Physique des Particules Avancées 2

Transcription

1 Physique des Particules Avancées Interactions Fortes et Interactions Faibles Leçon 1 Le mécanisme de Higgs ( enseignant Alessandro Bravar Alessandro.Bravar@unige.ch tél.: 9610 bureau: EP 06 assistant Alexander Korzenev Korzenev@mail.cern.ch tél.: bureau: EP 1 3 mai 018

2 Massive Gauge Bosons We have already seen that 1. massive gauge bosons break local gauge invariance of QED, QCD and EW theory. massive fermions break local gauge invariance of EW theory only (not QED nor QCD) Consider the QED Lagrangian with a massive photon (add the mass term m A m A m ) QED Under local U(1) gauge transformations, the fermion field Y and the photon field A m transform as and the photon mass term transforms as which is explicitly non invariant. Likewise the fermion mass term 1 1 e L y i m m y ey m A y F F m m A m A m m m m 4 y ( x) y ( x) e y ( x) A ( x) A ( x) A ( x) ( x) ie ( x) 1 m m m m e e m A m A m A m m A m A m A m m m m myy my [ 1 1 ] y m y Ry L y Ly R e mixes terms with different chiralities and is not invariant under SU() L U(1) Y transformations because of different transformation properties of the fermion fields y L and y R. Note that this terms is invariant under QED U(1) and QCD SU(3) local gauge transformations because the y L and y R fields transform in the same way.

3 Massive Gauge Bosons M W W m Why insist on rejecting terms like m? the theory is not renormalizable and looses all predictive power (meaningless) (for each order in the perturbative expansion introduce cutoffs specific to this order) (This is true also for the Fermi theory of weak interactions) Consider for instance a higher order contribution to the scattering of two fermions. To calculate the amplitude integrate over all propagators (loop) f f' q All possible 4-momenta q appear in the amplitude (internal lines), the momentum transfer q is not limited from above. In QED, this box diagram represent the exchange of photons (vacuum polarization). The propagator ig m /q makes the integral finite (all orders renormalized with e m and m m ). If the case the boson is massive, the propagator yields a divergent integral for large 4-momenta q. M 4 ~ d q (propagators) Introduce a cutoff on q? New parameter in the theory! For each new diagram a new set of cutoff parameters are required and the theory cannot be renormalized because of too many () cutoff parameters. Any theory breaking local gauge invariance is not renormalizable! 3 i gm qmq M qmq i q q M q M

4 Mechanical Analogy The following is usually presented as a mechanical example to develop an appreciation of this surprising phenomenon of symmetry braking. Imagine a one dimensional needle. Around its axis, the needle has no angular momentum. Then press the needle with a vertical force larger Than the elasticity limit of the needle. Initially nothing happens, the system remains in an equilibrium state (well, you will hurt yourself ). Starting from a quantum fluctuation, the lattice of the needle weakens in an unpredictable point, and the needle bends in a random direction. The cylindrical symmetry of the system has disappeared spontaneously. At the same time an angular momentum around the initial symmetry axis has appeared. The system has acquired a kind of mass. 4

5 Electromagnetic Analogy Gauge symmetry assures that a QFT gives finite results and requires massless fields. However real fields can be massive: it is possible that the fields do not have a real mass, but that their free propagation is slowed down by constraints connected to the space-time. Consider for instance the electromagnetic radiation propagating through a plasma of electrons. Because the plasma acts as a polarizable medium dispersion relation ne w e p n 1 1 0mew w with w p the plasma cutoff-frequency. Electromagnetic waves propagate in the plasma only for frequencies w > w p : Above this frequency EM waves propagate with a group velocity v g Rearranging gives E E p g E hw E p hw p v c v cn c 1 E E g p p v 1 1 c E Ep vg c mc with m Ep c Massless photons, which propagate through a plasma, behave as massive particles propagating in a vacuum with velocity v g! 5

6 The Higgs Mechanism Gauge invariance can coexists with massive boson fields, if the mass of these fields is in reality an artefact, a fake conclusions that we infer from their sub-luminal propagation. There are different possibilities how to realize the sub-luminal propagation of a particle, which in reality is massless. We will study the spontaneous symmetry braking or Higgs mechanism. The Higgs mechanism provides a mechanism for W +, Z 0, and W gauge bosons to acquire mass without violating the local gauge invariance principle. We will arrive at this result in several steps: 1. symmetry breaking for a real scalar theory. symmetry breaking for a complex scalar theory 3. symmetry breaking in a local gauge theory 4. symmetry breaking in a local SU() gauge theory 5. symmetry breaking of the electroweak gauge group SU() L U(1) Y Next, we will examine how fermions acquire mass via Yukawa couplings with the Higgs field. With the discovery of the Higgs boson (M H ~ 15 GeV) the Standard Model of particle physics is complete. 6

7 The Higgs Mechanism Propose a scalar spin-0 field h(x) that permeates the space-time (i.e. it is everywhere) with a non-zero vacuum expectation value (VEV) 0 hx ( ) 0 0. Massive gauge bosons propagating through the vacuum with a non-zero Higgs VEV correspond to massive particles. But what is the vacuum? 1) state of lowest energy (otherwise the system can always evolve to that state) ) state with no fields f = 0 (VEV = 0) state with fields f 0 (VEV 0) - the Higgs is electrically neutral but carries weak isospin T = 1/, T 3 = -1/ and weak hypercharge Y = 1 - the W ± and Z bosons couple to the weak isospin and hypercharge and acquire mass - the photon does not couple to the Higgs field and remains massless - the Higgs mechanism results in absolute predictions for the masses of gauge bosons - fermion masses are also ascribed to interactions with the Higgs field, however no predictions of masses are made (Higgs boson couplings are proportional to their mass) igmw g m igm Z g m ig M W m f 7

8 Real Scalar Fields We start with an idealized world containing only real scalar fields f(x), without fermions, without matter, characterized by the Lagrangian density in which T is the kinetic term (it depends only on the components of the field gradient), while V is a potential function. For the usual free scalar field, V(f) is a second-order function in f, m f /. The zero solution corresponds to the minimum of the potential function, and quantization of the free scalar field corresponds to the quantization of small oscillations around the equilibrium point at f = 0. The Lagrangian is symmetric w.r.t. the reflection transformation f f, and its lower state f = 0 is symmetric and stable. Now, let s introduce into V(f) a positive term of the form f 4 describing self-interactions of the fields (it can be also seen as an expansion of V(f) around f = 0): and the Lagrangian becomes Lf T m V f 1 m Lf m m f lf V ( f) m f lf lf 4 4 where we have introduced the term, which describes the self-interactions among the fields f, with constants m, l 0. This Lagrangian is symmetric for f f. f f 1 4 l f f 8

9 The comparison with the Lagrangian for free scalar fields of mass m (Klein-Gordon Lagrangian) LKG 1 m 1 m m f shows that L f describes a field of mass m (provided m > 0), which interacts with itself via a fourfold vertex with with a coupling constant l (f 4 theory). The potential density V f m f lf 4 has a minimum for f 0, dv d f f 0 0 and d V df f0 m 0 with the vacuum expectation value (VEV) 0 f 0 0 That s expected, since the potential energy should have a minimum in absence of fields. Note that there are no odd power terms, otherwise V would have no minimum. The interpretation changes drastically, if m < 0 ( imaginary m). The term proportional to f cannot be interpreted as a mass term, since the mass is an observable and must be real. This may suggest the introduction of supraluminal particles with imaginary mass (tachions). To us, however, what is important, is that the point f = 0 in not a minimum for the potential V. This Lagrangian, however, is still symmetric for f f. 9

10 d V The potential has a local maximum for f = 0, because 0. df Local minima appears for f min v with v, the vacuum expectation value VEV of the field, defined as m The VEV is therefore not zero. We call vacuum the state of minimal energy. (for m > 0 this is not possible because the solutions are imaginary, while the field is real) l f0 m lf 0 in two points (stable equilibrium), i.e. m 0 f 0 v 0 l m 0 m 0 +v Both minima have the same energy V 4 V f m l as a result of which the corresponding system has a doubly-degenerate vacuum state. 10

11 Perturbative Approach If we want to perform a perturbative calculation with this Lagrangian (m 0 and l 0), we must develop the field around a minimum of the energy. In this case the minimum corresponds to a finite value of the field f : f min v or f min v. The perturbative vacuum, therefore, does not correspond to a situation without fields, but it contains a field f(x) v for all the space-time points x. We have to choose a minimum among two possibilities, +v and v, we choose +v (arbitrary decision). To perform the perturbative calculation we transform f (and shift f to the minimum) as The initial Lagrangian can be rewritten in terms of the field h as The coefficient lv of the h term is now positive and the comparison of this Lagrangian with the Klein-Gordon Lagrangian allows us to interpret this term as a real mass term The Lagrangian can be finally rewritten as f x v h x 1 m 1 1 lv 4 4 Lh mh h h lvh lh lv mh lv m 1 m h h h h with V h lv h lh lv L m V h m h 4 4 The potential V describes three- and four-fold self-interactions among the fields h. 11

12 Spontaneous Symmetry Breaking 1. Find a non-symmetric solution of a theory which is otherwise symmetric.. The symmetric state does not coincide with the ground state 3. Identify a mechanism that evolves the system from the symmetric state to the non-symmetric state. With the transformation f h we lost the symmetry of the initial Lagrangian. L h is not invariant for h h (the term lvh 3 changes sign) different physics. This is due to the fact that we chose the minimum at +v. We could have also chosen the vacuum at v, but not both at the same time. Our choice has broken the symmetry of the system, and this symmetry breaking has given mass to the field f. The choice of the ground state among various possibilities, which are linked by a symmetry is what we call spontaneous symmetry breaking. The symmetry of the system is no longer evident in its ground state; more precisely, we should speak of hidden symmetry, since the system does not exhibit in its ground state the symmetry, which is well visible in the excited states of the system. One can ask himself how such a trivial transformation, which leaves the Lagrangian intact, can transform the massless field f into the massive filed h. With an exact calculation of solution for L f and L h one would find perfectly equivalent solutions. We are not calculating, however, an exact solution, but we approximate the solution with a perturbative expansion. The perturbative expansion is possible only for L h around a minimum of the potential energy h = 0 (i.e. f = v). Both Lagrangians describe a massive particle, but we know how to calculate only with L. 1 E E S E 0 symm. non-symm.

13 Mechanical Analogies In Nature, many examples systems exhibits spontaneous symmetry breaking. These systems are described by some functions (Lagrangians, Hamiltonians, eq. of motion) possessing some symmetry, while the real physical state of the system corresponding to a particular solution of the eq. of motion does not have this symmetry. That arises when the lowest symmetric state does not have the lowest possible energy and it is itself unstable (see e.g. magnetization). The actual cause of symmetry breaking may turn out to be an infinitesimal non-symmetric perturbation. A simple example is provided by the bottom of a bottle and a small ball inside it. The bottle has rotational symmetry around the vertical axis. Let the ball fall along this axis. When the ball reaches the bottom, the ball will not rest on the central protuberance and will roll down to the periphery. The final state has lost its initial rotational symmetry and the ball has acquired an angular momentum I. 13

14 Complex Scalar Fields A quite simple but physically profound generalization arises in the transition from the single-component field to an isotopic multiplet f = (f 1, f, f N ). For simplicity, let s consider a spin-0 complex field where f 1 and f are two spin-0 real fields, with the following Lagrangian density 1 and m < 0, i.e. without a real mass term and l > 0. The discrete reflection symmetry is replaced by the continuous symmetry of isotopic rotations. This Lagrangian is invariant under global phase transformations i f x f x e f x with a real constant. This Lagrangian is invariant under global gauge transformations of the U(1) group. Rewritten in terms of the f 1 and f fields, the Lagrangian becomes L 1 f f if 1 m Lf m f m f f l f f f f f f m f f l f f m m f m 1 1 m 1 1 The first two terms describe the kinetic energy density and the last two the potential. 14

15 As before, the potential has a local, unstable, maximum for f 0. Local minima occur for 1 v 0 f f f f This is the equation of a circle in the f 1 f plane. Now let s shift the field f by a constant vector f 0 = v, satisfying v = -m / l. We have the freedom to choose any minimum of the potential for our (perturbative) vacuum. Without loss of generality we choose f 1 v and f 0, such that m l where we introduced to new local fields h(x) and x(x) which are related to f(x) by the above expression. Inserting this new f(x) into the Lagrangian gives 1 f( x) v h( x) ix ( x) 1 m 1 m mx x v mh h h l h l hx lh lh x L lv v lx lv The terms O ~ 3 and O ~ 4 describe (self)interactions between the fields h and x. 15

16 Goldstone Theorem The Lagrangian is not anymore invariant under transformations of the h and x fields, because we chose a particular ground state. We can recognize a mass term for the field h with the good sign m h h m corresponding to a real mass (recall m < 0): the shifted component has acquired a mass. There is no mass term for the field x, even though L contains the kinetic term 1mx (there is no potential in the x direction : the other components correspond to massless degrees of freedom. The theory contains a massless scalar field x(x), referred to as Goldstone boson. This is an example of the Goldstone theorem, which predicts the appearance of a massless scalar field whenever we break spontaneously the continuous symmetry of a system. This property is invariably associated with the spontaneous breakdown of a continuous symmetry, because each of massless degrees of freedom correspond to an infinite degree of degeneracy of the vacuum state. (virtual motion of the ball within the circular hollow at the bottom of the bottle) In summary, the field f (field h) has acquired the mass m = v l, but also a new massless particle x has appeared. 1 lv h m h m h 16

17 Local Symmetry Breaking It seems that we are moving in the wrong direction: by giving mass to a field f by breaking spontaneously the symmetry of the system, we generated a Goldstone boson that we did not foresee. Let s study the symmetry breaking for a local gauge theory of the U(1) type: f To realize it in the Lagrangian we replace the m derivative by the covariant derivative with the gauge fields A m transforming as i ( x) e and the gauge invariant Lagrangian becomes D For m 0 this is exactly the QED Lagrangian for a scalar particle of mass m and charge e, enlarged by a self-interaction of the field f. f iea m m m 1 e A A ( x) m m m f f m f f l f f L iea iea F F m m 1 m m m m 4 However, for m 0, the term m f f cannot be interpreted as mass term. 17

18 As before, let m 0, this time with (x). For the ground state we chose i.e. a ground state with (x) 0. The Lagrangian becomes 1 We can identify the mass terms for the fields h and A, but not for x: f( x) v h( x) ix ( x) x x h h lh x x, h m m m m m L m m v e v Am A evam Fm F WW x 0 h l A As required, the gauge field A m has acquired mass, but at the price of introducing a massive field h and a Goldstone boson x. We succeeded in generating dynamically a mass for the gauge field, but we are not yet done because of the massless Goldstone boson. giving mass to A m 3 spin components ( 3) [we gained 1 d.o.f!] A m x m m v m ev The term m describes the interaction of a spin-1 field with a spin-0 field, in which the spin disappears and such terms cannot exist. We have the freedom to make a gauge transformation to eliminate it, by choosing the phase for the ground state. 18

19 The Goldstone boson corresponds to the extra (unphysical) d.o.f. in L. The field f(x) can be rewritten as A particular value for the phase (x) corresponds to a particular choice for the filed x(x). We chose a ground state with the fields h(x), q(x), and A m (x), all real, such that (note the different notation) with (x) = q(x) / v. ( )/ 1 1 f( x) v h( x) ix ( x) v h( x) e x f q ( )/ v h( x) e With these substitutions, the new Lagrangian becomes 1 i ev A A q ( x) m m m i x v i x v L h m h v h e v A A m vh h e A A m h ve A A m h F F m m l m l l m m m M H lv M A ev The field q(x) has disappeard in the new Lagrangian and the Goldstone boson has been reabsorbed. The new Lagrangian contains the scalar Higgs field h(x) and the massive gauge field A m. These fields interact between themselves, in particular the Higgs field h(x) Interacts with A m (x). 19

20 Local SU() Gauge Invariance Let us consider now SU() complex field doublets (no spin, scalars) The resulting Lagrangian is invariant under global phase transformations of the SU() symmetry group a it a / f( x) f( x) e f( x) with a a 3 component constant vector (i.e. the 3 Euler angles) and Pauli matrices t a as generators of the transformation. When imposing local gauge invariance we have to allow the complex fields to interact with vector gauge fields W m a introduced via the covariant derivative (see L3 and L11) with the W m a fields transforming as f ( x) 1 f1( x) if( x) f( x) f ( x) f3( x) if4( x) a a D ig W ( x) m m m m 1 g a a a b c W ( x) W ( x) ( x) i ( x) W ( x) m m m abc m with g the coupling constant of the fields f with the gauge fields W. The additional term (vector product between the phase and the fields) appears because SU() is non Abelian. t 0

21 The Lagrangian describing the fields f is given by a m m a m mf m f f f f m a t a a t 1 L ig W ig W a V W W 4 with the potential energy term V(f) (we chose this potential!) V f m f f l f f and the kinetic term for the 3 W gauge fields expressed in terms of field tensors W W W g W W a a a b c m m m abc m The vector products in the field transformations and tensors appear because the SU() group is non Abelian and the generators do not commute. For m 0 this Lagrangian describes 4 scalar particles, f 1, f, f 3, and f 4, with mass m, that interact among themselves and with the 3 gauge fields W m1, W m, and W m3. The gauge fields interact also between themselves like in QCD. 1

22 Local SU() Symmetry Breaking Again, we are interested in the case m 0 and l 0, because it allows us to break spontaneously the symmetry. The potential energy V(f) has minima for The ensemble of these minima is invariant under SU() transformations of the fields f. To apply a perturbative calculation, we have to choose a specific minimum, and this symmetry will not be visible anymore: The ground state corresponds to f0 ( x) v f v h( x) Thanks to the gauge invariance, all states can be described in this way. Of the 4 initial fields, only the Higgs field h(x) remains. 1 m l f f f f f f m v f f f f All fields can be expressed in terms of the field h(x) alone (apply an SU() transformation) a itq ( )/ a x v iq v iq q v q iq 1 f( x) e v h( x) iq 1 q v 1 iq 3 v v h( x) v h iq 3 l

23 After a small rotation in SU() we recover the initial form of the field f(x) expressed in terms of 4 real fields. The Lagrangian is invariant under these transformation by construction. In other words we can make disappear 3 out of 4 fields, q 1, q, and q 3. With an appropriate gauge transformation no Goldstone boson appears, the Lagrangian does not contain the fields q i. To obtain the mass terms, insert the vacuum state f 0 (x) in the Lagrangian. The term proportional to W (i.e. the mass term) is t a 3 1 g Wm Wm iw m 0 g v m m m Wm iwm W m v m ig W a f W W W 8 8 Comparison with a typical mass term for bosons shows that MW The three W gauge fields have eaten up 3 Goldstone bosons and have become massive: gauge bosons 3 d.o.f. (longitudinal polarization) The Lagrangian describes 3 massive boson fields W mi, vectors under Lorentz transformations, and a scalar field h(x), the Higgs boson. 1 In summary, in a theory that does not contain fermions we succeeded to transform the SU() locally invariant Lagrangian into a SU() locally invariant Lagrangian containing 3 massive vector bosons and one massive scalar via the spontaneous symmetry breaking The massless Goldstone bosons, associated with the symmetry breaking, are not present thanks to the appropriate choice of a gauge transformation. 3 gv

24 W + W W + W Scattering The W + W W + W scattering process is described at the lowest order by the following Feynman diagrams These diagrams alone violate unitarity already at relatively low energies (O ~ 1 TeV), because of W l W l W l W l amplitudes (l 3 rd polarization component of massive W s). violation of unitarity associated with the W boson being massive (3 rd component) It can be cancelled by the following diagrams involving a new scalar particle, the Higgs boson. This might work only if the couplings of the scalar particle are related to EW couplings. As shown by t Hooft (1975) locally gauge invariant theories are renormalizable: cancellation of infinities take place at all orders in the perturbative expansion with a finite number of renormalized quantities. breaking of the local gauge invariance of EW theory cannot be simply ignored! 4

25 The Standard Model Higgs Finally, we apply to Higgs mechanism to the Electroweak Lagrangian (L11): introduce 4 real scalar fields and an SU() U(1) gauge invariant Lagrangian for the scalar fields t a a Y LH f im g W g B V m m f f with the usual potential V f m f f l f f This additional term L H is invariant under combined transformations of the SU() U(1) group. To assure that, f(x) must belong to a multiplet of the group. The minimal choice is to arrange the 4 fields in an isospin doublet with weak hypercharge Y = 1 (Weinberg 1967) f ( x) 1 f1 if f( x) 0 f ( x) f3 if 4 With this minimal choice and the above Lagrangian we have specified the electro-weak sector of the Minimal Standard Model, i.e. the Glashow Weinberg Salam model. 5

26 Again, we are interested in the case m 0 and l 0, and we choose the following vacuum state f 0 (x) 1 0 m f v v f f f f l with v m /l. This ground state is characterized by T = 1/, T 3 = 1/, and Y = 1, and it is no longer invariant under SU() L or U(1) Y. On the other hand the electric charge of the field is Q = T 3 + Y/ = 0, such that the system is symmetric under U(1) Q with the generator Q. The charge of the state f 0 (x) is Qf 0 = 0, and for each phase (x) we have f ( x) f( x) e f ( x) f ( x) i ( x) Q This means that the U(1) Q symmetry is preserved, such that no mass is generated for the associated gauge boson, the photon. On the contrary, the symmetries SU() L and U(1) Y are broken: the W ± and Z 0 bosons have acquired mass. The mass is obtained by inserting the ground state f 0 (x) in the Lagrangian L H 0 1 g a g gw g B g W iw m 1 m m m m 0 i t W a i B m f g W iw gw g B v m m m m 1 v 3 m 3, m W W gb gw g m m m m B gw g v 8 8 0, m g v, m v 0 g ggw W W W, B 4 m 8 m m m gg g B 6

27 From the first term we obtain the mass of the W ± bosons MW 1 gv The second term in the mass expression is off diagonal in the (W 0, B) basis and can be written an v The matrix has two eigenvalues, one zero and one non-zero. The physical fields Z m and A m diagonalize the mass matrix so that the above expression must be identified with and we obtain (after normaliztion) 0 0, m 0 m m v 0 0 g Wm W ggw m B g Bm B gwm gb m 0 gw m gbm M Z M A Z m A m A gw gb g g Z gw gb g g 0 0 m m m m m m and mass M A 0 and Z v M g g v 0 f 0 46GeV By comparing this rotation which diagonalize the mass matrix with the transformation which rotates (W m3, B m ) in (A m, Z m ) one finds M M cosq W Z W 7

28 The Glashow Weinberg Salam Model In summary, the GWS model is based on the hypothesis that there exist two gauge fields. One of them W i has three components and corresponds to the adjoint representations of the group SU(), while the second one (B) has one component and the gauge group is the group U(1). The gauge group of the GWS model is the compact group SU() U(1). It contains two parameters, g and g. Four gauge particles are necessary for the description of three intermediate (massive) vector bosons and the photon. Of the four particles, W 1 and W are charged and W 0 and B are neutral. As the fields W 0 and B have the same quantum numbers, mixing between them can take place. The physical neutral vector particles, the photon and the Z boson are superpositions of the fields W 0 and B. To make the vector bosons massive, one uses the mechanism of spontaneous symmetry breaking, for which one introduces and auxiliary two-component complex (4 d.o.f.) scalar field f. Three of the four d.o.f. are used for providing via the Higgs mechanism a superfluous polarization state for each of the three vector components, while the fourth one leads to the physical massive Higgs boson. Thus the vector boson sector of the GWS model is based on the Lagrangian L IVB IVB 1 a m 1 m l 4 m a 4 m mf 4 f f L W W B B D v where the field f represents the isotopic doublet of the gauge group SU() of the complex fields and D m are the covariant derivatives. ig a ig a D f i t W YB f m m m m 8

29 Spontaneous symmetry breaking is realized when the second component of the field f is shifted by a real constant v: 0 f( x) f( x) v As a result of the shift, the term D m f gives the following contribution to the mass matrix which, after diagonalization, becomes and he mass terms are given by 1 v 0 Wm Wm gwm gb m g v 8 8 A gw gb g g Z gw gb g g 0 0 m m m m m m v v M g M g g M cosq M 0 W Z W W A The last term in L IVB can be rewritten as v v l l l F h h F h h 16 4 v 0 f 0 46GeV with F = f 1 + f + f 4, from which it follows that the filed h has mass M h = lv, while the components f i represent Goldstone fields. They can be removed by a gauge transformation, as a result of which three components of the non-abelian gauge field will acquire a superfluous polarization component (the Higgs effect). 9

30 Fermion Masses The mass term does not respect the SU() L U(1) Y gauge symmetry because y L is an SU() doublet and y R a singlet with different transformation properties. Therefore, a term which mixes left and right handed particles cannot be gauge invariant. Remarkably the Higgs mechanism can be used also to give masses to fermions, without the introduction of new particles. The same SU() L local gauge transformation on the Higgs field applies also to y L yf myy my [ 1 1 ] m y Ry L y Ly R f f I ig ( x) T f 0 L L L L y y I ig ( x) T y L L L y y y y I ig ( x) T L is invariant under SU() L transformations, and L Runder SU() L and U(1) Y is SU() L U(1) Y gauge invariant ( ) y fy g f y Lfy R y Rf y L y Lfy R y Rf y L with f the Higgs doublet. 30

31 To generate the mass term for the leptons we couple the Higgs field to the leptons as f 0 e Le ge e, e e 0 R erf, f L f e g e Yukawa coupling constant of the electron to the Higgs field, which is not predicted by the SM and which must be determined from the experiment. Yukawa coupling coupling of a fermion field with a scalar field (i.e. NpN) After spontaneous symmetry breaking of the Higgs doublet, 1 0 with the substitution f( x) the Lagrangian becomes v h( x) ge ge me Le veler erel heler erel Le meee eeh v L where the Yukawa coupling g e has been identified with the electron mass m e. Note that the neutrino remains massless. first term in L e : coupling of the electron to the Higgs field gives mass to e g m / v e e second term in L e : coupling between the electron and the Higgs boson 31

32 In addition to give mass to the electron (fermion) via the coupling of the left and right handed components to the Higgs field, the Higgs mechanism introduces also a direct interact between the fermion and the Higgs boson itself. Note that only the lower components of the weak isospin doublet are present! It can be used to generate the masses of the down quarks but not up quarks. To give masses to up quarks consider the conjugate doublet f C (recall the isospin transformation of antiparticles) 0 1 f3 if f 4 fc i f f1 if f which transforms exactly in the same way as f (property of SU(), not valid for SU(3) ). The Lagrangian describing the up quark mass is then given by gu gu mu Lu vulur urul hulur urul Lu muuu uuh v The gauge invariant mass terms for all fermions can be constructed from y fy y f y or y f y y f y L g L g f L R R L f L C R R C L with the Yukawa couplings of the fermions to the Higgs field g m / v where v = 46 GeV is the vacuum expectation value of the Higgs field. Note also that the coupling to the Higgs field is flavor conserving. f f 3

33 The Discovery of the Higgs Boson H H ZZ e e e e 33

34 Precision Tests of the Standard Model From LEP and the Tevatron, and now the LHC (+ lower energies!) have precise measurements that allows us to test the Standard Model predictions. M Z GeV For instance MW MZ cosqw measure sin q W predict M GeV M GeV W but measure W Very close, but not quite right. Have to include also radiative corrections (virtual loop diagrams) resulting in 0 M H MW MW amt bln MW with a, b exactly calculable coefficients (Feynman diagrams) Very high precision measurements sensitive to masses of particles inside virtual loops. The mass of the W boson depends also on the Higgs mass (albeit only logarithmically). 34

35 Predicting the Higgs Mass Before the discovery of the Higgs boson in 01 Measurements are sufficiently precise to have some sensitivity to the Higgs mass. Direct and indirect values of the top and W mass can be compared to predictions for different Higgs masses. Combined with direct searches, exclusion regions m H = 16 GeV 35

36 Properties of the Higgs Boson Neutral spin-0 particle Mass is a free parameter of the SM m lv GeV H The Higgs boson couples to all fermions with a coupling strength proportional to the fermion mass g m / v f f Feynman rules for the Yukawa interaction vertex mf mf i i g v M w The resulting matrix element is proportional to the mass of the particle which couples to the Higgs boson. determines dominant process for production and decay of the Higgs boson 36

37 Higgs Boson Decay The Higgs boson can decay to all SM particles provided that m H > m f. Because of the coupling, the Higgs boson decay predominantly to heaviest particles which are energetically allowed (largest BR). H bb and using Fermi s golden rule M M H scalar particle no polarization g mb 0 mb mb u pv p3 u v E M v v W m m b b M M 8E m H m bm H v v MW g H bb 3 m ~ (MeV) bmh O 3p MW Remember that also the masses run with q (not only S ), hence the masses appearing in are estimated at q = m H (m b = 3 GeV). The Higgs boson can decay also to g or via virtual top quark and W loops m v g b m H or H WW H ZZ and 37

38 Higgs Decay Branching Ratios Higgs boson couplings are proportional to the mass Higgs decays predominantly to heaviest particles which are energetically allowed Higgs decay branching ratios at 15 GeV (calculated) H bb H W W H tt H H gg cc H ZZ H 57.8% 1.6% 6.4% 8.6%.9%.7% 0.% m GeV H B.R. measurements are under way. 38

39 Higgs Boson Production in Hadronic Collisions Higgs production at the Large Hadron Collider LHC The LHC will collide 7 TeV protons on 7 TeV protons However underlying collisions are between partons Higgs production the LHC dominated by gluon-gluon fusion p 7 TeV p t t t H 0 7 TeV 1 1 pp HX ~ dx d x g( x ) g( x ) gg H Uncertainty in gluon PDFs lead to a ±5 % uncertainty in Higgs production cross section 39

40 Higgs Boson Production in e + e Annihilations Clear experimental signature ( Higgsstrahlung ), but small cross section with large backgrounds Need enough energy to produce a Z boson and the Higgs (ILC?): E s M M 17 GeV E 115 GeV beam Z H beam e + e mbb Z H 0 M Z H f f b b The only way to distinguish the two diagrams is from the invariant mass of the jets from the boson decay. main background to Higgssthralung e + e mbb e Z Z f f b b MZ 40

41 Tagging the Higgs Boson Decays One signature for the Higgs boson decay is the production of a b quark pairs. H 0 b b b b b q q b b b Each jet contains one b-hadron which decays weakly. Because V cb is small (~ 0.04) hadrons containing b quarks are relatively long-lived, i.e. t ~ s several mm long tracks with displaced verteces before decaying, and one can efficiently identify jets containing b quarks. 41

42 Concluding Remarks We covered almost all aspects of modern particle physics. The Standard Model, now completed with the discovery of the Higgs boson, is one of the greatest scientific triumphs of the late 0 th century. Many people contributed to this success with their insight and hard work. The Standard Model developed through close interplay of experiment and theory. Experiment is the ultimate test bench of any physical theory. Dirac Equation QFT Experiment Gauge Principle Higgs Mechanism The Standard Model Experimental Tests Experimental particle physics provides many precise measurements. The Standard Model describe almost all current data. Despite its great success, do not forget that it is just a model: a collection of beautiful theoretical ideas put together to fit with experimental data. There remains many issues / open questions 4

43 Standard Model Open Issues The Standard Model has too many free parameters ( = 6): m, m, m ; m, m, m ; m, m, m, m, m, m e m t u d c s t b 1 3 q, q, q, ; q, q, q, ; e, g, ; v, M ; q q CP CP W S H CP Why three generations? Why SU(3) C x SU() L x U(1) Y? Unification of Forces Origin of CP violation in early Universe and the matter antimatter asymmetry What is Dark Matter? Why is the weak interaction V A? Massive neutrinos Why are neutrinos so light? Where are the right-handed neutrinos? Higgs give rise to a huge cosmological constant Ultimately need to include gravity 43

44 Alternative Models: Extra Dimensions With the discovery of Higgs boson, alternative models to generate the bosons mass, have somehow lost interest. Nevertheless here we shall present one model based on extra dimensions beyond the four space-time dimensions. One can imagine that these additional dimensions are compacted, i.e. the fields are cyclic in the coordinates beyond x m. Let s consider a Universe with dimensions, where the additional dimension is confined to a radius R. To each space-time point a circle of radius R in the additional dimension is attached. Each field is confined to the periphery of this circle, which represents the additional coordinate z. The fact that particles do not disappear during propagation, requires the conservation of probability, i.e. the density of particles in a volume can change only when some particles cross the volume s border. For a generic field, that requires the conservation of the four-current j m : m m m j if f f f i.e. m j m = 0 in 4 dimensions. This implies that the fields must be cyclic in the additional dimension z: x, z x, z R f f p 44

45 Fields compatible with the four-current conservation can be decomposed as f / x, z f x e If the field is massless in dimensions, it evolves according to the Klein-Gordon equation with mass term Since the f k fields are orthogonal, each field f k must satisfy independently This is a Klein-Gordon equation for a massive filed in 4 dimensions with m = k/r. k How does a field evolve in a Kaluza-Klein vacuum? The small deviations on the loops in z prevent the field from propagating with speed c. The field behaves like a massive particle. The idea of additional dimensions also open the road to quantum gravity. Matter and ordinary interactions evolve on a 4-dimensional brane in the 4 + n dimensional space. The n additional dimensions are accessible to gravity only and gravity has no bosons carrying the force in the 4 dimensional space-time. k ikz R m z m k m z f 0 m fk( x, z) 0 k R m m k R fk ( x, z) 0 45