Transcript thhrgh
What is a fundamental particle??
No standard definition.
Formal definition : A particle which can not be described in terms of more fundamental particles.
Present day fundamental particles may not be fundamental particles of tomorrow.
LEPTONS (+ ANTI) QUARKS (+ANTI) e e u c t d s b BOSONS :
ɤ, W ± , Z 0 , GLUONS, Graviton (??)
1
Characteristics :
SPIN: Leptons & quarks : Spin ½, Bosons : spin 1 Graviton : spin 2 MASS: m e ≈0.51 MeV m µ m νe < 7eV m νµ ≈ 105 MeV < 270 KeV ∑ m (ν i ) ≤ 80 eV The speculation: m νe ≈ 10 -6 eV m τ m ντ ≈ 1.8 GeV < 31 MeV 2
Quark masses
quark
u d s c b t 5 MeV 10 MeV 150 MeV 1.5 GeV 5.0 GeV 175 GeV
3
310 MeV 316 MeV 450 MeV 1.6 GeV 5.2 GeV 175 GeV
charge
2/3 -1/3 -1/3 2/3 -1/3 2/3 3
Why two different sets of masses??
1. DEEP INELASTIC SCATTERING 2. LOW ENERGY PHENOMENON MASSES,SPECTROSCOPY Since quarks are not observed , therefore they enter as parameters in the calculations.
BOSON MASSES m( ɤ ) ≈ 0 < 3 X 10-28 eV
M(Z 0 )
≈
90 GeV m( gluons)
≈ m(
W ± )
≈
80 GeV 0 (no free gluons)
HOW DO QUARKS & LEPTONS GET MASSES??
NO ANSWERS, PRESENT SPECULATIONS 4
INTERACTIONS BETWEEN FUNDAMENTAL PARTICLES
Interaction between particles : Quantum & Relativistic Generalization of the concept of force P 1 / P 2 / In general exchange of any particle generates a force q P 1 P 2 We only observe P 1 , P 2 , P 1 / ,P 2 / exchange of virtual quantum 5
Different types of interactions
Electromagnetic (em) Weak Strong Gravitational em & weak best understood.
Quantum theory of gravity of least understood.
As ‘weak’ has been successfully unified with electromagnetic therefore we will discuss only em & strong here.
6
Electromagnetic interactions
The basic interaction is described by ieJ µ A µ e J µ = Ψ ɤ µ Ψ α = e 2 /4Π hc √α A µ e All em interactions are described by this prescription.
Theory is called QED.
7
LEVEL OF SUCCESS
{( g-2)/2} e QED µ = gµ B = (1159652.4 ± 0.4 ) x 10 S µ B = eh/2mc g Dirac -9 = 2 {( g-2)/2} e Ex = (1159652.4 ± 0.2 ) x 10 -9 e-e scattering ( Rutherford scattering ) e e
√α
e q
√α
e dσ/dΩ α α 2 /q 4 8
`
QED ALSO PERMITS SUCH PROCESSES SELF ENERGY VACUUM VERTEX CONTRIBUTION POLARIZATION CORRECTION 9
These effects are real were shown in the case of LAMB SHIFT.
HYDROGEN ATOM : non rel 2P 3/2 , 2P 1/2 , 2S 1/2 rel 2P 3/2 2P 1/2 , 2S 1/2 LAMB & RUTHERFORD (1947) 2S 1/2 2P 1/2 ∆ E LAMB ≈ 4 X 10 -6 eV = 1058 Mc/sec 10
Self energy contribution : 1017 Mc/sec Vacuum polarization Vertex correction : -27 Mc/sec : 68 Mc/sec Total : 1052 Mc/sec Showing that these effects are observable consequences. (accompanying a point particle).
Vacuum polarization effects are responsible for the increase of α at short distances.
11
Q eff = Q/Є In QED, the e + e electron.
Є ≥ 1 dielectric constant pairs are shielding the charge of an However, as we go towards the center of the electron, though the Q eff the charge stays constant.
is increasing but we consider that the coupling constant increases but α R (q 2 ) = α R (m 2 ) [ 1 + {α R (m 2 + O(α R 2 ) ] )/3Π} log (-q 2 /m 2 ) 12
QCD
Quantum field theory of q-q & q-q forces.
QCD very well understood in the deep inelastic limit but at low energy not at all understood.
q-q & q-q forces mediated by an octet of colored gluons (r,b,g) rb , rg ,bg ,br , gr, gb,(rr + bb – 2gg)/ √ 6, (rr-bb)/ √ 2 13
b r b r R r b Asymptotic freedom Quark confinement QCD inspired low energy phenomenology.
G 14
1.Write a lagrangian :L(
r
,∂
r
)
.
Identify the independent fields.
2.Use
Euler Lagrange Equations equations.
to find field 3.Impose commutation/anti commutation relations.
4.Identify the physical content 5.Find the Feynmann Rules
.
15
Notation :
x
µ
= (x
0
,x
1
,x
2
,x
3
) = (t,x,y,z)
x
µ
= (t,-x,-y,-z) A.B = A
µ
B
µ
= A
0
B
0
-A.B ∂
µ
= ∂/∂x
µ
= (∂/ ∂t, ∂/ ∂x, ∂/ ∂y, ∂/ ∂z)
16
Global Gauge Invariance :
Known as the Gauge Invariance of the First Kind or U(1) Gauge invariance . Consider a field function form, / = e i is real and constant. & a transformation of the If L( / ,∂ µ / ) = L( ,∂ µ ) then L is invariant under global gauge transformations .
17
Consequence:
Consider infinitesimal Gauge transformation: δ = / = iα α- infinitesimal 18
Using the Euler – Lagrange Equations : We obtain : 19
Define current density J µ : 20
Since δ L therefore ∂ µ J µ
δ
L = 0 , for global gauge invariance of L = 0
= -α∂
µ
J
µ , RESULT:
GLOBAL GAUGE INVARIANCE LEADS TO CONSERVED CURRENT OR A CONSERVED QUANTITY.
21
OBSERVATIONS :
Conservation of an additive quantum number corresponds to global gauge invariance.
The field and the derivative of the field transform in the same way. Example : / ∂ µ / = e = ∂ i µ e i 22
Local Gauge invariance:
In such transformations, itself is a function of x . For example in case of U(1), / = e i (x µ ) Consider again the Lagrangian L ( ,∂ ). In this case, way. and ∂ do not transform in the same Therefore L is not invariant under local gauge transformations .
23
Question : How can we make lagrangian invariant under such transformations??
Define another derivative called ‘COVARIANT DERIVTIVE’ such that D µ D µ / D µ D µ / = e i (x µ ) = e -i (x µ ) D µ D µ under the transformation: / = e i (x µ ) POSSIBLE IF : D µ & D µ = (∂ = (∂ + ieA - ieA µ µ ) ) 24
& If δ(D µ )= iα(D µ ) then invariance will exist and our task is accomplished.
Checking : δ(D µ )= 25
Thus : If we can find covariant derivative ,then we can always write a corresponding gauge invariant lagrangian .
26
Non Abelian Gauge Transformations
Consider SU(2) local gauge transformations for a field function , where, є·I = є 1 I 1 +є 2 I 2 +є 3 I 3 Є’s are real parameters depending upon x µ I 1, I 2 and I 3 are isospin matrices given by & 27
Let us consider the change in L infinitesimal transformation resulting from the If L = L ( ,∂ ), then we can write L Here 28
When є = є(x ), L is not invariant To restore gauge invariance , we employ triplet of gauge fields Define the covariant derivative, In analogy with the earlier definition.
29
What properties should A
µ
have in order that D
µ
Ψ under local non abelian gauge transformations transform in the same way as Ψ ?
we require that for infinitesimal є ( D ) = iє•I(D ) One can show that this is possible if
After having found the appropriate covariant derivative what is the corresponding term to -1/4 F µν F µν (K.E. term) ??
30
Again one can show that the required term is given by L
YM
YM stands for Yang-Mills So the Lagrangian L Is invariant under local non-abelian gauge transformations.
31
Comment :
The existence of the additional non abelian term –g A
X A
in E implies that in L YM we not only have ‘zeroth order’ term similar to ED but also terms of order g and g 2 respectively as ,
&
32
A µ 3 Now in limit L
YM
This shows that YM quanta must be zero mass vector boson. One can show that YM quanta or gauge fields A µ 3 form a triplet with charges +1,0,-1.
33
Generalization to other groups
The generalization to local SU(3) gauge invariance is straight forward.
If ½ I are the generators of SU(3) where i= 1,2,….8, then gauge transformations become, The covariant derivative is defined in a similar way & one can easily write the lagrangian. This invariance would lead to QCD lagrangian.
34
SPONTANEOUS SYMMETRY BREAKING
Let us consider Gauss’s Law while finding electric field at A. For that assume that symmetrical situations have symmetrical solutions. E(A) Spherical charge distribution Since small deviations always accompany a symmetrical situation, therefore we are not only making the above assumption but also that small deviations from the perfect symmetry will induce only small deviations from the symmetry solutions.
35
The last property is not always true as shown below :
SSB in a physical scenario
:
Consider the vibrations of a rod clamped as shown Figure (a) (b) 36
Let the deflections of the rod in the x and y directions as functions of Z given by X(Z) and Y(Z).
Condition for equilibrium will be given by, Here E = Young’s modulus and I = ¼ R 4 . 37
The boundary conditions of the system are X(0) =X(L) =0, X / (0) =X / (L) =0, X / = dX/dZ. Similar conditions on Y If we start with small force F, the equilibrium condition is that the rod is straight with solutions X(Z) =Y(Z) =0. The rod is stable against small perturbations from straightness 38
However if F is increased until it reaches a critical value F c, then the rod will become unstable against small perturbations from straightness and will bend (fig b). It is not difficult to show that such a solution is , X(Z) =const. Sin 2 Z/L, F c = 4 2 IE/L 2 When the rod is straight, the eigen frequencies of small oscillations are given by cos kL.cosh kL = 1, where, k= ω 1/2 (µ 0 /EI) 1/4 µ 0 is mass per unit length.
39
When the rod is bent, we may have small oscillations. However, the characteristic frequencies will be different and in particular the frequencies for oscillations in the plane of bending will not be same as in the perpendicular direction.
40
Conclusions & Comments
The original symmetry : Axial symmetry (x-y plane). For F For F F c , the system when perturbed infinitesimally, jumps to a new ground state in which the original symmetry is broken(rod is bent) 41 The original symmetry is not lost but hidden : In the sense that we can not predict the direction in the x-y plane in which the rod is going to be bent. They all correspond to the same energy. WE CALL SUCH A SYMMETRY TO BE SPONTANEOUSLY BROKEN . 42 Above example has all the characteristics of SSB : 1.There exist critical point which determines whether SSB will take place or not. 2.Symmetric solution becomes unstable. 3.Ground state becomes degenerate . 4.Ground state no longer respects the original symmetry of the system. QFT consequences of SSB are startling. 43 Consider a complex scalar field φ(x), corresponding lagrangian density is given by L where By inspection one can see that it is invariant under the global U(I) gauge transformations / = e i 44 Consider the corresponding Hamiltonian density, H =(∂ 0 Φ)∂ 0 Φ*) +(∂ i Φ)(∂ i Φ*) + V(Φ) Since the first two terms of H are positive, therefore the ground state of the system corresponds to minima of V(φ). For l >0, the position of minima depends on the sign of µ 2 45 (a) 2 0 (b) 2 <0 v =( 2 /2l) 1/2 For 2 0 the minima is at Φ = 0 but for 2 <0 , there is a whole circle of minima at the complex Φ – plane with radius v =( 2 /2l) 1/2 In this example we see that the critical point is 2 For 2 0, the symmetric solution is stable For 2 <0 , SSB occurs . = 0 46 L = Suppose we have 2 <0: To reach the stable solution we translate the field Φ and consider small oscillations around the translated point. (x)= (1/√2)[v+ 1 / (x)+i 2 / (x)] Substituting this in the Lagrangian we get, 47 Observation : 1. 2. The lagrangian does not contain any term proportional to 2 /2 . The present system after SSB, if considered as a QFT one , consists of two interacting scalar particles : 1 / (x) describes a particle of mass µ. 2 / (x) describes a massless particle. 48 To every generator of a spontaneously broken symmetry there corresponds a massless particle called Goldstone Boson . The vacuum of the theory does not obey the symmetry of the lagrangian. The lagrangian is invariant under φ e i (x µ ) φ But ground state is not. MIRACLE : when a gauge theory is spontaneously broken, the would be goldstone bosons decouple and disappear, the gauge bosons acquire mass. 49 D ieA . Photon interacting with a scalar particle L = where The Lagrangian is invariant under the local gauge transformations of the form e i (x µ ) A A / =A + i/e ∂ µ (x ). 50 SSB occurs for >0 and 2 <0. Consider the vacuum < >=v, v 2 /2 =-m 2 /2 . Next consider the oscillations about this vacuum Substituting in the lagrangian ,final lagrangian takes the form L 2 = -1/4 (F µν ) 2 + {e 2 v 2 /2 } A µ 2 -evA µ ∂ µ η + ………………. + ½ (∂ µ ζ) 2 + ½ (∂ µ η )- ½ (2λν 2 )ζ 2 51 Photon has acquired mass : earlier we had observed that local gauge invariance forbids photon to acquire mass.It is simply wrong. The gauge transformation has to be somewhat complicated. Degrees of freedom : Pre SSB : scalar (DOF) -2 vector (DOF) -2 Total DOF -4 Post SSB : scalar(DOF) -2 vector(DOF) –3 (A µ Total DOF – 5 has become massive) 52 No gauge transformation whatsoever can cause a change in the DOF . So we are wrong somewhere. L 2 can be transformed into L 3 , 53 • • • L 3 has Vector(DOF) -3 Scalar(DOF) -1 Total(DOF) –4 : same that we started with…… L 1 , L 2 , L 3 are equivalent. After SSB of Gauge theory, the goldstone boson has disappeared whereas photon has acquired mass. The above result can be repeated for nonabelian gauge theory. 54 A Yang Mills theory ,spontaneously broken , although contains massive vector bosons, remains renormalizable. 55 I. II. III. IV. V. VI. Choose a gauge group G {SU(2)xU(1) for Glashow, Weinberg & Salam model} Choose the field of elementary particles you want to introduce and their representations. Introduce enough scalars so that Higgs mechanism takes place. Write the most general renormalizable lagrangian invariant under G. Choose the parameters of Higgs scalars so that SSB takes place. Translate the scalars and rewrite the lagrangian in terms of translated fields. Look at the properties of the resulting model. If it resembles Physics PUBLISH IT. 56 L Y Unifies the electromagnetic and weak interactions. Limiting to the case of leptons, Observations from weak interactions 1.The lepton number is conserved separately, so it is natural to group the leptons into families of the same lepton number : ( e , e) : l 2.The weak interaction “sees” only a lepton & cannot distinguish between a neutrino and an electron. 3.Weak interaction prefers lepton to be left handed , ( electron left handed, e ; ( left handed) , ) : l case the system undergoes SSB : ; ( , ) : l . . 4.Must recognize the fact that a degeneracy, which might be there due to a given symmetry of the Lagrangian, might get lost in 57 . Shown by Glashow that the leptons seem to follow “weak” SU(2) and a “weak” hypercharge could be given to the leptons, defined by the Gell-Mann Nishizima relation Q = I 3 w + Y w /2. In the context of standard model, the neutrinos are mass less, therefore before SSB e L is also supposed to be mass less . 58 The left handed component of the electron and the neutrino are the I 3 w = - ½ & I 3 w = + ½ components of the weak isospin doublet l e with I w = 1/2 As the right handed component of electron has no weakly interacting partner , it must have I w = 0 (singlet) The neutrino and left handed electron are distinguished through weak hypercharge. 59 Fermion e , µ , τ e L , µ L ,τ L e R , µ R ,τ R u L , c L ,t L (d c ) L ,(s c ) L ,(b c ) L u R , c R ,t R (d c ) R ,(s c ) R ,(b c ) R 1/2 1/2 0 0 I w 1/2 1/2 0 I 3 w 1/2 -1/2 0 1/2 -1/2 0 0 -1 Y -1 -2 1/3 1/3 4/3 -2/3 0 Q 0 -1 2/3 -1/3 2/3 -1/3 60 Members of the weak isomultiplet have same Y For two generations (d are KM rotated states. c ) ,( s c ) are cabibbo “rotated” states and for three generations these Under an SU(2) L ( e, e) L ( transformation e, e) L = exp(i/2α .τ) ( e, The symmetry is imposed before SSB e) L 61 Restricting to the case of first generation of leptons L ( e , e ) L , L e L = ½(1 = ½(1 5 ) 5 , )e. R e R = ½(1+ 5 )e. By construction, the weak isospin projection I 3 and the weak hypercharge Y are commuting observables, [I 3 w ,Y w ] = 0. Henceforth superscript W will be dropped . 62 We now consider the (product) group of transformations generated by I and Y to be the gauge group SU(2) bosons : L U Y (1). To construct the corresponding gauge theory, introduce the gauge for SU(2) for U(1) Y L : b : A 1 ,b 2 ,b 3 ( b ) ; 63 L ( L = L gauge gauge + L leptons ) K.E. = -1/2 F l F l - 1/4 f f . • • • SU(2) gauge field U(1) gauge field f Matter term ( L) leptons µν μ =∂ ν A µ - ∂ µ A ν g : coupling constant for SU(2) L. g / /2 : coupling constant for U(1) y . 64 • The theory described by this L is not satisfactory. as it contains four massless gauge bosons while we require only one corresponding to the photon. As well as the local SU(2) invariance forbids a mass term for the electron. • Solution : Introduce scalars and achieve SSB so that electron & three gauge bosons acquire mass. 65 Introduce a doublet of complex scalars, = ( + , 0 ) , Y = 1. transforms as doublet under SU(2) L . L scalar 66 Introduce interaction of scalar with fermions, called Yukawa couplings L Yukawa symmetric under local SU(2) L U Y (1) gauge transformations and is Lorentz scalar. Next step generate SSB by considering breaking both SU(2) L invariance under U(1) EM charge . 2 < 0 and U(1) L but still preserving symmetry generated by electric 67 The vacuum expectation value of the scalar field is given by, < > 0 = (0 , v/√2) Check that < > 0 does not obey SU(2) x U(1). The vacuum is left invariant by a generator G, if , exp(i G)< > 0 = < > 0. For infinitesimal , G< > 0 = 0. 68 For the generators of SU(2) x U(1) ,we have 69 However , Q< > 0 This implies that the three gauge bosons (b b µ & A ) would still remain massless. ,A ) acquire mass whereas one (linear combination of 70 Now expand the Lagrangian about minimum of he Higgs’ potential, by considering, and transforming it to unitary gauge, e.g., 71 We now have to express the lagrangian in terms of u-gauge fields, omitting primes to avoid notational clutter & investigate the consequences : L Yukawa Thus the e m has acquired mass given by, e = G e v/ 2 72 The scalar term in the Lagrangian now reads, L scalar interaction terms. Observation :The field has acquired mass given as, M H 2 = -2 2 . This is the physical Higgs’ boson. Defining the charged Higgs’ boson as, 73 The mass terms of these can easily be found & one finds M W = gv/2. Defining the combinations ……….(a) One finds that, M Z 0 = (g 2 +g /2 ) v/2 = M W (1+g /2 /g 2 ). A remains massless gauge boson corresponding to surviving exp(iQ (x)) symmetry. 74 L w-l It corresponds to our low energy phenomenology of intermediate vector boson model, provided the coupling constant are identified as, g 2 /8 = G F M W 2 / 2. With this identification the parameter v is now determined as, F -1/2 . 75 The neutral gauge boson couplings to lepton fields are given by, L o-l = If we identify A as the photon field ,then gg / / (g 2 +g /2 ) = e. 76 To simplify the lepton – Z 0 weak mixing angle g / coupling, we introduce a = g tan W. Here W. is known as Wienberg angle. Equation (a) can be rewritten as, Z A µ = -A = A sin W cos +b 3 W + b 3 cos sin W , W . 77 Also, from the above expressions we can write, g = e/Sin W e g / =e/Cos W This leads to, M W 2 e, = (37.3 Gev/c 2 )/Sin 2 W . M Z 2 = (37.3 Gev/c 2 )/Cos 2 W . Also, M Z 2 ≥ M W 2 . 78 Now, L 0-l can be written as, L 0-l Z = Q e (e e)A - {(1/ √2 ) G F +(2Sin 2 W -1)e – (1/ √2) (G F M Z 2 M Z 2 / 2} 1/2 [2Sin 2 / 2) W e 1/2 (1 5 ) (1+ 5 )eZ (1 5 )eZ ] . 79 The standard model developed so had large number of immediate successes because of which it is now called SU(2) L X U(1) Y Electroweak theory . It is only after the observation of neutrino oscillations m ντ (consequently the fact that m νe ≠ m νµ ≠ ≠ 0) that we have the first clear cut signal of Physics Beyond Standard Model. 80 -i/√2 ( G F M Z 2 /√2) 1/2 ν ɤ λ (1 ɤ 5 ) ν Predictions of Neutral currents at the tree level : : One can easily check from Lo-l, that the following processes exist ν ν -i/√2 ( G F M Z 2 /√2) 1/2 ν ɤ λ (1 ɤ 5 ) ν Z 0 λ ν 81 -i/√2 ( G F M Z 2 (1+ ɤ 5 ) + L e (1 /√2) 1/2 ɤ 5 )] e e ɤ λ [R e R e L e = 2 sin 2 = 2 sin 2 ϴ W ϴ W - 1 Indicating that at high energies the electron-electron scattering should show the effects of Z 0 exchange. 82 In case the theory is extended to (ν µ also observe µ ), we should e ν µ ν µ e ν µ e Z 0 ν µ e e ν µ ν µ e ν µ e ν µ e 83 ν e e e ν e ν e e ν e e e ν e W ν e ( (charged current) ν e e e + Z 0 “ neutral current” + Z 0 e ν e 84 These diagrams represent the lowest order contributions to neutrino-electron elastic scattering. Neutral currents were first observed in quasi-elastic muon – neutrino – proton scattering ν µ + p ν µ + p + Π 0 85 The cross-section of the above reaction relative to the charged current version was measured to be σ (ν µ σ (ν µ +p ν µ +p µ +p+Π 0 +p+Π 0 ) ) = 0.51 ± 0.25 Showing them to be effects of the same order. 86 It needs to be mentioned that the reaction through ν µ + e µ + ν e which is forbidden in V-A theory , can proceed ν µ ν µ µ ν e dσ/dΩ ={ G F / 4Π 2 }. mass energy .(s-m µ ) 2 /s s: square of center of e e This process diverges with increasing energy. 87 In the IVB hypothesis the above process proceeds through ν µ ν µ W + W + e e This also diverges at energies > M 2 w 88 The cross-section of the above process in the Standard Model is predicted to be dσ/dΩ ={ G F 2 /Π}.(s-m µ ) 2 /s[1 +(s-m µ ) 2 /M W 2 ] At high energies it tends to G F M W 2 / Π . As the theory is renormalizable, so corrections can be calculated which are at a few % level. 89 In the polarized electron-deutron scattering, the experiment measures the difference in the scattering cross-section between left & right handed electrons off polarized deutron targets. The difference is of the order of 0.01% which is predicted correctly as well as its dependence on energy is also explained. 90 Another interesting parity violating experiment is concerning interaction of light with matter at low energies. The idea is to shine a beam of polarized light through vapour of metal atoms. The light is absorbed &re emitted by the various atomic electrons &, because of the effect of weak interaction between atomic nucleus & electrons, leads to a very slight but well defined rotation of the plane of polarization in a given direction. 91 ± The charged W ± boson have a mass M W = 80.33 ± 0.15 GeV Experimental decay rates to lepton pairs : Г ( W + e + ν ) = 224 ± 15 MeV Г (W+ µ + ν ) = 215 ± 19 MeV Г (W + Г (W + e + τν ) = 226 ± 27 MeV v) = Г (W ev) To lowest order ,the prediction is Г (W + e + v) = G F M W 2 / 6Π √ 2 = 226 ± 1 MeV for all processes 92 M z = 91.187 ± 0.007 GeV Partial decay widths : Expt. Г ( Z e Г (Z µ + µ + Г (Z τ + τ e- )=83.82 ± ) =83.83 ± ) =83.67 ± 0.30 MeV 0.39MeV 0.44 MeV To the lowest order, all are equal in Standard Model & the predicted value is Г ( Z e + e- ) = 83.4 MeV Again remarkable agreement between theory & experiment 93 The Standard Model predicts Г ( Z ν e ν e ) = Г ( Z ν µ ν µ ) = Г ( Z ν τ ν τ ) = G F M W 2 / 6Π √ 2 = 165.9 MeV Therefore, 3Г ( Z ν ν ) = 497.6 MeV Г(e + e invisible) = 498.3 ± 4.2 MeV Corresponding to neutrino as these are not recorded by detectors. Excellent agreement with expt. The list of agreement is long & agreement becoming better & better 94 Again restricting ourselves to one generation (u d) L (u R d R ) L (u B r B so color indices are ignored. ) L (u G d G ) L For weak interactions , color does not play any role , So, we start with L q R u R d = (u d) L =u R =d R = ½ (1+ ɤ 5 = ½ (1+ ɤ 5 ) u ) d Y (q L ) = 1/3 Y (u R ) = 1/3 Y (d R ) = -2/3 95 The complex doublet of scalars Φ =(Φ + Φ 0 ) Y(Φ) = +1 Φ = i σ 2 Φ* = (Φ 0 -Φ ) The only change in form from the theory of leptons is the interaction term L yukawa = - G u [ (L q Φ ) u R + u R (Φ + L q )] - G d [ (L q Φ ) d R + d R (Φ + L q )] which will generate upon spontaneous symmetry breaking the masses for up & down quark. 96 In a similar manner , the charged & neutral current interactions of quarks can be written as L w-q = -g/2 √ 2[u ɤ µ (1 ɤ 5 )d W µ + + d ɤ µ (1 ɤ 5 )uW µ ] L z -q = -g/4 cos ϴ w { u ɤ µ [ (1 ɤ 5 )τ 3 - 2x w Q ]u Z µ +d ɤ µ [ (1 ɤ 5 )τ 3 - 2x w Q ]d Z µ } For low energy phenomenology ,the hadronic charged current is described not by the quark doublet, rather by Cabbibo current : L q = (u d c ) L d c ϴ c = d cos ϴ c + s sin ϴ c : cabbibo angle 97 The resulting SU(2)L x U(1) model still has several defects s c = s cos ϴ c – d sin ϴ c is superfluous. The weak neutral current contains terms proportional to c c c µ 5 µ 5 which correspond to flavour changing (strangeness changing in present case) neutral currents. µ 98 For example , the decay K + ΠΠνν can be interpreted in terms of elementary decay s dνν. The decay is highly suppressed e.g. Г(K + Π + νν) Г(K + all) <10 -7 Implying that flavor changing neutral currents are highly suppressed. 99 An elegant solution to the above problem was proposed by Glashow – Illiopolous & Maiani. The key ingredient is the introduction of a new charmed quark c, the weak isospin partner of s c . In this theory the fundamental fermions are L e L u = (ν e e) L L µ = (u d c ) L L c = (ν µ = (c s µ) c ) L L R e R c = e R = c R ,R µ ,R s = µ R = s R 100 The weak current for the two generations now can be expressed as j µ = ½ (u c) ɤ µ (1 ɤ 5 ) U d s U = cos ϴ c sin ϴ c -sin ϴ c cos ϴ c With these modifications FCNC do not contribute, there will be a compensating term involving - sin ϴ c cos ϴ c effectively cancelling the term proportional to sin ϴ c cos ϴ c mentioned earlier. 101 The extension to the three generation case is straightforward, & in this case the charged current gets modified to j µ = ½ (u c t ) ɤ µ (1 ɤ 5 ) U d s b In the PDG representation c 12 c 13 U = -s 12 c 23 - c 12 s 23 s 13 e iδ s 12 s 23 - c 12 c 23 s 13 e iδ s c 12 12 c c - c 12 13 23 - s 12 s 23 - s 12 δ- CP violating phase s 23 s c 13 23 e s iδ 13 e iδ s s 13 c 23 e 23 c iδ 13 c 12 102WHAT HAPPENS TO THE ORIGINAL SYMMETRY??
Spontaneous breaking of a global symmetry
Goldstone theorem
Spontaneous breaking of a local gauge symmetry
Observation:
(HIGGS MECHANISM)
Theorem of ’t Hooft 1971
Model building
The Standard Model (SU(2)
x U(1)
gauge theory)
Q.What kind of symmetry one finds in the case of weak interactions of leptons??
The weak iso-spin & hypercharge assignments :
Observation
Comment :
Construction of Model
v = (
2 G
)
= 246 GeV
Consequences of Standard Model
Parity violating tests
The W
Boson
The Z Boson
The number of lepton families
Extension to quarks
sin ϴ
cos ϴ
[ d
ɤ
(1- ɤ
) s + s ɤ
(1- ɤ
) d ] Z
GIM Mechanism