What is a fundamental particle??

1. 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  u c t e- -  - d s b BOSONS :ɤ, W±, Z0, GLUONS, Graviton (??)

2. Characteristics : SPIN: Leptons & quarks : Spin ½, Bosons : spin 1 Graviton : spin 2 MASS: • me ≈0.51 MeV mµ ≈ 105 MeVmτ≈ 1.8 GeV • mνe < 7eV mνµ < 270 KeVmντ < 31 MeV • ∑ m (νi) ≤ 80 eV • The speculation: mνe≈ 10-6eV

3. Quark masses

4. 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 eVm(W± ) ≈ 80 GeV M(Z0 ) ≈ 90 GeVm( gluons) ≈ 0 (no free gluons) HOW DO QUARKS & LEPTONS GET MASSES?? NO ANSWERS, PRESENT SPECULATIONS

5. INTERACTIONS BETWEEN FUNDAMENTAL PARTICLES • Interaction between particles : Quantum & Relativistic Generalization of the concept of force In general exchange of any particle generates a force We only observe P1, P2 ,P1/ ,P2/ exchange of virtual quantum P1/ P2/ q P1 P2

6. 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.

7. Electromagnetic interactions • The basic interaction is described by ieJµAµ Jµ = ΨɤµΨ α = e2/4Πhc • All em interactions are described by this prescription. • Theory is called QED. e √α Aµ e

8. LEVEL OF SUCCESS • {( g-2)/2}eQED= (1159652.4 ± 0.4 ) x 10-9 µ = gµB S µB = eh/2mc gDirac = 2 {( g-2)/2}eEx = (1159652.4 ± 0.2 ) x 10-9 e-e scattering ( Rutherford scattering ) dσ/dΩαα2/q4 e e √α √α q e e

9. ` • QED ALSO PERMITS SUCH PROCESSES SELF ENERGY VACUUM VERTEX CONTRIBUTION POLARIZATION CORRECTION

10. These effects are real were shown in the case of LAMB SHIFT. HYDROGEN ATOM : non rel 2P3/2 , 2P1/2, 2S1/2 rel 2P3/2 2P1/2, 2S1/2 LAMB & RUTHERFORD (1947) 2S1/2 2P1/2 ∆ ELAMB ≈ 4 X 10-6eV = 1058 Mc/sec

11. Self energy contribution : 1017 Mc/sec • Vacuum polarization : -27 Mc/sec • Vertex correction : 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.

12. Q eff = Q/ЄЄ ≥ 1 dielectric constant • In QED, the e+e- pairs are shielding the charge of an electron. • However, as we go towards the center of the electron, though the Q eff is increasing but we consider that the coupling constant increases but the charge stays constant. αR (q2) = αR (m2) [ 1 + {αR (m2)/3Π} log (-q2/m2 ) + O(αR2) ]

13. 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

14. G • Asymptotic freedom • Quark confinement • QCD inspired low energy phenomenology. b r R r b R G r b

15. ABCD OF FIELD THEORY 1.Write a lagrangian :L(r,∂r) . Identify the independent fields. 2.UseEuler Lagrange Equations to find field equations. 3.Impose commutation/anti commutation relations. 4.Identify the physical content 5.Find the Feynmann Rules.

16. Notation : • xµ = (x0,x1,x2,x3) = (t,x,y,z) • xµ = (t,-x,-y,-z) • A.B = AµBµ = A0B0-A.B • ∂µ = ∂/∂xµ = (∂/ ∂t, ∂/ ∂x, ∂/ ∂y, ∂/ ∂z)

17. Global Gauge Invariance : Known as the Gauge Invariance of the First Kind or U(1) Gauge invariance. Consider a field function  & a transformation of the form, / = ei  is real and constant. If L(/,∂µ/) = L(,∂µ) then L is invariant under global gauge transformations .

18. Consequence: Consider infinitesimal Gauge transformation: δ = / -  = iαα- infinitesimal

19. Using the Euler – Lagrange Equations : We obtain :

20. Define current density Jµ :

21. δL = -α∂µJµ • Since δL = 0 , for global gauge invariance of L , therefore ∂µJµ = 0 RESULT: GLOBAL GAUGE INVARIANCE LEADS TO CONSERVED CURRENT OR A CONSERVED QUANTITY.

22. 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 : / = ei ∂µ/ = ∂µei

23. Local Gauge invariance: • In such transformations,  itself is a function of x. For example in case of U(1), / = ei(xµ)  Consider again the Lagrangian L(,∂). In this case,  and ∂ do not transform in the same way. Therefore L is not invariant under local gauge transformations .

24. Question : How can we make lagrangianinvariant under such transformations?? • Define another derivative called ‘COVARIANT DERIVTIVE’ such that • DµDµ/ = ei(xµ) Dµ • DµDµ/ = e-i(xµ) Dµ under the transformation: / = ei(xµ)  POSSIBLE IF : • Dµ = (∂+ ieAµ)  • & Dµ = (∂- ieAµ) 

25. & • If δ(Dµ)= iα(Dµ) then invariance will exist and our task is accomplished. • Checking : δ(Dµ)=

26. Thus : If we can find covariant derivative ,then we can always write a corresponding gauge invariant lagrangian.

27. Non Abelian Gauge Transformations • Consider SU(2) local gauge transformations for a field function , where, є·I = є1I1+є2I2+є3I3 Є’sare real parameters depending upon xµ & I1, I2 and I3 are isospin matrices given by

28. Let us consider the change in L resulting from the infinitesimal transformation If L = L (,∂), then we can write L Here

29. 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.

30. What properties should A µhave in order that DµΨ 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 FµνFµν (K.E. term) ??

31. Again one can show that the required term is given by LYM YM stands for Yang-Mills So the Lagrangian L Is invariant under local non-abelian gauge transformations.

32. Comment : • The existence of the additional non abelian term –g AX A in E implies that in LYM we not only have ‘zeroth order’ term similar to ED but also terms of order g and g2 respectively as , &

33. Aµ3 • Now in limit LYM This shows that YM quanta must be zero mass vector boson. One can show that YM quanta or gauge fields form a triplet with charges +1,0,-1. Aµ3

34. Generalization to other groups • The generalization to local SU(3) gauge invariance is straight forward. • If ½ Iare 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.

35. 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.

36. 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)

37. 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R4.

38. 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

39. However if F is increased until it reaches a critical value Fc, 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. Sin2 Z/L, Fc = 42IE/L2 • When the rod is straight, the eigen frequencies of small oscillations are given by coskL.coshkL = 1, where, k= ω1/2(µ0/EI)1/4 µ0 is mass per unit length.

40. 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.

41. Conclusions & Comments • The original symmetry : Axial symmetry (x-y plane). • For F<Fc, the ground state also has this symmetry. • For FFc, the system when perturbed infinitesimally, jumps to a new ground state in which the original symmetry is broken(rod is bent)

42. WHAT HAPPENS TO THE ORIGINAL SYMMETRY?? • 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 SPONTANEOUSLYBROKEN .

43. 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 SSBare startling.

44. Spontaneous breaking of a global symmetry • 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 / = ei

45. Consider the corresponding Hamiltonian density, H =(∂0Φ)∂0Φ*) +(∂iΦ)(∂iΦ*) + 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 µ2

46. (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 = 0 • For 2 0, the symmetric solution is stable • For 2 <0 , SSB occurs.

47. 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, L = 

48. Observation : 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.

49. Goldstone theorem • 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ei(xµ)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.

50. Spontaneous breaking of a local gauge symmetry • Photon interacting with a scalar particle • L= where The Lagrangian is invariant under the local gauge transformations of the form  ei (xµ) AA/=A+ i/e ∂µ(x ).