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Feynman Integrals Lecture 1 Stefan Weinzierl Institut für Physik, Universität Mainz Higgs Centre School for Theoretical Physics 2021 Stefan Weinzierl Feynman Integrals Higgs Centre School 2021 1 / 43

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Page 1: Feynman Integrals - Lecture 1

Feynman Integrals

Lecture 1

Stefan Weinzierl

Institut für Physik, Universität Mainz

Higgs Centre School for Theoretical Physics 2021

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Page 2: Feynman Integrals - Lecture 1

Section 1

Organisation

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Page 3: Feynman Integrals - Lecture 1

Organisation

Online course on Feynman integrals via zoom.

Daily schedule for this course:

9:00-11:00 Lecture followed by a

discussion session in the plenum

11:30-12:30 Tutorial

Slides and exercise sheets on the indico page.

Afternoon: Lectures by Matthew McCullough

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Page 4: Feynman Integrals - Lecture 1

Tutorials

For each lecture there will be an exercise sheet.

Tutorials: Small groups (5-7 people) in breakout rooms

Tutor: Yao Ma

Close-out at 12:20 in the plenum.

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Page 5: Feynman Integrals - Lecture 1

Advice

A summer school has a scientific part and a social part.

Talk to your fellow students in the tutorials.

Ask questions in the discussion session.

It’s your summer school!

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Page 6: Feynman Integrals - Lecture 1

Section 2

Motivation

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Page 7: Feynman Integrals - Lecture 1

Scattering amplitudes

We would like to make precise predictions for observables in scattering

experiments from quantum field theory.

Any such calculation will involve a scattering amplitude.

Unfortunately we cannot calculate scattering amplitudes exactly.

If we have a small parameter like a small coupling, we may use

perturbation theory.

We may organise the perturbative expansion of a scattering amplitude in

terms of Feynman diagrams.

Scattering amplitude = sum of all Feynman diagrams

Stefan Weinzierl Feynman Integrals Higgs Centre School 2021 7 / 43

Page 8: Feynman Integrals - Lecture 1

Scattering amplitudes

Example: Scattering amplitude for a 2 → 2-process:

p1

p2 p3

p4

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Page 9: Feynman Integrals - Lecture 1

Scattering amplitudes

We may compute the scattering amplitude within perturbation theory:

p1

p2 p3

p4

= + + ...

O(

g2)

+ + + ...

O(

g4)

+ + + ...

O(

g6)

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Page 10: Feynman Integrals - Lecture 1

Feynman diagrams

At leading order we (usually) have only tree diagrams. These do not pose

any conceptual problem.

But precise predictions require more terms in the perturbative expansion.

Beyond leading order we have loop diagrams and we have to compute

loop integrals.

Content of this course: How do we compute loop integrals?

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Page 11: Feynman Integrals - Lecture 1

Section 3

Basics

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Page 12: Feynman Integrals - Lecture 1

Introducing Feynman integrals

We may define Feynman integrals without knowledge of quantum field theory.

We just need:

Special relativity

Graphs

Stefan Weinzierl Feynman Integrals Higgs Centre School 2021 12 / 43

Page 13: Feynman Integrals - Lecture 1

Special relativity

Denote by D the number of space-time dimensions.

In our real world D equals 4 (one time dimension and three spatial dimensions), but it is

extremely helpful to keep this number arbitrary. We will always assume that space-time

consists of one time dimension and (D−1) spatial dimensions.

The momentum of a particle is a D-dimensional vector, whose first

component gives the energy E (divided by the speed of light c) and the

remaining (D−1) components give the components of the spatial

momentum, which we label with superscripts:

p =

(

E

c,p1, . . . ,pD−1

)

.

It is common practice to work in natural units, where

c = ~ = 1.

Stefan Weinzierl Feynman Integrals Higgs Centre School 2021 13 / 43

Page 14: Feynman Integrals - Lecture 1

Special relativity

In natural units we write p0 = E and

p =(

p0,p1, . . . ,pD−1)

.

A component of p is denoted by

pµ with 0 ≤ µ ≤ D−1

The index µ is called a Lorentz index.

Minkowski metric: gµν = diag(1,−1, . . . ,−1).

Minkowski scalar product of two momentum vectors pa and pb:

pa ·pb =D−1

∑µ=0

D−1

∑ν=0

a gµν pνb = p

µ

a gµν pνb .

In the last expression we used Einstein’s summation convention.

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Page 15: Feynman Integrals - Lecture 1

Special relativity

We represent a particle propagating in space-time by a line:

If we would like to indicate the direction of the momentum flow, we optionally

put an arrow:

p

The line is our first building block for a Feynman graph. A line for a particle

with momentum p and mass m stands for

p,m=

1

−p2 +m2.

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Page 16: Feynman Integrals - Lecture 1

Special relativity

Remark: If one follows standard conventions in quantum field theory the

propagator of a scalar particle with momentum p and mass m is given by

i

p2 −m2.

This differs by a factor (−i) from

1

−p2 +m2.

This is just a prefactor and easily adjusted.

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Page 17: Feynman Integrals - Lecture 1

Graphs

An unoriented graph consists of edges

and vertices, where an edge connects

two vertices.

A graph may be connected or

disconnected. We will mainly consider

connected graphs.

An oriented graph is a graph, where for

every edge an orientation is chosen. An

orientated edge is usually drawn with an

arrow line:

source sink

v1v2

v3

v4v5

e1

e2

e3 e4

e5

v6

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Page 18: Feynman Integrals - Lecture 1

Valency of a vertex

The valency of a vertex is the number of

edges attached to it.

A vertex of valency 0 is necessarily

disconnected from the rest of graph.

A vertex of valency 1 has exactly one

edge attached to it. This edge is called

an external edge. All other edges are

called internal edges.

A vertex of valency 2 is also called a dot.

v1v2

v3

v4v5

e1

e2

e3 e4

e5

v6

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Page 19: Feynman Integrals - Lecture 1

Momentum conservation

Choose an orientation for every edge.

E source(va) : set of edges, which have vertex va as source,

E sink(va) : set of edges, which have vertex va as sink.

At each vertex va of valency > 1 we impose momentum conservation:

∑ej∈E source(va)

qj = ∑ej∈E sink(va)

qj .

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Page 20: Feynman Integrals - Lecture 1

Dots

Consider a vertex of valency 2 inside a Feynman graph:

q,m q,m=

1

(−q2 +m2)2.

We get the same effect if we associate to each edge in addition to p and m a

number ν, corresponding to the power of the propagator:

q,m,ν=

1

(−q2 +m2)ν

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Page 21: Feynman Integrals - Lecture 1

The loop number

Consider a graph G with n edges, r vertices and k connected components.

The loop number l is defined by

l = n− r + k .

If the graph is connected we have

l = n− r +1.

The loop number l is also called the first Betti number of the graph or the

cyclomatic number.

Physics interpretation: If we fix all momenta of the external lines and if we

impose momentum conservation at each vertex, then the loop number is equal

to the number of independent momentum vectors not constrained by

momentum conservation.

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Page 22: Feynman Integrals - Lecture 1

Trees and forests

A connected graph of loop number 0 is called a tree.

A graph of loop number 0, connected or not, is called a forest.

If the forest has k connected components, it is called a k-forest.

A tree is a 1-forest.

Feynman graphs which are trees pose no conceptual problem.

Our focus in this lecture is on connected Feynman graphs, which are not trees,

e.g. Feynman graphs with loop number l > 0.

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Page 23: Feynman Integrals - Lecture 1

Notation

Consider a connected graph G with next external edges, nint internal edges

and loop number l :

External momenta: p1, . . . ,pnext

Internal momenta: q1, . . . ,qnint

Independent loop momenta: k1, . . . ,kl

qj =l

∑r=1

λjr kr +next−1

∑r=1

σjr pr , λjr ,σjr ∈ −1,0,1.

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Page 24: Feynman Integrals - Lecture 1

Example

p1

p2 p3

p4

q1 =

k1 + p2

q4 =

k1 + k2

q7 =

k2 + p3

q2 = k1 q5 = k2

q3 =

k1 + p1 + p2

q6 =

k2 − p1 − p2

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Page 25: Feynman Integrals - Lecture 1

Quiz

How many loops does this graph have?

(A) 5

(B) 6

(C) 11

(D) 16

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Page 26: Feynman Integrals - Lecture 1

Integration

We may consider the external momenta as input data, but what shall we

do with the l independent loop momenta?

Quantum field theory instructs us to integrate over the independent

loop momenta. Thus we include for every independent loop momentum

kr a D-dimensional integration

∫dDkr

iπD2

.

If one follows standard conventions in quantum field theory, the measure

is given by

∫dDkr

(2π)D.

The difference is a simple prefactor and one easily converts from one

convention to the other convention.

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Page 27: Feynman Integrals - Lecture 1

Integration domain

Naive expectation: We integrate each of the D components of kr along

the real axis from −∞ to +∞.

However, there will be poles on the real axis:

kr ,m=

1

−k2r +m2

Quantum field theory dictates us that the correct integration contour is:

Re(k0r)

Im(k0r)

−E~kr

E~kr

Feynman’s iδ-prescription:

q,m=

1

−q2 +m2 − iδ

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Page 28: Feynman Integrals - Lecture 1

Notation

For the space-time dimension D we write

D = 4−2ε.

Euler’s constant γE:

γE = limn→∞

(

− lnn+n

∑j=1

1

j

)

= 0.57721566490153286 . . .

In order to enforce that our Feynman integral is dimensionless, we

introduce an arbitrary parameter µ with mass dimension dim(µ) = 1 and

multiply by an appropriate power of µ.

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Page 29: Feynman Integrals - Lecture 1

The Feynman integral

Definition

The Feynman integral for a Feynman graph G with next external edges, nint

internal edges and l loops is given in D space-time dimensions by

I = elεγE(

µ2)ν− lD

2

∫ l

∏r=1

dDkr

iπD2

nint

∏j=1

1(

−q2j +m2

j

)νj,

where each internal edge ej of the graph is associated with a triple (qj ,mj ,νj),

qj =l

∑r=1

λjr kr +next−1

∑r=1

σjr pr , ν =nint

∑j=1

νj .

The coefficients λjr and σjr can be obtained from momentum conservation at

each vertex of valency > 1.

The integration contour is given by Feynman’s iδ-prescription.

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Page 30: Feynman Integrals - Lecture 1

Wick rotation

Minkowski scalar product:

k2 = k20 −k2

1 −k22 −k2

3 − ...

If the quarter-circles at infinity give no contribution:

∮dk0f (k0) = 0 ⇒

∞∫

−∞

dk0f(k0) =−

−i∞∫

i∞

dk0f(k0)

Change variables:

k0 = iK0,

kj = Kj , for 1 ≤ j ≤ D−1.

Final result:

∫dDk

iπD/2f (−k2) =

∫dDK

πD/2f (K 2),

with Euclidean scalar product K 2 = K 20 +K 2

1 +K 22 +K 2

3 + ....

Re k0

Im k0

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Page 31: Feynman Integrals - Lecture 1

The need for regularisation

Loop integrals in four space-time dimensions can be divergent!

∫d4k

(2π)4

1

(k2)2=

1

(4π)2

∞∫

0

dk2 1

k2=

1

(4π)2

∞∫

0

dx

x

This integral diverges at

k2 → ∞ (UV-divergence) and at

k2 → 0 (IR-divergence).

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Page 32: Feynman Integrals - Lecture 1

Dimensional regularisation

Replace the four-dimensional integral by an D-dimensional integral,

where D is now an additional parameter, which can be a non-integer or

even a complex number.

We consider the result of the integration as a function of D and we are

interested in the behaviour of this function as D approaches 4.

Concept closely related to: Consider a function f (z), which is defined for

any positive integer n ∈ N by

f (n) = n! = 1 ·2 ·3 · · · · ·n.

We would like to define f (z) for any value z ∈ C (except for a countable

set of isolated points, where f (z) is allowed to have poles).

Answer is well known and given by Euler’s gamma function

f (z) = Γ(z +1) .

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Page 33: Feynman Integrals - Lecture 1

Dimensional regularisation

A Feynman integral I has a Laurent expansion in ε:

I =∞

∑j=jmin

εj I(j),

where I(j) denotes the coefficient of εj .

For precision calculations we are interested in the first few terms I(jmin),

I(jmin+1), . . . of this Laurent series.

The exact number of required terms depends on the order of perturbation

theory we are calculating.

Let us stress that we would like to get the I(j) ’s, not necessarily I itself.

There are situations where a closed form expression for I is readily

obtained, but the Laurent expansion in ε is not immediate.

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Page 34: Feynman Integrals - Lecture 1

The tadpole integral

Let’s get our hands dirty: We start with a one-loop integral, which only

depends on the loop momentum squared (and no scalar product k ·p with

some external momentum).

The one-loop tadpole integral:

Tν = eεγE(

µ2)ν− D

2

∫dDk

iπD2

1

(−k2 +m2)ν .

After Wick rotation we have

Tν = eεγE(

µ2)ν− D

2

∫dDK

πD2

1

(K 2 +m2)ν .

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Page 35: Feynman Integrals - Lecture 1

Spherical coordinates

One radial variable K , D− 2 polar angles θ1, ..., θD−2 and one azimuthal angle θD−1:

K0 = K cosθ1

K1 = K sinθ1 cosθ2

...

KD−2 = K sinθ1...sinθD−2 cosθD−1

KD−1 = K sinθ1...sinθD−2 sinθD−1

The measure becomes

dDK = K D−1dKdΩD

Integration over the angles yields

∫dΩD =

π∫

0

dθ1 sinD−2 θ1...

π∫

0

dθD−2 sinθD−2

2π∫

0

dθD−1 =2πD/2

Γ(

D2

)

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Page 36: Feynman Integrals - Lecture 1

Euler’s gamma function

The gamma function is defined for Re(z) > 0 by

Γ(z) =

∞∫

0

e−t tz−1dt .

It fulfils the functional equation

Γ(z + 1) = z Γ(z).

For positive integers n it takes the values

Γ(n+ 1) = n! = 1 ·2 ·3 · ... ·n.

The expansion around z = 1 reads

Γ(1+ ε) = exp

(

−γEε+∞

∑n=2

(−1)n

nζnεn

)

, ζn =∞

∑j=1

1

jn.

ζn is called a zeta value. Laurent expansion around z = 0:

Γ(ε) =1

ε− γE +O(ε).

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Page 37: Feynman Integrals - Lecture 1

Euler’s beta function

Euler’s beta function is defined for Re(z1)> 0 and Re(z2)> 0 by

B (z1,z2) =

1∫

0

tz1−1(1− t)z2−1dt,

or equivalently by

B (z1,z2) =

∞∫

0

tz1−1

(1+ t)z1+z2dt.

The beta function can be expressed in terms of Gamma functions:

B (z1,z2) =Γ(z1)Γ(z2)

Γ(z1 + z2).

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Page 38: Feynman Integrals - Lecture 1

The tadpole integral

Performing the angular integrations for our tadpole integral we obtain

Tν =eεγE

(

µ2)ν− D

2

Γ(

D2

)

∞∫

0

dK 2

(

K 2)

D2−1

(K 2 +m2)ν .

We substitute t = K 2/m2 and obtain

Tν =eεγE

Γ(

D2

)

(

m2

µ2

)D2−ν ∞∫

0

dtt

D2−1

(t +1)ν .

The remaining integral is just Euler’s beta function

∞∫

0

dtt

D2−1

(1+ t)ν =Γ(

D2

)

Γ(

ν− D2

)

Γ(ν).

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Page 39: Feynman Integrals - Lecture 1

The tadpole integral

Tadpole integral:

(

D,m2

µ2

)

=eεγEΓ

(

ν− D2

)

Γ(ν)

(

m2

µ2

)D2−ν

.

We are interested in the Laurent expansion of Feynman integrals. With

D = 4−2ε, ν = 1 and L = ln(m2/µ2) we obtain

T1 (4−2ε) =m2

µ2eεγEΓ(−1+ ε)e−εL

=m2

µ2

[

−1

ε+(L−1)+

(

−1

2L2 −

1

2ζ2 +L−1

)

ε

]

+O(

ε2)

.

The pole at ε = 0 originates from the ultraviolet divergence of tadpole integral

with ν = 1 in four-space time dimensions.

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Page 40: Feynman Integrals - Lecture 1

The tadpole integral

Tadpole integral:

(

D,m2

µ2

)

=eεγEΓ

(

ν− D2

)

Γ(ν)

(

m2

µ2

)D2−ν

.

The result is valid for any D, so we may as well expand it around two

space-time dimensions. Just for fun, let’s do it:

ε T1 (2−2ε) = eεγEΓ(1+ ε)e−εL

= 1−Lε+

(

1

2L2 +

1

2ζ2

)

ε2 +O(

ε3)

.

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Page 41: Feynman Integrals - Lecture 1

Weights

Define a weight as follows:

Any algebraic expression in the kinematic variables has weight 0.

The dimensional regularisation parameter has weight (−1).

L = ln(m2/µ2) has weigth 1, ζn has weight n.

The weight of a product is the sum of the weights of its factors.

We observe

T1(4−2ε) contains terms of mixed weight.

ε T1(2−2ε) only involves terms of weight zero.

We call ε T1(2−2ε) to be of uniform weight.

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Page 42: Feynman Integrals - Lecture 1

Dimensional shift relations

We may show

Tν (D) = νTν+1 (D+2) .

This is an example of a dimensional shift relation, relating integrals in D and

D+2 space-time dimensions.

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Page 43: Feynman Integrals - Lecture 1

Master integrals

Write D = [D]− 2ε and set

J1 = ε T1 (2− 2ε) = eεγEΓ(1+ ε)e−εL.

It is not too difficult to show that

Tν (D) =Γ(

ν− [D]2+ ε)

Γ(ν)Γ(1+ ε)

(

m2

µ2

)

(

[D]2 −ν

)

J1.

For ν ∈ N and [D] even, the prefactor is always a rational function in ε and m2. Thus

we may express Tν(D) as a rational coefficient times J1.

Later on we will see, that this generalises as follows: We may express any member of

a family of Feynman integrals as a linear combination of Feynman integrals from a

finite set. The Feynman integrals from this finite set are called master integrals. A

master integrals, which is of uniform weight, is called a canonical master integral.

Thus J1 is a canonical master integral.

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