Quantum Field Theory - CAS · Klein paradox To illustrate this feature we will study Klein™s...

37
Quantum Field Theory Ling-Fong Li (Institute) Chapter 1 Introduction 1 / 37

Transcript of Quantum Field Theory - CAS · Klein paradox To illustrate this feature we will study Klein™s...

Quantum Field Theory

Ling-Fong Li

(Institute) Chapter 1 Introduction 1 / 37

Chapter 1 IntroductionNecesscity of field theory in relativistic systemSchrodingerSchrodinger equation ⇒ conservation of particle number.Hψ = i h ∂ψ

∂t ⇒ ddt

∫d 3xψ†ψ = 0 →

∫d 3x (ψ†ψ) indep of time

If H is hermitian, H = H †. Then number of particles is conserved and no particle creation orannihilation.

Canonical commutation relation gives uncertainty relation,

[x , p] = −i h, ⇒ 4x4p > h

Fromp2c2 +m2c4 = E 2

get

4E = p4pE

c2 > p hc2

E4x or 4x > pcE(hc4E )

To avoid new particle creation we require 4E 6 mc2. Then we get a lower bound on 4x

4x > pcE

hmc

= (vc)(

hmc)

(Institute) Chapter 1 Introduction 2 / 37

For relativistic particlevc≈ 1, then

4x > ( hmc) Compton wavelength

⇒ Particle can not be confined to a interval smaller than its Compton wavelength

(Institute) Chapter 1 Introduction 3 / 37

Klein paradoxTo illustrate this feature we will study Klein’s paradox in the context of the Klein-Gordonequation given by (

∂2

∂t2−∇2 −m2

)ψ (x , t) = 0

Let us consider a square potential with height V0 > 0 as shown in the figure,

(Institute) Chapter 1 Introduction 4 / 37

A solution to the wave equation in regions I and II is given by

ψI (x , t) = e−iEt−ip1x + R e−iEt+ip1x

ψII (x , t) = Te−iEt−ip2x

where

p1 =√E 2 −m2, p2 =

√(E − V0)2 −m2

The constants R(reflection) and T(transmission) are computed by matching the twosolutions across the boundary x = 0. The conditions ψI (t , 0) = ψII (t , 0) and∂xψI (t , 0) = ∂xψII (t , 0) give

1+ R = T , (1− R ) p1 = Tp2

Solve for R and T

T =2p1

p1 + p2, R =

p1 − p2p1 + p2

if E > V0 +m both p1 and p2 are real and there are both transmitted and reflected wave.If E < V0 +m and E > m, then p2 is imaginary and p1 real, we get a reflected wave,transmitted wave being exponentially damped within Compton wavelength inside thebarrier and there is total reflection.when V0 > 2m and m < V0 − E then both p1 and p2 are real and there are both reflectedand transmitted waves . This implies that there is a nonvanishing probability of findingthe particle at any point across the barrier with negative kinetic energy (E −m − V0 < 0)!This result is known as Klein’s paradox. This result can only be understood in terms ofparticle creation at sudden potential step.

(Institute) Chapter 1 Introduction 5 / 37

Gauge Theory—Quantum Field Theory with Local SymmetryGauge principleAll fundamental Interactions are descibed in terms of gauge theories;

1 Strong Interaction-QCD;gauge theory based on SU(3) symmetry

2 Electromagnetic and Weak interaction-gauge theory based on SU (2)× U (1) symmetry

3 Gravitational interaction-Einstein’s theory-gauge theory of local coordinate transformation.

(Institute) Chapter 1 Introduction 6 / 37

Nature UnitsIn high energy physics, convienent to us the natural unit

h = c = 1

=⇒ many formulae simplified. Recall that in MKS units

h = 1.055× 10−34J sec

Thus } = 1 implies that energy has same dimension as (time)−1. Also

c = 2.99× 108m/sec

so c = 1 =⇒ time and length have the same dimension. In this unit, at the end of thecalculation one needs put back the factors of h and c to get the right dimension for the physicalquantities in the problem. For example, the quantity me can have following different meaningsdepending on the contexts;

1 Reciprocal length

me =1hme c

=1

3.86× 10−11cm

(Institute) Chapter 1 Introduction 7 / 37

2 Reciprocal time

me =1h

me c2

=1

1.29× 10−21 sec

3 energyme = me c2 = 0.511 Mev

4 momentumme = me c = 0.511 Mev/c

The following conversion relations

h = 6.58× 10−22Mev − sec hc = 1.973× 10−11Mev − cm

are quite useful in getting the physical quantities in the right units.Example:

(Institute) Chapter 1 Introduction 8 / 37

1 Thomson cross section

σ =8πα2

3m2e

Cross section has dimension of (length)2 and the only quantity here with dimension is meFirst compute this in natural unit,

σ =8πα2

3m2e=

8× 3.143× (0.5Mev )2

×(1137

)2=(1.78× 10−3

)(Mev )−2

Then we multiply this by(hc = 1.973× 10−11Mev − cm

)2 to convert this into cm2σ =

(1.78× 10−3

)(Mev )−2

(1.973× 10−11Mev − cm

)2= 6.95× 10−25cm2

(Institute) Chapter 1 Introduction 9 / 37

2 Decay rate of W−bosonIn Standard Model the decay rate for W − → eν is given by

Γ(W − → eν) =GF√2

M 3W6π

where MW = 80.4Gev/c2 is the mass of W−boson and GF = 1.166× 10−5Gev−2 is theweak coupling constant. We will first compute the rate in Gev and then convert to(sec)−1 , the unit for decay.

Γ(W − → eν) =GF√2

M 3W6π

=

(1.166× 10−5Gev−2

)× (80.4Gev )3√

26π= 0.227Gev

Then we divide by } to get the right unit

Γ(W − → eν) =0.228Gev

6.58× 10−22Mev − sec = 3.5× 1023/ sec

(Institute) Chapter 1 Introduction 10 / 37

3 Neutrino cross sectionFor quasi-elastic neutrino scattering νµ + e → µ+ νe , the cross section at low energies isof form

σ = 2G 2FmeE

where E is the energy of the neutrino. We want to find σ for E = 10Gev . Again, innatural unit.

σ = 2×(1.166× 10−5Gev−2

)2 × (.5Mev ) (10Gev ) = 1.34× 10−12Gev−2

Now we use hc = 1.973× 10−11Mev − cm to convert

σ = 1.34× 10−12Gev−2(1.973× 10−11Mev − cm

)2= 5.2× 10−40cm2

Note that this is a very small cross section which means that neutrino hardly interactswhen encounter matter with many electrons. Another feature is that in these lowenergies, this cross section increases with energies.

(Institute) Chapter 1 Introduction 11 / 37

4 Conversion of Newton constant

GN = 6.67× 10−11m3kg−1sec−2

to some energy scale, Planck scale. Use

hc = 3.16× 10−26Jm

we can write

GN = 6.67× 10−11(m2kgs2

)mkg 2

= 6.67× 10−11J mkg 2

Then we get for the combination

(hcGN) = 3.16× 10−26Jm × 1

6.67× 10−11Jm kg2 = 4.73× 10−16(kg )2

Use √hcGN

= 2.176× 10−8(kg )

and

1kg c2 = 1kg × (3× 108m/sec )2 = 9× 1016m2kgs2

= 9× 1016J or 1kg = 9× 1016J/c2

(Institute) Chapter 1 Introduction 12 / 37

we have

=⇒√hcGN

= 1.96× 109J/c2

Use the conversion factor

1Gev = 1.6× 10−10J =⇒ 1J =11.6× 1010Gev

we finally get

mp ≡√hcGN

=1.96× 109

1.6× 1010Gev = 1.225× 1019Gev/c2

This is the usual statement that the energy associated with gravity (Planck scale) is~1019Gev . Another way to express the result is

GN = 6.07× 10−39( hc )(Gev/c2)−2

(Institute) Chapter 1 Introduction 13 / 37

Review of Special RelativityBasic principles of special relativity :

1 The speed of light : same in all inertial frames.

2 Physical laws: same forms in all inertial frames.

Lorentz transformation—relate coordinates in different inertial frame

x ′ =x − vt√1− v 2

y ′ = y , z ′ = z , t ′ =t − vx√1− v 2

⇒t2 − x 2 − y 2 − z 2 = t ′2 − x ′2 − y ′2 − z ′2

Proper time τ2 = t2 −−→r 2 invariant under Lorentz transfomation.Particle moves from

→r1(t1) to

→r2(t2) .The speed is

|−→v | = 1|t2 − t1 |

√(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2

For |−→v | = 1,(t1 − t2)2 = |−→r1 −−→r2 |2

this is invariant under Lorentz transformation ⇒ speed of light same in all inertialframes .

(Institute) Chapter 1 Introduction 14 / 37

Another form of the Lorentz transformation

x ′ = coshω x − sinhω t , y ′ = y , z ′ = z , t ′ = sinhω x − coshω t

wheretanhω = v

For infinitesmal interval (dt , dx , dy , dz), proper time is

(dτ)2 = (dt)2 − (dx )2 − (dy )2 − (dz )2

(Institute) Chapter 1 Introduction 15 / 37

Minkowski space,xµ = (t , x , y , z ) = (x 0, x 1, x 2, x 3), 4− vector

Lorentz invariant product can be written as

x 2 = (x0)2 − (x1)2 − (x2)2 − (x3)2 = xµx νgµν

where

gµν =

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

Define another 4-vector

xµ = gµνx ν = (t ,−x 1,−x 2,−x 3) = (t ,−−→r )

so thatx 2 = xµxµ

For infinitesmal coordinates

(dx )2 = (dxµ)(dxµ) = dxµdx νgµν = (dx 0)2 − (d−→x )2

(Institute) Chapter 1 Introduction 16 / 37

Write the Lorentz transformation as

xµ → x ′µ = Λµνx

ν

For example for Lorentz transformation in the x−direction, we have

Λµν =

1√1−β2

−β√1−β2

0 0−β√1−β2

1√1−β2

0 0

0 0 1 00 0 0 1

Write

x ′2 = x ′µx ′νgµν = Λµα Λν

β gµνx αx β

then x 2 = x ′2 impliesΛµ

α Λνβ gµν = gαβ

and is called pseudo-orthogonality relation.

(Institute) Chapter 1 Introduction 17 / 37

Energy and MomentumStart from

dxµ = (dx 0, dx 1, dx 2, dx 3)

Proper time is Lorentz invariant and has the form,

(dτ)2 = dxµdxµ = (dt)2 − (d−→xdt)2(dt)2 = (1−→v

2)(dt)2

4− velocity ,

uµ =dxµ

dτ= (

dx 0

dτ,d−→xdτ

)

there is a constraint

uµuµ =dxµ

dxµ

dτ= 1

Note that−→u = d−→x

dτ=d−→xdt(dtdτ) =

1√1− v 2

−→v ≈ −→v , for v � 1

4− velocity =⇒ 4−momentum

pµ = muµ = (m√1− v 2

,m−→v√1− v 2

)

(Institute) Chapter 1 Introduction 18 / 37

For v � 1,

p0 =m√1− v 2

= m(1+12v 2 + ...) = m +

m2v 2 + ..., energy

−→p = m−→v 1√1− v 2

= m−→v + ... momentum

pµ = (E ,−→p )

Note that

p2 = E 2 −−→p 2 = m2

1− v 2 [1− v2 ] = m2

(Institute) Chapter 1 Introduction 19 / 37

Tensor analysisPhysical laws take the same forms in all inertial frames, if we write them in terms of tensors inMinkowski space.Basically, tensors are

tensors ∼ product of vectors

2 different types of vectors,x ′µ = Λµ

νxν, x ′µ = Λ ν

µ xν

multiply these vectors to get 2nd rank tensors,

T ′µν = ΛµαΛν

βTαβ, T ′µν = Λ α

µ Λ βν Tαβ, T ′µν = Λµ

αΛ βν T

αβ

In general,

T′µ1 ···µnν1 ···νm = Λµ1

α1 · · ·ΛµnαnΛ β1

ν1 · · ·Λβm

νm Tα1 ···αnβ1 ···βm

transformation of tensor components is linear and homogeneous.

(Institute) Chapter 1 Introduction 20 / 37

Tensor operations; operation which preserves the tensor property

1 Multiplication by a constant, (cT ) has the same tensor properties as T2 Addition of tensor of same rank3 Multiplication of two tensors

4 Contraction of tensor indices. For example,T µαβγµ is a tensor of rank 3 while T µαβγ

ν is atensor or rank 5. This follows from the psudo-orthogonality relation

5 Symmetrization or anti-symmetrization of indices. This can be seen as follows. SupposeT µν is a second rank tensor,

T ′µν = ΛµαΛν

βTαβ

interchanging the indices

T ′νµ = ΛναΛµ

βTαβ = Λν

βΛµαT

βα

ThenT ′µν + T ′νµ = Λµ

αΛνβ

(T αβ + T βα

)symmetric tensor transforms into symmetric tensor. Similarly, the anti-symmetric tensortransforms into antisymmetic one.

6 gµν, and εαβγδ have the property

Λµα Λν

β gµν = gαβ, εαβγδ det (Λ) = ΛαµΛβ

ν Λγρ Λδ

σεµνρσ

gµν, and εαβγδ transform in the same way as tensors if det (Λ) = 1.

Example: M µν = xµpν − x νpµ, F µν = ∂µAν − ∂νAµ 2nd rank antisymmetric tensor.

(Institute) Chapter 1 Introduction 21 / 37

Note that if all components of a tensor vanish in one inertial frame they vanish in all inertialframe. Suppose

f µ = maµ

Definetµ = f µ −maµ

then tµ = 0 in this inertial frame. In another inertial frame,

t ′µ = f µ′ −ma′µ = 0

we get

f µ′ = ma′µ

Thus physical laws in tensor form are same in all inertial frames .

(Institute) Chapter 1 Introduction 22 / 37

Action principle: actual trajectory of a partilce minimizes the actionParticle mechanicsA particle moves from x1 at t1 to x2 at t2. Write the action as

S =∫ t2

t1L(x , x ) dt L : Lagrangian

For the least action, make a small change x (t),

x (t)→ x ′(t) = x (t) + δx (t)

with end points fixed

i .e . δx (t1) = δx (t2) = 0 initial conditions

Then

δS =∫ t2

t1[∂L∂x

δx +∂L∂x

δ(x )] dt

Note that

δx = x ′(t)− x (t) = ddt[δ(x )]

Integrate by parts and used the initial conditions

δS =∫ t2

t1[∂L∂x

δx +∂L∂x

ddt(δx )] dt =

∫ t2

t1[∂L∂x− ddt(

∂L∂x)]δx dt

(Institute) Chapter 1 Introduction 23 / 37

Since δx (t) is arbitrary, δS = 0 implies

∂L∂x− ddt(

∂L∂x) = 0 Euler-Lagrange equation

Conjugate momentum is

p ≡ ∂L∂x

Hamiltonian is ,H (p, q) = px − L(x , x )

Consider the simple case

md 2xdt2

= − ∂V∂x

Suppose

L =m2(dxdt)2 − V (x )

then∂L∂x=ddt(

∂L∂x), ⇒ − ∂V

∂x= m

d 2xdt2

Hamiltonian

H = px − L = m2(x )2 + V (x ) where p =

∂L∂x= mx

is just the total energy.

(Institute) Chapter 1 Introduction 24 / 37

Generalizationx (t)→ qi (t), i = 1, 2, ..., n

S =∫ t2

t1L(qi , qi ) dt

Euler-Lagrange equationsddt(

∂L∂qi)− ∂L

∂qi= 0 i = 1, 2, ..., n

pi =∂L∂qi

, H = ∑ipi qi − L

Example: harmonic oscillator in 3-dimensionsLagrangian

L = T − V = m2(x12 + x22 + x32)−

mw 2

2(x 21 + x

22 + x

23 )

=m2

(d→xdτ

)2− mw

2

2

(→x)2

and∂L∂xi

= −mw 2xi ,∂L∂xi

= mxi

(Institute) Chapter 1 Introduction 25 / 37

Euler-Langarange equationm

..x i = −mw 2xi

same as Newton’s second law.Remarks:

We need action principle for quantization

In action principle formulation, the discussion of symmetry is simpler

Can take into account the constraints in the coordinates in terms of Lagrange multiplers

(Institute) Chapter 1 Introduction 26 / 37

Field TheoryField theory ∼ limiting case where number of degrees of freedom is infinite. qi (t)→ φ(−→x , t).Action

S =∫L(φ, ∂µφ) d 3xdt L : Lagrangian density

Variation of action

δS =∫[∂L∂φ

δφ+∂L

∂(∂µφ)δ(∂µφ)] dx 4 =

∫[∂L∂φ− ∂µ

∂L∂(∂µφ)

]δφ dx 4

Use δ(∂µφ) = ∂µ(δφ) and do the integration by part. then δS = 0 implies

=⇒ ∂L∂φ= ∂µ(

∂L∂(∂µφ)

) Euler-Lagrange equation

Conjugate momentum density

π(−→x , t) =∂L

∂(∂0φ)

and Hamiltonian densityH = πφ− L

Generalization to more than one field

φ(−→x , t)→ φi (−→x , t), i = 1, 2, ..., n

(Institute) Chapter 1 Introduction 27 / 37

Equations of motion are∂L∂φi

= ∂µ(∂L

∂(∂µφi )) i = 1, 2, ..., n

and conjugate momentum

πi (−→x , t) =

∂L∂(∂0φi )

Hamiltonian density isH = ∑

iπi φi − L

(Institute) Chapter 1 Introduction 28 / 37

Symmetry and Noether’s TheoremContinuous symmetry =⇒ conservation law, e.g. invariance under time translation

t → t + a, a is arbitrary constant

gives energy conservation. Newton’s equation for a force derived from a potential V (→x , t) is,

md 2→x

dt2= −

→∇V (→x , t)

Suppose V (→x , t) =V (

→x ), then invariant under time translation and

md−→xdt·(d 2−→xdt2

)= −

(d−→xdt

)· −→∇V = − d

dt[V (−→x )]

Orddt[12m(d−→xdt)2 + V (−→x )] = 0, energy conservation

Similarity, invariance under spatial translation

−→x → −→x +−→a

gives momentum conservation and invariance under rotations gives angular momentumconservation. Noether’s theorem : unified treatment of symmetries in the Lagrangian formalism.

(Institute) Chapter 1 Introduction 29 / 37

Particle mechanicsAction in classical mech

S =∫L(qi , qi ) dt

Suppose S is invariant under a continuous symmetry transformation,

qi → q ′i = fi (qj ) ,

For infinitesmal changeqi → q ′i ' qi + δqi

The change of S

δS =∫[

∂L∂qi

δqi +∂L∂qi

δqi ] dt where δqi →ddt(δqi )

Using the equation of motion,∂L∂qi

=ddt(

∂L∂qi)

we can write δS as

δS =∫[ddt(

∂L∂qi)δqi +

∂L∂qi

ddt(δqi )] dt =

∫[ddt(

∂L∂qi

δqi )] dt

Thus δS = 0 ⇒ddt(

∂L∂qi

δqi ) = 0

(Institute) Chapter 1 Introduction 30 / 37

This can be written as

ordAdt

= 0, A =∂L∂qi

δqj

A is the conserved charge.

Note if δL 6= 0 but changes by a total time derivative δL =ddtK ,we still get the conservation

law in the form,ddt(

∂L∂qi

δqi −K ) = 0

because the action is still invariant. For example, for translation in time, t −→ t + ε,

q (t + ε) = q (t) + εdqdt

, =⇒ δq = εdqdt

Similarly,

δL =dLdt

The conservation law is thenddt(

∂L∂qi

δqi − L) = 0

OrdHdt

= 0, with H =∂L∂qi

qi − L

(Institute) Chapter 1 Introduction 31 / 37

Example: rotational symmetry in 3-dimensionaction

S =∫L(xi , xi ) dt

Suppose S is invariant under rotation,

xi → x ′i = Rij xj , RRT = RT R = 1 or RijRik = δjk

For example, a rotation around z−axis,

Rz (θ) =

cos θ sin θ 0− sin θ cos θ 00 0 1

For infinitesmal rotations, we write

Rij = δij + εij , |εij | � 1

Orthogonality requires,

(δij + εij )(δik + εik ) = δjk =⇒ εjk + εkj = 0 i , e , εjk is antisymmetric

For example, for θ � 1,

Rz (θ)→ 1+

0 θ 0−θ 0 00 0 0

(Institute) Chapter 1 Introduction 32 / 37

and ε12 = θ corresponds to infinitesmal rotation about z -axis. Similarly ε23 corresponds torotation about x -axis and ε31 rotation about y -axis.For the general case, we can compute the conserved quantities as

J =∂L∂xi

εij xj = εijpi xj

More explicitly,

J = ε12 (p1x2 − p2x1) + ε23 (p2x3 − p3x2) + ε13 (p1x3 − p1x3)

If we writeε12 = −θ3, ε23 = −θ1, ε31 = −θ2

orεij = −εijk θk

thenJ = −θk εijkpi xj = θk Jk Jk = εijk xipj

More explicitly,

J1 = (x2p3 − x3p2) , J2 = (x3p1 − x1p3) , J3 = (x1p2 − x2p1) ,

Hence Jk can be identified with k -th component of the usual angular momentum.If we write εij = −εijk θk

J = −θk εijkpi xj = −θk Jk Jk = εijk xipj

Jk k-th component of angular momentum.

(Institute) Chapter 1 Introduction 33 / 37

Field TheoryStart from the action

S =∫L(φ, ∂µφ) d 4x

Symmetry transformation,φ(x )→ φ′(x ′),

which includes the change of coordinates,

xµ → x ′µ 6= xµ

Infinitesmal transformation

δφ = φ′(x ′)− φ (x ) , δx ′µ = x ′µ − xµ

need to include the change in the volume element

d 4x ′ = Jd 4x where J =

∣∣∣∣ ∂(x ′0, x ′1, x ′2, x ′3)∂(x0, x1, x2, x3)

∣∣∣∣J :Jacobian for the coordinate transformation. For infinitesmal transformation,

J = | ∂x′µ

∂x ν| ≈ |g µ

ν +∂(δxµ)

∂x ν| ≈ 1+ ∂µ(δxµ)

(Institute) Chapter 1 Introduction 34 / 37

we have used the relation

det(1+ ε) ≈ 1+ Tr (ε) for |ε| � 1

Thend 4x ′ = d 4x (1+ ∂µ(δxµ))

change in the action is

δS =∫[∂L∂φ

δφ+∂L

∂(∂µφ)δ(∂µφ) + L∂µ(δxµ)] dx 4

Define the change of φ for fixed xµ,

δφ(x ) = φ′(x )− φ(x ) = φ′(x )− φ′(x ′) + φ′(x ′)− φ(x ) = −∂µφ′δxµ + δφ

or δφ = δφ+ (∂µφ)δxµ

Similarly,δ(∂µφ) = δ(∂µφ) + ∂ν(∂µφ)δx ν

Operator δ commutes with the derivative operator ∂µ,

δ(∂µφ) = ∂µ(δφ)

(Institute) Chapter 1 Introduction 35 / 37

Then

δS =∫[∂L∂φ(δφ+ (∂µφ)δxµ) +

∂L∂(∂µφ)

(δ(∂µφ) + ∂ν(∂µφ)δx ν) + L∂µ(δxµ)] dx 4

Use equation of motion∂L∂φ= ∂µ(

∂L∂(∂µφ)

)

we get∂L∂φ

δφ+∂L

∂(∂µφ)δ(∂µφ) = ∂µ ∂L

∂(∂µφ)δφ+

∂L∂(∂µφ)

∂µ(δφ) = ∂µ[∂L

∂(∂µφ)δφ]

Combine other terms as

[∂L∂φ(∂νφ) +

∂L∂(∂µφ)

∂ν(∂µφ)]δx ν + L∂ν(δx ν) = (∂νL)δx ν + L∂ν(δx ν)

= ∂ν(Lδx ν)

Then

δS =∫dx 4∂µ[

∂L∂(∂µφ)

δφ+ Lδxµ]

and if δS=0 under the symmetry ransformation, then

∂µJµ = ∂µ[∂L

∂(∂µφ)δφ+ Lδxµ] = 0 current conservation

(Institute) Chapter 1 Introduction 36 / 37

Simple case: space-time translationHere the coordinate transformation is,

xµ → x ′µ = xµ + aµ =⇒ φ′(x + a) = φ(x )

thenδφ = −aµ∂µφ

and the conservation laws take the form

∂µ[∂L

∂(∂µφ)(−aν∂νφ) + Laµ] = −∂µ(Tµνaν) = 0

where

Tµν =∂L

∂(∂µφ)∂νφ− gµνL energy momentum tensor

In particular,

T0i =∂L

∂(∂0φ)∂iφ

andPi =

∫dx 3T0i momentum of the fields

(Institute) Chapter 1 Introduction 37 / 37