ Tn−m(xm+1, . . . , xn),
if {x1, . . . , xm} < {xm+1, . . . , xn}.
We are now ready to sketch the game plan for the inductive construction
of the time ordered products Tn.
34

The Inductive Construction of Tn(x1, . . . , xn).
1. Assume Tm(x1, . . . , xm) for 1 ≤ m ≤ n − 1 are known.
2. Construct advanced and retarded distributions as follows
A ′ (x1, . . . , xn) =
˜Tn1 (X)Tn−n1 (Y, xn) (5.7)
P2
R ′ (x1, . . . , xn) =
P2
Tn−n1 (Y, xn) ˜ Tn1 (X) (5.8)
where P2 is the set of all partitions {x1, . . . , xn−1} = X ∪ Y , X = ∅
and n1 = |X| ≥ 1.
3. Include the empty set ∅ as follows
A(x1, . . . , xn) =
˜Tn1 (X)Tn−n1 (Y, xn)
P 0 2
= A ′ n(x1, . . . , xn) + Tn(x1, . . . , xn) (5.9)
R(x1, . . . , xn) =
Tn−n1 (Y, xn) ˜ Tn1 (X)
4. The difference is
5. Now
P 0 2
= R ′ n(x1, . . . , xn) + Tn(x1, . . . , xn). (5.10)
Dn = R ′ n − A ′ n = Rn − An. (5.11)
Tn = Rn − R ′ n = An − A ′ n. (5.12)
We thus need to determine either Rn or An. This will be done by
investigating the support properties of the distributions.
35

5.2 Example - In the Hilbert Space Setting of
Quantum Mechanics
Before going on we illustrate the significance of the table by applying it to
the more transparent theories of Hilbert spaces and quantum mechanics.
In quantum mechanics we substitute distributions by functions and replace
x by t.
Now, assuming T1(t) is given we want to construct T2(t). We follow the
rules given from the box on the previous page and construct the advanced
and retarded functions.
and
A ′ 2(t1, t2) = ˜ T1(t1)T1(t2) = −T1(t1)T1(t2) (5.13)
R ′ 2(t1, t2) = T1(t2) ˜ T1(t1) = −T1(t2)T1(t1), (5.14)
where the last inequalities follow from (5.1). Then
and
A2(t1, t2) = A ′ 2(t1, t2) + T2(t1, t2) (5.15)
R2(t1, t2) = R ′ 2(t1, t2) + T2(t2, t1)
From Theorem 5.1 we know that if ti > tj, then
T2(ti, tj) = T1(t1)T1(tj).
Hence A2 vanishes for t1 > t2 and R2 vanishes for t1 < t2.
Now
D2 = R2 − A2 = R ′ 2 − A ′ 2
is known from equations (5.13) and (5.14). The T2 cancel out since they
are symmetric.
Finally, R2 and A2 can be determined by their support properties. Let
Θ(t) be the Heaviside function
Then
Θ(x) =
1, t ≥ 0,
0, t < 0.
A2(t1, t2) = Θ(t2 − t1)D2(t1, t2)
= Θ(t2 − t1)(T1(t1)T1(t2) − T1(t2)T1(t1)).
Hence A2 is uniquely determined up to its value at t1 = t2.
36

Now from equation (5.15)
T2(t1, t2) = A2(t1, t2) − A ′ 2(t1, t2)
= Θ(t2 − t1)(T1(t1)T1(t2) − T1(t2)T1(t1)) + T1(t1)T1(t2)
= Θ(t1 − t2)T1(t1)T1(t2) − Θ(t2 − t1)T1(t2)T1(t1)
=: T {T1(t1)T1(t2)}.
FIXME: draw parallel to S-matrix with usage of Schrödinger equation.
37

5.3 Splitting of Distributions
In our previous example we saw that the splitting of Dn into advanced and
retarded parts could be done by use of the Heaviside function. This was
because we were dealing with operators in Hilbert space. But in our
setting we are considering distributions which cannot simply be multiplied
with discontinuous step-functions. In this section we consider the
properties of the advanced and retarded distributions.
First a technicality.
Theorem 5.3. Let Y = P ∪ Q, P = ∅, P ∩ Q = ∅, |Y | = n1 ≤ n − 1 and
x /∈ Y .
If {Q, x} > P ,|Q| = n2, then we have
If {Q, x} < P , then we have
R ′ n1+1(Y, x) = −Tn2+1(Q, x)Tn1−n2 (P ). (5.16)
A ′ n1+1(Y, x) = −Tn1−n2 (P )Tn2+1(Q, x). (5.17)
Proof. Equation (5.16) is proved in [6]. Therefore we only prove equation
(5.17).
A ′ n1+1(Y, x) =
˜Tn3 (X)Tn−n3 (Y ′ , x),
where P2 is the set of all partitions Y = X ∪ Y ′ such that X = ∅.
Now let
and
Now since
causality implies
P2
Y ′ = Y1 ∪ Y2, where Y1 = Y ′ ∩ P and Y2 = Y ′ ∩ Q
X = X1 ∪ X2, where X1 = X ′ ∩ P and X2 = X ∩ Q
Q ∪ {x} < P, Y2 < Y1, X2 < X1 and Y1 > Y2 ∪ {x},
A ′ n1+1(Y, x) =
˜Tn3 (X)Tn−n3 (Y ′ , x)
P2
=
˜T (X2) ˜ T (X1)T (Y1)T (Y2, x), (5.18)
P 0 4
subscripts are omitted for simplicity. P 0 4
form
is the set of all partitions of the
P 0 4 : P = X1 ∪ Y1, Q = X2 ∪ Y2 and X1 ∪ X2 = ∅.
38

However, for x2 = ∅ we can have X1 = ∅. From Equation (5.3) it follows
that
˜T (X1)T (Y1) = 0. (5.19)
P 0 2
As a consequence, in (5.18) only terms with X2 = ∅ and hence Y2 = Q
remain. That is, remembering that T0(∅) = 1 = ˜ T0(∅),
A ′
n1+1(Y, x) = ˜T (X1)T (Y1) T (Q, x),
P2
where P2 is set set of all partitions such that P = X1 ∪ Y1 and X1 = ∅.
Including ∅ would give 0 according to (5.19), hence
A ′
n1+1(Y, x) = ˜T (X1)T (Y1) − ˜
T (∅)T (P ) T (Q, x)
as wanted.
P 0 2
Fixme: investigate the meaning of this.
= (0 − T (P ))T (Q, x) = −T (P )T (Q, x),
Corollary 5.4 (Support property 1). The supports of A and R are
respectively advanced and retarded, that is
and
Proof. Equation (5.9) gives us
An1+1(Y, x) = A ′ n1+1 + Tn(Y, x)
supp An1+1(Y, x) ⊂ Γ − n1+1 (x)
supp Rn1+1(Y, x) ⊂ Γ + n1+1 (x)
= −Tn1−n2 (P )Tn2+1(Q, x) + Tn1+1(P ∪ Q, x)
= −Tn1−n2 (P )Tn2+1(Q, x) + Tn1−n2 (P )Tn2+1(Q, x) = 0.
2. follows the same way from (5.10).
The corollary explains why A and R are called advanced and retarded
distributions. An+1(X, x) vanishes if there exists a point x ′ ∈ X with
x ′ > x, i.e if P exists. On the other hand Rn+1(X, x) vanishes if there
exists a point x ′ ∈ X with x ′ < x.
Theorem 5.5 (Support property 2). If n ≥ 3 then
supp Dn(x1, . . . , xn−1, xn) ⊆ Γ − n (xn) ∪ Γ + n (xn). (5.20)
39

A proof of this is found in [6] page 167. For n ≤ 2 this property has to be
verified explicitly.
It is here the difficult part of the method of Epstein and Glaser described
in [6] lies, that is in decomposing
such that
Dn(x1, . . . , xn) = Rn(x1, . . . , xn) − An(x1, . . . , xn)
supp Rn ⊆ Γ + n−1 (xn) and supp An ⊆ Γ − n−1 (xn).
40

Chapter 6
Regularly Varying Functions
Before going on in the splitting process we need to consider some
properties of a class of functions which will become important.
Definition 6.1. A positive function ρ(t) is called regularly varying at
t = 0, if it is measurable in (0, t0] for some t0 > 0, and there exists ω ∈ R
such that the limit
ρ(at)
lim = aω
t→0 ρ(t)
exists for all a > 0. In this case ω is called the order of ρ. If ω = 0 the
function is said to be slowly varying.
(6.1)
Note that a regularly varying function always can be reduced to be slowly
varying. That is, let
ρ(t) = t ω r(t). (6.2)
Then (6.1) gives
r(at)
lim = 1, (6.3)
t→0 r(t)
for all a > 0.
The following representation theorem will be important for our use.
Theorem 6.2. Let r : (0, t0] → R+ be a slowly varying function. Then
there exists some t1 ∈ (0, t0) such that
t1 h(s)
r(t) = exp η(t) +
t s ds
holds for all 0 < t < t1. Here η is bounded and measurable on (0, t1], and
η → c for t → 0 (|c| < ∞). Further h(t) is continuous on (0, t1] and
h(t) → 0 for t → 0.
Before proving the theorem we will need some lemmas.
41

Lemma 6.3. Let f : [x0, ∞) → R be a measurable function such that
f(x + k) − f(x) → 0 for x → ∞ (6.4)
for all k ∈ R. 1 Let I = [a, b] ⊂ R be a closed interval then
Proof. Let I = [0, 1]. Assume
sup |f(x + k) − f(x)| → 0 for x → ∞.
k∈I
sup |f(x + k) − f(x)| 0
k∈I
for x → ∞. (6.5)
Then there exists an ɛ > 0 and a sequence (xn, kn) where xn → ∞ and
kn ∈ I such that |f(xn + kn) − f(xn)| ≥ ɛ for all n ∈ N.
Now consider the measurable sets
Fixme: are they?
and
Un = {s ∈ [0, 2]||f(xm + s) − f(xm)| < ɛ/2 for all m ≥ n}
Vn = {t ∈ [0, 2]||f(xm + km + t) − f(xm + km)| < ɛ/2 for all m ≥ n}.
By (6.4) these sets are clearly monotonically increasing towards [0, 2]
hence, letting m denote the Lebesgue measure, there exists an N ∈ N such
that m(Un) > 3/2 and m(Vn) > 3/2 for n, m ≥ N.
Consider the translation V ′ N = VN + kN, then since m(V ′ N ) > 3/2 and
UN, V ′ N ⊂ [0, 3] the two sets must intersect in some point k. Now since
k ∈ UN
|f(xN + k) − f(xN)| < ɛ/2
and since k − kN ∈ VN
|f(xN + k) − f(xN + kN)| = |f(xN + kN + k − kN) − f(xN + kN)| < ɛ/2.
Hence
|f(xN + kn) − f(xN)| = |f(xN + k) − f(xN) − (f(xN + k) − f(xN + kN))|
≤ |f(xN + k) − f(xN)| + |f(xN + k) − f(xN + kN)|
< ɛ/2 + ɛ/2 = ɛ,
contradicting (6.5). This proves the theorem for I = [a, b] where a = 0 and
b = 1.
1 For k < 0 we may only consider x large enough that f be defined. This is no problem,
though, since we are only concerned about the limit x → ∞.
42

Now we want to use this to show it for arbitrary a, b ∈ R+. Let
Then for a fixed,
g(x) := f((b − a)x).
f(x + k) − f(x) = f(x + a + k − a) − f(x + a) + f(x + a) − f(x)
x + a k − a
x + a
= g + − g + f(x + a) − f(x)
b − a b − a b − a
= g(y + h) − g(y) + f(x + a) − f(x), (6.6)
where y and h are defined in the obvious way. Now since k ∈ [a, b] we see
that h ∈ [0, 1]. Further y → ∞ is equivalent to x → ∞. Now by (6.4),
f(x − a) − f(x) → 0 for x → ∞. And by the proof above for I = [0, 1]
Hence by (6.6)
as claimed.
sup |g(y + h) − g(y)| → 0 for y → ∞.
h∈I
sup
k∈[a,b]
|f(x + k) − f(x)| → 0 for x → ∞.
Lemma 6.4. Let f(x) = ln r(e −x ), where r is the slowly varying function
defined by (6.2). Then f satisfies (6.4) and there exists a constant x0 such
that f is bounded on every interval [a, b] where x0 ≤ a.
FIXME: x0 comes from lemma 6.3 and tells us what the domain of f is.
Proof.
lim (f(x + k) − f(x)) = lim
x→∞ x→∞ (ln r(e−(x+k) ) − ln r(e x ))
r(e
= ln lim
x→∞
−(x+k) )
= ln lim
t→0
= ln 1 = 0,
r(e −x )
since (6.3) holds for r(t). Therefore f satisfies (6.4).
By Lemma 6.3 there exists a such that
for all x ≥ a and for all k ∈ [0, 1].
r(at)
r(t) , substituting t = e−x and a = e −k
|f(x + k) − f(x)| < 1 (6.7)
43

We want to show by induction that for x ∈ [a + m − 1, a + m],
|f(x)| ≤ |f(a)| + m. (6.8)
For m = 1, let x ∈ [a, a + 1] then we may write x = a + k for some
k ∈ [0, 1]. Then
|f(a + k)| = |f(a + k) − f(a) + f(a)| ≤ |f(a + k) − f(a)| + |f(a)|
≤ |f(a)| + 1.
In the (n + 1)th step, let x ∈ [a + n, a + n + 1]. Then there exists a
kn ∈ [0, 1] such that x = a + n + kn and
|f(x)| = |f(a + n + kn)| = |f(a + n + kn) − f(a + n) + f(a + n)|
≤ |f(a + n + kn) − f(a + n)| + |f(a + n)|
< |f(a)| + n + 1
by the induction hypothesis. This concludes the proof by induction.
Clearly if x satisfies (6.8), then it is also satisfied by all y ∈ [a, a + m] and
we have proven that f is bounded on the interval [a, b].
Corollary 6.5. Let f(x) = ln r(e −x ). Then there exists a constant x0 such
that f is integrable on every interval [a, b] where x0 ≤ a.
Proof. This follows directly from the definition of integrability since f is
measurable and bounded on [a, b] by Lemma 6.4.
Lemma 6.6. Let a ≥ x0 from Lemma 6.4. Then f(x) = ln r(e−x ) can be
represented for all x ≥ a as
f(x) = g(x) +
x
a
h(s)ds, (6.9)
where g and h are measurable and bounded on every interval [a, b] and
g(x) → g0 (|g0| ≤ ∞) and h(x) → 0 for x → ∞.
Proof. Let x ≥ a we may write f(x) on the form (6.9) by letting
g(x) :=
x+1
FIXME: maybe write out the above
By (6.4) h(x) → 0 for x → ∞.
x
f(x) − f(s)ds +
a+1
h(x) := f(x + 1) − f(x)
44
a
f(s)ds

By Lemma 6.4 the second integral of g(x) is bounded. Consider the first
integral
x+1
x
f(x) − f(s)ds =
≤
1
0
1
by (6.7). Hence g is bounded. Further
lim k(x) = lim
x→∞ x→∞
=
0
1
0
f(x) − f(x + k)dk, substituting s = x + k
|f(x) − f(x + k)|dk < 1,
1
0
f(x) − f(x + k)dk
lim (f(x) − f(x + k))dk = 0,
x→∞
by (6.4) using the Dominated Convergence Theorem with (6.7). We
conclude that
as claimed.
g(x) →
a+1
a
f(s)ds =: g0 for x → ∞, (6.10)
Lemma 6.7. There exists a∗ such that for all x ≥ a∗ x
f(x) = g ∗ (x) +
a ∗
h ∗ (s)ds
where g ∗ and h ∗ are measurable and bounded on every interval [a ∗ , b] and
g ∗ (x) → g ∗ 0 (|g∗ 0 | ≤ ∞) and h∗ (x) → 0 for x → ∞. In fact, h ∗ is continuous
on every interval [a ∗ , b].
FIXME: reformulate.
Proof. Let f ∗ (x) = x
a h(s)ds. Then by (6.9) and (6.10)
For all k > 0,
f(x) − f ∗ (x) = g(x) → g0 for x → ∞.
f ∗ (x + k) + f ∗ x+k
(x) =
a
x+k
=
=
=
x
x+k
x
k
0
h(s)ds +
h(s)ds
a
x
h(s)ds
f(s + 1) − f(s)ds
f(t + x + 1) − f(t + x)dt,
45

substituting t := s − x. We may write
f(t + x + 1) − f(t + x) = [f(x + t + 1) − f(x)] − [f(x + t) − f(x)].
Both terms on the right hand side converge uniformly in t towards 0 on
[0, k] as x → ∞ by Lemma 6.3. Hence
f ∗ (x + k) − f ∗ k
(x) =
0
f(x + t + 1) − f(x + t)
≤ k sup |f(x + t + 1) − f(x + t)| → 0
t∈[0,k]
for x → ∞. The same way it can be shown to hold for k < 0 as well.
Hence f ∗ satisfies (6.4).
Note that f ∗ is continuous in any interval [a, b], because given ɛ > 0,
choose δ = ɛ/ sup [a,b] |f(s + 1) − f(s)|. Then
x0
|f(x0) − f(x)| = |
≤
x x0
x
f(s + 1) − f(s)ds|
|f(s + 1) − f(s)|ds
≤ |x0 − x| sup
[a,b]
|f(s + 1) − f(s)|
< δ sup |f(s + 1) − f(s)| < ɛ.
[a,b]
Now we want to construct a representation of f ∗ .
First note that since f ∗ is continuous it is measurable.
Thus, by Lemma 6.3 we may conclude that
sup |f
[c,d]
∗ (x + k) − f ∗ (x)| → 0 for x → ∞.
Further, since by Lemma 6.4 f is bounded on every closed interval so is f ∗ ,
because
sup
[c,d]
|f ∗ (x)| = sup |
x
[c,d] a
x
≤ sup
[c,d]
a
f(s + 1) − f(s)ds|
|f(s + 1) − f(s)|ds
≤ (d − a) sup |f(t + 1) − f(t)| < ∞.
[a,d]
Since f ∗ is measurable and bounded on closed intervals, it is integrable.
By Lemma 6.6 f ∗ has the following representation for x ≥ a ∗ ,
FIXME: what is a ∗ ?.
46

where
and
k ∗ (x) =
f ∗ (x) = k ∗ x
(x) +
a∗ h ∗ (s)ds,
x+1
f
x
∗ (x) − f ∗ a∗ +1
(s)ds +
a∗ h ∗ (x) = f ∗ (x + 1) − f ∗ (x).
Note that h∗ is continuous, since f ∗ is continuous.
Now (6) implies
f(x) = g(x) + f ∗ (x) = g(x) + g ∗ x
(x) +
a∗ Finally, let g ∗ (x) := g(x) + k ∗ (x).
f ∗ (s)ds
h ∗ (s)ds
Now we are finally ready to prove the representation theorem.
Proof of Theorem 6.2. Let f(x) := ln r(e−x ). By Lemma 6.7 f has the
following representation
That is
ln(r(e −x )) = g ∗ x
(x) +
a∗ ln r(t) = g ∗ (− ln t) +
− ln t
a ∗
h ∗ (s)ds.
h ∗ (s)ds.
Letting ν(t) := g ∗ (− ln(t)), substituting s := − ln t ′ and h(t ′ ) := h ∗ (− ln t ′ ),
Thus,
where t1 = e −a∗
t
ln r(t) = η(t) −
e−a∗ h(t ′ )
dt′
t ′
t1 h(t
r(t) = exp(η(t) +
t
′ )
t ′ dt′ ),
as claimed.
Note that the representation of 6.2 only holds for t < t1 but since we are
only concerned with the limit t → 0 we will usually not have any problem
using the representation theorem. The following theorem will be important
later when we shall split distributions.
Corollary 6.8. Let ρ be a regularly varying function and ɛ > 0, then
there exist some S0(ɛ) and non-negative constants C and C ′ such that
C ′ t ω+ɛ ≤ ρ(t) ≤ Ct ω−ɛ , for 0 ≤ t ≤ s0(ɛ). (6.11)
47

Proof. Reduce ρ to be slowly varying
ρ(t) = t ω r(t) = t ω t1
exp η(t) +
t
h(s)
s ds
, for 0 < t < t1,
where we substitute r by the representation given by Theorem 6.2. Since
h(s) → 0 for s → 0 there exists s0(ɛ) ≤ t1 such that |h(s)| ≤ ɛ for s ≤ s0(ɛ).
So if 0 < t ≤ s0(ɛ), then
s0(ɛ)
t
h(s)
s ds
≤
s0(ɛ)
t
h(s)
s ds ≤
s0(ɛ)
t
ɛ
s ds = ɛ ln s0(ɛ) − ɛ ln t
s0(ɛ)
= ɛ ln .
t
Hence
ρ(t) = t ω s0(ɛ) h(s)
exp η(t) +
t s ds
≤ t ω
s0(ɛ)
exp η(t) + ɛ ln
t
= t ω−ɛ e η(t) s ɛ 0(ɛ)
≤ Ct ω−ɛ ,
where C = s ɛ 0 (ɛ) sup 0

Chapter 7
Splitting of Numerical
Distributions
7.1 The Singular Order of a Distribution
We expect the operator-valued distributions to be expanded in terms of
free fields as follows1 Dn(x1, . . . , xn) =
:
ψ(xj)d k n(x1, . . . , xn)
ψ(xl) ::
A(xm) :,
k
j
where the numerical distributions d k n ∈ S ′ (R 4n ) have the causal support
property (5.20). We have to assume them to be tempered in order to use
the Fourier transformation. Our aim is to split these distributions
where
d k n(x) = rn(x) − an(x),
supp rn ⊂ Γ + n−1 (xn) and supp an ⊂ Γ − n−1 (xn).
We have assumed translational invariance thus we may assume xn = 0 and
thus only consider
d(x) := d(x1, . . . xn−1, 0) ∈ S ′ (R m ), m = 4(n − 1).
In the Hilbert space setting of our earlier example we would do the
splitting as
rn(x) = χn(x)d k n(x), where χn(x) =
n−1
j=1
l
Θ(x 0 j − x 0 n) =
m
n−1
1 Note that the double dots denote normal ordering. See Definition 4.4
2 We write it like this to show what it would be if xn = 0.
49
j=1
Θ(x 0 j) 2 .

In our setting this is not defined, but it might help us get an idea of what
to do.
The discontinuity plane of χn(x) is
Pχn = {x = (x1, . . . , xn−1, 0)|x 0 j = 0 for all j}.
Now if y ∈ supp d ⊂ Γ + n (0) ∪ Γ− n (0), then y2 j
then y0 j = 0 for all j hence
−
3
(y i j) 2 ≥ 0
i=1
≥ 0 for all j. If also y ∈ Pχn,
This is only possible if yi = 0 for i = 1, 2, 3. Hence the intersection is the
origin, that is, supp d ∩ Pξn = {0}.
We now define some tools to investigate the singularity at the origin.
Definition 7.1. The distribution d(x) ∈ S ′ (R m ) is said to have a
quasi-asymptotics d0(x) at x = 0 with respect to a positive continuous
function ρ(δ), δ > 0, if the limit
exists in S ′ (R m ) 3 .
lim
δ→0 ρ(δ)δmd(δx) = d0(x) = 0 (7.1)
The equivalent definition in momentum space reads
Definition 7.2. The distribution ˆ d(p) ∈ S ′ (R m ) has quasi-asymptotics
ˆd0(p) at p = ∞ if
exists for all ˇ φ ∈ S(R m ).
lim
δ→0 ρ(δ)〈 ˆ d( p
δ ), ˇ φ(p)〉 = 〈 ˆ d0, ˇ φ〉 = 〈0, ˇ φ〉 (7.2)
We should show these definitions are indeed equivalent. This follows since
δ m 〈d(δx), φ(x)〉 = 〈d(x), λδφ(x)〉
= 〈 ˆ d(p), (λδφ)ˇ(p)〉
= 〈 ˆ d(p), δ m (λ 1/δ ˇ φ)(p)〉, by (7.4)
= 〈 ˆ d( p
δ ), ˇ φ(p)〉, (7.3)
by Proposition 1.12 and using
(λδφ)ˇ(p) = (2π) −m
e ix·p
x
φ dx = (2π)
δ
−m δ m
e iy·δp φ(y)dy
= δ m λ 1/δ ˇ φ(p). (7.4)
3 An explanation of why we require d0 = 0 is found after Definition 7.4
50

Let’s show that quasi-asymptotics do in fact say something about the
origin, only.
Say, d(x) = d1(x) + d2(x) where supp d1 is compact containing {0} and
supp d2 is bounded away from {0}. Then
lim
δ→0 ρ(δ)δm 〈d2(δx), φ0(x)〉 = lim ρ(δ)〈d2(x), φ0(
δ→0 x
)〉 = 0,
δ
for all φ0 ∈ C ∞ 0 (Rm ) due to Proposition 1.12 and the support properties of
d2. Since C ∞ 0 is dense in S the distribution also vanishes on S, hence
lim
δ→0 ρ(δ)δm 〈d1(δx), φ(x)〉 = 〈d0, φ〉 for all φ ∈ S.
FIXME: why does K0 need to be compact?
Lemma 7.3. Let ρ be the positive continuous function of definitions 7.1
and 7.2. Then there exists ω ∈ R such that
for all a > 0.
Proof. By Proposition 1.12
ρ(aδ)
lim
δ→0 ρ(δ) = aω , (7.5)
lim
δ→0 ρ(δ)〈 ˆ d( p
δ ), (λ1/a ˇ φ)(p)〉 = a −m lim ρ(δ)〈
δ→0 ˆ d( p
aδ ), ˇ φ(p)〉
= a −m lim
δ→0
= a −m lim
δ→0
ρ(δ)
ρ(aδ) ρ(aδ)〈 ˆ d( p
ρ(δ)
ρ(aδ) 〈 ˆ d0(p), ˇ φ(p)〉,
since the limit
lim
δ→0 ρ(aδ)〈 ˆ d( p
aδ ), ˇ φ(p)〉 = 〈 ˆ d0(p), ˇ φ(p)〉
exists. Thus the following is defined
aδ ), ˇ φ(p)〉
ρ(aδ)
ρ0(a) := lim
δ→0 ρ(δ) = a−m 〈 ˆ d0(p), ˇ φ(p)〉
〈 ˆ d0(p), ˇ . (7.6)
φ(ap)〉
Note that ρ0 is continuous, since λ 1/a : S → S is continuous, and that
ρ(aδ)
ρ0(a)ρ0(b) = lim
δ→0 ρ(δ) lim
ρ(bδ) ρ(abδ)
= lim
δ→0 ρ(δ) δ→0 ρ(bδ) lim
ρ(bδ) ρ(abδ)
= lim
δ→0 ρ(δ) δ→0 ρ(δ)
Now define a function by f(s) = ln(ρ0(e s )). Then
= ρ0(ab).
f(s) + f(t) = ln(ρ0(e s )) + ln(ρ0(e t )) = ln(ρ0(e t )ρ0(e s )) = ln(ρ0(e s+t ))
51
= f(s + t).

Let n, m ∈ N then
and
Hence
mf( n n
n
) = f( ) + · · · + f(
m m m ) = f(m
m terms
n
) = f(n)
m
f(n) = nf(1).
f( n n
) =
m m f(1).
Since ρ0 is continuous f(x) = xf(1) for all x > 0.
Thus
ρ0(e x ) = e f(x) = e xf(1) ,
that is, substituting x = ln a,
Let ω := f(1).
ρ0(a) = e f(1) ln(a) ln af(1)
= e = a f(1) .
Note that (7.5) is the condition of a regularly varying function.
We regard ω as an indication of how singular a causal distribution is near
{0}. This leads us to define:
Definition 7.4. The distribution d ∈ S ′ (R m ) is called singular of order ω,
if it has a quasi-asymptotics d0(x) at x = 0, or its Fourier transform has
quasi-asymptotics ˆ d0(p) at p = ∞, respectively, with power-counting
function ρ(δ) satisfying
for each a > 0.
ρ(aδ)
lim
δ→0 ρ(δ) = aω ,
Note that d0 = 0 in (7.1) and the corresponding requirement in (7.2),
respectively, are required to make sure that the singular order is uniquely
defined. If we skipped the requirement any ω ′ ≥ ω could be chosen as
singular order.
Fixme: It this explained well enough?
Lemma 7.5. The distributions d0 and ˆ d0 are homogeneous of degree
−(m + ω) and ω respectively.
Proof. From (7.6) and (7.5),
Hence by Proposition 1.12
a −ω 〈 ˆ d0(p), ˇ φ(p)〉 = a m 〈 ˆ d0(p), ˇ φ(ap)〉.
〈 ˆ d0( p
a ), ˇ φ(p)〉 = a m 〈 ˆ d0(p), ˇ φ(ap)〉 = a −ω 〈 ˆ d0(p), ˇ φ(p)〉. (7.7)
52

Thus the condition of Definition 1.14 is satisfied by ˆ d0.
Further from (7.3)
and
hence by (7.7)
a m 〈d0(ax), φ(x)〉 = 〈d0(x), φ( x
a )〉 = 〈 ˆ d0( p
a ), ˇ φ(p)〉.
〈 ˆ d0(p), ˇ φ(p)〉 = 〈d0(x), φ(p)〉
〈d0(ax), φ(x)〉 = a −(m+ω) 〈d0(x), φ(x)〉.
The condition of Definition 1.14 is satisfied by d0.
The singular order now gives us a way to distinguish two different
situations in the splitting process. We investigate the splitting first in the
case where ω < 0 and then for ω ≥ 0.
53

7.2 Case I: Negative Singular Order
7.2.1 Existence
In this subsection we will show that there exists a decomposition of d into
an advanced part a and a retarded part r.
Choose ɛ > 0 such that ω + ɛ < 0 then t ω+ɛ → ∞ for t → 0 − hence by
(6.11) ρ(t) → ∞ for t → 0. Hence by (7.1)
lim
δ→0
〈d(x), φ(x
δ
)〉 = lim
δ→0 δm 〈d0(x), φ(x)〉
〈d(δx), φ(x)〉 = lim
= 0. (7.8)
δ→0 ρ(δ)
Let v = (v1, . . . , vn−1) ∈ Γ + n−1 (0) be an arbitrary but fixed vector and
define a hyperplane by
PH := {x ∈ M n−1 n−1
|v · x := 〈vj, xj〉 = 0}.
Lemma 7.6. All products satisfy
and
j=1
〈vj, xj〉 ≥ 0 if x ∈ Γ + n−1 (0) (7.9)
〈vj, xj〉 ≤ 0 if x ∈ Γ − n−1
(0). (7.10)
Proof. For all j, (v0 j )2 − 3 i=1 (vi j )2 ≥ 0 hence v0 j ≥
3
i=1 (vi j )2 since
v0 j ≥ 0. Now if x ∈ Γ+ n−1 (0), then for all j
〈vj, xj〉 = v 0 j x 0 j −
3
i=1
v i jx i j ≥ v 0 j x 0 j −
3
i=1
If on the other hand x ∈ Γ − n−1 (0), then x0 j
〈vj, xj〉 = v 0 j x 0 j −
3
i=1
(v i j )2
v i jx i j ≤ v 0 j x 0 j − x 0 j
≤ x 0 j
FIXME: read this through carefully.
3
Thus PH splits causal support (5.20) of d.
54
i=1
3
i=1
(x i j )2 ≥ v 0 j x 0 j − v 0 j x 0 j = 0
≤ 0 for all j, hence
3
i=0
(v i j )2 − x 0 j
(v i j )2
3
i=0
(v i j )2 = 0

Now choose a monotonous function χ0 ∈ C∞ (R) such that
⎧
⎪⎨ 0, t ≤ 0,
χ0(t) = s ∈ [0, 1),
⎪⎩
1,
0 < t < 1,
t ≥ 1.
The following theorem gives us a retarded distribution.
Theorem 7.7. The limit
exists.
v · x def
lim χ0( )d(x) = θ(v · x)d(x) (7.11)
δ→0 δ
In order to prove this we need the following lemma.
Lemma 7.8. Let a1 > 1, then
ψ0( x
)d(x) :=
δ
uniformly in a ≥ a1.
χ0(a
v · x v · x
) − χ0(
δ δ )
d(x) → 0 for δ → 0
Proof. Let U be a neighborhood of Γ + (0) ∪ Γ − (0). Choose a function
ψ1 ∈ C ∞ (R m ) such that
Then, by (5.20)
⎧
⎪⎨
ψ1(x) =
⎪⎩
0, Rm \ U.
〈ψ0( x
δ
1, x ∈ Γ + (0) ∪ Γ − (0),
s ∈ (0, 1], U \ Γ + (0) ∪ Γ − (0),
x x
)d(x), φ(x)〉 = 〈ψ0( )d(x), ψ1(
δ δ )φ(x)〉
= 〈φ(x)d(x), ψ0( x
δ
for any φ ∈ S(R m ), since ψ0ψ1 ∈ C ∞ 0 (Rm ).
By (7.1)
dδ def
= ρ(δ)δ m d(δx) → d0
for δ → 0. Further
in C ∞ (R m ) since
φ(δx) → φ(0) for δ → 0
D α (φ(δx) − φ(0)) = δ |α| (D α φ)(δx) − D α φ(0).
Therefore by [12] Theorem 6.1.8
x
)ψ1( )〉, (7.12)
δ
φ(δx)dδ → φ(0)d0 for δ → 0. (7.13)
55

Therefore φ(x)d(x) also has quasi-asymptotics of order ω < 0 at x = 0
with respect to ρ.
Hence by (7.8) and (7.12)
〈ψ0( x
)d(x), φ(x)〉 → 0 for δ → 0 (7.14)
δ
for all φ ∈ S(R m ).
Now for fixed a1 > 1 the convergence is uniform on the compact interval
1 ≤ a ≤ a1.
We want to show the uniformity of the limit in a ′ ≥ a1 > 1.
To this end, we claim that
Claim.
χ0
n v · x
v · x
n−1
a − χ0 =
δ
δ
j=0
ψ0
j x
a
δ
We prove this by induction. It holds for n = 1 by definition. Assume it
holds for n = k − 1. Then,
k v · x
v · x
χ0 a − χ0
δ
δ
k v · x
k−1 v · x
k−1 v · x
v · x
= χ0 a − χ0 a + χ0 a − χ0
δ
δ
δ
δ
k−1 x
k−2
j x
= ψ0 a + ψ0 a .
δ
δ
j=0
Which proves the claim.
We write a ′ = a n , for a suitable n and 1 ≤ a ≤ a1. Now the problem is
translated into showing uniformity in n. By (7.13) we may choose δ0 such
that for δ < δ0
|ρ(δ)〈φ(x)d(x), ψ1(x/δ)ψ0(x/δ)〉| ≤ |〈φ(0)d0(x), ψ1(x)ψ0(x)〉| + 1 = K.
Note that δ/a j ≤ δ < δ0. Thus
From (7.5) we know that
Now since a −ω > a −ω
1
Since δ/a q < δ for all q ≥ 1
|〈φ(x)d(x), ψ1(a j x/δ)ψ0(a j x/δ)〉| ≤ K/ρ(δ/a j )
ρ(δ/a)
ρ(δ) .
for ω ≤ 0 we may assume, that for δ < δ0
ρ(δ/a) ≥ a −ω
1 ρ(δ).
ρ(δ/a j ) ≥ a −ω
1 ρ(δ/aj−1 ) ≥ a 2ω
1 ρ(δ/a j−2 ) ≥ a −jω
1
56
ρ(δ).

Thus
n−1
|
j=0
〈φ(x)d(x), ψ1(a j x/δ)ψ0(a j n−1
x/δ)〉| ≤
But ρ(δ) → ∞ Hence the convergence is uniform.
We now prove the theorem.
K/ρ(δ/a j )
j=0
≤ Kρ(δ) −1
∞
j=0
a jω
1
= Kρ(δ) −1 (1 − a ω 1 ) −1 .
Proof of Theorem 7.7. Another way to state the theorem is that the
limit in (7.11) exists for all sequences δn → 0. This is done by showing
〈χ0( v·x)d(x),
φ(x)〉 is a Cauchy sequence. Consider
δn
v · x v · x
v · x v · x
χ0( ) − χ0( ) = χ0(an,m ) − χ0( ), (7.15)
δm
δn
where an,m = δn/δm. Assuming δm < δn, we know from Lemma 7.8 that
for all φ and for all ɛ > 0 there exists some δ0 > 0 such that for all δ < δ0
|〈(χ0(an,m
v · x
δ
δn
δn
v · x
) − χ0( ))d(x), φ(x)〉| < ɛ
δ
Hence given φ and ɛ > 0 we may choose δ0 and N such that δn < δ0 for all
n, m ≥ N
v · x v · x
|〈(χ0( ) − χ0( ))d(x), φ(x)〉| < ɛ.
δm
δn
Hence 〈χ0( v·x)d(x),
φ(x)〉 is a Cauchy sequence.
δn
Now suppose δn and δn ′ are two sequences converging towards 0. Then as
above, with an,n ′ = δn ′/δn in (7.15),
v · x
v · x
〈χ0( )d(x), φ(x)〉 − 〈χ0( )d(x), φ(x)〉 → 0
δn
as n → ∞. Hence the limit exists.
Now we may define
Definition 7.9.
δn ′
v · x
r(x) = lim χ0( )d(x) = θ(v · x)d(x).
δ→0 δ
Similarly there exists an advanced distribution.
57

Corollary 7.10. The limit
exists.
−v · x
lim χ0( )d(x) := −a(x)
δ→0 δ
Proof. This follows by the proof of Lemma 7.8 and Theorem 7.7 with χ0(t)
substituted by χ0(−t).
FIXME: think about this.
Note that by (7.9) and (7.10),
and
Hence
v · x
supp r ⊂ supp(lim χ0(
δ→0 δ )) ⊂ Γ+ n−1 (0)
supp a ⊂ supp(lim
δ→0 χ0(
−v · x
)) ⊂ Γ
δ
− n−1 (0).
r(x) − a(x) = Θ(v · x)d(x) − Θ(−v · x)d(x) = d(x). (7.16)
These support properties allows us to apply a and r to discontinuous
test-functions as follows
〈r, φ〉 = 〈r(x), Θ(v · x)φ(x)〉 and 〈r, (1 − Θ)φ〉 = 0
〈a, φ〉 = 〈a(x), (1 − Θ)(v · x)φ(x)〉 and 〈a, Θφ〉 = 0.
Thus by (7.16) we may split d in a retarded and an advanced part, simply
by multiplication with Θ,
and
Note that by (7.1)
and
〈d, Θφ〉 = 〈r, Θφ〉 − 〈a, Θφ〉 = 〈r, φ〉
〈d, (1 − θ)φ〉 = 〈r, (1 − Θ)φ〉 − 〈a, (1 − Θ)φ〉 = −〈a, φ〉.
〈d0, Θφ〉 = lim
δ→0 ρ(δ)δ m 〈d(δx), Θφ(x)〉 = lim
δ→0 ρ(δ)〈d(x), Θλδφ(x)〉
= lim ρ(δ)〈r(x), φ(x/δ)〉
δ→0
= lim ρ(δ)δ
δ→0 m 〈r(δx), φ(x)〉
〈d0, (1 − Θ)φ〉 = lim
δ→0 ρ(δ)δ m 〈d(δx), (1 − Θ)φ(x)〉 = lim
δ→0 ρ(δ)〈a(x), φ(x/δ)〉
We see that r and a have same singular order as d.
58
= lim
δ→0 ρ(δ)δ m 〈a(δx), φ(x)〉

7.2.2 Uniqueness
We still need to prove that the decomposition of d is unique. To this end
we need the following result
Theorem 7.11. Let u ∈ D ′ (R n ) with support supp u = {0}. Then there
exists a N ∈ N such that
where cα ∈ C.
u =
|a|≤N
cα∂ α δ,
This theorem is a consequence of the following lemmas.
Lemma 7.12.
1
k! ∂k t f(xt) =
|α|=k
x α
α! f (α) (xt) (7.17)
Proof. The first we note that by the chain rule, we may write the left hand
side as
1
k! ∂k t f(xt) = 1
n
k
∂
xi f(y)
k! ∂yi
i=1
y=xt
It is this form we will use when proving (7.17) by induction. That is, we
want to prove
1
n
∂
k xi f(y) =
k! ∂yi
xα α! ∂αf(y) i=1
59
|α|=k

It is obvious for k = 0 and k = 1. Now, assume it holds for k = m. Then
1
n
(m + 1)!
= 1
n
m + 1
j=1
= 1
n
m + 1
= 1
m + 1
= 1
m + 1
= 1
m + 1
= 1
m + 1
=
j=1
n
i=1
xj
xj
xi
∂
∂yj
∂
∂yj
∂
∂yi
m+1
f(y)
1
n
m!
x
xj
j=1 |α|=m
α
α!
n
j=1 |α|=m
|α|=m j=1
|β|=m+1 j=1
n
|β|=m+1 j=1
|α|=m
i=1
xi
∂
∂yi
x α
α! ∂α f(y)
∂
∂
∂yj
α f(y)
m
f(y)
xα+(0,...,0,1,0,...,0) ∂
α!
α+(0,...,0,1,0,...,0) f(y)
n
x
(αj + 1)
α+(0,...,0,1,0,...,0)
(α + (0, . . . , 0, 1, 0, . . . , 0))! ∂α+(0,...,0,1,0,...,0) f(y)
n
x β
β! ∂β f(y)
x
βj
β
β! ∂βf(y), where β = α + (0, ..., 0, 1, 0, ...0)
where only the jth entry of (0, . . . , 0, 1, 0, . . . , 0) is non-zero and where we
have used
(α + (0, . . . , 0, 1, 0, . . . , 0))! = (α1, α2, . . . , αj + 1, . . . , αn)!
Finally, rename β = α.
= α1!α2! · · · (αj + 1)! · · · αn!
= (αj + 1)α!.
Lemma 7.13. Let f ∈ C ∞ 0 (Rn ) and t ∈ R, then
f(x) ≤
for any N ≥ 1.
|α|≤N−1
xα α! f (α)
N N
(0) + |x|
α!
|α|=N
sup |f (α) (x ′ )|
′
|x | < |x| ,
60

Proof. Let F : t ↦→ f(xt). By application of Taylor’s formula [10] to F ,
f(x) = F (1) =
=
N−1
k=0
N−1
k=0
=
1
k! ∂k t F (0) +
|α|≤N−1
|α|=k
1
(N − 1)!
xα α! f (α)
(0) + N
xα α! f (α) (0) +
|α|=N
1
(1 − t)
0
N−1 ∂ N t F (t)dt
1
N−1
(1 − t)
xα α! f (α) (xt)dt
0
|α|=N
x α
α! fα(x), (7.18)
where fα(x) = N 1
0 (1 − t)N−1 f (α) (xt)dt and where we have used (7.17).
Now, since |x α | ≤ maxj |xj| N ≤ |x| N when |α| = N, we can make the
following estimate
As wanted.
|α|=N
xα α! fα(x)
N 1
≤ |x|
α!
|α|=N
fα(x)
N
= |x|
|α|=N
N
≤ |x|
|α|=N
N
≤ |x|
|α|=N
1
α! N
1
(1 − t)
0
N−1 f (α) (xt)dt
N
α! sup |(1 − t)
t∈[0,1]
N−1 f (α) (xt)|
N
α! sup |f (α) (x ′ )| |x ′ | < |x| .
Lemma 7.14. Let u ∈ D(R n ) with support supp u = 0, then there exists
an N ≥ 0 such that 〈u, φ〉 = 0 for all φ ∈ C ∞ 0 (Rn ) and ∂ α φ(0) = 0 when
|α| ≤ N.
Proof. Let ψ ∈ C ∞ 0 (Rn ) be defined such that
Let ɛ ∈ (0, 1). Then
⎧
⎪⎨ 1, |x| < 1/2,
ψ(x) = s ∈ [0, 1],
⎪⎩
0,
1/2 ≤ |x| ≤ 1,
|x| > 1.
〈u, φ〉 = 〈u(x), φ(x)ψ(x/ɛ)〉, for all φ ∈ C ∞ 0 (R n ),
since supp u ⊂ {x ∈ R n ||x| < ɛ/2} =: U and ψ(x/ɛ) = 1 on U.
If x > 1, then x/ɛ > x > 1, hence the map x ↦→ φ(x)ψ(x/ɛ) is supported in
the unit-ball. From now on we may therefore assume that |x| ≤ ɛ.
61

By compactness of the unit-ball and the definition of distributions there
exists a real number C ≥ 0 and a non-negative integer N such that
|〈u, φ〉 ≤ C
α≤N
sup |∂ α (φ(x)ψ(x/ɛ))|, for all φ ∈ C ∞ 0 (R n ). (7.19)
If ∂ α φ(0) = 0 when |α| ≤ N and |β| ≤ N, we may apply Lemma 7.13 to
∂ β φ as follows
|∂ β φ(x)| ≤ |x| N−β+1
≤ ɛ N+1−β
By Leibniz’ formula,
|γ|=N+1−β
|γ|=N+1−β
∂ α (φ(x)ψ(x/ɛ))) =
α=β+γ
=
α=β+γ
N + 1 − β
γ!
N + 1 − β
γ!
sup |∂ γ (x ′ )| |x ′ | < |x|
sup |∂ γ (x)| |x| < 1 . (7.20)
α!
β!γ! ∂β φ(x)∂ γ (ψ(x/ɛ))
α!
β!γ! ɛ−|γ| ∂ β φ(x)(∂ γ ψ)(x/ɛ). (7.21)
Combining (7.20) and (7.20) together with the fact that ∂ γ ψ(x/ɛ has
compact supported contained in the unit-ball, we see that there exists
constants Cα for each α ≤ N which is independent of ɛ such that
FIXME: (start) Could say that supp ∂ γ compact (in unit-ball) so attains
its maximum value. For a fixed α there is only finitely many possible γ, so
the maximum of these supremums gives us the estimate. Fixme: (end)
Plugging this into (7.19),
|∂ α (φ(x)ψ(x/ɛ)) ≤ Cαɛ N+1−|β| e −|γ| = Cαɛ N+1−|α|
|〈u, φ〉| ≤ C
|α|≤N
letting K be the constant. Finally, let ɛ → 0.
FIXME: maybe this could be better.
Cαɛ N+1−|α| ≤ Kɛ,
Proof of Theorem 7.11. Pick N and ψ as in Lemma 7.14.
By (7.18) every φ ∈ C ∞ 0 (Rn ) can be written on the form
φ = ψ(x)
|α|≤N
62
x α
α! ∂α φ(0) + φ ′

with N and ψ as in the lemma. Then φ ′ ∈ C ∞ 0 (Rn ) and ∂ α φ ′ (0) = 0 for
|α| ≤ N. Hence, by Lemma 7.14 〈u, φ ′ 〉 = 0. Thus
〈u, φ〉 = 〈u(x), ψ(x)
|α|≤N
= 〈u(x), ψ(x)
=
|α|≤N
=
|α|≤N
with cα = (−1) |α| 〈u(x),ψ(x)xα 〉
α! .
|α|≤N
x α
α! ∂α φ(0) + φ ′ 〉
x α
α! ∂α φ(0)〉
〈u(x), ψ(x)xα 〉
∂
α!
α φ(0)
cα∂ α δ,
Now to prove the decomposition d = r − a is unique assume there are two
solutions {r1, a1} and {r2, a2} to the splitting problem. That is,
d = r1 − a1 = r2 − a2 hence r1 − r2 = a1 − a2.
The support of the differences is {0}
FIXME: Is it??
hence by Theorem 7.11
r1 − r2 = a1 − a2 =
cα∂ α δ. (7.22)
|α|≤N
We want to investigate each term of the sum. From (8.6) (FIXME: maybe
better to place this equation before) we know that ˆ δ = 1, hence by
Theorem 5.17 in [2]
Hence,
ρ(δ)〈 cα∂ α δ( p
δ ), ˇ φ(p)〉 = ρ(δ)cαi |α| 〈
∂ α δ = i |α| D α δ = i |α| p αˆ δ = i |α| p α
p
α ,
δ
ˇ φ(p)〉 = ρ(δ)
δ |α| 〈 cα∂ αδ(p), ˇ φ(p)〉.
Thus ρ(δ) = δ |α| for dα(x) = cα∂ α δ(x) by (7.2) and then clearly the
singular order is ωα = |α|.
For r1 − r2 =
|α|≤N cα∂ α δ we then must have |α| = ωα ≤ ω for all α.
This implies
r1 − r2 =
|α|≤ω
cα∂ α δ (7.23)
Further in our case ω < 0. Hence all the cα must vanish in (7.22). This
shows r1 = r2 and a1 = a2. Hence the splitting is unique.
63

7.3 Case II: Positive Singular Order
Choose positive ɛ < 1 then since C ′ δ ɛ−1 = C′ δ ω+ɛ
δ ω+1
≤ ρ(δ)
δ ω+1 by (6.11),
ρ(δ)
→ ∞ for δ → 0
δω−1 Hence if we choose a multi-index b such that |b| = ω + 1, then
lim
δ→0 〈d(x)xb , ψ(x/δ)〉 = lim δ
δ→0 m+ω+1 〈d(δx)x b , ψ(x)〉
= lim
δ→0
δ m+ω
ρ(δ) 〈d0(x), x b ψ(x)〉 = 0, (7.24)
by (7.1).
We want to show the splitting can be done for test-functions φ satisfying
Definition 7.15. We define 4
D a φ(0) = 0 for |a| ≤ ω.
S ω (R m ) = {φ ∈ S(R m )|D α φ(0) = 0 for all α with |α| ≤ ω}
This is the subspace of S(R m ) of test-functions which vanish up to order ω
at 0.
FIXME: show ∞ > sing order(d) = ω ≥ 0 ⇒ d ∈ S ′ω (R m ).
Definition 7.16. We define operators W (ω,w) : S(R n ) → S ω (R n ) by
(W (ω,w)φ)(x) := φ(x) − w(x)
for each w(x) ∈ S(R n ) such that
ω
|a|=0
w(0) = 1 and D α w(0) = 0 for 1 ≤ |a| ≤ w.
x a
a! (Da φ)(0) (7.25)
Recognizing Taylor’s formula given by (7.18) we note that W is actually a
modified Taylor subtraction operator which projects S(Rm ) into the space
of test-functions which vanish up to order ω at 0.
Note that we may write
(W φ)(x) =
x b ψb(x)
|b|=ω+1
by (7.18)
We now want to apply θ from Theorem 7.7.
4 If ω is not an integer use [ω] the largest integer such that [ω] < ω. The reason I
haven’t written it like this straight away, is that ω always seems to be an integer in QED.
64

Lemma 7.17. Let d (b) (x) = x b d(x), then d (b) has quasi-asymptotics
x b d0(x) with respect to ρ (b) (δ) = δ −|b| ρ(δ), and it has singular order ω − |b|.
Proof. The quasi-asymptotics follows from
lim
δ→0 ρ(b) δ m d (b) (δx) = lim
delta→0 ρ(b) (δ)δ m+|b| x b d(δx) = lim
δ→0 ρ(δ)δ m x b d(δx)
The singular order is given by
for all a > 0.
= x b d0(x)
ρ
lim
δ→0
(b) (aδ)
ρ (b) (δa)
= lim
(δ) δ→0
−|b| ρ(aδ)
δ−|b| = aω−|b|
ρ(δ)
By the previous lemma x b d(x) has singular order ω − |b|. Hence it is
negative for b ≥ ω + 1. Thus we may define
Definition 7.18.
and
〈r(x), φ〉 := 〈d, θ(v · x)W φ〉
a(x) := d(x) − r(x).
By construction supp r ⊂ Γ + n−1
x ∈ Γ + n−1 (0) \ {0} by (7.25), since a test-function φ ∈ S supported in
Γ + n−1 (0) \ {0} satisfies Dαφ(0) = 0 for all α.
Like for the case ω < 0, r and a have the same singular order as d. This
follows from the equations
(0). Further r(x) = d(x) for
lim ρ(δ)〈r(x), φ(x/δ)〉 = lim ρ(δ)〈d(x), (θW φ)(x/δ)〉
δ→0 δ→0
= 〈d0(x), θW φ(x)〉.
In contrast to the case where ω < 0 the splitting is not unique for w ≥ 0
because of the dependence on w. Again the support of the difference
between two solutions is {0}, hence by (7.23)
r1 − r2 =
Cα∂ α δ
|α|≤ω
The coefficients Cα have to be fixed by additional normalization conditions.
65

Chapter 8
Application to QED
8.1 Using the Game Plan
In this section we will use the game plan from the end of Section 5.1 to
find D2 from T1 and then show how to find T2.
In QED 1
T1(x) = ie : ψ(x)γ µ ψ(x) : Aµ(x),
By (5.1) this means that
˜T1(x) = −T1(x) = −ie : ψ(x)γ µ ψ(x) : Aµ(x).
We now want to construct an advanced distribution according to (5.7).
This is done using Theorem 4.6 remembering that the only contractions
existing in QED are given by equations (4.1), (4.2) and (4.3).
1 [6] page 183
66

A ′ 2(x1, x2) = ˜ T1(x1)T1(x2) = −T1(x1)T1(x2)
= e 2 γ µ
abγν cd : ψa(x1)ψb(x1) :: ψc(x2)ψd(x2) : Aµ(x1)Aν(x2)
: ψa(x1)ψb(x1)ψ c(x2)ψd(x2) :
= e 2 γ µ
ab γν cd
= e 2 γ µ
ab γν cd
+ : ψ a(x1) | ψb(x1)ψ c(x2) | ψd(x2) :
+ : | ψ a(x1)ψb(x1)ψ c(x2)ψd(x2) | :
+ : | ψ a(x1) | ψb(x1)ψ c(x2) | ψd(x2) | :
×
: ψa(x1)ψb(x1)ψ c(x2)ψd(x2) :
: Aµ(x1)Aν(x2) : + : | Aµ(x1)Aν(x2) | :
+ 1
i S(+)
bc (x1 − x2) : ψa(x1)ψd(x2) :
+ 1
i S(−)
da (x2 − x1) : ψb(x1)ψc(x2) :
− S (+)
bc (x1 − x2)S (−)
da (x2
− x1)
×
: Aµ(x1)Aν(x2) : +gµνiD (+)
0 (x1 − x2)
. (8.1)
From (5.8) we see that the retarded distribution is given by (8.1) simply by
substituting x1 ↔ x2. For convenience we also interchange the indices
µ ↔ ν, a ↔ c and b ↔ d, arriving at
R ′ 2(x1, x2) = T1(x2) ˜ T1(x1) = −T1(x2)T1(x1)
: ψa(x1)ψb(x1)ψ c(x2)ψd(x2) :
= e 2 γ µ
ab γν cd
− 1
i S(+)
da (x2 − x1) : ψb(x1)ψc(x2) :
− 1
i S(−)
bc (x1 − x2) : ψa(x1)ψd(x2) :
− S (+)
da (x2 − x1)S (−)
bc (x1
− x2)
×
: Aµ(x1)Aν(x2) : +gµνiD (+)
0 (x2 − x1)
67
. (8.2)

Finally we can calculate the difference D(x1, x2) by (5.11).
D(x1, x2) = R ′ n(x1, x2) − A ′ n(x1, x2)
= e 2 γ µ
abγν
cd ψa(x1)ψb(x1)ψ c(x2)ψd(x2) : gµνi(D (+)
0 (x2 − x1) − D (+)
0 (x1 − x2))
− 1
S
i
(+)
da (x2 − x1) + S (−)
da (x2
− x1) : ψb(x1)ψc(x2) :: Aµ(x1)Aν(x2) :
− gµν S (+)
da (x2 − x1)D (+)
0 (x2 − x1) + S (−)
da (x2 − x1)D (+)
0 (x1 − x2) : ψb(x1)ψc(x2) :
− 1
i (S(−)
bc (x1 − x2) + S (+)
bc (x1 − x2))ψa(x1)ψd(x2) :: Aµ(x1)Aν(x2) :
− gµν S (−)
bc (x1 − x2)D (+)
0 (x2 − x1) + S (+)
bc (x1 − x2)D (+)
0 (x1 − x2) : ψa(x1)ψd(x2) :
+ S (+)
bc (x1 − x2)S (−)
da (x2 − x1) − S (−)
bc (x1 − x2)S (+)
da (x2
− x1) : Aµ(x1)Aν(x2) :
+ 1
i gµν
S (−)
bc (x1 − x2)S (+)
da (x2 − x1)D (+)
0 (x2 − x1)
. (8.3)
− S (+)
bc (x1 − x2)S (−)
da (x2 − x1)D (+)
0 (x1 − x2)
This formula may be considered as a correspondent to the Feynman rules.
Each term can be treated separately and represents different scattering
scenarios, where the field operators create and annihilate particles. Like
the Feynman rules, they may be illustrated by graphs. But notice the huge
difference that while all our terms are well-defined distributions, the
Feynman rules are ill-defined for closed loops, which here are represented
by terms with products of the propagators S and D.
Lets us consider the first term. First we note that
D + 0 (x1 − x2) − D + 0 (x2 − x1) is in fact the Jordan-Pauli distribution for
mass m = 0, where the full Jordan-Pauli distribution is given by 2
D(x) =
sgn x0
2π (δ(x2 ) − Θ(x 2 ) m
2 √ x 2 J1(m √ x 2 )).
Note that for the 1-dimensional δ-distribution,
Therefore
〈δ(δ 2 x 2 ), φ(δ 2 x 2 )〉 = δ −2 〈δ(x 2 ), φ(x 2 )〉.
D0(δx) =
sgn x0
2π δ(δ2x 2 sgn x0 δ(x
) =
2π
2 )
δ2 Hence if we choose ρ(δ) = δ 2−n we get the quasi-asymptotics
2 [6] page 89.
lim
δ→0 ρ(δ)δn sgn x0
D0(δx) =
2π δ(x2 ).
68

and for each a > 0,
ρ(aδ)
ρ(δ)
= a2−n
Thus in R 4 , ω = −2. Since the singular order is negative the splitting may
be done by multiplication by the Θ-function (7.16). Hence the retarded
part of the first term of (8.3) is
R2(x1, x2) = −ie 2 : ψ(x1)γ µ ψ(x1)ψ(x2)γ ν ψ(x2) : ΘD0(x1 − x2)
Further R ′ (x1, x2) is given by (8.2) by inspecting each term. By (5.12)
T2(x1, x2)
= R2(x1, x2) − R ′ (x1, x2)
= −ie 2 : ψ(x1)γ µ ψ(x1)ψ(x2)γ ν ψ(x2) :
ΘD0(x1 − x2) + D (+)
0 (x2 − x1) .
It is not the aim of this project to introduce Feynman propagators but the
reader with knowledge of these might notice that the expression in
brackets is actually a Feynman propagator describing the exchange of a
photon between electrons.
Figure 8.1: Photon exhange.
8.2 The Adiabatic Limit
In the introduction of the S-matrix we used test-functions g ∈ S(R 4 ),
so-called switching functions. As the name infers they switch off
long-range interaction to prevent infrared divergences 3 . Of course this is
not a good model and we need to consider the so-called adiabatic limit
g → 1 to take long-range interaction like the Coulomb-potential into
account. We show how we may carry out the adiabatic limit. In practise it
can me done by taking the so-called scaling limit.
Let g0 ∈ S(R 4 ) be a fixed test-function such that g0(0) = 1. Then we let
g(x) := g0(ɛx) and take the scaling limit, that is, we let ɛ → 0.
Calculations are in practise usually done in momentum space. With this in
mind we note that
ˆg(k) =
g(x)e −ix·k d 4 x =
g0(ɛx)e −ik·x d 4 x = 1
ɛ
k
ˆg0( ). (8.4)
4 ɛ
3 By infrared divergence we simply mean a divergence due to physical phenomena at
very long distances or because of contributions from objects with very small energy
69

Theorem 8.1. Identify ˆg(k) with the distribution it induces, then
Proof. For all φ ∈ S(R 4 )
lim〈ˆg,
φ〉 = lim
ɛ→0 ɛ→0 〈g, ˆ φ〉 = lim
ɛ→0
lim
ɛ→0 ˆg(k) = (2π)4δ(k).
g(k) ˆ φ(k)d 4
k = lim
ɛ→0
g0(ɛk) ˆ φ(k)d 4
k
= ˆφ(k)d 4 k, (8.5)
by the Dominated Convergence Theorem.
Now note that generally
〈 ˆ δ, φ〉 = 〈δ, ˆ φ〉 = ˆ
φ(0) = φ(x)dx = 〈1, φ〉. (8.6)
Therefore ˆ δ = 1. Similarly
〈 ˇ δ, φ〉 = 〈δ, ˇ φ〉 = ˇ φ(0) = (2π) −n
hence
φ(x)dx = (2π) −n 〈1, φ〉,
〈ˆ1, φ〉 = (2π) −n 〈F( ˇ δ), φ〉 = (2π) −n 〈δ, φ〉. (8.7)
That is, ˆ1 = (2π) nδ. It then follows from (8.5), that
lim〈ˆg,
φ〉 =
ɛ→0
as wanted.
Now by (8.4)
= 1
ɛ 4n
= 1
ɛ 4n
ˆφ(k)d 4 k = 〈1, ˆ φ〉 = 〈ˆ1, φ〉 = (2π) 4 〈δ, φ〉,
d 4 k1 · · · d 4 kn ˆ T (k1, . . . , kn)ˆg(k1) · · · ˆg(kn)
d 4 k1 · · · d 4 kn ˆ T (k1, . . . , kn)ˆg0(k1/ɛ) · · · ˆg0(kn/ɛ)
d 4 k1 · · · d 4 kn ˆ T (ɛk1, . . . , ɛkn)ˆg0(k1) · · · ˆg0(kn),
by substituting ki by ɛki. Finally, we may calculate in the usual way and
in the end take the limit ɛ → 0 without problems.
70

Chapter 9
The Microlocal Approach -
A Condition on the Wave
Front Set.
In this section I will use the notation T ∗ (X) even though X will be an
open subset of R n (and thus T ∗ (X) = X × R n ). I use this notation to
emphasize that the results can be generalized to manifolds 1 . As the results
are based on [4] where we only considered the case where X was open in
R n it would not make sense to consider manifolds.
The Method of Epstein and Glaser may be generalized to work on
manifolds using microlocal analysis.
Definition 9.1. Let Ψ ∈ H. Then the operator-valued function
Ψ(f) def
= : W (f) : Ψ, where W (f)φ def
= exp(iφ(f) ∗∗ ),
is said to be infinitely often differentiable at f = 0 if for each n,
1. For all h ∈ D(M) and all t ∈ R the map t ↦→ Ψ(th) is n times
norm-differentiable at 0.
2. The derivatives at 0 define symmetrical operator-valued distributions.
With Ψ as above we define a distribution
: φ(x1) · · · φ(xn) : def
=
∂n in : W (f) :
∂f(x1) · · · ∂f(xn)
1 In fact, globally hyperbolic manifolds. See [3] page 7.
71
f=0

in the sense that if h ∈ D(M), then
〈: φ(x1) · · · φ(xn) :, Ψ〉(h n
) = : φ(x1) · · · φ(xn) : Ψ h(x1) · · · h(xn) (9.1)
= i −n
∂nΨ ∂f(x1) · · · ∂f(xn) h(x1) · · · h(xn)
We want to generalize this.
Definition 9.2. The microlocal domain of smoothness D is defined by
where
D = {Ψ ∈ H|Φ(f) is infinitely often differentiable at f = 0
∂nΨ and for all n ∈ N, WF
∂f n
(T ∗ M n )± def
={x, k) ∈ T ∗ M|ki ∈ V − (0)}.
⊂ (T ∗ M n )−},
Now we may use the idea of equation (9.1) to define an operator-valued
distribution on D.
Definition 9.3. We call the operator-valued distribution : φ(x1) · · · φ(xn) :
on D defined by (9.1) a Wick monomial. 2
Now we would like to generalize the Wick monomial to a polynomial. That
is, for arbitrary indices l = (l1, . . . , ln) we would like to make sense to
To this end we need the following definition.
: φ l1 (x1) · · · φ ln (xn) : . (9.2)
Definition 9.4. A partial diagonal ∆l1,...,lj , where l1 + · · · + lj = n is a
subset of M n on the form
∆l1,...,lj (M) = {(x1, . . . , x1),
. . . , (xj, . . . , xj)|xi
∈ M, i = 1 . . . , j} M
l1 times
lj times
j .
Now we want to define the polynomial (9.2) as a restriction of ∂ n Ψ
∂f1···∂fn to
arbitrary partial diagonals. This is done in practice by a pullback. 3
Define the partial diagonal map
by
δn,l : M n → ∆l1,...,ln (M)
δn,l : (x1, . . . , xn) ↦→ (x1, . . . , x1,
. . . , xn, . . . , xn)
l1 times
ln times
2 Note that there are different definitions of a Wick monomial and polynomial in the
literature. The definitions used here are not the most used ones.
3 Remember that we considered the pullback of a distribution by a function in Sec-
tion 4.2 of [4].
72

Now by Theorem 4.5 from [4] we may define a Wick polynomial as an
operator-valued distribution on D defined as the pullback of a Wick
monomial by the a diagonal map δn,l. 4
First we calculate the set of normals of the partial diagonal map.
The transpose of the Jacobi matrix of the partial diagonal map is
δ ′ n,l (x1, . . . , xn) t =
⎛
∂x1
⎜
⎝
(δn,l)1
⎞t
. . . ∂xn(δn,l)1
.
. ..
⎟
. ⎠
∂x1 (δn,l)n
⎛
1 . . . 1
⎜ 0 . . . 0
⎜
= ⎜ .
⎝ 0 . . .
. . . ∂xn(δn,l)n
0 . . .
1 . . . 1 0 . . .
0 1 . . . 1 0
⎞
0
0 ⎟
. ⎟ ,
⎟
0 ⎠
0 . . . 0 1 . . . 1
where row j contains lj ones. Hence
δ ′ n,l (x1, . . . , xn) t η =
⎛
⎜
⎝
η1 + . . . + ηl1
ηl1+1 + . . . + ηl1+l2
.
ηl1+...+ln−1+1 + . . . + ηl1+...+ln
⎞
⎟
⎠ .
Note that the matrix product is independent on x. The set of normals is
Nδn,l = {(δn,l(x), η) ∈ ∆l(M) × R n |δ ′ n,l (x)t η = 0}
= {(y, η) ∈ ∆l(M) × R n li+1
| ηli+j = 0 for all i = 0, . . . , n − 1},
Where we have set l0 := 0.
j=1
Definition 9.5. By a Wick polynomial : φ l1 (x1) · · · φ ln (xn) : we mean the
operator-valued distribution
when it exists, that is, if
.
: φ l1 (x1) · · · φ ln (xn) : def
= δn,l ∗ : φ(x1) · · · φ(xn) :
Nδn,l ∩ WF (: φ(x1) · · · φ(xn) :) = ∅.
4 Note that we proved the theorem exactly for the case of a diagonal map.
73

Note that the definition agrees with Definition 9.3, since the restriction to
the total diagonal ∆ n,(1,...,1)(M) simply returns the monomial itself.
Translational invariance has been an important ingredient in the
development of the method of Epstein-and Glaser. Unfortunately
translations are not well-defined on general manifolds and have to be
replaced by parallel transport. Therefore it is one of the key problems one
has to face in the task of creating a microlocal formulation of the method
(and thus make the method work on curved spaces) to find a suitable
condition of smoothness to replace the translational invariance.
The condition is related to graph theory. Letting G denote the set of all
graphs, we define
and
Gn := {G ∈ G|G is non-oriented with vertices {1, . . . , n}}.
E G := {e|e is an edge of G}.
Further given e ∈ E G connecting i and j with i < j we say that i := s(e)
and j := r(e) are the source and range of e, respectively.
Definition 9.6. Let G ∈ Gn and M a manifold. An immersion (x, γ, k) is
a triplet consisting of
1. A map x mapping the vertices v of G to x(v) ∈ M
2. a map γ mapping the edges e ∈ E G to null-geodesics γ(e) ⊂ M
connecting x s(e) and x r(e)
3. A map k mapping edges e ∈ E G to future directed covector fields
k γ(e) := k(e), which are coparallel to the tangent vector ˙γe of the
null-geodesic.
Along with symmetry of Tn and causality we expected the S-matrix to
have we have to add one more property:
74

Microlocal Spectrum Condition.
For the numerical distribution tn ∈ D ′ (M n ), n ≥ 2 of any time-ordered
product,
WF (tn) ⊂ Γn,
where
Γn =
(x1, k1; . . . ; xn, kn) ∈ T ∗ M n \ {0}|∃g ∈ Gn and
an immersion (x, γ, k) of G, in which ke is
future directed whenever x s(e) /∈ J − (x r(e)) and
such that ki =
m:s(m)=i
km(xi) −
n:r(n)=i
kn(xi) ,
where t and s runs over all curves terminating and starting at xi,
respectively.
where
J − (x) := {y ∈ M|y < x and there exists γ causal, connecting x and y}
is the set of all points in M in the past of x that can be connected with x
by a causal curve.
The microlocal spectrum condition may be seen as a replacement of the
spectrum condition (Axiom 4iii of the Wightman Axioms). Further it
ensures that the wave front set has the following property.
Figure 9.1: Illustration of Lemma 9.7.
Lemma 9.7. Let (x, k) ∈ Γn then there exists a pair (xm, km) such that
km /∈ V + (0).
75

Proof. Let xm be a maximal point, that is xm /∈ J − (xi) for all i. Note that
we may write
km = −k1,m − · · · − km−1,m + km+1,m + · · · + kn,m,
some of which might be zero, but not all since xm is connected to at least
> 0 and
one point. Now since xm is maximal, k0 1,m , . . . , k0 m−1,m
k0 m+1,m , . . . k0 n,m < 0.
The spectrum condition is important for the construction of the Scattering
matrix, as the following theorem will witness.
Theorem 9.8. If the spectral condition is satisfied, then
Tn(x1, . . . , xn) : φ l1 (x1) · · · φ ln (xn) : (9.3)
is a well-defined operator-valued distribution for any n and any choice of
indices l1 . . . , ln on D. FIXME: dense invariant?
Proof. Let Ψ ∈ D. Then by Definition 9.5
: φ l1 (x1) · · · φ ln (xn) : Ψ
is an operator-valued distribution. By Definition 9.2
Thus FIXME: tjeck det følgede:
WF (: φ(x1) · · · φ(xn) : Ψ) ⊂ (T ∗ M)−.
WF (: φ l1 (x1) · · · φ ln (xn) : Ψ) = WF (δn,l ∗ : φ(x1) · · · φ(xn) : Ψ)
= δn,l ∗ WF (: φ(x1) · · · φ(xn) : Ψ)
⊂ M n × V ×n
− ,
where we have used that WF (f ∗ u) ⊂ f ∗ WF (u) by Theorem 4.2 in [4].
Now remember our main result Theorem 4.10 of [4], that the product of
two distributions u and v exists unless
(x, k) ∈ WF (u) and (x, −k) ∈ WF (v)
for some (x, k).
But in our case the spectral condition insures that WF (Tn) ⊂ Γn and
therefore by Lemma 9.7 there doesn’t exist a (x1, k1, . . . , xn, kn) ∈ Γn with
ki ∈ V ×n
+ for all i. Thus the product (9.3) is well-defined.
76

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