Weak interaction. Chapter The Fermi s theory Birth of a new interaction

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1 Chapter 6 Weak interaction Few references: F. Halzen and D. Martin, Quarks & Leptons, Wiley, (984) chapter. Carlo Giunti and Chung W. Kim Fundamentals of Neutrino Physics and Astrophysics, Oxford University press (007) chapter 7 and 8. So, far we have studied the electromagnetic and the strong interaction as examples of gauge theories. This chapter presents the last interaction described by the standard model of particle physics, the weak interaction. We will follow an historical approach, trying to explain the evolution of its theoretical description. The gauge theory of the weak interaction will be described later, in the next chapter, within the context of the unification of the electromagnetism and the weak interaction. 6. The Fermi s theory 6.. Birth of a new interaction In 934, Fermi proposed his theory to describe the so-called decay: n! pe where the neutrino, not yet discovered, had been previously postulated by Pauli to explain the observed continuous energy spectrum of the electron. By analogy with the nuclear decay where a photon is emitted N! N Fermi assumed that the electron and neutrino, were created in the process leading to the graph: n e p 59

2 60 Weak interaction This graph is an example of a so-called four-fermion contact interaction (no propagation of bosons). While in QED, the Lagrangian for p! p (the proton being in a nucleus) is: L = e(ū p µ u p )A µ Fermi replaced the proton current by a current with the transition n! p and the photon A µ by a current expressing the creation of the electron and neutrino: L = G F (ū p µ u n )(ū e µ u ) the charge e being replaced by a constant, the Fermi s constant [9]: G F = GeV (6.) consistent with the decay rate. A new coupling constant meant that a new force was born. The Fermi coupling constant is so small that the probability of interaction is necessarily small, leading to naming this new interaction: the weak force. Moreover, the currents in the latter Lagrangian are charged while in QED they are always neutral. So far, so good. Perturbation theory based on this Lagrangian described pretty well the -decay. 6.. The + and + mystery Many experiments were made to better understand the property of the weak interaction. At that time (in the 950s), two particles called + and + (the latter was not the now well-known lepton of the third family) had very similar properties: same mass and same lifetime (within experimental errors). However they decayed via the weak interaction di erently: +! + 0 +! + + They were considered as two di erent particles because according to the decays above, they have opposite parities (see section.3.4.) which were assumed to be conserved for all interactions. Indeed, knowing that the intrinsic parity of pions is -, the parity of + is: ( + )= ( + ) ( 0 ) (spatial) =( )( )( ) l=0 =+ where l = 0 because the + had spin-0 as well as the pions and thus by angular momentum conservation (J = L + S), l is necessarily 0. In the case of +, for the spatial parity one has to consider the orbital momentum l between the first s and the orbital momentum l between the third and the barycenter of the first s. Since the + had also a spin 0, necessarily 0=l + l ) l = l and thus (spatial) =( ) l = +. Hence: ( + )= ( + ) ( + ) ( ) (spatial) =( )( )( )(+) = In 956, two physicists, Yang and Lee suggested that the + and + were the same particle, the K + in our modern nomenclature, and that parity is not conserved by weak interactions. A real revolution in physics! Nature distinguishes left and right! Indeed, your left hand appears as a right hand in a mirror. The non-conservation of parity induces that the two situations lead to di erent physics measurements (with particles not your hands!). P.Paganini Ecole Polytechnique Physique des particules

3 The Wu s experiment: parity is violated The Wu s experiment: parity is violated After such hypothesis, the challenge was to prove it experimentally. Yang and Lee published a paper suggesting several experiments. The first experiment was led by Wu using a source of Cobalt polarized thanks to an external magnetic field. The Cobalt has the following decay: 60 Co! 60 Ni e and the total angular momenta are respectively J(Co) = 5 and J(Ni ) = 4. The field was large enough and the temperature low enough (about 0.0 K!) to be sure that the spin of the Cobalt was properly aligned. Now recall that the angular momentum is a pseudovector (or equivalently named axial vector) meaning that contrary to usual vectors as the position or the momentum, it remains unchanged under a parity transformation. Indeed, for the orbital angular momentum: ~L = ~r ~p parity! ~ L 0 = ~r ~p = ~ L The spin S, the intrinsic angular momentum, is thus a pseudovector as well as the total J = L+S. Now, if parity were conserved, we should find the same rate of electrons emitted with an angle and an angle, the angle being measured with respect to the Nuclei angular momentum (or the B field) axis (see figure 6.). However Wu and her team measured an asymmetry which indicated a clear sign of parity violation! Electrons were more likely to be emitted at an angle = than = 0. Figure 6.: Schema of the Wu s experiment. From [6]. With J(Co) = 5 and J(Ni ) = 4, we have J =, the projection of the spin of the system e on an axis is thus necessarily and hence both the electron and the neutrino have S z =+/. Since they are emitted back to back (the Nitrogen stays (almost) at rest as was the Cobalt), they have opposite helicity. The electron being more likely emitted in the opposite direction of its spin, left handed helicity is favored, while for the anti-neutrino, we expect the right-handed helicity. Recall that when the masses are neglected, both helicity and chirality are equivalent which is reasonably the case here. 6. Modification of Fermi s theory: the V A current 6.. Parity conservation in QED and QCD The parity violation signed the death of Fermi s theory. Indeed, the current involved in the Lagrangian has the same structure as the one in QED (by construction): Physical Review 04:54 j µ = eū µ u = eu 0 µ u P.Paganini Ecole Polytechnique Physique des particules avancée

4 6 Weak interaction We saw in section.3.4., that the parity operator is ˆP = under a parity transformation the current becomes: 0 when applied on spinors. Hence, and thus: j µ = e( 0 u ) 0 µ 0 u = eu µ 0 u j 0 parity! eu 0 0 u = eū 0 u = j 0 j i parity! eu i 0 u = eu 0 i u = eū i u = j i We have just checked that j µ is a vector (polar vector more precisely), namely the temporal component is not a ected by parity transformation while the spatial components are inverted. Therefore, the amplitude being made with the product of currents j µ and jµ will give: M = j µ j µ = j 0 j 0 + j i j i parity! j 0 j 0 +( j i )( j i )=j µ j µ We do have invariance of the QED amplitude under parity transformation. Similarly, in QCD there is just an additional Gell-Mann matrix which remains unchanged under parity transformation. QCD is thus parity invariant. 6.. The V A current How should we modify the QED current to include parity violation? We have just seen that electrons with left-handed helicity were favored in the Wu s experiment. We could think that inserting an helicity projector would be enough. However, in section.3.3.5, we saw that helicity is not Lorentz invariant. We cannot have a weak interaction that would depend on the frame! Relativistic fermions are described by spinors, so a covariant current is necessarily of the form: j = O where O is an operator having a 4 4 matrix representation. The constraints of Lorentz invariance and hermiticity reduce the possibilities of O to these 5 combinations: O =, 5, µ, µ 5, µ µ Let us consider the case O = µ 5. A parity transformation of such current gives: j µ =ū µ 5 u parity! ( 0 u ) 0 µ 5 0 u = u µ 0 5 u where we have used the anticommutation property.58. The time and spatial components read: j 0 parity! ū 0 5 u = j 0 j i parity! +ū µ 5 u = j i The spatial components are not a ected meaning that j µ is an axial vector (pseudo). This behavior is opposite to that of the (polar) vector. Clearly, an amplitude based on the products of axial vectors remains invariant under parity transformation. However, the parity violation is obtained by a mixture of vector current j µ V and axial vector jµ A : j µ V =ū µ u, j µ A =ū µ 5 u, j µ = c V j µ V c A j µ A =ū µ (c V c A 5 )u P.Paganini Ecole Polytechnique Physique des particules

5 The V A current 63 where the minus sign is purely conventional (it will be more clear in few lines). Let us see how an amplitude based on the product of currents of this type is transformed by parity: M = c V j µ V c A j µ A c 0 V j0 Vµ c 0 A j0 Aµ = c V c 0 V jµ V j0 Vµ + c Ac 0 A jµ A j0 Aµ c V c 0 A jµ V j0 Aµ c A c 0 V jµ A j0 Vµ M parity! M 0 = c V c 0 V j0 V j0 V 0 + c Ac 0 A ( j0 A )( j0 A0 ) c V c 0 A j0 V ( j0 A0 ) c Ac 0 V ( j0 A )j0 V 0 c V c 0 V ( ji V )( j0 Vi )+c Ac 0 A ji A j0 Ai c V c 0 A ( ji V )j0 Ai c A c 0 V ji A ( j0 Vi ) = c V j µ V + c Aj µ A c 0 V j0 Vµ + c0 A j0 Aµ 6= M if c A or c 0 A 6= 0 and c V 6= 0 or c 0 V 6=0 Conclusion: a weak current based on the di erence (or addition) of a vector current and an axial vector current, thus the name V A does permit parity violation. Let us come back to the decay: n! pe. We then expect a current between the (outgoing) electron and the anti-neutrino of this kind: j µ e =ū e µ (c V c A 5 )v (6.) Wu s experiment showed that the electron was mainly produced with a left-handed helicity, which in the low mass limit, is equivalent to a left-handed chirality state. But, in section.3.4.3, we saw that the left-handed chirality projector for fermions (equ..6) is: Hence the left-handed electron reads: P L = ( 5 ) u el = ( 5 )u e ) ū el = ( 5 )u e = ( 5 )u 0 e = u e =ū e ( + 5 ) ( 5 ) 0 where we used the 5 properties of anticommutation and ( 5 ) = 5. Hence if only left(right) handed components of (anti)fermions are involved in weak interaction 3,acurrentlike: j µ e = P L u e µ P L v =ū e ( + 5 ) µ ( 5 )v =ū e µ ( 5 ) ( 5 )v =ū e µ ( 5 )v (using PL = P L) has precisely the form required for parity violation: see equ. 6. with c V = c A =. This is precisely what was proposed in 958 by Feynman, Gell-Mann, Sudarshan and Marshak. The amplitude of the decay 4 then reads: M = p 4 apple µ G F ū apple p ( 5 )u n ū e µ ( 5 )v where the factor 4 and p are purely conventional to keep the original value of the Fermi s constant. 3 Recall that P L for antifermions projects on the right component! 4 Historically a factor.6 was put in front of the 5 of the n! p current to match the experimental rate. We know now that it comes from the composite nature of the neutron. There is no such factor when the decay is transposed to quark level: d! ue. Hence we have omitted it here. P.Paganini Ecole Polytechnique Physique des particules avancée

6 64 Weak interaction 6..3 Evidence for V A current Consider the decays of the charged pion. Since it is the lightest meson, it cannot decay via the strong interaction. And since its quarks content is ūd of respective charge -/3 and /3, these quarks cannot annihilate into a photon (the photon would carry an electric charge). Thus, it must decay via the weak interaction into lighter objects namely:! e e,! µ µ Since the phase space available in the electron channel is much larger than in the muon channel, we would expect naively a branching ratio larger for the former than the latter. However, experimentally, the measurement gives: (! e e ) 4 =.3 0 (! µ µ ) The opposite of our naive expectation! The explanation is the following: the pion has a spin 0, thus the charged lepton and the anti-neutrino have an opposite spin projection. In the pion rest frame, they fly back to back and hence, they have necessarily the same helicity. Now, we know that the neutrino has a tiny mass (because of the neutrino oscillation phenomenon). Thus, we can safely consider that its helicity state corresponds to its chirality state. But, because of the V A structure of the weak interaction, we know that only right-handed anti-neutrino can interact. Consequently, its helicity state is also right-handed, meaning that the charged lepton state has also a right-handed helicity state. However, V A predicts that only the left-handed chirality state of charged lepton interacts. So we have a situation in which only the left-handed chirality projection of a right handed helicity state is involved. Now consider the right-handed helicity state (equation.45 for fermion) and project it on left-handed chirality state: ( 5 ) + = B 0 0 p 0 0 A E + m cos ~p B e i sin C cos A e i sin = p E + m cos e i sin ~p E+m cos ~p E+m ei sin We see clearly that if the mass of the charged lepton was 0 (and thus ~p = E), the projection would give 0, and hence the decay of the pion would not occur. Since m e or m µ are however very small, the decay is naturally suppressed explaining the long lifetime of the charged pion ( s). And since m e m µ, the decay must be much more suppressed in the electron channel than in the muon one explaining the ratio of the decay widths above The four-fermion contact interaction failure The four-fermion contact interaction with V A currents described well the weak interaction processes known in the 950s-960s, namely at low energy. Rapidly, it was realized that this theory was only an approximation, ie an e ective theory in the low energy range. For example, consider the following neutrino electron interaction µ e! e µ : P.Paganini Ecole Polytechnique Physique des particules C A

7 The four-fermion contact interaction failure 65 µ (k) µ (k 0 ) e (p) e (p 0 ) The matrix element is then: M = p 4 apple µ G F ū apple e ( 5 )u e ū µ µ ( 5 )u µ = G F p ū e µ ( 5 )u e ūµ µ ( 5 )u µ (6.3) We wish to calculate the cross-section of this process. The calculation is very similar to e µ! e µ. But instead of µ, we need to replace it by µ ( 5 ). The spin average requires attention. The neutrino mass is so tiny that we will consider that only one helicity state (left-handed) is possible. Hence, only the spins of incoming electron has to be averaged. Using equality 3.50 replacing µ! µ ( 5 ) and neglecting all masses, it comes: M = G F Tr /p µ ( 5 )/p 0 ( 5 ) Tr /k µ ( 5 )/k 0 ( 5 ) Let us evaluate the traces: Tr /p µ ( 5 )/p 0 ( 5 ) = Tr /p µ /p 0 Tr /p µ /p 0 5 Tr /p µ 5 /p 0 + Tr /p µ 5 /p 0 5 = p p 0 4(g µ g g g µ + g g µ ) p p 0 4i µ p p 0 4i µ where the following properties were used: 3.56, 3.60, the anticommutation of 5, and ( 5 ) =. Hence, using the cyclic permutation µ = µ = µ : Tr /p µ ( 5 )/p 0 ( 5 ) = 8(p µ p 0 p.p 0 g µ + p p 0µ + ip p 0 µ ) Tr /k µ ( 5 )/k 0 ( 5 ) = 8(k µ k 0 k.k 0 g µ + k k 0 µ + ik k 0 µ ) (6.4) Now we can make the product of the traces, denoted in what follows by Tr Tr. Notice that µ g µ =0sinceg µ 6= 0 only when µ = for which µ = 0. Tr Tr = 64( p.k p 0.k 0 +p.k 0 p 0.k +ip p 0 k µ k 0 µ + ip p 0 k kµ 0 µ + ip µ p 0 k k 0 µ + ip p 0µ k k 0 µ p p 0 k k 0 µ µ ) In the first term of the second line, we can change the dummy indices µ $. Doing so, we have µ = µ = µ = µ. Hence the first terms of the second line will cancel. And similarly with the third and fourth terms of the second line. Now, using properties 3.6: µ µ = µ µ = ( ) P.Paganini Ecole Polytechnique Physique des particules avancée

8 66 Weak interaction And thus: Tr Tr = 64 p.k p 0.k 0 +p.k 0 p 0.k +p p 0 k k 0 ( ) (6.5) = 56 p.k p 0.k 0 Because of 4-momentum conservation, p + k = p 0 + k 0 ) p.k = p 0.k 0 where the masses are neglected. The square of the center of mass energy is s =(p + k) ' p.k. Hence, the matrix element is simply reduced to: M = G F 56s 4 = 6 G F s (6.6) Now using formula.86 for the di erential cross-section and.54 for the momentum in the rest-frame ~p = ~ p 0 = s/: d d = ~ p 0 64 s M ~p = G F s 4 ) = G F s After this lengthy calculus, we see that there is clearly a problem: the cross section is proportional to the square of the center of mass energy! Hence, it is divergent for s!! In our jargon, we say that the unitarity is violated, meaning that the probability of observing this reaction is greater than! Clearly the theory cannot be used at high energy: it is an e ective theory, i.e. an approximation of a more general theory, only valid in the low energy regime. There is another (related) problem with the four-fermion contact interaction: it can be shown that it not renormalizable. Thus any calculation to higher order diverges. 6.3 A more modern weak interaction 6.3. The W boson In our modern interpretation of the interactions, the forces are mediated by vector bosons. Since in the decay, we have currents that are vector-like (for Lorentz transformation), the W boson responsible for the weak interaction, must be a spin boson as the photon. However, there is an obvious di erence: the weak current is charged! A charge is transferred between the neutrino and the electron for instance. Therefore, the W boson must be charged: The W and its charge conjugate W + are the intermediate vector bosons of the weak interaction with charged current. The decay now corresponds to the graph below: d u W e At each vertex, only left(right)-handed chirality for (anti)fermions is involved. It would be tempting to imagine that the W s are just charged photons. If it would be the case, we should be able to see this kind of Compton process producing W bosons as easily as photons: P.Paganini Ecole Polytechnique Physique des particules

9 The charged current of leptons 67 W e It is not the case. However, nothing forbids this process. Why don t we detect weak charged bosons as easily as photon in detectors? The logical explanation is that they simply have a large mass suppressing their production at low energy. A natural consequence of their mass is the short range of the weak interaction. Experimentally, the W ± was discovered at CERN in 983 in the UA experiment, and few days later by the UA experiment, both operating at the SPS, a proton-anti-proton collider with p s = 540 GeV (at that time). The channel p p! W X! e X (and charge conjugate) was used. Nowadays, the W property is well known and it mass has been measured with a good accuracy thanks to the Tevatron collider (near Chicago) and more recently to the LHC at CERN. Its mass is [9]: M W = ± 0.03 GeV/c (6.7) How can we change the photon propagator to accommodate for the W mass? It can be shown that it is [7, p. 50]: W propagator = i g µ + q µ q /M W q M W (6.8) which can be compared to the usual photon propagator ig µ /q.well,settingm W! 0does not recover the photon propagator. The reason is related to the fact that with a massive spin boson, the 3 polarizations are allowed contrary to the photon which has only polarizations. And this is the longitudinal polarization that is responsible for the term q µ q /MW The charged current of leptons Looking closely at the amplitude squared 6.6 which is dimensionless, we conclude that G F has the dimension GeV. It is not dimensionless as its analog in QED. In complete analogy with QED, let us define g 5 w the universal weak coupling constant, the analog of e in QED. The charged current for leptons is then straightforward: j µ e! e = g w µ p ū e ( 5 )u e (6.9) where for convenience, the factor p is used so that it will match the electroweak definition that we will see later. The reaction considered in the section 6..4, µ e! e µ is now described by the graph: 5 the w of g w stands for weak, not for W boson. P.Paganini Ecole Polytechnique Physique des particules avancée

10 68 Weak interaction µ (k) µ (k 0 ) for which the amplitude is: e (p) e (p 0 ) g µ +q µq /M W q M W im = j µ e! e i h i = p gw ū µ e ( 5 )u e i j µ!µ g µ +q µq /M W q M W h gw p ū µ ( 5 )u µ i (6.0) When q M W, we should recover the Fermi s e ective theory. In this approximation, the amplitude becomes: h i h i M = p gw ū µ e ( 5 gµ )u gw e MW p ū µ ( 5 )u µ ū e µ ( 5 )u e ūµ µ ( 5 )u µ = g w 8M W Comparing this expression to 6.3, we can immediately identify: G F p = g w 8M W (6.) Now, let us come back to the general case without assuming q M W. Neglecting the masses of the fermions, and applying the Dirac s equations in the momentum space to e and e gives: (/p m)u e ' /pu e =0, ū e ( /p 0 m) ' ū e /p 0 =0 Since q = p p 0,thetermq µ q due to the W propagator in 6.0 contracted with j µ e! e becomes: h i h i p gw ū µ e ( 5 )u e q µ q = p gw ū e (/p /p 0 ) ( 5 )u e q h gw i = p ū e /p ( 5 )u e q h i = p gw ū e ( + 5 )/pu e q =0 Hence, the amplitude is simply reduced to: h i h i M = p gw ū µ e ( 5 g )u µ gw e MW p q ū µ ( 5 )u µ = ū e µ ( 5 )u e ūµ µ ( 5 )u µ g w 8(M W q ) Compared to 6.3, we see that now, the amplitude does not diverge anymore thanks to the quadratic term /q. (The calculation of the cross-section is very similar to the previous case). If g w is universal, we expect by definition the same coupling for the 3 families of leptons e µ,, and the W e µ ± namely: P.Paganini Ecole Polytechnique Physique des particules

11 The charged current of leptons 69 e µ i gw p µ ( 5 ) W i gw p µ ( 5 ) W i gw p µ ( 5 ) W e µ The vertex factor has been simply deduced by analogy with QED. Indeed in QED, the current is j µ = qū µ u with a vertex factor iq µ. For weak interaction, j µ = p gw ū µ ( 5 )u and hence a vertex factor G F = p g w 8M W i gw p µ ( 5 ). Can we test that g w = g e w = g µ w = g w or equivalently = G e F = Gµ F = G F? Consider for example the muon decay µ! e e µ. One can show that [9, p. 63] (neglecting masses of the final state): = µ!e e µ = G F m5 µ µ 9 3 Assuming non universal constants, one would have found: = µ!e e µ = Ge F Gµ F m5 µ µ 9 3 Similarly for the decays:!e e = Ge F G F m5 9 3!µ µ = Gµ F G F m5 9 3 But the partial decay-width is related to the branching-ratio and the lifetime via:!e e =Br(! e e ) = Br(! e e ) Hence combining the muon and tau lifetime: µ = 9 3 G e F Gµ F m5 µ G e F G F m5 9 3 Br(! e e ) = G F G µ F m Now, these quantities are well known with a good accuracy [9]: m µ 5 Br(! e e ) µ =.97034() 0 6 s m µ = (4) GeV = (90.6 ±.0) 0 5 s m = ± 0.6 GeV Br(! e e ) = (7.8 ± 0.04)% so that: G F G µ F =.004 ± A value compatible with. A similar combination yields a ratio of coupling constants electron/muon compatible with as well. Hence, the weak coupling constant is universal for all leptons. P.Paganini Ecole Polytechnique Physique des particules avancée

12 70 Weak interaction The charged current of quarks The Cabibbo angle It seems logical to assume that the quarks current is identical to the lepton one, or in other words, that the coupling constants are the same. However, compare the values of the Fermi constant measured in decay and obtained from muon lifetime measurement: G F =.36(3) 0 5 GeV G µ F =.6637() 0 5 GeV There is a slight, but statistically significant, discrepancy between the two. In addition, in the early time of the weak interaction, there was another indication that the quark sector behaves di erently: the decay K +! µ + µ was observed: s µ u µ + meaning that a transition u! s was possible. Compared to the corresponding pion decay (so u d! µ + µ )with S = 0, the kaon decay S = is suppressed by a factor 0. Do we have to give up the universality of weak interaction? In 963, the physicist Cabibbo postulated that the strength G F of the weak interaction was shared between S = 0 transition and S =. The sharing was described by an angle c,thecabibbo angle, so that: S=0 / G F cos c and S= / G F sin c Empirically, the factor 0 required an angle c ' 3 or sin c ' 0.3. After the discovery of quarks, the reinterpretation of the Cabibbo angle was the following: the quark that couples to the u-quark in the weak interaction is a mixture of d and s quarks: d 0 i = cos c di +sin c si d 0 is the weak eigenstates whereas d and s are the mass eigenstates (the ones that propagate). This is the mass state that matters for the QCD interaction. In terms of vertex factor: ū ū i cos c g w p µ ( 5 ) W i sin c g w p µ ( 5 ) W d s The GIM mechanism In 970, the physics community was puzzled by the very small branching ratio of the decay K 0! µ + µ, about which was supposed to be described by a so-called box diagram: P.Paganini Ecole Polytechnique Physique des particules

13 The charged current of quarks 7 s µ sin c cos c u µ d µ + By symmetry with the leptonic sector where families were already known, Glashow, Iliopoulos and Maïani postulated the existence of a fourth quark, the charm so that another box diagram was possible with a quark c instead of a u in the box. Since the c quark can couple to the s or d quark, one can use the linear combination of d and s orthogonal to the d 0 to define the s 0 state so that: d 0 cos c sin s 0 = c d (6.) sin c cos c s and hence the W boson couples c quark to s 0 and u quark to d 0 : u d 0 c s 0 The u and d 0 quarks are respectively the counterpart of the e and e (and c, s 0 play the role of µ, µ ). The new diagram is thus: s µ in the leptonic sector cos c sin c c µ d µ + The amplitude of the first diagram is thus proportional to gw 4 cos sin while the second is proportional to gw 4 cos sin (the power 4 is coming from the 4 vertices including the muons ones). Hence, if the u and c quarks had the same mass, these diagrams would perfectly cancel (one has to add the amplitudes since we have the same initial/final states). Since m c m u,the overall contribution is simply strongly suppressed with respect to just considering the original diagram above. That s the so-called GIM mechanism, GIM standing for Glashow, Iliopoulos and Maïani. The three physicists predicted that the c quark had to have a mass between to 3 GeV to explain the measured rate of the K 0! µ + µ decay. Four years later, in 974, the charm was discovered as a constituent of the J/ meson made of a pair c c. Its mass is m c '.3 GeV! The CKM matrix Nowadays, 6 quarks have been discovered. The Cabibbo-Kobayashi-Maskawa (CKM) matrix, V CKM is an extension of GIM matrix 6. to the 3 families: 0 d V ud V us V ub s 0 A V cd V cs V cb sa (6.3) b 0 V td V ts V tb b P.Paganini Ecole Polytechnique Physique des particules avancée

14 7 Weak interaction where d 0,s 0,b 0 are the weak eigenstates, and d, s, b the mass eigenstates. The CKM matrix is a unitary matrix (V V = ) and hence the following equalities hold: X V ik = X i k V ik =, X k V ik V jk = X k V ki Vkj =0(i6= j) (6.4) The charge rising current with the 3 generations is then: 0 j µ cc+ = g w p (ū, c, t) µ d ( 5 )V sa (6.5) b with a V CKM element being for example with d! u: V ud. The charge lowering weak current is obtained with the hermitician conjugate: 0 j µ cc =(j µ cc+ ) = g w p ( d, s, b)v CKM µ u ( 5 ca (6.6) t with: 0 V CKM Vud Vus Vub Vcd Vcs Vcb This time the matrix element is for example with u! d: Vud. The vertex factor is then: Vtd Vts Vtb A d i u i i gw p µ ( 5 )V uj d i u j i gw p µ ( 5 )V u i d j d j W + where u i is an up-type quark (charge +/3) u i=,,3 = u, c, t and d i a down-type quark (charge -/3) d i=,,3 = d, s, b. What is the number of free parameters in the CKM unitary matrix? Any 3 3 unitary matrix can be decomposed as the product of 3 rotations (3 real parameters, the angles ij=,3,3 )plus one complex parameter interpreted as a phase. The common choice is then: V CKM 0 c 3 s 3 0 s 3 c 3 0 W c 3 0 s 3 e i 0 0 s 3 e i 0 c 3 0 c s 0 s c 0A 0 0 c c 3 s c 3 s 3 e i s c 3 c s 3 s 3 e i c c 3 s s 3 s 3 e i s 3 c 3 A (6.7) s s 3 c c 3 s 3 e i c s 3 s c 3 s 3 e i c 3 c 3 where c ij = cos ij and s ij =sin ij. If the third generation of quarks did not mix with the first one, the angle would be exactly c. The mixing is however very small, and hence ' c. These parameters have to be measured, they are not predicted by the theory. The CKM matrix P.Paganini Ecole Polytechnique Physique des particules

15 The neutral current 73 elements can be more precisely determined by a global fit that uses all available experimental measurements and imposes constraints as assuming three generations and the unitarity of the matrix (i.e. V ub + V cb + V tb = ) [9]: V exp CKM = (5) 0.53(7) 3.47(6) (7) (6) 4() (6) (.) (45) A (6.8) As expected, the dominant decay is always within a given family u $ d, c $ s and t $ b, the matrix being almost diagonal. Mixing between the first and third family is almost zero The neutral current In the mid-sixtees, Glashow, Salam and Weinberg developed a theory to unify the weak interaction and the electromagnetism. This theory, called electroweak will be described in the next chapter, but it predicted the existence of a neutral boson the Z 0 that mediates the weak interaction via neutral current (and hence no exchange of electric charge). Hence reactions of this kind were predicted: µ e! µ e, µ N! µ X (6.9) the former occurring via the diagram: µ µ Z 0 Notice that a W cannot be exchanged, this diagram: µ e e µ W e being forbidden because of a violation of lepton number (or in other words because the W couples only to leptons of the same family). In 973, the processes 6.9 were observed at CERN in the Gargamelle experiment (a big bubble chamber that can be seen now in the CERN garden!). The coupling to the Z 0 respects the lepton number, meaning that in: Z 0! f f the fermion f is necessarily produced with its anti-fermion. Such coupling Z 0! e µ inducing a Flavor Changing Neutral Current (FCNC) are thus forbidden (but actively looked for as a sign of new physics). Moreover, the measured cross section of: µ N! µ X P.Paganini Ecole Polytechnique Physique des particules avancée e

16 74 Weak interaction related at the parton level to reaction µ q! µ q was about a third larger than the corresponding pure charged current reaction: µ N! µ + X suggesting, as the electroweak theory predicts (see next chapter) that the structure of the neutral current is di erent than the one of the charged current. In fact, the experimental data ruled out apurev A structure [30], and justified a neutral current: j µ NC / q µ (c V c A 5 )q with c V 6= c A 6=, meaning that not only the left handed chirality state is involved in the neutral current but also the right handed chirality state! The Z 0 boson has been observed for the first time at CERN in 983, still in the UA and UA experiments few months after the discovery of the W ±. It was observed in its leptonic decay: p p! Z 0 X! l + l X where l = e or µ. Later on, it has been extensively studied at the LEP e + e is now very well known [9]: collider. Its mass M Z = ± 0.00 GeV/c (6.0) explaining again the weakness of the weak interaction. In the next chapter, more details will be given Strength of the weak interaction The very small value of G F ' GeV compared to =/37 explains why the weak interaction is qualified as weak at low energy (where G F is valid). However, injecting the W mass value 6.7 into 6. allows to estimate g w : g w = s 8M W G F p =0.65 ) w = g w 4 ' 9 Finally w is larger than! Intrinsically, the weak force is thus stronger than the electromagnetism. Its weakness, observed at low energy, is simply due to the large mass of the W boson. At high energy, the weak interaction becomes stronger than the electromagnetism. At LHC for instance, it is more likely to interact via the weak interaction than the electromagnetism Failure of the model There is a major problem with the weak theory described so far: it is not renormalizable! Such high order (O(g 4 w)) process µ µ! µ µ corresponding to the diagram: 6 At very large energy, the distinction between the interactions become irrelevant. See the electroweak model in the next chapter. P.Paganini Ecole Polytechnique Physique des particules

17 The neutrinos case 75 µ W + µ µ µ + µ W µ is divergent! Even if this process cannot be detected, it is a problem for the theory! In fact, one can show that this problem is related to the fact that the W or Z boson in the theory exposed in this chapter are not quanta of a gauge field. What is missing here is a gauge invariance to make the theory renormalizable. The next chapter is dedicated to this topic. 6.4 The neutrinos case 6.4. The PMNS matrix Until the 990s, the neutrinos were assumed to be massless. In these conditions, the solutions of the Dirac s equation simplify. Looking back to the coupled equations.33, setting m = 0 and being interested by the E = + ~p solution gives: ~p u a ~. ~p u b =0 ~. ~p u a ~p u b =0 where u a and u b are -components spinors of the wave function: ua (x) = e ip.x Let us define the following linear combination: u b u L = u a u b u R = u a + u b We see that they satisfy two independent equations: ~. ~p ~p u L = u L = u R ~. ~p ~p u R Since ~. ~p ~p is the helicity projector (equivalent to chirality in the massless assumption), we see that u L,u R are respectively the left-handed and right-handed spinors. Let us define the wave functions L/R(x) =u L/R e ip.x so that the original wave function (x) can be expressed as: (x) = ur +u L u R u L e ip.x = R(x)+ R(x) L (x) L(x) Since (x) satisfies the Dirac equation i µ (x) = 0, it comes with L/R (x): 0 R (x)+@ 0 L (x)+ i R (x) i L (x) =0 0 R (x)+@ 0 L (x) i R (x) i L (x) =0 P.Paganini Ecole Polytechnique Physique des particules avancée

18 76 Weak interaction which decouples into: 0 i ) L(x) = i ) R(x) = 0 These two equations known as the Weyl equations 7 describe independently the evolution of a massless left-handed particle and a massless right-handed particle. Thus, the massless lefthanded and the right-handed particles can be considered as two di erent particles. This applies to L and R. Since only L is sensitive to the interactions, there was no way (even in principle) to detect a R. Conclusion: the R did not have any physical meaning and was then absent from the Standard Model. However, in the late 990s, several experimental facts suggested that neutrinos are massive. As we will see in the next section, massive neutrinos lead to a phenomena called neutrino oscillations, where neutrinos can change their flavour while propagating. The very first hints of the oscillations came from a large deficit (50% to 70%) in the number of e coming from the sun (produced by the nuclear reaction 4p +e! 4 He + e ) with respect to the standard solar model predictions. It turns out that because of their oscillation, the fraction of e reaching the earth is only the third of the total flux of neutrinos (assuming 3 flavors). These last years, neutrino oscillations have been confirmed by observing directly the change of flavor of neutrinos coming from a nuclear reactor or neutrino beam. Such oscillations (i.e flavour changing) are explained by assuming that neutrinos eigenstates of the weak interaction di er from the mass eigenstates (the eigenstates of the free-propagation of the neutrinos). This scenario is then similar to the quark case of the previous section. After the quark mixing, we now have the neutrino mixing! However, there is an historical di erence: quarks were discovered first through their strong interaction. The usual labels u, d etc names the strong or equivalently the mass eigenstates 8. Only later, it was realized that weak interaction acts on other states. In the case of neutrinos, the reverse occurred: since neutrinos are only sensitive to the weak interaction and are not directly observed, the usual neutrinos e, µ, are by definition the ones associated to the charged leptons that are used to detect them: those neutrinos are thus weak eigenstates. Only recently, we consider that they di er from the mass eigenstates. But contrary to the quarks case, there is no way to detect directly the mass eigenstates. Anyway, as for the quarks, there is a unitary matrix similar to CKM, called PMNS for Pontecorvo-Maki-Nakagawa-Sakata that relates the mass eigenstates,, 3 to the weak eigenstates e, µ, : µ A = V 3 V e V e V e3 A V µ V µ V µ3 V V V 3 3 A (6.) with V PMNS having a similar parametrization 6.7 in terms of mixing angles, 3, 3 and a phase as the CKM matrix. However, the components of the neutrino matrix is known with a very poor accuracy. The phase is totally unconstrained and a crude estimation (combining 7 We would have found these equations more simply by using another representation of the Dirac matrices called the chiral representation. 8 The quarks cannot be seen as free particles, they always feel the strong interactions. Thus the propagation of free quarks (pure mass eigenstates) does not really make sense. Therefore, what is called mass eigenstate is by definition the eigenstate of the kinetic and strong interaction parts of the Hamiltonian. P.Paganini Ecole Polytechnique Physique des particules

19 Neutrino oscillations formalism 77 various sources of data) of the absolute value of the components of the matrix gives [7]: V PMNS A The di erence of this matrix with respect to CKM is striking: it is not diagonal at all, all elements having pretty large values 9. One may wonder why the mixing is applied to the neutrinos and not to the charged leptons which are considered with a definite mass. All charged leptons have the same quantum numbers, the only di erence being their masses. Thus, the flavor of a charged lepton is determined by its mass. However, the mass di erence between charged leptons is very large. Consequently, the mass measurement uncertainties is small enough to discriminate the 3 masses and moreover, the large di erence of masses yields very distinct experimental signature (very fast decay of the, Bremsstrahlung of the light e etc.). Therefore, there is no ambiguity in the assignment of the nature of the particle. Hence, charged leptons with a definite flavor are, by definition, particles with definite mass. On the other hand, neutrinos are never detected directly: they are identified via the charged leptons produced by the weak interaction which define the flavor of the neutrino. Therefore, flavor neutrinos are not required to have a definite mass and the mixing implies that they are superpositions of neutrinos with definite masses Neutrino oscillations formalism Let us denote by =e,µ, the neutrinos with a given flavor and k=,,3, the neutrinos mass eigenstates. The charged current transition producing a neutrino from a charged lepton l (and coupled to the W field, see equ. 6.9) can be expressed as function of the k : j µ l! = gw p u µ ( 5 )u l = gw p P 3 k= V ku k µ ( 5 )u l = gw p P 3 k= V k u k µ ( 5 )u l Consequently, the created neutrino with flavor is described by the state: i = 3X V k ki (6.) k= the states being normalized: h i =, h i k i = ik. The flavor state is expressed as a linear combination of the mass eigenstates. Let us consider that a neutrino k created in the space-time point (0,~0) has an energy E k and a momentum ~p with: The evolution of the state k in space-time is simply: E k = ~p + m k (6.3) k (x, t)i = e i(e kt ~p. ~x ) k i 9 Recently, thanks to neutrino oscillation experiments, the angle 3 has been measured with a good accuracy giving sin ( 3) =0.09 ± 0.07 [8]. P.Paganini Ecole Polytechnique Physique des particules avancée

20 78 Weak interaction k being an eigenstate of the free hamiltonian. Let us assume that the 3 mass eigenstates propagate in the same direction, for example ~p = p~e x where ~e x is a unit vector in the x-axis direction. Then, the state at a time t and a distance L along the x-axis is given by: (L, t)i = X k V k e i(e kt pl) k i (6.4) Now, because of equ. 6., we have: 0 0 e i µ ia = ia i 3 i and since V PMNS is a unitary matrix: V PMNS V PMNS =(V PMNS) t V PMNS =) (V PMNS ) t V PMNS =) (V PMNS) =(V PMNS ) t and hence: i V e V µ V e ia V e V µ V µ ia 3 i V e3 V µ3 V 3 i so that k i = P V k i. Conclusion, equ. 6.4 can be re-written: (L, t)i = X X k V k e i(e kt pl) V k i where only flavor neutrinos appear. The probability to observe the transition! is thus: P! (L, t) = h (L, t)i = P k V k e i(e kt pl) V k = P k,j V k V kv j V j e i(e kt pl) e i(e jt pl) For ultra-relativistic neutrinos, t L and E k E j = E p thus E k,j t pl =(E k,j p)l = E k,j p E k,j + p L = m k,j E L Hence we conclude: P! (L) = X k,j V k V kv j V j e i m kj E! L (6.5) where: m kj = m k m j (6.6) Clearly, P! (0) = since the unitarity of the PMNS matrix imposes P P k V k V k = j V jv j =. In other words, there is no change of flavor without propagation. For very short distances, close to the point of creation of the neutrino, the probability converges towards. In order to observe a change of flavor between the creation of neutrinos and their detection in an experiment located at L, three conditions have to be fulfilled: the corresponding PMNS matrix elements must not be 0, P.Paganini Ecole Polytechnique Physique des particules

21 Neutrino oscillations formalism 79 neutrinos must be massive. their masses must not be degenerated. It is convenient to introduce the oscillation length and decompose formula 6.5 as: P! (L) = X k L osc kj V k V k +Re X k>j = 4 E m kj V k V kv j V j exp! i L L osc kj (6.7) (6.8) The first term does not depend on L. Very far from the production point of the neutrino where L L osc kj, the probability will converge toward this constant value because of the finite resolution in the measurement of E and L [7]. Even if there is a transition from one flavor to another (mixing), it is not an oscillation. The second term depending on L is the oscillation: the probability of a transition from one flavor to another does depend on the distance (or equivalently on the time). Using standard units, the oscillation length reads: L osc E(GeV) kj (km) = 47 m kj (0 ev ) (6.9) Global constraints on various sources of data (solar neutrinos, atmospheric neutrinos etc.) give an estimation of m O(0 4 )ev and m 3 m 3 O(0 3 )ev. Typical neutrinos experiments trying to observe the neutrinos oscillations use either (anti-) neutrinos ( e ) from nuclear plants (with typical energy of the order of few MeV) or neutrinos from accelerators ( µ with few GeV produced by decay). Consequently, the oscillation length is necessarily large (macroscopic), ranging typically from one kilometer (Double-CHOOZ, Daya Bay, Reno experiments etc) to several hundreds of kilometers (TK, OPERA etc.). This is the typical distance between the source of neutrinos and their detection to be able to observe the oscillation phenomena. Notice that the sign of m 3 and m 3 is still unknown. The mass hierarchy could be m <m <m 3 (so called normal hierarchy) or m 3 <m <m (inverted hierarchy). Coming back to the oscillation formula 6.8, we see that when 6=, P! can be di erent from P!.Indeed, Re " V k V kv j V j exp!# " i L L osc 6= Re V k V kv j V j exp kj!# i L L osc kj as soon as the elements of the PMNS are complex. In addition, if we assume that the CPT symmetry is conserved by the weak interaction, the CPT transformation of the reaction! being! 0, we would expect P! = P! and thus a complex PMNS matrix implies P! 6= P! ) P! 6= P!. But the reaction! is just the CP transformation of! (remember the anti-neutrinos are only right handed and neutrinos left-handed). Conclusion, the complexity of the PMNS matrix implies that CP is violated in the neutrino sector. 0 The only interacting (anti-)neutrinos are left (right) handed so the CPT transformation of! L! L C =) L! L P =) R! R T =) R! R! P.Paganini Ecole Polytechnique Physique des particules avancée

22 80 Weak interaction Considerations on the measurement of neutrinos masses and the oscillations phenomena We have seen that the k neutrinos have by definition a definite mass. What about the? Consider the 4-momentum operator ˆP µ. By definition, we have: ˆP µ ˆPµ k i = m k ki And the average value of the mass squared is as expected h k ˆP µ ˆPµ k i = m k. Hence, according to equ. 6.: ˆP µ ˆPµ i = X m k V k ki k Showing that is not a mass eigenstate as announced. The average mass squared is however: hm i = h ˆP µ ˆPµ i = X k m k V k h k i = X k m k V k The same result would have been obtained by using the development of (x, t)i (from 6.4). On average, we would measure hm i which does not depend on t or L (and so on the distance of the detector), but event by event we would measure m k with a probability V k. The measurement supposes that the mass can be measured directly. Let us take for instance the pion decay +! µ + µ. Assuming an ideal detector, able to measure perfectly the 4-momentum of the µ +, according to formula.5: m k = m + m µ m qm µ + ~p µ where ~p µ is the muon momentum in the pion rest frame. If the 4-momentum of the pion is also perfectly measured, then one can determine ~p µ from the 4-momentum of the µ + measured in the lab. But, beware that the measurement forces the superposition to collapse on the massive neutrino whose mass has been measured. In other words, there is no more oscillation. Indeed after the collapse, the state is k (x, t)i = k i e i(e kt ~p. ~x ) and the probability to detect the state i (via weak interaction) is: P! k! = h k (x, t)i h k i = V k V k which does not depend anymore on t or L. As soon as the measurement is accurate enough to determine m k with an error less than m ij and thus determine unambiguously which k was involved, oscillation cannot exist anymore. What is the reason for the destruction of the oscillation pattern? It relies on the uncertainty principle. Coming back to our example of the pion decay, the more accurate the pion momentum is measured, the more uncertain the point where the neutrino has been created would be. Thus the uncertainty on L can even exceed the oscillation length L osc kj, and so washing out the oscillation pattern. A rigorous proof of this explanation, would require to use a more complicated formalism than the one we used based on plane-wave approximation. It is necessary to apply wave packets treatment which introduces the coherence length L coh [7] related to the size of the neutrinos wave-packets. When this size is larger that L osc kj, or when L is larger than Lcoh, the oscillation is suppressed. By the way, the wave-packet treatment eliminates idealizing assumptions like assuming that all k have the same momentum during the propagation. P.Paganini Ecole Polytechnique Physique des particules

23 Lepton number: conserved or not conserved? Lepton number: conserved or not conserved? As a conclusion for this section about neutrinos, let us give few considerations about individual lepton number conservation. Because of neutrinos oscillations, a neutrino produced with a given flavor can be detected with another flavor. Hence, there is clearly no conservation of the individual lepton number for neutrinos. But what about charged leptons? Consider for instance the following diagram describing the muon decay µ! e (the photon is here only to conserve the 4-momentum): W µ µ e Because of the neutrino mixing (symbolized by the cross), this reaction is possible, leading to muon and electron lepton numbers violation. However, the probability for such reaction is proportional to ( m /M W ) [9]. Since m. 0 3 ev, this decay is virtually impossible to observe: a branching ratio of the order of 0 56 is expected! Hence, even if strictly speaking the individual lepton number for charged lepton is also violated, it remains in all practical cases conserved. Beware that at the vertex level, the lepton number is always conserved. Such diagram does not exist in the standard model: e W µ The source of the individual lepton number violation must always be due to a neutrino oscillation. However, the total lepton number remains always conserved: if a lepton is present in the initial state (of any flavor), there must be a lepton in the final state (but not necessarily with the same flavor). e 6.5 What is violated by weak interaction? 6.5. Charge conjugation C We already know that the weak interaction violates the parity. What about the charge conjugation? Consider the charged pion decay:! µ µ Since the pion has a spin 0, the spin of µ and µ is opposite and since in the rest-frame of the pion, they are produced back-to-back, they necessarily have the same helicity. Considering the neutrino massless, its chirality state coincides to its helicity state and thus because of weak P.Paganini Ecole Polytechnique Physique des particules avancée

24 8 Weak interaction interaction property, µ is necessarily in the right-handed helicity (=chirality) state µr.therefore, µ is also in the right-handed helicity state µ R. Now let us transform every particles in its antiparticle (and vice-versa):! µ R µr charge conjugation! +! µ + R µ R Is this last reaction possible? No obviously since µ only interacts as a left-handed particle! Hence, weak interaction violates both P and C transformations CP transformation Continuing with the previous example, if we apply in addition a parity transformation:! µ R µr charge conjugation! +! µ + R µ R parity! +! µ + L µ L which is perfectly possible and gives a similar rate as the original reaction. So weak interaction seems to preserve the CP symmetry. Actually not, but only in rare cases. The first case where it was observed is in the K 0 K 0 system in 956. The K 0 =(d s), K 0 = ( ds) are very easily produced by strong interaction. For instance: K + + n! K 0 + p, K + p! K 0 + n Notice that the strangeness is conserved (strong interaction) since K + =(u s). In order to decay to lightest mesons (pions), they have to decay via the weak interaction, since this is the only interaction that does not conserve the quark flavour (and hence the strangeness). But the K 0 s, eigenstates of the strong interaction, have a peculiarity: they can transform spontaneously to their anti-particle and vice versa: K 0 $ K 0 thanks to weak couplings. We say that they oscillate via the two box diagrams: d W s d u; c; t s u; c; t u; c; t W + W s W + d s ū; c; t d Since K 0 and K 0 are permanently oscillating, they do not have a definitive lifetime, and the physical states that propagates are a linear combination of the two. Those states are eigenstates of the total hamiltonian which includes the strong hamiltonian and the weak hamiltonian (since the original K 0 are sensitive to both interactions). These states are called K L and K S where the L and S stand for Long and Short. This appellation emphasizes the large di erence between their (definite) lifetimes: KL ' s, KS ' 9 0 s The masses of the K L and K S are almost equal, about 498 MeV/c but di er however by only MeV/c. What is the relationship with CP violation? Consider the K 0 parity: K 0 = d s ( ) l = ()( )( ) 0 = ) P K 0 i = K 0 i P.Paganini Ecole Polytechnique Physique des particules