Particle Physics. Dr. Yazid Delenda. Department of Physics, Faculty of Sciences University of Batna - Algeria.

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1 UEF 01/013 Particle Physics Dr. Yazid Delenda Department of Physics, Faculty of Sciences University of Batna - Algeria. yazid.delenda@yahoo.com Lecture notes available at Contents I Foundations of particle physics 1 Classification of particles and forces Weak Interactions 8 3 Electroweak interactions in the Standard Model 33 4 Hadron Spectroscopy 38 II Gauge theories in particle physics and the standard model 58 1

2 5 Synopsis 58 6 Gauge field theory 59 7 Non-Abelian gauge invariance 64 Part I Foundations of particle physics 1 Classification of particles and forces 1.1 Elementary particles and fundamental forces Elementary particles: are point-like particles with zero-dimension and no spacial extensions or internal structure or excited states. Examples include: the electron, the photon. Note that the proton (p) and neutron (n) are not considered as elementary particles because they have an internal structure (made of quarks) and dimensions ( fm). There are three types of elementary particles: 1. Force carriers: also known as gauge bosons, of spin-0, spin-1 or spin-. For example: the photon is the electromagnetic force carrier.. Leptons: are spin-half (fermions) particles, for example the electron and the electron neutrino. 3. Quarks: are the constituents of (for example) the proton and neutron. They are fermions of fractional electric charge (1/3 or /3). Note that quarks cannot be observed as free particles. They are confined into hadrons (confinement property). Hadrons: are particles composed of quarks. Thus they are not elementary and they have dimensions (of order fm) and excited states. Examples of hadrons: the proton (p), the neutron (n) with spin-half, and the pions (π ± and π 0 ) with spin-1 or 0, and many others. Fundamental Forces There are four fundamental forces of nature: 1. Electromagnetic forces (EM): are responsible for the binding of electrons and protons in atoms, and they lead to the process of photon emission from atoms and nuclei.. Weak Interactions: are responsible for electron emission and neutrino emission (βdecay) from a nucleus - see nuclear physics course. 3. Strong interactions: are responsible for the binding of quarks in hadrons and are also responsible for the binding of protons and neutrons into the nuclei. 4. Gravity: is so weak that we ignore it (at microscopic level).

3 Summary There are four types of particles: three of them are elementary (leptons, quarks, gauge bosons) and one of them is composed (Hadrons). There are three fundamental interactions (strong, weak, EM) + gravity. 1. Leptons and conservation laws Leptons are elementary spin-half particles with no strong interactions. There are six known leptons which are grouped into three generations: ( ) ( ) ( ) νe νµ ντ, µ, τ so we have three charged leptons: the electron, the muon µ and the tauon τ, all of negative charge Q = e = C, and three neutrinos: electron neutrino ν e, muon neutrino ν µ and tauon neutrino ν τ which are electrically neutral and have very small mass (order ev/c ). Hence the neutrinos do not interact electromagnetically (only weak interactions). In addition there are six anti-particles: ( ) νe e +, ( νµ µ + ), ( ντ τ + e + is the anti-electron (also known as the positron), ν e is the anti-electron neutrino, µ + is the anti-muon, ν µ is the anti-muon neutrino, τ + is the anti-tauon and ν τ is the anti-tauon neutrino. ) 1..1 Conservation laws for leptons Lepton numbers: are conserved quantum numbers associated with each generation. They include the electron number L e, the muon number L µ and the tauon number L τ : L e = N( )+N(ν e ) N(e + ) N( ν e ) L µ = N(µ )+N(ν µ ) N(µ + ) N( ν µ ) L τ = N(τ )+N(ν τ ) N(τ + ) N( ν τ ), where N( ) is the number of electrons present in a state, etc. For example a state which has just an electron has L e = +1, L µ = L τ = 0. A state with just an anti-muon has L µ = 1 and L e = L τ = 0. Non-leptons (p, n, γ, π,...) have all lepton numbers equal to zero. Lepton numbers are conserved in all known interactions Examples In electromagnetism we have the vertices: but not the vertex: Consider the weak interaction: ν µ +n µ +p L µ =

4 µ µ and the vertex µ Here L µ = 1 in the initial state and L µ = 1 in the final state. Thus L µ is conserved and the reaction is readily observed with intense high energy ν beams. Examples of unobserved reactions: where both L e and L µ are violated violates L µ conservation. ν µ +n +p L µ = L e = ν µ +p µ + +n L µ = No lepton-number violating interactions have ever been observed despite extensive searches 1.. The first generation ( νe L e = 1 ) ( νe e + ) L e = 1 (Historically) the electron neutrinos are emitted in nuclear-β decays (via weak interactions): (Z,A) (Z +1,A)+ + ν e and in neutron decays: (Z +1,A) (Z,A)+e + +ν e n p+ + ν e 4

5 Neutrinos are usually not detected experimentally. Their existence is inferred by Pauli to satisfy the energy and angular momentum conservation. The neutrino mass is inferred from observed electron energies. The best results from Tritium decay: H 3 He ν e (pnn ppn + + ν e ) which gave an experimental bound on the neutrino mass: m νe < 3 ev/c m e (i.e. m νe m e ) and thus often assumed to be zero. Note that neutrino oscillations require that m ν 0. Neutrino detection we use the inverse-β decay: or ν e +n +p ν e +p e + +n to detect neutrinos emitted in β-decay. This experiment is typically very difficult because the cross-section for this reaction is very small. For solar neutrinos the energy of a neutrino is of order E ν 1 MeV, and this cross-section is σ m on protons. The mean-free-path of these neutrinos (l = 1/(σρ) light years in iron). However with massive detectors this can be done ( 5 interactions/sec in 1m 3 of Fe) 1..3 More generations The charged leptons have masses: ( νµ µ ), ( ντ τ L µ = 1 L τ = 1 L e = L τ = 0 L e = L µ = 0 m e = 0.511MeV/c m µ = 106MeV/c ) m τ = 1777MeV/c (1) and life times: τ e > s(stable) τ µ = s τ τ = s () 1..4 Electromagnetic interactions of leptons EM interactions are identical to those of the electron provided the masses are taken into account, for example the µ is much more penetrating than the electron because it is heavier. 5

6 1.3 Hadrons (Baryons and Mesons) Quarks Quarks are fundamental spin-half fermions. They have weak, strong and EM interactions (c.f. leptons have no strong interactions). Quarks are not observed as isolated free particles, instead they exist in multi-quark bound states, known as hadrons. There are six types of quarks (quark flavours) grouped in three generations: ( ) ( ) ( ) ( u c t +,,, with charges e ) 3 d s b 1e 3 they are respectively up (u) down (d) charm (c) strange (s) top (t) bottom (b), and there are six anti-quarks ( ) ( ) ( ) ( c,,, with charges 3 ū d s t b e ) + 1e 3 The masses of these quarks are inferred from properties of hadrons. Their approximate masses are respectively (in GeV/c ): ( ) ( ) ( ) ,, The topquark istoo heavy to formstable hadrons, andits mass isinferred fromits decay products. These masses cannot be observed directly because they are never isolated as free particles Hadrons Hadrons are composed particles (not elementary) with integer charge. They can either be fermions or bosons. There are more than 00 known types of hadrons and they exist in two types: Baryons (and anti-baryons): are qqq bound states. Examples include: Proton (p = uud) with mass 938 MeV/c and charge +1e and spin-1/. Neutron (n = udd) with mass 940 MeV/c and neutral charge and spin-1/. Anti-proton (ūū d) with same mass as proton and charg1e and spin-1/. Ω (Omega particle = sss) with electric charg1e and mass 167 MeV/c and spin 3/. and many others. Since there are three spin-half constituents in baryons they must be fermions. Mesons are q i q j bound states. Examples include: Pion (π + = u d) with mass 140 MeV/c and charge +1e and spin-0. Neutral-pion (π 0 = uū or d d) with mass 135 MeV/c and neutral charge and spin-0. B-zero (B 0 = d b) with mass 5.79 GeV/c and neutral charge and spin-0. Since Mesons are composed of two quarks they are bosons of spin-0 or spin-1. 6

7 1.3.3 Quark numbers There are six quark numbers related to quark contents of any physical state. Most importantly we have: Strangeness S = N s = [N(s) N( s)] Charm C = N c = [N(c) N( c)] Beauty B = N b = [ N(b) N( b) ] Truth T = N t = [N(t) N( t)] Baryon number B = 1 3 [N(q) N( q)] Other numbers N u = [N(u) N(ū)] and N d = [ N(d) N( d) ] In the baryon number N(q) is the total number of all quarks and N( q) is the total number of anti-quarks. For baryons the baryon number is +1, for anti-baryons the baryon number is -1 and for Mesons the baryon number is zero. The total hadronic electric charge for a hadron is: Q H = 3 [N u +N c ] 1 3 [N d +N s +N b ] In a given state the total electric charge is the sum of hadronic and leptonic charges: Q = Q L +Q H. Note that the truth quantum number = 0 for all known hadrons (because it decays quickly before forming a hadron) so we ignore it from now on. Both the strong and EM interactions conserve all quark numbers because a gluon or a photon can only produce a pair of particle-anti particle. However in weak interactions individual quark numbers are not conserved except the Baryon and total charge quantum numbers. Example: the n-decay: which effectively is: n p+ + ν e d u+ + ν e conserves all lepton numbers (as it should be) and conserves charge and Baryon number B. However this process violates the conservation of N u and N d (weak interaction) Strange particles Strange hadrons are those with S = (N(s) N( s)) 0 (i.e. contains strange particles). Lightest ones K + (u s) with mass 494 MeV/c and K 0 = d s with mass 498 MeV/c, both having strangeness S = +1; and their antiparticles K = sū (494 MeV/c ) and K 0 = s d (498 MeV/c ) with strangeness S = 1. 7

8 The lightest strange baryon is Λ 0 = uds with mass 1116 MeV/c with strangeness S = 1. Note that the strange particle is a bit heavier than the lightest quarks. Weak Interactions.1 Classification Like QED and QCD, the force carriers of weak interactions are spin-1 bosons that couple to quarks and leptons. Force carriers of weak interactions are three intermediate vector bosons: W + and W with mass 80 GeV/c, and Z 0 with mass 91. GeV/c. Since these gauge bosons are very massive the weak interaction is a very short range force(recall that range 1/M) of order 10 fm. Unlike the strong, electromagnetic and gravitational forces, the weak interaction is not responsible for any kind of bound systems (the bound systems for the aforementioned forces are successively: hadrons and nuclei, atoms and molecules, astronomical objects). The first dedicated experiment to study vector bosons is the SPS proton-anti proton collider at CERN: p+ p W + +X l + +ν l +X as shown in the figure p+ p W +X l + ν l +X p+ p Z 0 +X l +l + +X p p q q W ±,Z 0 From the quark point of view the above processes are essentially quark anti-quark annihilation processes: u d W +, d+ū W u+ū Z 0, d+ d Z 0 Hence the signature of a W boson is a lepton with large momentum emitted at wide angles with large missing transverse momentum carried out by the neutrino. If the weak boson has zero transverse momentum then by momentum conservation the missing transverse momentum must equal to the transverse momentum of the lepton. The signature of the Z 0 boson is a pair of leptons in the final state with large transverse momentum. Therefore the mass of the Z 0 boson is the invariant mass of the leptons. The weak interactions are classified in two types: Charged current reactions: Before the electroweak theory was formulated all observed weak processes were charged current reactions (such as β decay) mediated by W + or W bosons. 8

9 Neutral current reactions: The electroweak theory predicted the neutral current reactions caused by the Z 0 boson. The charged current reactions can be classified into three types: Purely leptonic process e.g. Purely hadronic process e.g. semi-leptonic process e.g. µ + ν e +ν µ Λ π +p n p+ + ν e Purely leptonic processes Recall that the electromagnetic interactions can be built from the basic interactions: γ γ e + e + which give the following basic vertices: e + e + γ e + γ γ γ γ e + e + γ e + In a similar way, leptonic weak interaction processes can be built from a certain number of reactions corresponding to basic vertices: 9

10 ν l l ν l l + W ± W ± For example, from the left-hand vertex we can derive the following vertices: ν l W l + W + l + l W + ν l ν l W l ν l l W ν l ν l W + l + Similarly eight other diagrams can be built by replacing particles with anti-particles. Weak interactions always conserve lepton quantum numbers Diagram-wise this conservation is guaranteed by the fact that at each vertex, there is one arrow pointing in and one pointing out. The arrow pointing-in in the initial state always represents a particle (l or ν l ) and the arrow pointing out in the initial state represents an anti-particle (l + or ν l ). In the final state the situation is the opposite. Note that processes shown above are virtual, so that two or more have to be combined to conserve energy. However processes like l + +ν l W + and W l + ν l do not violate energy conservation if M W > M l +M(ν l ). The leptonic vertices are characterized by the corresponding strength parameter α W independently on lepton type involved. The strength of the weak interaction is comparable with the e.m. one analogues to electron-electron scattering by photon exchange. e.g. muon decay: ν µ µ W ν e 10

11 Since W bosons are very heavy, interactions like the above can be approximated by a zero-range interaction: ν µ µ G F ν e Taking into account spin effects, the relation between α W and G F in zero-range approximation is: G F = g W 8M W = 4πα W 8M W where g W is the coupling constant in W-vertices, by definition. Purely Hadronic and semi-leptonic weak interactions Constituent quarks emit or absorb W bosons: From the Lepton-quark symmetry the corresponding generations of quarks and leptons have identical weak interactions: ( ) ( ) νe u,etc (3) d For example the neutron beta decay is: g W ν e n d d u g ud W u u d p The basic vertices are: u d ū ū c s c s g ud g ud g cs g cs W ± W ± W ± W ± 11

12 The corresponding coupling constants do not change upon exchange of quarks/leptons: g ud = g cs = g W For example, an allowed reaction is: π µ + ν µ (dū µ + ν µ ) However, some reactions do not comply with the lepton-quark symmetry: K µ + ν µ (sū µ + ν µ ) Example: The dominant quark diagram for Λ decay is: dū π Λ s u d W d uu p To solve the contradiction, the quark mixing hypothesis was introduced by Cabibbo. We shall return to quark mixing towards the end of this course.. Beta decay: Fermi theory The weak interaction is responsible for the β decay of atomic nuclei, which involves the transformation of a proton to a neutron or vice versa, and which allows nucleus to reach most stable p/n ratio. Examples: 10 C 10 B +e + +ν e 14 O 14 N +e + +ν e where one proton converts into a neutron from Carbon 10 nucleus to Boron 10 nucleus, or from Oxygen 14 nucleus to Nitrogen 14 nucleus (β + decay). The atomic number remains the same but the charge of the nucleus is reduced: p n+e + +ν e whichforfreeprotonisforbiddenbyparticlemasses 1,thusenergyinputisneededinthesereactions. However the crossed reaction: n p+ + ν e is energetically allowed and is the reason for instability of neutrons (β decay). 1 m p = 938 MeV/c and m n = 940 MeV/c, neutron being heavier. 1

13 The reaction can also happen by electron capture (a.k.a. K-capture) where a proton captures an electron from the K-shell and converts into a neutron and electron neutrino: p+ n+ν e Historically the existence of the neutrino was inferred by conservation of momentum, since the emitted β particle had a spectrum of energies. Problem: Using the relativistic energy-momentum relation, show that in the decay of a particle of mass M at rest into two-particles of masses m 1 and m, the two particles must have a definite energy. Pauli therefore proposed in 1930 that an additional particle should be emitted besides the electron in the decay of the neutron. This particle, nowadays called the anti-neutrino should carry no electric charge and have a low mass. Fermi s explanation of β-decay (193) was inspired by the structure of the electromagnetic interaction. Recall that the invariant amplitude for electromagnetic electron-proton scattering: γ p p is M = (ieū p γ µ u p ) igµν q ( ieū e γ ν u e ) = i(eū p γ µ u p ) 1 q ( eū eγ µ u e ) where we have treated the proton as a structure-less Dirac particle with charge +e and that of the electron is e. M is the product of the electron and proton electromagnetic currents, together with the propagator of the exchanged photon. We define an electron and proton electromagnetic current of the form: ejµ e = eū eγ µ u e, ejµ p = +eū pγ µ u p in which case the matrix element becomes: im = e q jp µ jµ e By analogy with the current-current form, Fermi proposed that the invariant amplitude for β-decay be given by: G F (ū n γ µ u p )(ū νe γ µ u e ) where G F is the weak coupling constant which remains to be determined by experiment; G F is called the Fermi constant. 13

14 p n ν e Note the charge-raising or charge-lowering structure of the weak current. We speak of these as the charged weak currents. However there is one point which was not foreseen by Fermi, that the neutrinos are lefthanded particles and violate parity, so the Lagrangian must not allow right-handed neutrinos to be created, only left handed ones are allowed. To account for this we just make the replacement γ µ γ µ (1 γ 5 )/ (see next section). We also add the factor 4/ to G F (to keep the original definition of G F which did not include γ 5 ). The matrix element for beta-decay becomes: im = G F (ū n γ µ (1 γ 5 )u p )(ū νe γ µ (1 γ 5 )u e ) which can be written in the form: im = 4G F j µ eνj pn µ with the charge-raising weak current: j eν µ = ū ν e γ µ1 γ 5 u e This is charge-raising because in the sense of Feynmann diagrams the transition ν e involves raising the charge by one unit (from -1 to 0), while the charge-lowering current: involves lowering the charge by one unit. j pn µ = ū 1 γ 5 nγ µ To estimate G F we compute the transition amplitude: u p T fi = i(π) 4 δ 4 (p p p n p p ν )M which can be calculated by making the assumption that the low energy interaction enables us to use non-relativistic spinors in which γ 1 ; γ ; and γ 3 do not contribute. Note that we have made the assumption that the proton and neutron are structure-less and we ignored strong interactions. This is feasible since the beta decay is a very low energy reaction (point-like interaction) which also means strong interactions are also ignorable here. Summing over spin we obtain the energy spectrum of the emitted positron: dγ dp e = G F π 3 p e (E E 0 ) γ µ alone in the Lagrangian does not violate parity in QED, however the term γ µ (1 γ 5 )/ does. Recall γ 5 = iγ 0 γ 1 γ γ 3. 14

15 where E 0 is the energy released to the lepton pair, E 0 = E ν +E e. Thus, if from the observed positron spectrum we plot 1 p e dγ dp e as a function of E e we should obtain a linear plot with end point E 0. This is called the Kurie plot. It can be used to check whether the neutrino mass is indeed zero. A non-vanishing neutrino mass destroys the linear behavior, particularly for E e near E 0 1 dγ p e dp e E 0 E e µ ν 0 We can crudely integrate the differential rate above, putting p e = E e, i.e. neglecting the electron mass. We also neglect corrections due to the atomic electrons, which distort the electron wave function. In this way we obtain Γ = 1 τ = G F E5 0 30π 3 We can thus extract the value of G F from experiment, G F = 10 5 m p This value can be obtained from the measured β-decay lifetimes of several different nuclei. Experimental evidence thus supports the existence of the neutrino and the validity of the effective Fermi interaction. Including the above mentioned corrections, as well as radiative corrections, and using the so called super allowed Fermi transitions the very precise value below is obtained: G F (βdecay) = GeV which has dimensions of inverse-mass. Note that the value for G F from muon decay is slightly different (1.136)..3 Inverse-beta decay: neutrino interactions The inverse-β decay process is one in which a neutrino maybe detected by being absorbed by an atomic nucleus: ν e +p n+e + ν e +n p+ However this reaction requires a very large detector in order to detect a significant number of neutrinos because of the small cross-section of this interaction. Antineutrinos were first detected in the 1950s near a nuclear reactor. Two scintillation detectors were placed next to the cadmium targets. Antineutrinos with an energy above the threshold of 15

16 1.8 MeV caused charged current interactions with the protons in the water, producing positrons and neutrons. The resulting positron annihilation with electrons in the detector material created photons with an energy of about 0.5 MeV. Pairs of photons in coincidence could be detected by the two scintillation detectors above and below the target. The neutrons were captured by cadmium nuclei resulting in gamma rays of about 8 MeV that were detected a few microseconds after the photons from a positron annihilation event. Many experiments are established to detect neutrinos from various sources, such as solar neutrinos, atmospheric neutrinos, etc. Neutrino detectors are often built underground in order to isolate the detector from cosmic rays and other background radiation..4 Neutrino helicity Consider the Dirac equation for free neutrinos with energy and momentum p µ = (E, p), (assuming them to be massless for now). We look for solutions of the standard form: ( ) χ ψ = i(et p. x) (4) φ where χ and φ are -component spinors which only depend on momentum. Substituting in the Dirac equation: i ψ t = ( i α. +βm)ψ (5) recall that γ µ = β(1, α), with ( ) 0 σ α = (6) σ 0 in the Dirac representation and σ are the pauli matrices. We find the coupled equations: ( ) ( )( ) χ 0 σ. p χ E = φ σ. p 0 φ so χ = σ. p E φ, φ = σ. p E χ (8) These equations can be decoupled by defining the component spinors: (7) ν R = χ+φ, ν L = χ φ which leads to the decoupled equations: ( σ.ˆp)ν R = +ν R, ( σ.ˆp)ν L = ν L (9) where ˆp = p/e = p/p (massless particle) is the unit vector along the momentum axis of the particle. The operator: H = 1 σ.ˆp 16

17 is known as the helicity operator, that is, the helicity operator is the projection of the spin of a particle onto its momentum axis. We therefore see that ν L is an eigenfunction of H with eigenvalu1/, so it corresponds to a solution to the Dirac equation with negative helicity, while ν R corresponds to positive helicity. In other words ν L describes a left-handed neutrino while ν R describes a right-handed one. The operator H commutes with the Hamiltonian and therefore the helicity is a good quantum number. It is not, however a Lorentz-invariant quantity for a massive particle. If a particle of given helicity moves with a velocity β < 1 we can overtake it and find its helicity flipped around. Note also that under the parity operator (spacial inversion), it may easily be shown that these components transform to one another. The operators O ± = (1±γ 5 )/, where γ 5 = iγ 0 γ 1 γ γ 3, are projection operators. They satisfy the following relations: O ± = O ± In the Dirac representation we have: O + O = O O + = 0 γ 5 = where each bloc is a dimensional matrix. ( The result of application of these operators on the wave function ψ yields O ψ = ψ L O + ψ = ψ R where ψ L has a negative helicity and ψ R has a positive helicity. A particle with negative helicity has the spin antiparallel to the direction of motion and is called a left-handed particle. Similarly ψ R has positive helicity and corresponds to a right-handed particle. An experiment on the weak decay of Cobalt-60 nuclei carried out by Chien-Shiung Wu and collaborators in 1957 demonstrated that parity is not a symmetry of the universe. The experiment studied r-transitions of polarized cobalt nuclei: ) 60 Co 60 Ni + + ν e Thenuclear spinsinasampleof 60 Cowerealignedbyanexternalmagneticfield, andanasymmetry in the direction of the emitted electrons was observed. The asymmetry was found to change sign upon reversal of the magnetic field such that electrons prefer to be emitted in a direction opposite to that of the nuclear spin. The essence of the argument is sketched in the figure. The observed correlation between the nuclear spin and the electron momentum is explained if the required J z = 1 is formed by a right-handed anti-electron neutrino, ν R, and a left-handed electron, L. The cumulative evidence of many experiments is that indeed only ν R (and ν L ) are involved in weak interactions. The absence of the mirror image states, ν R and ν L, is a clear violation of parity invariance. Also, charge conjugation (C) invariance is violated, since C transforms a ν L state into a ν L state. However, the γ µ (1 γ 5 ) form leaves the weak interaction invariant under the combined CP operation. (10) 17

18 ν e z-axis + J z = 5 J z = 4 J z = 1 {}}{ 60 Co 60 Ni + + ν e For example: Γ(π + µ + +ν L ) Γ(π + µ + +ν R ) = 0, Γ(π + µ + +ν L ) Γ(π µ + ν L ) = 0, but: Γ(π + µ + +ν L ) = Γ(π µ + ν R ), In this example, ν denotes a muon neutrino. P violation C violation CP invariance.5 V-A interaction How do we build a theory of weak interactions with parity violation? The most general form of the matrix element we can write is: M (ū ψf Ôu ψi )(ū φ,f Ôu φi ) (11) where Ô is a combination of γ matrices. It turns out that there are only 5 bilinear covariant expressions that can be formed by the gamma matrices: Name Symbol Current Number of components Effect under parity Scalar S ψψ 1 + Vector V ψγ µ ψ 4 (+,,, ) Tensor T ψσ µν ψ 6 Axial-vector A ψγ µ γ 5 ψ 4 (+,+,+,+) Pseudo-Scalar P ψγ5 ψ 1 where σ µν = i(γ µ γ ν γνγ µ )/. Youcanshow, forexample, thatthevector current ψγ µ ψ transformsundertheparityoperation: P = on the coordinates as shown in the table. 18

19 A series of experiments through the end of the 1950 s lead to a new form of the effective weak interaction: M β = G F [ū n γ µ (1 γ 5 )u p ][ū νe γ µ (1 γ 5 )u e ] The factor 1/, introduced for historical reasons, maintains the value of the Fermi constant G F. The ūγ µ u and ūγ µ γ 5 u transform, under Lorentz transformations of the coordinates, respectively as a vector (V) and an axial vector (A) 3 : ψγ µ ψ Λ µ ν ψγ ν ψ, vector transformation from which the name V-A. ψγ µ γ 5 ψ det(λ)λ µ ν ψγ ν γ 5 ψ, Axial vector transformation Parity violation comes from the fact that the behaviour of the vector and axial vector currents under a parity transformation are different. As you can see from the table, the vector current flips sign under parity whereas the axial vector does not. The interference between these two terms creates the parity violation. One can see this schematically by remembering that what we observe is usually the square of the amplitude. Suppose the amplitude is pure V-A. Then: M (V A)(V A) = VV +AA AV If we apply a parity transformation then the sign of the V term flips, but the sign of the A term does not: P{ M } ( V)( V)+AA A( V) = VV +AA+AV so we see that the cross-term has flipped sign. The V-A interaction actually violates parity maximally as both currents have the same strength. Parity is not just violated in a small percentage of interactions, it is violated in all of them. The form of the interaction suggests that we put it in the form of a current-current interaction in analogy with electromagnetism. We write the effective lagrangian as: where where for instance: The corresponding matrix element: L = G F J µ + (x)jµ+ (x)+h.c. J + µ = ( ν e ) µ +( pn) µ ( ν e ) µ = ψ(ν e )γ µ (1 γ 5 )ψ( ) ( ν e ) µ = ū νe γ µ (1 γ 5 )u e The superscript + reminds us that the current is a charge raising current, corresponding to the transitions n p and ν in beta decay. The two currents are taken at the same space time point x. 3 Thesearetermedsobecauseafour-vectorunderparitygetsitsspacialcomponentsreversed(suchasmomentum) while axial vector (or pseudovector) is unchanged, such as angular momentum. Under parity these components also change as described. 19

20 The presence of the factor 1 γ 5 in the current requires that all fermions participating inaweak process be left-handed and all anti-fermions be right-handed. For neutrino which are massless we expect neutrino to always have negative helicity and anti-neutrino to have positive helicity. This does not preclude the possibility of the existence of a neutrino with right-handed helicity. It can be shown, however, that the probability of generating a neutrino with right-handed helicity is proportional to (m ν /E ν ) and is therefore almost impossible. We know that the mass of the neutrino is of order of a few ev. For a neutrino with energy of, say, 10 MeV the probability of emitting a wrong sign neutrino is around The V-A form of the weak interaction has been verified experimentally both for neutrinos in β-decay, which we call electron-neutrinos or ν e as well as for neutrinos form π µν decays, called muon-neutrinos or ν µ. These experimental results have greatly contributed to establishing the V-A interaction..6 Weak-currents and allowed transitions The leptonic contribution ū(ν e )γ µ (1 γ 5 )u(e), contains terms that resemble the electromagnetic current j µ (x) = ψ(x)γ µ ψ(x). By analogy with the electromagnetic current, we therefore introduce the weak leptonic current: j α (x) = ū(ν e )γ α (1 γ 5 )u(e)+ū(ν µ )γ α (1 γ 5 )u(µ)+ū(ν τ )γ α (1 γ 5 )u(τ) = j α (e) +jα (µ) +jα (τ). To describe the mutual weak interaction of leptons we postulate that each leptonic hierarchy interacts with itself as well as with each of the other two. The following diagrams are some examples for such possible processes: Neutrino-electron scattering: ν e ν e j (e) α j α (e) = [ū e γ α (1 γ 5 )u νe ][ū νe γ α (1 γ 5 )u e ], Muon decay: ν µ µ ν e 0

21 j (e) α jα (µ) = [ū eγ α (1 γ 5 )u νe ][ū νµ γ α (1 γ 5 )u µ ], and muon production in muon-neutrino-electron scattering: ν µ µ ν e On the other hand, a process like: j (µ) α j α (e) = [ū µγ α (1 γ 5 )u νµ ][ū νe γ α (1 γ 5 )u e ]. ν µ ν µ is not allowed. This means that ν µ and can interact only via the creation of a muon, which is an immediate consequence of the specific form of the currents j µ (i), allowing for a neutrino converting into a charged lepton (or vice versa!), but prohibiting an interaction without a conversion of particles. This property of the interaction is usually expressed by calling the currents by charged currents (more accurate by charged transition currents) since the charge of the particle of a particular leptonic hierarchy changes by one unit. In the electromagnetic current the charge of the particle does not change, it is therefore called a neutral current. We shall later see that neutral currents also appear in the context of the gauge theory of weak interaction..7 Weak bosons and Fermi couplings We now know that the weak interaction is mediated by two massive gauge bosons: the charged W ± and the neutral Z 0. The propagator term for the massive boson is: 1 M W,Z q where q is the square of the 4-momentum. If we assume that the Fermi theory is the low energy limit of the weak interaction, then we can estimate the intrinsic coupling at high energy. In the Fermi limit, the coupling factor appears to be G F /. At low energies, with MW,Z q, the propagator term reduces to just 1/MW and we can make the identification: G F = g w 8M W 1

22 g W g W G F / Coupling G F 1 1 MW q MW Coupling g W 8M W This allows us to compare the intrinsic couplings of the weak interaction with the electromagnetic interaction. Experimentally the mass of the W boson is 80.4 GeV and the Fermi constant is GeV. We get a weak coupling factor of g w = Now, remember that the electromagnetic interaction coupling factor is the square root of the fine structure constant, we have: EM coupling: α EM = 1 137, Weak coupling: α W = g W 4π = The factor 1/8 comes from two 1/ factors from the insertion of (1 γ 5 )/ (projector operators) and two 1/ factors coming from the original definition of g w. In fact the weak interaction is, intrinsically, about 4 times stronger than the electromagnetic interaction. What makes the interaction so weak is the large mass of the relevant gauge bosons. In fact at very high energies, where q MW, the weak interaction is comparable in strength to the electromagnetic interaction. At high energies the mass of the W-boson suppresses the total cross section and stops it going to infinity..8 Muon and Pion decay.8.1 Muon decay The muon decay: µ (p) (p )+ ν e (k )+ν µ (k), is the model reaction for weak decays. The particle four-momenta are defined in the above equation. According to Feynman rules, the Feynman diagram must be drawn using only particle lines; and so the outgoing ν e is equivalent to an incoming ν e. The invariant amplitude for muon decay: ν µ (k) µ (p) G F ν e (k ) (p ) is: M = G ][ ] F [ū s ν µ (k)γ α (1 γ 5 )u S µ(p) ū S e (p )γ α (1 γ 5 )u s ν e (k ),

23 where s,s,s,s are the spins of the corresponding particles. The matrix element squared is: M = G F ][ ] [ū sνµ (k)γ α (1 γ 5 )u Sµ (p) ū S µ (p)γ β(1 γ 5 )u s ν µ (k) [ ][ ] ū S e (p )γ α (1 γ 5 )u s ν e (k ) ū s ν e (k )γ β (1 γ 5 )u S e (p ). Performing the summation over final-state spins and averaging over initial-state spins to obtain: we use the fact that: M = 1 M, s,s,s,s u s (p)ū s (p) = p+m, where m is the mass of the corresponding particle, we arrive at: s M = 1 G F Tr[γ α(1 γ 5 )( p+m µ )γ β (1 γ 5 ) k]tr [ γ α (1 γ 5 ) k γ β (1 γ 5 )( p +m e ) ]. Exercise: Prove the following relations: and Tr[γ α (1 γ 5 ) pγ β (1 γ 5 ) k]tr [ γ α (1 γ 5 ) k γ β (1 γ 5 ) p ] = 56(p.k )(k.p ), Hence we may write: Tr[γ α (1 γ 5 )γ β (1 γ 5 ) k] = Tr [ γ α (1 γ 5 )γ β (1 γ 5 ) p ] = 0. M = 64G F (p.k )(k.p ). The muon decay rate can now be obtained using: dγ = 1 E M dq, where E is the energy of the incoming muon, and the invariant phase space of the decay products is dq = d3 p d 3 k d 3 k δ (4) (p p k k ), (π) 3 E (π) 3 ω(π) 3 ω (π)4 over which we shall integrate. Using the easily-derived relation: d 3 k ω = d 4 kθ(ω)δ(k ). we can turn the 3-momentum integration over k into 4-momentum integration. We thus have: dγ = 3G F (π) 5 E (p.k )(k.p ) d3 p E d 3 k ω d4 kθ(ω)δ(k )δ (4) (p p k k ). 3

24 Performing the integration over d 4 k using the Dirac-δ (4) function forces k = p p k in the integrand (which means ω = E E ω ), with k 0 = ω, p 0 = E, k 0 = ω and k 0 = ω. Thus: dγ = G F π 5 E (p.k )((p p k ).p ) d3 p E d 3 k ω Θ(E E ω )δ((p p k ) ). Now since m µ > 00m e, we can safely neglect the mass of the electron. One can show that in the muon rest frame, where p = (m µ,0,0,0): (p.k )((p p k ).p ) = (p.k )(p k ).p = (p.k )(p k ) / = m µ ω (m µ m µω )/, where we treat all particles massless except the muon. In the third equality we used the relation for the massless neutrino: k = p k p k = 0 = (p k p ) = (p k ) (p k ).p, and we used p = 0 (massless electron). Thus: (p k ).p = (p k ) / = m µ/ m µ ω /. The decay rate in the muon rest frame is (E = m µ ): dγ = G F π 5 m µ d 3 p E d 3 k ω m µω (m µ m µ ω )Θ(m µ E ω )δ(m µ m µ E m µ ω +E ω (1 cosθ)), where θ is the angle between p and k. We have the freedom to fix the polar angle of say p and multiply by a factor to account for the averaging π 0 sinθ =, and so we can express the phase space d 3 p d 3 k by: 4πE de πω dω dcosθ. where we integrated over the azimuthal angles (giving π factors). Thus we have: dγ = G F π 3 E de ω dω dcosθω (m µ m µω )Θ(m µ E ω ) 1 ( m µ m µ E m µ ω E ω δ E ω where we used the relation: δ(m µ m µe m µ ω +E ω (1 cosθ)) = 1 E ω δ ( m µ m µ E m µ ω E ω ) +(1 cosθ), ) +(1 cosθ), in order to performthe integration over the opening angleθ between the emitted and ν e. Hence: dγ = G F dω dcosθω (m π 3dE µ m µ ω )Θ(m µ E ω ) ( m µ m µ E m µ ω δ E ω ) +(1 cosθ). 4

25 The integration over θ is straightforward (gives 1) forcing cosθ = 1+(m µ m µe m µ ω )/E ω, but sinc1 < cosθ < 1 then: 1 < 1+(m µ m µe m µ ω )/E ω < 1. Thus we obtain the following constraints on E and ω together with the step function above: 0 E + ω 1 m µ m µ E ω, m µ m µ E +ω m µ, where we respectively subtracted 1, then multiplied by E ω /m µ in the first inequality. We can rewrite these constraints as: E + ω E ω 1 m µ m µ m µ m µ, 1 m µ E +ω m µ. From the first inequality we may write: ) E (1 ω 1 (1 ω Now we either have m µ m µ m µ ). 1 ω m µ 0 thus E 1 m µ, which means ω m µ / and E m µ /, which implies ω +E m µ, which contradicts the second inequality in the above; or we have: ω m µ thus E 1 m µ, so we just take the last inequality. Hence the above coupled inequality is equivalent to: 0 E 1 m µ, 1 m µ E ω 1 m µ. We are now in position to perform the energy integration over ω to obtain the energy spectrum of the emitted electron: dγ de = G F π 3m µ mµ ω (m µ ω )dω = G F E mµ Hence we finally arrive at: 8π 3m4 µ 1 x(1 x)dx = G F 1 E /m µ 48π µ[x (3 x)] 1 3m4 1 E /m µ. ) dγ = G F de 1π µe (3 4 E. 3m m µ 5

26 This prediction is in excellent agreement with the observed electron spectrum. Finally, we calculate the muon decay rate: Γ = 1 τ = mµ/ 0 de dγ de = G F m5 µ 19π 3. Inserting the measured muon lifetime τ = sec, we can calculate the Fermi coupling G F. We find: G F = 10 5 /m N. Comparison of the values of G F obtained in muon decay and beta decay supports the assertion that the weak coupling constant is the same for leptons and nucleons, and hence universal. It means that nuclear β-decay and the decay of the muon have the same physical origin. Indeed, when all corrections are taken into account, G β and G µ are found to be equal to within a few percent: G µ = (1.1663±0.0000) 10 5 GeV, G β = (1.136±0.003) 10 5 GeV. (1) The reason for the small difference is quark mixing..8. Pion decay Can we now also understand the lifetime of the π ± -Mesons? To be specific, we take the decay: π (q) µ (p)+ ν µ (k), which is depicted as follows: π (q) { d ū q W ν µ (k) k p qf π µ (p) The amplitude for the pion decay is of the form: M = G F (...) α ū µ (p)γ α (1 γ 5 )u νµ (k), where (...) α represents the weak quark current. It is tempting to write it as ū u γ α (1 γ 5 )u d, but this is incorrect since the ū, d quarks are not free states (so cannot be described by Dirac free spinors), but are quarks bound into a π meson. We know, however, that M is Lorentz invariant, so that (...) α must be a vector or axial-vector, as indicated. Also The π-meson is spin-less, so that 6

27 q α is the only four-vector available to construct the quark current (no spin structure and therefore no γ matrices. We therefore have: (...) α = q α f(q ) q α f π, where f is a function of q since it is the only Lorentz scalar that can be formed from q, but q = m π and f(m π ) f π, is a constant. The amplitude becomes: M = G F (p α +k α )f π [ū µ (p)γ α (1 γ 5 )u νµ (k)] = G F f π m µ [ū µ (p)(1 γ 5 )u νµ (k)]. Here, we have used ku νµ (k) = 0 and ū µ (p)( p m µ ) = 0, the Dirac equations for the neutrino and muon, respectively. The invariant amplitude-squared is: M = G F f πm µ [ū µ (p)(1 γ 5 )u νµ (k)][ū νµ (k)(1+γ 5 )u µ (p)]. Summing over final state spins (no initial state spins since the pion is spin-less) we obtain: M = G F f π m µ Tr[(1 γ 5 ) k(1+γ 5 )( p+m µ )], where we use s u ν µ (k)ū νµ (k) = k and S u µ(p)ū µ (p) = p + m µ, with s and S the spins of the neutrino and the muon respectively. Exercise: Show that: In its rest frame, the π-decay rate is Tr[(1 γ 5 ) k(1+γ 5 )( p+m µ )] = 8(p.k). dγ = 1 d 3 p d 3 k M m π (π) 3 E (π) 3 ω (π)4 δ (4) (q p k). Hence: dγ = G F f πm µ p 8m π π (p.k)d3 E d 3 k ω δ(4) (q p k) In the pion rest frame ( k = p), hence: The decay rate thus is given by: Γ = G F f π m µ 8m π π p.k = Eω k. p = Eω + k = ω(e +ω). d 3 p d 3 k E ω ω(e +ω)δ(m π E ω)δ (3) ( p+ k), where we used q = (m π, 0vivhyvbbbb,n), i.e. k = p in the pion rest frame. The d 3 p integration is taken care of by the δ (3) function and imposing k = p in the remaining integrand, and, since there is no angular dependence, we are left with only the integration over dω: Γ = G F f π m µ 8m π π 4π ω dω(1+ω/e)δ(m π E ω), 7

28 where the 4π is the leftover from angular integrations. Now since k = p we have k = ω = p = E m µ, thus E = ω +m µ. Then we write: Γ = G F f πm µ ω 1 ( ) 1+ δ ω πm +m π 1+m µ/ω µ +ω m π dω. To perform this integration we use the fact that: δ[f(ω)] = δ(ω ω 0 )/ f ω, ω=ω0 where ω 0 is the zero 4 of the function f(ω) = ω +m µ +ω m π, which can easily be found: and we have: ω 0 = m π m µ m π f ω = 1 +1 f 1 1+m µ ω = 1+. /ω ω=ω0 1+m µ/ω0 Now we can easily perform the ω integration using the Dirac-δ function, which forces ω = ω 0 in the rest of the integrand. We note here that the term 1+ 1, 1+m µ /ω 0 in the integrand cancels exactly with 1/ f ω, ω=ω0 and we are left with only the ω 0 term in the integrand, hence: Therefore, finally we obtain: Γ = G F f π m µ πm π ω 0. Γ = 1 τ = G F 8π f πm π m µ ( ) 1 m µ. m π Taking the universal value of G F = 10 5 m N obtained from β or µ-decay and assuming that f π = m π, (a guess which at least guarantees the correct dimension), we indeed obtain the π lifetime sec. If we repeat the calculation for the decay mode π ν e, we obtain the same result above with the replacement m µ m e. Therefore: Γ(π ν ( ) ( ) e ) Γ(π µ ν µ ) = me m π m e = , m µ m π m µ 4 In case of many zeros we must sum over all of them. 8

29 where the numerical value comes from inserting the particle masses. The charged π prefers (by a factor of 10 4 ) to decay into a muon, which has a similar mass, rather than into the much lighter electron. This is quite contrary to what one would expect from phase-space considerations, so some dynamical mechanism must be at work. The pion is spinless, and so, by the conservation of angular momentum, the outgoing lepton pair ( ν e ) must have J = 0. As the ν e has positive helicity, the is also forced into a positive helicity state. But recall that this is the wrong helicity state for the electron. In the limit m e = 0, the weak current only couples negative helicity electrons, and hence the positive helicity coupling is highly suppressed. π ν e Thus, in the π-decay, the (or µ ) is forced by angular momentum conservation into its wrong helicity state. This is much more likely to happen for the µ than for the relatively light, in fact, 10 4 times more likely. Experiment confirms this result, which is a direct consequence of the 1 γ 5 or left-handed structure of the weak current. Problem Predict the ratio of the K ν e and K µ ν µ decay rates. Given that the lifetime of the K is, τ = s and the K µν, branching ratio is 64%, estimate the decay constant f K. Comment on your assumptions and on your result..9 Neutral weak currents We have noted that there are no scattering processes of the form ν µ ν µ or ν µ ν µ in the framework of Fermi s theory with V-A coupling. One therefore has to carefully investigate experimentally whether such scattering occurs in nature. These experiments are extremely difficult, because the expected cross sections (if any) lie in the range to 10 0 barn. Only with the high neutrino currents in modern accelerators (Fermilab near Chicago, CERN-SPS) and high neutrino energies (several hundred GeV) did such experiments become practical at all. In fact many such processes were observed; the best experimental values for the cross sections are: 1 E νµ σ(ν µ ν µ ) = (1.45±0.6) 10 4 cm /GeV 1 E νµ σ( ν µ ν µ ) = (1.3±1.0) 10 4 cm /GeV The existence of such so-called neutral currents can therefore be regarded as being firmly established. Here the name neutral current has the following origin. If one starts from the conservation of electron and muon numbers separately, the only possible interpretation of the scattering process ν µ ν µ is that at the interaction point the incoming electron turns into the outgoing electron and the incoming ν µ neutrino turns into the outgoing ν µ neutrino. The 9

30 obvious method to implement this process in our theory is therefore to supplement the leptonic current by expressions of the form: ū µν γ α (1 γ 5 )u νµ ū e γ α (g V g A γ 5 )u e Here we have made use of the fact that in any event neutrinos must have negative helicity. The former current does not contain a charged particle at all, that is it is really neutral, while in second the charge of the particle is conserved, which one also somewhat sloppily refers to as neutral (in this sense the electromagnetic current ūγ µ u is neutral for all particles!). Important is the fact that the incoming particle changes its charge in the charged transition currents (carried by W ± boson) but this is not the case with neutral currents (carried by Z 0 boson)..10 Discovery of W ± and Z 0 bosons Consider the e + annihilation into µ + µ process at PETRA (Hamburg). In QED The electron e + µ γ µ + and positron collide head-on so the centre of mass frame is the Lab frame: µ + θ e + µ Exercise: Show that the QED differential cross-section for this process in the high energy limit (E CM m µ ) is: dσ dω = α (1+cos θ) 4ECM Thus QED predicts symmetric angular distribution dσ/dω as a function of cos θ that looks like: 30

31 .0 dσ d cosθ Notethatwehavenotincludedtheproductionofe + becauseithasmorechannels(t-channel). At low energies (13.8 GeV M Z 0 90 GeV) we have the possible processes: e + µ e + µ γ + Z 0 virtual µ + µ + where the Z 0 is produced off-shell. The interference term in the amplitude-squared between these two processes causes an asymmetry in the differential angular distribution. The asymmetry is dσ d Pure QED Γ Z 0 interferenc cosθ measured by (backward-forward asymmetry). A µµ = N(µ back) N(µ forward) N(µ back)+n(µ forward) The measured asymmetry confirms the existence of Z 0 and predicts the mass of the Z 0 around 93 GeV as this symmetry is restored at 93 GeV where the Z 0 is produced on-shell (real particle). 31

32 In 1983 the first discovery (production) of Z 0 boson achieved in p p collider at CERN. 450 GeV u d Z GeV ū d ū The energy available for the cross-section was small so it took a long time (only few Z 0 bosons produced). In 1989 the LEP collider (CERN) produced the Z 0 which decayed into hadrons hadrons σ(e + hadrons) e + Z 0 q Γ Z 0 q E CM hadrons M Z The predicted mass and width of the Z 0 boson was: M Z 0 Γ Z 0 = ( ±0.001)GeV = (.495±0.003)GeV which gives a lifetime of about 10 5 s (recall E t = τ = /Γ) Note that having too much or too less energy would produce the Z 0 as a virtual particle. The reason for this is that all the energy and momentum goes into creating the Z 0 boson, and by momemtnum conservation it has zero 3-momentum (i.e. the Z 0 is created at rest). So the centre of mass energy must equal to the mass of the created particle (E = p +m ) = m, so m = E, so the mass of the created particle is less than the actual mass and the particle is therefore not real (off-shell) (this is allowed by the Heisenberg uncertainty. 3

33 3 Electroweak interactions in the Standard Model The Z 0 boson is neutral, it couples to leptons and quarks: Z 0 ν l Z 0 l + ν l l Z 0 q q For example the electron proton scattering in QWD (range m): Z 0 u u d u u d Experimental test of flavour conservation at Z 0 vertex: considering two possible processes changing strangeness: ν µ ν l K + s u W + µ + K + s u Z 0 ν l uū π 0 u d π + Allowed Forbidden K + π 0 +µ + +ν µ 33

34 and K + π 0 +ν l + ν l The Measured upper limit on the ratio of the decay rates K + π 0 +ν l + ν l to K + π 0 +µ + +ν µ is: l Γ(K+ π 0 +ν l + ν l ) < 10 7 Γ(K + π 0 +µ + +ν µ ) Thus the Z 0 boson does not change (carry) lepton number or quark number (i.e. no FCNC - flavour changing neutral current). Some of these properties are demonstrated by the photon (ep scattering in EM interactions): γ u u d u u d Here the range is [recall range can be found using E t. For weak interactions E = 100 GeV t = range/c, so range m] Comparing vertices involving γ, W ± and Z 0, one can conclude that they all are governed by the same coupling constant g e. Weak interactions only become comparable to EM interactions if distances are of order m, a concept which leads to EM unification. The unification condition establishes relation between coupling constants (α em = e /4πǫ 0 ): where θ W is the weak mixing angle, or Weinberg angle: e ǫ 0 = g W sinθ W = g Z cosθ W (13) cosθ W = M W M Z. 3.1 Divergences in weak interactions The theory of weak interactions only by means of W ± bosons leads to infinities. Examples of processes which lead to divergences include: A good theory (such as QED) must be renormalisable: all expressions can be made finite by reexpressing them in a finite number of physical parameters (e, m e and in QED). Introduction of Z 0 boson fixes the problem and cancel the divergences: 34

35 e + γ W + e + W + µ + ν e ν µ W W µ e + Z 0 W + e + Z 0 µ + e + γ µ + W Z 0 µ µ Z 0 µ µ 3. Weinberg-Salam model Renormalisable theories are gauge invariant theories. A Gauge transformation is a certain alteration of a quantum field variables that leave basic properties of the field unchanged i.e. a symmetry transformation. There are several forms of gauge invariance corresponding to different interactions. In QED, the Lagrangian must be must be invariant under the phase transformation of the field. φ e iα(x) φ Here α is an arbitrary continuous function. If a particle is free, then the transformed field can not be a solution of the Dirac equation. Gauge principle: to keep the invariance condition satisfied, a minimal field should be added to the Lagrangian, i.e., an interaction term should be introduced. In QED, the transition from one electron state to another with different phase,, demands the emissions (or absorption) of a photon: +γ. More generally, transformations like: ν e, ν e,,ν e ν e lead via gauge principle to interactions ν e W, ν e W +, W 0, ν e ν e W 0 with W +, W and W 0 are corresponding spin-1 gauge bosons. While W + and W are well-known charged currents, W 0 has not been identified. Electroweak unification regards both Z 0 and γ as mixtures of W 0 and yet another neutral boson B 0 : ( ) ( )( ) γ cosθw sinθ Z 0 = W B 0 sinθ W cosθ W W 0 (14) 35