1interaccion Debil en Ingles

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Fenomenologa de las Interacciones Fundamentales(La Interaccion Debil. Unificacion Electrodebil)

Agustn NietoDepartamento de Fsica

Universidad de Oviedo

Enero Mayo 2012


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Outline

Spontaneous Symmetry Breaking and The Higgs Mechanism

Beta Decay. Fermis Model

Parity Violation

Fermis Model Revisited

The GWS Model. Electroweak Unification


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Massive Vector Bosons and Gauge Symmetry

The weak interaction is mediated by massive vector (spin 1) bosons.

Under an infinitesimal gauge transformation:

Aa Aa + 1g

a + fabcAb

c

a mass term for a gauge field (such as m2AaAa) isnt invariant:

AaAa AaAa + 2

g(a) A

a +

1

g2(a) (

a)

How can we construct a gauge theory with a massive gauge field?

For the moment, lets just study a few non-gauge scalar theories.


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The Baby ModelThe Lagrangian

Lets consider a field theory for a scalar field described by the Lagrangian:

L() = 12

()() +

1

22 2

44

where 2 > 0.

Two comments:

This Lagrangian is symmetric under the transformation. If the sign in front the 2-term were instead of +, this Lagrangian

would describe a scalar particle of mass . But ...... with +, the2-term isnt a mass term anymore.


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The Baby ModelThe Potential

The Lagrangian describes a system with potential:

V() =

1

22 2 +

44

V()

vv

It has:

Two minima at = v with v =2/. A maximum at = 0.

in L() represents fluctuations around = 0, but ...... it is unstable.


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The Baby ModelThe Vacuum and Spontaneous Symmetry Breaking

The Lagrangian has to be written in terms of a field that represents

fluctuations around the vacuum (a minimum):

(x) = v + (x)

Now, (x) represents quantum fluctuations around the minumum at = v.

We substitute in L() and arrive atL() = 1

2()(

) 2 2 v3 4

4 + const.

The 2-term now has the correct sign for a mass term. Therefore,

represents a particle of mass m =22. This way of generating mass terms is called Spontaneous Symmetry

Breaking (SSB).


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The Baby ModelThe Hidden Symmetry

The Lagrangian:

L() = 12

2 2 +

44

shows a symmetry

that is hidden in

L() = 12

()() 2 2

2 3 4

4

More precisely the symmetry is responsible for the mass and coupling

constants not being independent (the 3 term has coupling constant

2).

But L() gives the correct particle content.


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The Child ModelThe Lagrangian

Lets consider a field theory with a complex scalar field which is described

by the Lagrangian:

L(, ) = ()() + 2 ()2

with 2

> 0.

This theory has a global phase symmetry

(x) ei (x)

The quadratic has the wrong sign for a mass term.


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The Child ModelA Story of Two (Real) Fields

Since is a complex field, it can be written in terms of 2 real fields:

=1

2(1 + i2)

In terms of 1 and 2, the Lagrangian is

L(1, 2) = 12

(1)(1) +

1

2(2)(

2) +

+1

22 (21 +

22)

4(21 +

22)

2


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The Child ModelThe Potential

The potential

V(1, 2) = 12

2 (21 + 22) +

4

(21 + 22)

2

has a circle of minima in the (1, 2) plane of radius v such that

21 + 22 = v

2 with v2 = 2/

1

2

V

circle of minimaof ra ius v


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The Child ModelThe Vacuum. Spontaneous Symmetry Breaking

As in the baby model, we want to write the Lagrangian in terms of fields

that represent fluctuations around a minimum (the vacuum).

We can choose any point in the minima circle; for simplicity, we take

(1, 2) = (v, 0)

and write

(x) =

1

2v + (x) + i(x)where now (x) and (x) describe the vacuum fluctuations.

L(, ) = 12

()() +

1

2()(

) 2 2

+ cubic and quartic terms in and

+ const.

represents a particle of mass m =

22,

is a massless scalar which is known as a Goldstone boson.


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The Teen ModelSpontaneous Symmetry Breaking

Lets consider a field theory with N real scalar fields a (with a = 1, . . . , N):

L(a) = 12

(a)(a) +

1

22 aa

4(aa)

2

Fluctuations around, lets say (v, 0, . . . , 0), give the SSB Lagrangian:

L(, i) = 12

(i)(i) + 1

2()(

) 2 2

+ cubic and quartic terms in and i

+ const.

which describes:

A scalar of mass m =22. N 1 massless Goldstone bosons.


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The U(1) Gauge TheoryThe Lagrangian

Lets consider the a U(1) gauge theory described by the Lagrangian

L(, A) = 14

FF + (D)

(D) + 2 ()2

where 2 > 0, F = A A and D = i g A.

This Lagrangian is invariant under U(1) (Abelian) gauge transformations:

(x) (x) ei(x) (x)

1 + i(x)

A(x) A(x) + 1g

(x)

The quadratic term has the wrong sign for a mass term.


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The U(1) Gauge TheoryThe Vacuum

As in the child model we can set = (1 + i2)/

2.

The potential V = 2 + ()2 has minima at 21 + 22 = v2 wherev2 = 2/: a circle of radius v in the (1, 2) plane.

Now, expanding the Lagrangian around (1, 2) = (v, 0) with

(x) =1

2

v + (x) + i(x)

we get ...

Th U( ) G Th


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The U(1) Gauge TheorySpontaneous Symmetry Breaking

...

L(,, A) = 14

FF + 12

()() + 12

()()

2 2 + 12

g2v2 AA gv A()

+ cubic and quartic terms in and

+ const.

We obtain:

A scalar with mass m =

22.

A massless Goldstone boson .

A gauge field A with mass mA = gv. We have found a way of gettingmassive gauge bosons without breaking gauge symmetry !!.

The gauge symmetry which is apparent in L(, A) is hidden in L(,, A)and it gives the particle content.

Th U(1) G Th


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The U(1) Gauge TheoryWait a Minute!

BUT, something must be wrong.

L(, A) has 4 degrees of freedom:d.o.f

(complex scalar) 2A (massless vector) 2

TOTAL 4

L(,, A) has 5 degrees of freedom:d.o.f

(real scalar) 1

(real scalar, Goldstone) 1A (massive vector) 3

TOTAL 5

and, yet, they describe the same physical system.

Th U(1) G Th


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The U(1) Gauge TheoryEating up Goldstone Bosons

Actually, we can see

(x) =

1

2 v + (x) + i(x)as an infinitesimal gauge transformation of a field (v + )/

2 with gauge

parameter /(v + ):

(x) = v + 2 1 + i v + 12v + H(x) ei(x)If we now write the gauge field as a gauge transformed field

A(x) = B(x) +1

g

v +

B(x) + 1

g(x)

doesnt appear in the gauge invariant Lagrangian L(, B) = L(H, B). TheGoldstone boson has been eaten up.

Note that:

H(x)

(x) and (x)

(x)/(v + (x))

The Higgs Mechanism


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The Higgs Mechanism

Given a gauge invariant theory the gauge fields can get a mass by adding

a scalar Lagrangian LHiggs:L(A, ) = 1

4FF

+ LHiggs

where

LHiggs = (D)

(D) + 2

()2

and expanding around the vacuum state v = v/2 (which is left invariant bygauge transformations) by setting

(x) =1

2

v + H(x)

ei(x) and A(x) = B(x) +1

g(x)

This procedure is known as the Higgs mechanism. Here, we implicitly choosethe unitary gauge.

H is the Higgs field or particle. Sometimes also is called the Higgs.

Higgs Mechanism The Lagrangian


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Higgs Mechanism - The Lagrangian

The Lagrangian:

L = 14

GG +

1

2g2(v + H)2BB

+1

2

(H)(H) +

1

2

2 (v + H)2

4

(v + H)4

Note that the Higgs has mass mH =

22 and the gauge boson mB = gv.

In order to obtain the broken theory, it is enough to substitute

(x)

1

2(v + H).

Outline


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Outline



Parity Violation



Beta Decay


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Beta Decay

In beta decay an atomic nucleus A is transformed into a slightly lighter

nucleus B with the emission of an electron:

A B+ e

Some examples:

40K 40Ca + e64Cu 64Zn + e

3H 3He + e

The underlying process involves the transformation of a neutron to a proton:

n p+ + e

Beta Decay


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Beta DecayExperiments

Using elementary relativistic kinematics, the electron energy in the

center-of-mass frame should be:

Ee =m2A m2B + m2e

2mA

However, experiments (Lise Meitner and Otto Hahn, 1911) show that the

energy of the emitted electrons vary considerably and this formula only gives

the maximum electron energy.

Emax

E

d

dE

Beta Decay


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Beta DecayThe Third Particle

Pauli (1930) suggested that the experimental results could be understood

if the beta decay, instead of producing 2 particles, produces 3 particles.

The third particle must be electrically neutral because it doesnt leaveany trace in detectors.

Its mass must be very small because some electrons can almost reach

the maximum energy Ee.

The new particle is called neutrino and is represented by the symbol (actually the particles produced in beta decay are electron antineutrinos).

Typical beta decay processes:

40K 40Ca + e + e64Cu 64Zn + e + e

3H 3He + e + eThe underlying process:

n p+ + e + e

Beta Decay


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Beta DecayFermis Model

Fermi (1932) proposed a theory of beta decay inspired by the structure of

the electromagnetic interaction that incorporated the neutrino.

In modern terms, Fermis model for the beta decay n p e e isdescribed by the Lagrangian:

L-decay = GF

2 p(x)n(x) e(x)e(x)+ h.c.where fermion fields are named by their particle names and GF is the Fermi

coupling constant.

Recall that the electromagnetic interaction (for instance, for an electron

and a structureless proton) is described by:

Lem = e A(x) e(x)e(x) + e A(x) p(x)p(x)

Beta Decay


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Beta DecayFermis Model

The electromagnetic ep ep process and the crossed beta decayep ne are represented by the Feynman diagrams:

e

p

e

p

e

p

e

n

We see that:

the electromagnetic process is mediated by a particle (a photon).

the interaction of the Fermi model is at a point.

Beta Decay


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Beta DecayMore Decays

Other processes involving neutrinos where discovered.

Pion decay: + + + Muon decay: e + e +

These processes, like beta decays, were found to have comparatively long

lifetime.

The concept of a distinctive class of interactions, the weak interactions,

began to emerge.

It is natural to hope that all weak interaction phenomena are described by an

interaction with a universal coupling constant GF.

Beta Decay


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y(Charge-rising) Weak Currents

Defining the electromagnetic current: Jem = e(x)e(x) + p(x)p(x)the electromagnetic interaction ca be written as

Lem = e A Jem

In a similar way, we can define the hadron charge-rising weak current:

Jcw,h = p(x)n(x)

and the lepton charge-rising weak current:

J

cw,l = e(x)

e(x)

Note that (Jcw,h) = n(x)p(x) and (Jcw,l)

= e(x)e(x) are thecorresponding charge-lowering currents.

Then, the Lagrangian for n p e e is

L-decay = GF2

(Jcw,l)

Jcw,h + h.c.

Outline


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Parity Violation



5, , and A Basis for 4 4 Matrices


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We define the 5 matrix:

5

i0123

which has the properties:, 5

= 0 ,

5

= 5 ,

52

= I

In the Pauli-Dirac (Bjorken-Drell) representation of the Dirac matrices:

5 =0 1

1 0

We also define:

i

2 (

) =

i

2 , It can be shown that matrices I, , 5, 5, and are a basis for thespace of 4 4 matrices: Every 4 4 matrix can be written as a linearcombination of these 16 matrices.

Spinor Bilinears


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For a spinor , in total there are 16 products: i j with i,j = 1, 2, 3, 4.

These products can be added together in various linear combinations to

construct quantities with distinct behavior under Lorentz transformations:

Scalar: (1 component)

Pseudoscalar: 5 (1 component)

Vector: (4 components)

Axial vector: 5 (4 components)

Tensor: (6 components)

Under a proper Lorentz transformation (det = +1), scalars andpseudoscalars transform as scalars, vectors and axial vectors transform as

vectors, and tensors transform as antisymmetric tensors.

Chiral Spinors


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For a Dirac spinor we can construct two chiral spinors:

L 12

1 5) (left-handed)

R 12

1 + 5) (right-handed)

NOTES:

= L + R.

5L = L and 5R = +R. Chiral spinors may have a definite helicity.

If u is a solution of Diracs equation, uL and uR are also solutions of it.

Parity


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Definition

Parity is the space inversion operation:

P : x = (x0, x) xP = (x0,x)

A vector transforms as

V VP = V

with

=

1

11

1

Note that

1

= . Also, the parity transformation when applied twicegives the identity:

P2 = I

Therefore, the eigenvalues of the parity transformation are 1.

Parity


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Spinors

Under a Lorentz transformation , spinors transform as

Swhere S and are related by the identity:

S1S =

Under the parity transformation:

S10S = 0

S1iS = i

Since 000 = 0 and 0i0 = i, we can take S = 0. Here is anarbitrary phase which can be taken as = 1. Then,

S = 0

and

0

ParityS C S


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Spinor Bilinears and Chiral Spinors

Under the parity transformation, spinor bilinears transform as follows:

Scalar: Pseudoscalar: 5 5

Vector:

0 = 0

i = i

Axial vector: 5 05 = 0i5 = i

Chiral spinors transform as:

L 0RR 0L

We see that parity takes right-handed states into left-handed states (and

the other way around). This means that any process that treats

right-handed and left-handed states in a different way will violate parity.

ParityI t i i P it


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Intrinsic Parity

Spinors that represent fermions (antifermions) at rest (

p = 0) areeigenvectors of the parity operator 0 with eigenvalue +1 (1). However,

fermions and antifermions which are not at rest dont have parity defined.

According to Quantum Field Theory

The parity of a fermion is opposite to that of its antiparticle.

The parity of a boson is the same as its antiparticle.

If we just want to study parity conservation, all we need is to assign parity

conventionally and use the fact that the electromagnetic and strong

interactions respect parity to make further assignments.

We set the parity of quarks to be positive (P = +1), the parity of antiquarksto be negative (P = 1), and the parity of the photon negative (P = 1).

ParityComposite S stems


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Composite Systems

The parity of a composite system in its ground state is the product of the

parities of its constituents.

Baryons have positive parity: P = (+1)3 = +1

Antibaryons have negative parity: P = (

1)3 =

1

Mesons have negative parity: P = (1)(+1) = 1

For excited states, there is an extra factor (1)L, where L is the orbitalangular momentum.

In general, mesons have parity P = (

1)L+1.

Parity ViolationFermis Model Only has Vectors


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Fermi s Model Only has Vectors

Fermis guess of the form of the weak current for the beta decay

L-decay = GF2 (J

cw,l) Jcw,h + h.c.

with

Jcw,h = p(x)n(x) and Jcw,l = e(x)

e(x)

is a very specific choice from among the different Lorentz invariant

amplitudes that can be constructed using all the possible bilinear quantities:

Scalar:

Pseudoscalar: 5

(Polar) Vector:

Axial Vector:

5

There is a priori no reason to use only vectors. The presence of

pseudoscalar or axial vector terms would automatically violate parity and in

the 1930s there were no good reason to include such terms.

Parity ViolationThe tau theta Puzzle


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The tau-theta Puzzle

The tau-theta puzzle.

Two strange mesons (called and ) appeared to be identical in everyrespect except that one of them decayed into two pions and the other one

into three pions which are states of opposite parity:

+ + + 0 (P = +1)

+ + + 0 + 0+ + + +

(P = 1)

It seemed most peculiar that two otherwise identical particles should carry

different parity.

T.D. Lee and C.N. Yang (1956) suggested that and are really the sameparticle (now known as the K+), and parity is simply not conserved in one of

the decays.

Parity ViolationThe Experiment


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The Experiment

Since at the time there was no parity tests for the weak interaction, Lee

and Yang proposed a test.

The experiment was carried out by C.S. Wu (1957).

Radioactive 60Co undergoes beta decay, and the direction of the emitted

electrons is recorded.

What she found was that most of them came out in the north direction (the

direction of the nuclear spin).

This was the first experimental evidence of parity violation in weak

interactions.

Parity ViolationPion Decay


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Pion Decay

I will not discuss Wus experiment here. Instead, I will study the simpler

pion beta decay:

+

+

+

We will analyze what one would expect if parity were preserved and we will

see what actually happens.

The spin of the pion is 0 and its parity 1 (JP = 0). Therefore, in thecenter-of-mass frame, the angular momentum L and spin S of the outgoingparticles must be equal L = S (for otherwise their sum wouldnt give J = 0).On the other hand, since both + and are spin 1/2 particles, S = 0, 1.

The lowest energy state will be the one with L = S = 0:

|S = 0, ms = 0 =1

2 |S = + 12 , S = 12 |S = 12 , S = + 12 The z-axis for spins is taken in the direction of the outgoing particles which in

the center-of-mass frame come out back to back. Since spins are oppositely

aligned, if the neutrino has positive helicity, the muon also has positive

helicity and the other way around.

Parity ViolationPion Decay


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Pion Decay

Note that in state |S = 0, ms = 0 half the neutrinos have positive helicityand the other half has negative helicity. It we make a parity transformation,

z z which is equivalent to changing signs in the states in|S = 0, ms = 0, nothing changes (except for an irrelevant global sign): thisstate is parity invariant.

However, experiments tell us that the weak interaction violates parity and

there are only neutrinos with negative helicity. So, if the system is an state

|S = + 12 , S = 12 which describes neutrinos with negative helicity, after aparity transformation, the system must be in state |S = 12 , S = + 12 . Thestate is not parity invariant.

Only left-handed neutrinos are found experimentally.

Parity ViolationModifying Fermis Model


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y g

The only essential change requiered in Fermis original proposal:

L-decay = GF2

(Jcw,l)

Jcw,h + h.c.

is the replacement of by (1

5) in the hadron charged weak current:

Jcw,h = p(1 5) n = 2 pL (1 5) nL

and in the lepton charged weak current:

Jcw,l = e (1 5) e = 2 eL (1 5) eL

Outline


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Parity Violation



Fermis Model RevisitedMuon Decay


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The muon decay ee

e

e

is described by the Lagrangian

L-decay = GF2

[e(1 5)e] [(1 5)] + h.c.

which can be written as

L-decay =

GF

2(Jcw,l)

Jcw,l

where we have defined the lepton charged weak current:

Jcw,l = e(1 5)e+ (1 5)

In terms of Jcw,l

,L-decay has two additional terms that couple two electrons to two es and two s to two s; these terms dont

contribute to the muon decay, but they would contribute, for instance, to Compton-like processes such as ee ee.

Fermis Model RevisitedQuarks


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There is another important modification that was (necessarily) missing in

Fermis theory.

By the mid 1970s it was clear that protons, neutrons and other hadrons

are made out of quarks.

The (non-leptonic) fundamental weak interaction must involve

quarks instead of hadrons.

Fermis Model RevisitedBeta Decay


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Since the neutron quark content is udd and the proton is uud, Fermis

theory of beta decay, at fundamental level, must describe the decay

d u ee:d

e

e

u

L-decay = GF2

(Jcw,e)

Jcw,u + h.c.

with the electron charged weak current:

Jcw,e = e (1 5) eand the quark charged weak current:

Jcw,u = u(1 5) d

Fermis Model RevisitedPion Decay


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Since + ud, the pion decay + + + is ud +:

d

u

+

This process will be described by the Lagrangian

L-decay = GF2

(Jcw,) Jcw,u + h.c.

where Jcw, = (1 5).

Fermis Model RevisitedSummary


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All these decays (, and +) plus other processes can be summarized in

the Lagrangian

LFermi = GF2

(Jcw) Jcw

where Jcw is the charged weak current defined as

J

cw= e

(1

5) e+ (1

5) + u(1

5) d

A theory for the weak interactions has to include flavor mixing and 3

generations of leptons and quarks. This implies that the charged weak

current has to be modified:

Jcw = (1 5) e + u (1 5) V d

where V is the CKM matrix and = 1, 2, 3 represents the generation.

Fermis Model RevisitedBeyond Fermis Model: The GWS Model


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Today, the Fermi model is understood as a low energy approximation of

the GWS model for the electroweak theory.

In the GWS model there is a heavy particle (the W) which couples to the

charged weak current.

In the low energy limit (p2 M2W) its momentum can be neglected, thepropagator becomes a constant (proportional to 1/M2W) and the interactionseems to take place at a point.

Fermis Model RevisitedFermis Model vs. GWS Model

The amplitude for the muon decay e :


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The amplitude for the muon decay e e :

e

e

e

e

Fermis Model GWS Model

In Fermis model:

iM = i GF2

u

(1 5)u

ue(1 5)ve

In the GWS model:

iM = ig222u(15)u ip2 M2W + ippM2W ue(15)ve

In the low energy limit (p2 M2W):

iM = i

g

2

2MW2

u

(1 5)uue(1 5)ve

Fermis Model RevisitedFermis Model vs. GWS Model


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Comparing results, we find a relation between GF and g:

GF2

=g2

8M2W

The Fermi constant can be obtained by measuring the muon lifetime and

comparing with the value computed in Fermis model from i

M:

(exp.) 2.197 03 106 s , (comp.) = 192 3

7

G2Fm5c4

We obtain:

GF (comp.)

1.163 8

105 GeV (c)3

which is not very different from the value given in the Review of Particle

Physics:

GF 1.166 37 105 GeV2 (c)3

Outline


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Parity Violation



The GWS ModelIntroduction


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The Glashow-Weinberg-Salam (GWS) model provides a way of unifyingweak and electromagnetic interactions.

It is based on a SU(2) U(1) gauge field theory that isspontaneously broken to a theory that containselectromagneticand

weak interactions.

This model shows perfect agreement of theory and experiments.

There is though a missed piece: the Higgs boson.

One may try other models, but none of them is as consistent with

experiments as the GWS.

The GWS ModelMotivation


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The motivation of the model goes back to the 1960s particle physics which

we will not discuss here. Instead, we recall that:

The weak interaction is mediated by three massive vector bosons: W

and Z0 with masses MW and MZ respectively.

The electromagnetic interaction is mediated by the photon which is

massless. Quarks and leptons (3 families each) feel the weak interaction.

Neutrinos dont feel the electromagnetic interaction (their electric charge

is zero).

Neutrinos are almost massless. For simplicity, we will assume that

neutrinos are massless. The weak interaction violates parity in such a way that (for massless

neutrinos), they are all left-handed.

The GWS ModelThe Players


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A theory for the weak and electromagnetic interactions must involve:

leptons: L, eL, eR (3 families)

quarks: uL, uR, dL, dR (3 families)

vector bosons: W, Z0

photon:

The GWS model is based on breaking the full SU(2) U(1) group. Lefthanded leptons L and eL are put into an SU(2) doublet (fundamentalrepresentation) and eR is put into a singlet representation of SU(2):

L

LeL

, eR

Similarly, for quarks, we put uL and dL in a doublet and uR and dR in an SU(2)singlet:

qL

uLdL

, uR , dR

The GWS ModelSome Comments


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All L, eL, eR, uL, uR, dL, and dR are Dirac fields.

Quarks also are in the fundamental representation of SU(3)color but it is

not important now.

The other families are included in the same way, but for now we only

consider the first family.

The GWS ModelGauge Transformations


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Under a gauge SU(2) U(1) transformation:L ei

a(x)a

2+i(x)

Y2 L

eR ei(x)Ye

2 eR

qL

eia(x)

a

2+i(x)

Yq2 qL

uR ei(x)Yu

2 uR

dR ei(x)Yd

2 dR

where a= 1, 2, 3 and the as are the Pauli matrices.

The constants Ys are called hypercharges.

The GWS ModelCovariant Derivatives


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The covariant derivatives are:

DL =

i g a

2Wa i g Y

2B

L

DeR =

i g Ye

2B

eR

DqL = i g a2 Wa i g Yq2 B qLDuR =

i g Yu

2B

uR

DdR =

i g Yd

2B

dR

Wa and B are the SU(2) and U(1) gauge bosons, respectively.

g is the coupling constant for SU(2) and g for U(1).

The GWS ModelThe Higgs Field


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In order to break the SU(2)U(1) gauge symmetry we need a Higgs fieldwhich we take as a SU(2) doublet

12

Under a gauge transformation:

eia(x)a

2+i(x) I

2

The covariant derivative of is:

D = i g a

2Wa

i g

I

2B

The GWS ModelThe Lagrangian (1/2)


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The GWS model is described by the following SU(2) U(1) gaugeinvariant Lagrangian:

LGWS = LG + LH + LF + LHF

LG is the purely gauge part of the the Lagrangian:

LG = 14

WaW

a 14

BB

where

Wa = Wa Wa + gabc WbWc

B = B B LH contains the Higgs field:

LH =

D

D

+ 2 2This is the term that spontaneously breaks the symmetry.

The GWS ModelThe Lagrangian (2/2)


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LFcontains the kinetic terms for fermions and their interactions with the

gauge fields:

LF = L i/DL + eR i/DeR + qL i/DqL + uR i/DuR + dR i/DdR

LHFdescribes the interaction of the Higgs field with fermions:

LHF = fe LeR fu qLuR fd qLdR + h.c.where, in the second term, we have used

= 2

1instead of .

The GWS ModelComments on the Lagrangian


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Comments:

In LF, mass terms such as (eLeR + eReL) would violate gauge symmetrybecause left handed and right handed fields transform in different

representations of SU(2).

Also note that eLeL = eReR = 0.

So there are no mass terms for fermions.

Just on gauge invariance grounds there is no reason for not using instead of in LHF, but otherwise (after symmetry breaking) the u quarkwould remain massless.

In order for LHF to be gauge invariant, the hypercharges must be related:

Y = Ye + 1Yq = Yu 1 = Yd + 1 Yu = Yd + 2

The GWS ModelBreaking the Symmetry


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The Higgs potential in LH:

2 + 2has minima at

=v2

2with v +

2

We choose the vacuum at

v = 120

v

To obtain the broken theory in the unitary gauge, in LGWS, we take:

(x) = v+H =12 0v + H(x)

H(x) represents the fluctuations around the vacuum. Its the Higgs field.

The GWS ModelVector Bosons and The Photon (1/2)

We will eventually substitute v+H in the Higgs Lagrangian LH but we start


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We will eventually substitute v+H in the Higgs Lagrangian LH, but we startby substituting v in the covariant derivative terms

D

D

to get

mass terms for gauge fields. We find:

Dv

Dv

=vg

2

2W+ W

+

v2

8

W3 B g2 g gg g g2

W3B

where we have defined

W 1

2W1 i W2 To diagonalize the matrix, we define:

AZ

sin w cos wcos w

sin w

W3B

The weak mixing angle (or Weinberg angle) w is defined as:

tan w g

g

The GWS ModelVector Bosons and The Photon (2/2)


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Then,

Dv

Dv

= M2W W+ W

+

v2

8

A Z0 0

0 g2 + g2A

Z

= M2W W+ W

+

1

2M2Z ZZ

where

MW v2

g and MZ v2g2 + g2

are the masses of the W and Z vector bosons, respectively.

Note thatMW

MZ= cos w

The field A remains massless; this a hint about A describing the photon.

The GWS ModelNew Covariant Derivatives

N l t it th i t d i ti


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Now, lets write the covariant derivatives

D = i ga

2 W

a

i gY

2 B

in terms of the new vector fields W , Z, and A.

Since

g1

2W1 + g

2

2W2 =

g

2 0 W+

W

0 and

g3

2W3+g

Y2

B =g g

g2 + g2

T3+

Y

2

A+

g T3 cos wg Y

2sin w

Z

where T

3

= 3

/2. Defining

e g g

g2 + g2= g cos w = g sin w

The GWS ModelThe Broken Symmetry


67/91

We can write

D = i eQ A ig2

0 W+W 0

ig

cos w

T3 Qsin2 w

Z

where,

Q = T3 +Y

2with T3 =

3

2

We see that A is the vector field associated to the generator Q of the

U(1) symmetry of the broken theory.

The covariant derivative for the broken theory is

D = ieQAWe say that the SU(2)L U(1)Y theory is broken into U(1)em.

The GWS ModelComments on Q


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The operator Q is:

Q = 12Y + 1 0

0 Y 1 for doublets.Q =

Y

2for singlets.

for singlets.

Then, the coupling of singlets and doublet components to A is e times a

number which we call Qf.

So, if A represents the photon, eQf is the electric charge!

Qf =Y/2 for a singlet(Y + 1)/2 for a doublet upper component

(Y 1)/2 for a doublet lower component

The GWS ModelLH


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Substituting = v+H in LHLH =

D

D

+ 2 2we get

LH = MW + g2 H2

W+ W + 12MZ + g2cos w H2

ZZ + LHiggs

with

LHiggs = 12

(H)(H) 1

2M2HH

2 + The dots represent terms that depend on the Higgs potential explicitly.

The GWS ModelLHF

S b i i i L


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Substituting v+H in LHF

LHF =

fe LeR

fu q

LuR fd qLdR + h.c.we find

LHF =

me +g me2MW

H

ee

mu +g mu2MW

H

uu

md +g md2MW

H

dd

where ee = eLeR + eReL (similarly for u and d). The fermion masses are

m v2

f

which can also be written as

LHF =fmf + g mf2MW H f f

where f = , e, u, d. We can use this expression for neutrinos no matterwhether they are massless or massive.

The GWS ModelLF (1/2)

In order to determine the coupling of the particles, we have to write


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LF = L i/DL + eR i/DeR + qL i/DqL + uR i/DuR + dR i/DdR

in terms of W , Z, and A.

The result can be summarized as follows:

LF = L i/ L + e i/e+ u i/u+ d i/d+ e AJ

em +

g

2cos wZJ

nw +

g

22W+ J

cw +

g

22W Jcw

where e i/e = eL i/eL + eR i/eR (similarly for u and d).

The currents Jem, Jcw, and J

nw:

The electromagnetic current is defined by (with f = , e, u, d)

Jem

f

Qf ff =

f

Qf

fL

fL + fRfR

= ee+ 2

3uu 1

3dd

The GWS ModelLF (2/2)

The charged weak current is defined by


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The charged weak current is defined by

Jcw (1 5) e+ u(1 5) d= 2 L eL + 2 uL dL

The neutral weak current is defined by

Jnw

f

f

cfV cfA 5

f

=1

2 (1 5)+ eceV ceA 5e

+ u

cuV cuA 5

u+ d

cdV cdA 5

d

In these expressions,

cfV = T3

f 2 Qf sin2 wcfA = T3f

It is convenient to summarize here the values of Qf and T3

f :

f u d eQf +2/3 1/3 0 1T3f 1/2

1/2 1/2

1/2

The GWS ModelLG

The purely gauge part of the the Lagrangian,

1 1


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LG = 14

WaW

a 14

BB

must be written in terms of W , Z, and A.

LG = 14

FF 1

4ZZ

12

(W)(W+)

+ ie

FW+ W

(W)W+A (W+)AW

+ igcos w ZW+W (W)W+Z (W+)ZW

W+ W

e2 AA++ e gcos w (AZ + ZA) + g

2 cos2 w ZZ +g2

2W+ W

where

F = A AZ = Z Z

(W) = W W

e = g sin w

The GWS ModelMore Generations Introduction


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We have studied the standard electroweak unification of one family

(generation) of fermions.

However, the electroweak interaction involves more than one generation and

we will have to modify the GWS model to include them.

All experiments are consistent with the number of generations being 3.

For one generation, the GWS model was described by the Lagrangian:

LGWS = LG + LH + LF + LHFWhen we add more generations:

LGand

LHdo not change.

LF and LHF have to be modified.

The GWS ModelMore Generations LF

LF contains the kinetic terms for fermions and gives rise to the interactionb f i d b T i l d i i


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terms between fermions and vector bosons. To include more generations in

LF we just have to add a copy per family:

LF = L

i/DL

+ eR

i/DeR

+ qL

i/DqL

+ uR

i/DuR

+ dR

i/DdR

where is a family index that runs from 1 to 3 with

L

= L

eL

= eeL ,

L ,

L

eR

=

eR , R ,

R

qL

=

uL

dL

=

u

d

L

,

c

s

L

,

t

b

L

u

R

= uR , cR , tR

dR

=

dR , sR , b

R

We have used primes for the fields because, as we will see next, they maynot correspond to the physical particles.

The GWS ModelLHF Fermion Mass

LHF contains the interactions of fermions with themselves and the Higgsfield and gauge invariance allows family changing terms So in general it


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field and gauge invariance allows family changing terms. So, in general, it

can be written as

LHF = fe L eR fu qL uR fd qL dR + h.c.where and are family indices.

After spontaneous symmetry breaking, LHF gives rise to the fermion massterms. Taking = v:

LF,mass = v f

e2

eL

eR v f

u2

uL

uR v f

d2

dL

dR

+ h.c.

If we define:

me v f

e2

, mu v f

u2

, and md v f

d2

,

and use matrix notation, we can write

LF,mass = eLmeeR uLmuuR dLmddR + h.c.

The GWS ModelThe Mass Basis

In general m isnt either symmetric or Hermitian Therefore m cannot be


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In general m isn t either symmetric or Hermitian. Therefore m cannot be

diagonalized by a unitary transformation.

However, m can be diagonalized by a biunitary transformation:

m = SRm

SL

where SL and SR are unitary matrices and m is diagonal with positive

eigenvalues. Then,

Lm

R + h.c. = LmR + RmL = m

where

L SLL , R SRRWe conclude that

LF,mass = eme e umu u dmd d= e me e e mu u d md d

The GWS ModelLHF


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Substituting = v+H in LHF we get the fermion-Higgs interaction terms(in addition to the mass terms):

LHF = e me +

gme2MW

H e

u

mu +gmu2MW

H

u d

md +gmd2MW

H

d

where I have used MW = vg/2.

The GWS ModelLHF

We have to write LF in the mass basis. Since, all of the terms in LF onlyi l it t i i th f il j t t i


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involve a unit matrix in the family space, we can just put primes:

LF = L i/ L + e i/e + u i/u + d i/d+ e AJ

em +

g

2cos wZJ

nw +

g

2

2W+ J

cw +

g

2

2W

Jcw

where the electromagnetic current is

Jem = ee +2

3 uu

1

3 dd

the neutral weak current is

Jnw =1

2 (1 5) + e

ceV ceA 5

e

+ u cuV cuA 5u + d cdV cdA 5 d

and the charged weak current is

Jcw = u (1 5) d + (1 5) e

The GWS ModelLHF: Electromagnetic and Neutral Weak Currents


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Since

= and

5 = 5

it is clear that the kinetic, the electromagnetic current terms Jem, and neutral

weak currents terms Jnw for u, d, and e dont change in the mass basis; i.e,

we can just drop the primes for these fields in these terms.

The electromagnetic current:

Jem = ee+ 23

uu 13

dd

The neutral weak current:

J

nw = J

nw, neutrino + e

ceV ceA 5e+ u

cuV cuA 5

u+ d

cdV cdA 5

d

The GWS ModelLHF: Charged Weak Currents (Leptons)

Lets consider the leptonic part of the charged weak current:


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Jcw-l = (1

5) e = 2 L

eL

which can be written as

J

cw-l = 2 0LS

L

(SeL )

eL = 2 L

eL

where I have defined:

L

SeL (SL )

0L = SeL

L

We use L as the definition of the neutrino physical states.

Since SeL is unitary, we can just drop the primes to get the kinetic term:

L i/ L

and the neutral weak term:

Jnw, neutrino =1

2 (1 5)

The GWS ModelLHF: Charged Weak Currents (Quarks)


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Lets consider the quark part of the charged weak current:

Jcw-q = u (1 5) d

= 2 uL dL

= 2 uLSuL

(SdL )dL

= 2 uL SuL (S

dL )

dL

Then,

Jcw-q = 2 uL V dL

where we have defined the Cabbibo-Kobayashi-Maskawa (CKM) matrix V:

V

SuL (S

dL )

(by convention, the mixing is assigned to the Q = 1/3 quarks)

The GWS ModelLF, LHF Summary

Now we can write the fermion part of the electroweak Lagrangian:

/ / gme


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LF + LHF = L i/ L + e

i/me gme

2MWH

e

+ u i/mu gmu2MW

H u + d i/md gmd2MW

H d+ e AJ

em +

g

2cos wZJ

nw +

g

2

2W+ J

cw +

g

2

2W

Jcw

where the electromagnetic current:

Jem = ee + 23

uu 13

d

d

the neutral weak current:

Jnw =1

2

(1

5) + e c

eV

ceA

5

e

+ u

cuV cuA 5 u + d cdV cdA 5 dand the charged weak current:

Jcw = (1 5) e + u (1 5) V d

The GWS ModelFeynman Rules Propagators

The full set of electroweak Feynman rules is rather lengthy. Here we just

write some useful rules.


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Photon propagator (A):

p =

i

p2 + i

(1 )ppp2

Z Boson propagator (Z):

p = ip2 M2Z+ i

pp

M2Z

W Boson propagator (W ):

p =

i

p2 M2W+ i

pp

M2W

Fermion propagator (f = e, u, d):

p=

i

/pmf + i =

i (/p+mf)

p2 m2f + i

The GWS ModelFeynman Rules Vertices involving W, Z, and (1/2)

Feynman rules for vertices involving the vector bosons (W, Z, and thephoton):


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kp

q

A

W+

W

ie

(p q) + (q k) + (kp)

kp

q

Z

W+

W

ig cos w

(pq)+(qk) +(kp)

A

A

W+

W

= ie22

The GWS ModelFeynman Rules Vertices involving W, Z, and (2/2)

Z Z


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Z Z

W+

W

=ig2 cos2 w

2

A Z

W+

W

=ie g cos w

2

W+ W

W+

W

= ig2 2

+

The GWS ModelFeynman Rules Fermions (1/2)

Feynman rules for vertices involving fermions:


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Feynman rules for vertices involving fermions:

f f

A

= ieQf

f f

Z

=ig

2cos w(cfV c

fA

5)

Here f = , e, u, d.

The GWS ModelFeynman Rules Fermions (2/2)


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e

W+

e

W

=ig

22(1 5)

d u

W

+

u d

W

=ig

22

(1 5)V

Here , = 1, 2, 3 represent the family number.

The GWS ModelFeynman Rules Higgs (1/2)

Feynman rules involving the Higgs field (H):


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Feynman rules involving the Higgs field (H):

p = ip2 M2

H+ i

H

Z Z

=igMZ

cosw

H H

Z Z

=ig2

2 cos2 w

The GWS ModelFeynman Rules Higgs (1/2)

H


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W W+

= igMW

H H

W

W+

=

ig2

2

f f

H

= ig mf

2MW

The GWS ModelMassive Neutrinos


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Experiments show that neutrinos have (a small) mass.

How to modify the GWS model?

Add a mass term for neutrinos.

The e--W-vertex involves a CKM-like matrix V instead of :

e

W

=ig

22

(1 5)V

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