THE STANDARD MODEL OF PARTICLE PHYSICS

Transcription

1 THE STANDARD MODEL OF PARTICLE PHYSICS P.J. Mulders Department of Theoretical Physics, Faculty of Sciences, Vrije Universiteit, 08 HV Amsterdam, the Netherlands Lectures given at the BND School Center de Krim, Texel, Netherlands 9-30 September 005 September 005 (version.3)

2 Preface In these lectures I will mainly discuss the symmetries and concepts underlying the Standard Model that is so successful in describing the interactions of the elementary particles, the quarks and leptons. I will assume a basic knowledge of field theory. As an additonal help in understanding these notes, I suggest students to use the Introductory quantum field theory notes found under mulders/lectures.html#master or consult text books such as those given below.. L.H. Ryder, Quantum Field Theory, Cambridge University Press, M.E. Peshkin and D.V. Schroeder, An introduction to Quantum Field Theory, Addison-Wesly, M. Veltman, Diagrammatica, Cambridge University Press, S. Weinberg, The quantum theory of fields; Vol. I: Foundations, Cambridge University Press, 995; Vol. II: Modern Applications, Cambridge University Press, C. Itzykson and J.-B. Zuber, Quantum Field Theory, McGraw-Hill, 980.

3 Contents Gauge theories Spontaneous symmetry breaking 5 3 The Higgs mechanism 9 4 The standard model SU() W U() Y 0 5 Family mixing in the Higgs sector and neutrino masses 5 A kinematics in scattering processes 0 B Cross sections and lifetimes C Unitarity condition 4 D Unstable particles 6 3

4 Gauge theories Abelian gauge theories Consider a theory that is invariant under global gauge transformations or gauge transformations of the first kind, e.g. in the Klein-Gordon theory for a scalar field, describing a spinless particle the transformation φ(x) e i eλ φ(x), () in which the U() phase involves an angle e Λ, independent of x. Gauge transformations of the second kind or local gauge transformations are transformations of the type φ(x) e i eλ(x) φ(x), () i.e. the angle of the transformation depends on the space-time point x. The lagrangians for free particles (e.g. Klein-Gordon, Dirac) are invariant under global gauge transformations and corresponding to this there exist a conserved Noether current. Any lagrangian containing derivatives, however, is not invariant under local gauge transformations, φ(x) e i eλ(x) φ(x), (3) φ (x) e i eλ(x) φ (x), (4) µ φ(x) e i eλ(x) µ φ(x) + i e µ Λ(x) e i eλ(x) φ(x), (5) where it is the last term that spoils gauge invariance. A solution is the one known as minimal substitution in which the derivative µ is replaced by a covariant derivative D µ which satisfies To achieve invariance it is necessary to introduce a vector field A µ, The required transformation for D µ then demands D µ φ(x) e i eλ(x) D µ φ(x). (6) D µ φ(x) ( µ + i ea µ (x))φ(x), (7) D µ φ(x) = ( µ + i ea µ (x))φ(x) e i eλ µ φ + i e ( µ Λ) e i eλ φ + i ea µ ei eλ φ = e i eλ ( µ + i e(a µ + µλ) ) φ e i eλ ( µ + i ea µ ) φ. (8) Thus the covariant derivative has the correct transformation behavior provided A µ A µ µ Λ, (9) the behavior which is familar as the gauge freedom in electromagnetism described via for massless vector fields and the (free) lagrangian density L = (/4)F µν F µν. Replacing derivatives by covariant derivatives and adding the (free) part for the vector fields to the original lagrangian produces a gauge invariant lagrangian, L(φ, µ φ) = L(φ, D µ φ) 4 F µνf µν. (0) The field φ is used here in a general sense standing for any possible field. As an example consider the Dirac lagrangian, L = i ( ψγ µ µ ψ ( µ ψ)γ µ ψ ) M ψψ.

5 Minimal substitution µ ψ ( µ + i ea µ )ψ leads to the gauge invariant lagrangian L = i ψ ψ M ψψ e ψγ µ ψ A µ 4 F µνf µν. () We note first of all that the coupling of the Dirac field (electron) to the vector field (photon) can be written in the familiar interaction form L int = e ψγ µ ψ A µ = e j µ A µ, () involving the interaction of the charge (ρ = j 0 ) and three-current density (j) with the electric potential (φ = A 0 ) and the vector potential (A), e j µ A µ = e ρφ + e j A. The equation of motion for the fermion follow from δl δ( µ ψ) = i γµ ψ, δl δψ = i / ψ M ψ e/a ψ giving the Dirac equation in an electromagnetic field, For the photon the equations of motion follow from (i/d M) ψ = (i/ e/a M) ψ = 0. (3) δl δ( µ A ν ) = F µν, δl δa ν = eψγ ν ψ, giving the Maxwell equation coupling to the electromagnetic current, µ F µν = j ν, (4) where j µ = e ψγ µ ψ. This latter current is the conserved current that is obtained for the Dirac lagrangian using Noether s theorem. Non-abelian gauge theories Quantum electrodynamics is an example of a very successful gauge theory. The photon field A µ was introduced as to render the lagrangian invariant under local gauge transformations. The extension to non-abelian gauge theories is straightforward. The symmetry group is a Lie-group G generated by generators T a, which satisfy commutation relations [T a, T b ] = i c abc T c, (5) with c abc known as the structure constants of the group. For a compact Lie-group they are antisymmetric in the three indices. In an abelian group the structure constants would be zero (for instance the trivial example of U()). Consider a field transforming under the group, φ(x) e i θa (x)l a φ(x) inf. = ( + i θ a (x)l a ) φ(x) (6) where L a is a representation matrix for the representation to which φ belongs, i.e. for a threecomponent field φ under an SO(3) or SU() symmetry transformation, φ e i θ L φ φ θ φ. (7)

6 The complication arises (as in the abelian case) when one considers for a lagrangian density L(φ, µ φ) the behavior of µ φ under a local gauge transformation, U(θ) = e i θa (x)l a, φ(x) U(θ)φ(x), (8) µ φ(x) U(θ) µ φ(x) + ( µ U(θ)) φ(x). (9) Introducing as many gauge fields as there are generators in the group, which are conveniently combined in the matrix valued field W µ = W a µ L a, one defines and one obtains after transformation D µ φ(x) ( µ ig W µ ) φ(x), (0) D µ φ(x) U(θ) µ φ(x) + ( µ U(θ)) φ(x) ig W µ U(θ)φ(x). Requiring that D µ φ transforms as D µ φ U(θ) D µ φ, i.e. D µ φ(x) U(θ) µ φ(x) ig U(θ) W µ φ(x), one obtains or infinitesimal W µ = U(θ) W µ U (θ) i g ( µu(θ)) U (θ), () W a µ = W a µ c abc θ b W c µ + g µθ a = W a µ + g D µθ a. It is necessary to introduce the free lagrangian density for the gauge fields just like the term (/4)F µν F µν in QED. For abelian fields F µν = µ A ν ν A µ = (i/g)[d µ, D ν ] is gauge invariant. In the nonabelian case µ W a ν ν W a µ does not provide a gauge invariant candidate for G µν = G a µνl a, as can be checked easily. Expressing G µν in terms of the covariant derivatives provides a gauge invariant definition for G µν with G µν = i g [D µ, D ν ] = µw ν ν W µ ig [W µ, W ν ], () and thus It transforms like The gauge-invariant lagrangian density is now constructed as G a µν = µ W a ν ν W a µ + g c abc W b µw c ν, (3) G µν U(θ) G µν U (θ). (4) L(φ, µ φ) L(φ, D µ φ) Tr G µνg µν = L(φ, D µ φ) 4 Ga µνg µν a (5) with the standard normalization Tr(L a L b ) = (/)δ ab. Note that the gauge fields must be massless, as a mass term M W W a µ W µ a would break gauge invariance. QCD, an example of a nonabelian gauge theory As an example of a nonabelian gauge theory consider quantum chromodynamics (QCD), the theory describing the interactions of the colored quarks. The existence of an extra degree of freedom for each species of quarks is evident for several reasons, e.g. the necessity to have an antisymmetric wave function for the ++ particle consisting of three up quarks (each with charge +(/3)e). With the 3

7 quarks belonging to the fundamental (three-dimensional) representation of SU(3) C, i.e. having three components in color space ψ r ψ = ψ g, ψ b the wave function of the baryons (such as nucleons and deltas) form a singlet under SU(3) C, color = 6 ( rgb grb + gbr bgr + brg rbg ). (6) The nonabelian gauge theory that is obtained by making the free quark lagrangian, for one specific species (flavor) of quarks just the Dirac lagrangian for an elementary fermion, L = i ψ/ ψ m ψψ, invariant under local SU(3) C transformations has proven to be a good candidate for the microscopic theory of the strong interactions. The representation matrices for the quarks and antiquarks in the fundamental representation are given by F a = λ a F a = λ a for quarks, for antiquarks, which satisfy commutation relations [F a, F b ] = i f abc F c in which f abc are the (completely antisymmetric) structure constants of SU(3) and where the matrices λ a are the eight Gell-Mann matrices. The (locally) gauge invariant lagrangian density is with L = 4 F a µν F µν a + i ψ/dψ m ψψ, (7) D µ ψ = µ ψ ig A a µ F aψ, F a µν = µ A a ν ν A a µ + g c abc A b µa c ν. Note that the term i ψ/dψ = i ψ/ ψ + g ψ/a a F a ψ = i ψ/ ψ + j µ a A a µ with jµ a = ψγ µ F a ψ describes the interactions of the gauge bosons A a µ (gluons) with the color current of the quarks (this is again precisely the Noether current corresponding to color symmetry transformations). Note furthermore that the lagrangian terms for the gluons contain interaction terms corresponding to vertices with three gluons and four gluons due to the nonabelian character of the theory. The Gell-Mann matrices are the eight traceless hermitean matrices generating SU(3) transformations, i λ = λ = i λ 3 = i λ 4 = λ 5 = λ 6 = i λ 7 = i λ 8 = i 3 4

8 Feynman rules for QCD For writing down the complete set of Feynman rules it is necessary to account for the gauge symmetry in the quantization procedure. This will lead (depending on the choice of gauge conditions) to the presence of ghost fields. In the axial gauge n µ A a µ = 0 gauge fields are not needed. From the Lagrangina of QCD, L = 4 F a µν(x)f µνa (x) + ψ(x)(/d m)ψ(x) λ (nµ A a µ), (8) including a gauge fixing term that assures n µ A a µ = 0 one reads off the Feynman rules. The propagators are derived from the quadratic terms ( ) k i δ i, α j, β ij = i δ ij (/k + m) βα /k m + iɛ βα k m + iɛ [ k i δ ab µ a ν b k g µν + (n + λ ) k µ k ν + iɛ k (k n) kµ n ν + k ν n µ ]. k n The vertices are derived from the interaction terms in the lagrangian, µ,a i g (γ µ ) βα (F a ) ji i, α ρ,c j, β g c abc [(p q) ρ g µν + (q r) µ g νρ + (r p) ρ g ρµ ] µ, a ν,b µ,b ν, c ρ,d σ, e i g c abc c ade (g µρ g νσ g µσ g νρ ) +i g c abd c ace (g µν g ρσ g µσ g νρ ) +i g c abe c acd (g µρ g νσ g µν g ρσ ). Spontaneous symmetry breaking In this section we consider the situation that the groundstate of a physical system is degenerate. Consider as an example a ferromagnet with an interaction hamiltonian of the form H = i>j J ij S i S j, which is rotationally invariant. If the temperature is high enough the spins are oriented randomly and the (macroscopic) ground state is spherically symmetric. If the temperature is below a certain critical temperature (T < T c ) the kinetic energy is no longer dominant and the above hamiltonian prefers a lowest energy configuration in which all spins are parallel. In this case there are many possible groundstates (determined by a fixed direction in space). This characterizes spontaneous 5

9 symmetry breaking, the groundstate itself appears degenerate. As there can be one and only one groundstate, this means that there is more than one possibility for the groundstate. Nature will choose one, usually being (slightly) prejudiced by impurities, external magnetic fields, i.e. in reality a not perfectly symmetric situation. Nevertheless, we can disregard those perturbations and look at the ideal situation, e.g. a theory for a scalar degree of freedom (a scalar field) having three (real) components, φ φ = φ, φ 3 with a lagrangian density of the form L = µ φ µ φ m φ φ 4 λ( φ φ). (9) }{{} V ( φ) The potential V ( φ) is shown in fig.. Classically the (time-independent) ground state is found for a constant field ( φ = 0) and the condition V φ = 0 ϕ c ϕ c = 0 or ϕ c ϕ c = m λ F, ϕ c the latter only forming a minimum for m < 0. In this situation one speaks of spontaneous symmetry breaking. The classical groundstate appears degenerate. Any constant field ϕ c with length φ = F is a possible groundstate. The presence of a nonzero value for the classical groundstate value of the field will have an effect when the field is quantized. A quantum field theory has only one nondegenerate groundstate 0. Writing the field φ as a sum of a classical and a quantum field, φ = ϕ c + φ quantum where for the (operator-valued) coefficients in the quantum field one wants 0 c = c 0 = 0 one has 0 φ quantum 0 = 0 and 0 φ 0 = ϕ c. (30) Stability of the action requires the classical groundstate ϕ c to have a well-defined value (which can be nonzero), while the quadratic terms must correspond with non-negative masses. In the case of degeneracy, therefore a choice must be made, say 0 φ 0 0 = 0. (3) F V( φ) F φ Figure : The symmetry-breaking potential in the lagrangian for the case that m < 0. 6

10 The situation now is the following. The original lagrangian contained an SO(3) invariance under (length conserving) rotations among the three fields, while the lagrangian including the nonzero groundstate expectation value chosen by nature, has less symmetry. It is only invariant under rotations around the 3-axis. It is appropriate to redefine the field as ϕ φ = ϕ, (3) F + η such that 0 ϕ 0 = 0 ϕ 0 = 0 η 0 = 0. The field along the third axis plays a special role because of the choice of the vacuum expectation value. In order to see the consequences for the particle spectrum of the theory we construct the lagrangian in terms of the fields ϕ, ϕ and η. It is sufficient to do this to second order in the fields as the higher (cubic, etc.) terms constitute interaction terms. The result is L = ( µϕ ) + ( µϕ ) + ( µη) m (ϕ + ϕ ) m (F + η) 4 λ (ϕ + ϕ + F + η + F η) (33) = ( µϕ ) + ( µϕ ) + ( µη) + m η (34) Therefore there are massless scalar particles, corresponding to the number of broken generators (in this case rotations around and axis) and massive scalar particle with mass m η = m. The massless particles are called Goldstone bosons. Realization of symmetries In this section we want to discuss a bit more formal the two possible ways that a symmetry can be implemented. They are known as the Weyl mode or the Goldstone mode: Weyl mode. In this mode the lagrangian and the vacuum are both invariant under a set of symmetry transformations generated by Q a, i.e. for the vacuum Q a 0 = 0. In this case the spectrum is described by degenerate representations of the symmetry group. Known examples are rotational symmetry and the fact that the the spectrum shows multiplets labeled by angular momentum l (with members labeled by m). The generators Q a (in that case the rotation operators L z, L x and L y or instead of the latter two L + and L ) are used to label the multiplet members or transform them into one another. A bit more formal, if the generators Q a generate a symmetry, i.e. [Q a, H] = 0, and a and a belong to the same multiplet (there is a Q a such that a = Q a a ) then H a = E a a implies that H a = E a a, i.e. a and a are degenerate states. Goldstone mode. In this mode the lagrangian is invariant but Q a 0 0 for a number of generators. This means that they are operators that create states from the vacuum, denoted π a (k). As the generators for a symmetry are precisely the zero-components of a conserved current Jµ(x) a integrated over space, there must be a nonzero expectation value 0 Jµ a (x) πa (k). Using translation invariance and as k µ is the only four vector on which this matrix element could depend one may write 0 J a µ(x) π b (k) = f π k µ e i k x δ ab (f π 0) (35) for all the states labeled by a corresponding to broken generators. Taking the derivative, 0 µ J a µ(x) π b (k) = f π k e i k x δ ab = f π m π a ei k x δ ab. (36) 7

11 If the transformations in the lagrangian give rise to a symmetry the Noether currents are conserved, µ Jµ a = 0, irrespective of the fact if they annihilate the vacuum, and one must have m π a = 0, i.e. a massless Goldstone boson for each broken generator. Note that for the fields π a (x) one would have the relation 0 π a (x) π a (k) = e i k x, suggesting the stronger relation µ Jµ a (x) = f π m π a πa (x). Chiral symmetry An example of spontaneous symmetry breaking is chiral symmetry breaking in QCD. Neglecting at this point the local color symmetry, the lagrangian for the quarks consists of the free Dirac lagrangian for each of the types of quarks, called flavors. Including a sum over the different flavors (up, down, strange, etc.) one can write L = ψ(i/ M)ψ, (37) where ψ is extended to a vector in flavor space and M is a diagonal matrix, ψ u m u ψ = ψ d, M = m d.... (Note that each of the entries in the vector for ψ is a 4-component Dirac spinor). This lagrangian density then is invariant under unitary (vector) transformations in the flavor space, (38) ψ e i α T ψ, (39) which for instance including only two flavors form an SU() V symmetry (isospin symmetry) generated by the Pauli matrices, T = τ/. The conserved currents corresponding to this symmetry transformation are found directly using Noether s theorem (see chapter 6), Using the Dirac equation, it is easy to see that one gets V µ = ψγ µ T ψ. (40) µ V µ = i ψ [M, T ] ψ. (4) Furthermore µ V µ = 0 [M, T ] = 0. From group theory (Schur s theorem) one knows that the latter can only be true, if in flavor space M is proportional to the unit matrix, M = m. I.e. SU() V (isospin) symmetry is good if the up and down quark masses are identical. This situation, both are very small, is what happens in the real world. This symmetry is realized in the Weyl mode with the spectrum of QCD showing an almost perfect isospin symmetry, e.g. a doublet (isospin /) of nucleons, proton and neutron, with almost degenerate masses (M p = MeV/c and M n =939.6 MeV/c ), but also a triplet (isospin ) of pions, etc. There exists another set of symmetry transformations, socalled axial vector transformations, ψ e i α T γ 5 ψ, (4) which for instance including only two flavors form SU() A transformations generated by the Pauli matrices, T γ 5 = τγ 5 /. Note that these transformations also work on the spinor indices. The currents corresponding to this symmetry transformation are again found using Noether s theorem, Using the Dirac equation, it is easy to see that one gets A µ = ψγ µ T γ5 ψ. (43) µ A µ = i ψ {M, T} γ 5 ψ. (44) 8

12 In this case µ A µ = 0 will be true if the quarks have zero mass, which is approximately true for the up and down quarks. Therefore the world of up and down quarks describing pions, nucleons and atomic nuclei has not only an isospin or vector symmetry SU() V but also an axial vector symmetry SU() A. This combined symmetry is what one calls chiral symmetry. That the massless theory has this symmetry can also be seen by writing it down for the socalled lefthanded and righthanded fermions, ψ R/L = ( ± γ 5)ψ, in terms of which the Dirac lagrangian density looks like L = i ψ L / ψ L + i ψ R / ψ R ψ R Mψ L ψ L Mψ R. (45) If the mass is zero the lagrangian is split into two disjunct parts for L and R showing that there is a direct product SU() L SU() R symmetry, generated by T R/L = ( ± γ 5) T, which is equivalent to the V-A symmetry. This symmetry, however, is by nature not realized in the Weyl mode. How can we see this. The chiral fields ψ R and ψ L are transformed into each other under parity. Therefore realization in the Weyl mode would require that all particles come double with positive and negative parity, or, stated equivalently, parity would not play a role in the world. We know that mesons and baryons (such as the nucleons) have a well-defined parity that is conserved. The conclusion is that the original symmetry of the lagrangian is spontaneously broken and as the vector part of the symmetry is the well-known isospin symmetry, nature has choosen the path SU() L SU() R = SU() V, i.e. the lagrangian density is invariant under left (L) and right (R) rotations independently, while the groundstate is only invariant under isospin rotations (R = L). From the number of broken generators it is clear that one expects three massless Goldstone bosons, for which the field (according to the discussion above) has the same behavior under parity, etc. as the quantity µ A µ (x), i.e. (leaving out the flavor structure) the same as ψγ 5 ψ, i.e. behaves as a pseudoscalar particle (spin zero, parity minus). In the real world, where the quark masses are not completely zero, chiral symmetry is not perfect. Still the basic fact that the generators acting on the vacuum give a nonzero result (i.e. f π 0 remains, but the fact that the symmetry is not perfect and the right hand side of Eq. 44 is nonzero, gives also rise to a nonzero mass for the Goldstone bosons according to Eq. 36. The Goldstone bosons of QCD are the pions for which f π = 93 MeV and which have a mass of m π 38 MeV/c, much smaller than any of the other mesons or baryons. 3 The Higgs mechanism The Higgs mechanism occurs when spontaneous symmetry breaking happens in a gauge theory where gauge bosons have been introduced in order to assure the local symmetry. Considering the same example with rotational symmetry (SO(3)) as for spontaneous symmetry breaking of a scalar field (Higgs field) with three components, made into a gauge theory, where L = 4 G µν G µν + D µ φ D µ φ V ( φ), (46) D µ φ = µ φ ig W a µ L a φ. (47) Since the explicit (conjugate, in this case three-dimensional) representation (L a ) ij = i ɛ aij one sees that the fields W µ and G µν also can be represented as three-component fields, D µ φ = µ φ + g Wµ φ, (48) G µν = µ Wν νwµ + gw µ W ν. (49) 9

13 The symmetry is broken in the same way as before and the same choice for the vacuum, ϕ c = 0 φ 0 0 = 0. F is made. The difference comes when we reparametrize the field φ. We have the possibility to perform local gauge transformations. Therefore we can always rotate the field φ into the z-direction in order to make the calculation simple, i.e. φ = 0 0 φ 3 = 0 0 F + η. (50) Explicitly one then has D µ φ = µ φ + g Wµ φ = gf W µ + g W µ η gf W µ g W µ η µ η, which gives for the lagrangian density up to quadratic terms L = 4 G µν G µν + D µ φ D µ φ m φ φ λ 4 ( φ φ) = 4 ( µ W ν ν Wµ ) ( µ W ν ν W µ ) g F ( W µ W µ + W µ W µ ) + ( µη) + m η +..., (5) from which one reads off that the particle content of the theory consists of one massless gauge boson (W 3 µ), two massive bosons (W µ and W µ with M W = gf ) and a massive scalar particle (η with m η = m. The latter is a spin 0 particle (real scalar field) called a Higgs particle. Note that the number of massless gauge bosons (in this case one) coincides with the number of generators corresponding to the remaining symmetry (in this case rotations around the 3-axis), while the number of massive gauge bosons coincides with the number of broken generators. One may wonder about the degrees of freedom, as in this case there are no massless Goldstone bosons. Initially there are 3 massless gauge fields (each, like a photon, having two independent spin components) and three scalar fields (one degree of freedom each), thus 9 independent degrees of freedom. After symmetry breaking the same number (as expected) comes out, but one has massless gauge field (), massive vector fields or spin bosons ( 3) and one scalar field (), again 9 degrees of freedom. 4 The standard model SU() W U() Y The symmetry ideas discussed before play an essential role in the standard model that describes the elementary particles, the quarks (up, down, etc.), the leptons (elektrons, muons, neutrinos, etc.) and the gauge bosons responsible for the strong, electromagnetic and weak forces. In the standard model one starts with a very simple basic lagrangian for (massless) fermions which exhibits more symmetry than observed in nature. By introducing gauge fields and breaking the symmetry a more complex lagrangian is obtained, that gives a good description of the physical world. The procedure, however, implies certain nontrivial relations between masses and mixing angles that can be tested experimentally and sofar are in excellent agreement with experiment. The lagrangian for the leptons consists of three families each containing an elementary fermion (electron e, muon µ or tau τ ), its corresponding neutrino (ν e, ν µ and ν τ ) and their antiparticles. 0

14 As they are massless, left- and righthanded particles, ψ R/L = ( ± γ 5)ψ decouple. For the neutrino only a lefthanded particle (and righthanded antiparticle) exist. Thus L (f) = i e R / e R + i e L / e L + i ν el / ν el + (µ, τ). (5) One introduces a (weak) SU() W symmetry under which e R forms a singlet, while the lefthanded particles form a doublet, i.e. L = ν e with I e W = { +/ L and I3 W = / and Thus the lagrangian density is R = e R with I W = 0 and I 3 W = 0. L (f) = i L/ L + i R/ R, (53) which has an SU() W symmetry under transformations e i α T, explicitly L SU()W e i α τ/ L, (54) R SU()W R. (55) One notes that the charges of the leptons can be obtained as Q = IW 3 / for lefthanded particles and Q = IW 3 for righthanded particles. This is written as Q = I 3 W + Y W, (56) and Y W is considered as an operator that generates a U() Y symmetry, under which the lefthanded and righthanded particles with Y W (L) = and Y W (R) = transform with e iβyw /, explicitly L U()Y e i β/ L, (57) R U()Y e i β R. (58) Next the SU() W U() Y B µ, symmetry is made into a local symmetry introducing gauge fields W µ and D µ L = µ L + i g W µ τ L i g B µ L, (59) D µ R = µ R i g B µ R, (60) where W µ is a triplet of gauge bosons with I W =, I 3 W = ± or 0 and Y W = 0 (thus Q = I 3 W ) and B µ is a singlet under SU() W (I W = I 3 W = 0) and also has Y W = 0. Putting this in leads to L (f) = L (f) + L (f), (6) L (f) = i Rγ µ ( µ ig B µ )R + i Lγ µ ( µ i g B µ + i g W µ τ)l L (f) = 4 ( µ W ν ν Wµ + g W µ W ν ) 4 ( µb ν ν B µ ). In order to break the symmetry to the symmetry of the physical world, the U() Q symmetry (generated by the charge operator), a complex Higgs field φ = φ+ φ 0 = (θ + iθ ) (θ 4 iθ 3 ) (6)

15 with I W = / and Y W = is introduced, with the following lagrangian density consisting of a symmetry breaking piece and a coupling to the fermions, where L (h) = L (h) + L (h), (63) L (h) = (D µ φ) (D µ φ) m φ φ λ (φ φ), }{{} V (φ) L (h) = G e (LφR + Rφ L), and D µ φ = ( µ + i g W µ τ + i g B µ )φ. (64) The Higgs potential V (φ) is choosen such that it gives rise to spontaneous symmetry breaking with ϕ cϕ = m /λ v /. For the classical field the choice θ 4 = v is made. Using local gauge invariance θ i for i =, and 3 may be eliminated (the necessary SU() W rotation is precisely e i θ(x) τ ), leading to the parametrization φ(x) = 0 v + h(x) (65) and ( ig W ) µ iwµ D µ φ = (v + h) ( µ h i gw 3 ) µ g B µ (v + h). (66) Up to cubic terms, this leads to the lagrangian L (h) = ( µh) + m h + g v [ (W 8 µ ) + (Wµ )] + v ( gw 3 8 µ g ) B µ +... (67) = ( µh) + m h + g v [ (W + 8 µ ) + (Wµ )] + (g + g ) v (Z µ ) +..., (68) 8 where the quadratically appearing gauge fields that are furthermore eigenstates of the charge operator are W ± µ = ( W µ ± i W µ), (69) Z µ = g W 3 µ g B µ g + g cos θ W W 3 µ sin θ W B µ, (70) A µ = g W 3 µ + g B µ g + g sin θ W W 3 µ + cos θ W B µ, (7) and correspond to three massive particle fields (W ± and Z 0 ) and one massless field (photon γ) with MW = g v 4, (7) MZ = g v 4 cos θ W = M W cos, θ W (73) Mγ = 0. (74)

16 The weak mixing angle is related to the ratio of coupling constants, g /g = tan θ W. The coupling of the fermions to the physical gauge bosons are contained in L (f) giving L (f) = i eγ µ µ e + i ν e γ µ µ ν e g sin θ W eγ µ e A µ + g (sin θ W e R γ µ e R cos θ cos θ W e L γ µ e L + ) ν eγ µ ν e Z µ W + g ( νe γ µ e L Wµ + e L γ µ ν e W + ) µ. (75) From the coupling to the photon, we can read off e = g sin θ W = g cos θ W. (76) The coupling of electrons or muons to their respective neutrinos, for instance in the amplitude for the decay of the muon ν µ νµ µ e W µ = e ν e ν e is given by i M = g (ν µγ ρ µ L ) i g ρσ +... k + MW (e L γ σ ν e ) g i 8 MW (ν µ γ ρ ( γ 5 )µ) (eγ σ ( γ 5 )ν e ) }{{}}{{} (j (µ) L )ρ (j (e) L )ρ (77) i G F (j (µ) L ) ρ (j (e) L )ρ, (78) the good old four-point interaction introduced by Fermi to explain the weak interactions, i.e. one has the relation G F = g e 8 MW = = 8 MW sin θ W v. (79) In this way the parameters g, g and v determine a number of experimentally measurable quantities, such as e /4π /37, (80) G F = GeV, (8) sin θ W = 0.3, (8) M W = 80.4 GeV, (83) M Z = 9.98 GeV. (84) The coupling of the Z 0 to fermions is given by g/ cos θ W multiplied with I 3 W ( γ 5) sin θ W Q C V C A γ 5, (85) 3

17 with C V = I 3 W sin θ W Q, (86) C A = I 3 W. (87) From this coupling it is straightforward to calculate the partial width for Z 0 into a fermion-antifermion pair, Γ(Z 0 ff) = M Z g 48π cos (CV + C θ A). (88) W For the electron, muon or tau, leptons with C V = / + sin θ W 0.05 and C A = / we calculate Γ(e + e ) 78.5 MeV (exp. Γ e Γ µ Γ τ 83 MeV). For each neutrino species (with C V = / and C A = / one expects Γ(νν) 55 MeV. Comparing this with the total width into (invisible!) channels, Γ invisible = 480 MeV one sees that three families of (light) neutrinos are allowed. Actually including corrections corresponding to higher order diagrams the agreement for the decay width into electrons can be calculated much more accurately and the number of allowed (light) neutrinos turns to be even closer to three. The masses of the fermions and the coupling to the Higgs particle are contained in L (h). With the choosen vacuum expectation value for the Higgs field, one obtains L (h) = G e v (e L e R + e R e L ) G e (e L e R + e R e L ) h = m e ee m e v eeh. (89) First, the mass of the electron comes from the spontaneous symmetry breaking but is not predicted (it is in the coupling G e ). The coupling to the Higgs particle is weak as the value for v calculated e.g. from the M W mass is about 50 GeV, i.e. m e /v is extremely small. Finally we want to say something about the weak properties of the quarks, as appear for instance in the decay of the neutron or the decay of the Λ (quark content uds), u u d s - W e - - W e - ν e ν e n pe ν e d ue ν e, Λ pe ν e s ue ν e. The quarks also turn out to fit into doublets of SU() W for the lefthanded species and into singlets for the righthanded quarks. A complication arises as it are not the mass eigenstates that appear in the weak isospin doublets but linear combinations of them, where u d d s b L = L c s L V ud V us V ub V cd V cs V cb V td V ts V tb t b, L This mixing allows all quarks with IW 3 = / to decay into an up quark, but with different strength. Comparing neutron decay and Λ decay one can get an estimate of the mixing parameter V us in the socalled Cabibbo-Kobayashi-Maskawa mixing matrix. Decay of B-mesons containing b-quarks allow d s b L (90) 4

18 estimate of V ub, etc. In principle one complex phase is allowed in the most general form of the CKM matrix, which can account for the (observed) CP violation of the weak interactions. This is only true if the mixing matrix is at least three-dimensional, i.e. CP violation requires three generations. The magnitudes of the entries in the CKM matrix are nicely represented in a socalled Wolfenstein parametrization V = λ λ λ 3 A(ρ i η) λ λ λ A λ 3 A( ρ i η) λ A + O(λ4 ) with λ 0.3, A 0.8 and ρ 0.3 and η The imaginary part i η gives rise to CP violation in decays of K and B-mesons (containing s and b quarks, respectively). 5 Family mixing in the Higgs sector and neutrino masses The quark sector Allowing for the most general (Dirac) mass generating term in the lagrangian one starts with L (h,q) = Q L φλ d D R D R Λ d φ Q L, Q L φ c Λ u U R U R Λ u φc Q L (9) where we include now the three lefthanded quark doublets in Q L, the three righthanded quarks with charge +/3 in U R and the three righthanded quarks with charges /3 in D R, each of these containing the three families, e.g. U R = ( ) u R c R t R. The Λu and Λ d are complex matrices in the 3 3 family space. The Higgs field is still limited to one complex doublet. Note that we need the conjugate Higgs field to get a U() Y singlet in the case of the charge +/3 quarks, for which we need the appropriate weak isospin doublet φ c = ( φ 0 φ ) = ( v + h 0 For the (squared) complex matrices we can find positive eigenvalues, Λ u Λ u = V u G u V u, and Λ d Λ d = W d G d W d, (9) where V u and W d are unitary matrices, allowing us to write ). Λ u = V u G u W u and Λ d = V d G d W d, (93) with G u and G d being real and positive and W u and V d being different unitary matrices. Thus one has L (h,q) = D L V d M d W d D R D R W d M d V d D L, U L V u M u W u U R U R W u M u V u U L (94) with M u = G u v/ (diagonal matrix containing m u, m c and m t ) and M d = G d v/ (diagonal matrix containing m d, m s and m b ). One then reads off that starting with the family basis as defined via the left doublets that the mass eigenstates (and states coupling to the Higgs field) involve the righthanded states UR mass = W uu R and DR mass = W d D R and the lefthanded states UL mass = V u U L and DL mass = V d D L. Working with the mass eigenstates one simply sees that the weak current coupling to the W ± becomes U L γ µ γ 5 D L, U mass L γ µ γ 5 V u V d DL mass, i.e. the weak mass eigenstates are D L = Dweak L = V u V d D mass L the unitary CKM-matrix introduced above in an ad hoc way. = V CKM D mass L, (95) 5

19 The lepton sector (massless neutrinos) For a lepton sector with a lagrangian density of the form in which L (h,l) = LφΛ e E R E R Λ eφ L, (96) L = ( NL E L ), is a weak doublet containing the three families of neutrinos (N L ) and charged leptons (E L ) and E R is a three-family weak singlet, we find massless neutrinos. As before, one can write Λ e = V e G e W e and we find L (h,l) ( = M e EL V e W e E R E R W e V e ) E L, (97) with M e = G e v/ the diagonal mass matrix with masses m e, m µ and m τ. The mass fields ER mass = W e E R, EL mass = V e E L and the neutrino fields NL mass = V e N L are also the states appearing in the W -current, i.e. there is no family mixing and the neutrinos are massless. The lepton sector (massive Dirac neutrinos) In principle a massive Dirac neutrino could be accounted for by a lagrangian of the type L (h,l) = LφΛ e E R E R Λ e φ L, Lφ c Λ n N R N R Λ n φc L (98) with three righthanded neutrinos added to the previous case, decoupling from all known interactions. Again we continue as before now with matrices Λ e = V e G e W e and Λ n = V n G n W n, and obtain L (h,l) = E L V e M e W e E R E R W e M e V e E L, N L V n M n W n N R N R W n M n V n N L. (99) We note that there are mass fields ER mass = W e E R, EL mass = V e E L, NL mass = V n N L and NR mass = W n N R and the weak current becomes E L γ µ γ 5 N L = EL mass γ µ γ 5 V e V n NL mass. Working with the mass eigenstates for the charged leptons we see that the weak eigenstates for the neutrinos are N weak V e N L with the relation to the mass eigenstates for the lefthanded neutrinos given by or N mass L = U N weak L with U = V n V e. The lepton sector (massive Majorana fields) L = N L = N weak L = V e V n N mass L, (00) The simplest option is to add in Eq. 97 a Majorana mass term for (lefthanded) neutrino mass eigenstates, L mass,ν = ( ML NL c N L + ML N L NL c ), (0) but this option is not attractive as it violates the electroweak symmetry. The way to circumvent this is to introduce as in the previous section righthanded neutrinos with for the righthanded sector a mass term M R, L mass,ν = ( MR N R NR c + M R N R c N ) R. (0) For neutrinos as well as charged leptons, the right- and lefthanded species are coupled through Dirac mass terms as in the previous section. Thus (disregarding family structure) one has two Majorana neutrinos, one being massive. For the charged leptons there are no Majorana mass terms (it would break the U() electromagnetic symmetry) and left- and righthanded species combine to a Dirac fermion. Moreover, if the Majorana mass M R M D one obtains in a natural way tiny neutrino masses. This is called the seesaw mechanism. 6

20 The seesaw mechanism Consider (for one family N = n) the most general Lorentz invariant mass term for two independent Majorana spinors, Υ and Υ (satisfying Υc = Υ and as discussed in chapter 6, Υ c L (Υ L) c = Υ R and Υ c R = Υ L). We use here the primes starting with the weak eigenstates. Actually, it is easy to see that this incorporates the Dirac case by considering the lefthanded part of Υ and the righthanded part of Υ as a Dirac spinor ψ. Thus Υ = nc L + n L, Υ = n R + nc R, ψ = n R + n L. (03) As the most general mass term in the lagrangian density we have L mass = ( ML n c L n L + ML n L n c ( L) MR n R n c R + MR n c R n R) ( MD n c L nc R + MD ) ( n L n R MD n R n L + MD n c R L) nc (04) = ( n c L n R ) ( ) ( ) M L M D nl M D M R n c + h.c. (05) R which for M D = 0 is a pure Majorana lagrangian and for M L = M R = 0 and real M D represents the Dirac case. The mass matrix can be written as ( ) M M = L M D e iφ M D e iφ (06) M R taking M L and M R real and non-negative. This choice is possible without loss of generality because the phases can be absorbed into Υ and Υ (real must be replaced by hermitean if one includes families). This is a mixing problem with a symmetric (complex) mass matrix leading to two (real) mass eigenstates. The diagonalization is analogous to what was done for the Λ-matrices and one finds U M U T = M 0 with a (unitary) matrix U, which implies U M U = U M U = M 0 and a normal diagonalization of the (hermitean) matrix MM, U (MM ) U = M 0, (07) Thus one obtains from ( MM = ML + M D M D ( M L e iφ + M R e +iφ) M D ( M L e +iφ + M R e iφ) MR + M D ), (08) the eigenvalues M / = [ M L + M R + M D ] ± (ML M R ) + 4 M D (ML + M R + M LM R cos(φ)), (09) and we are left with two decoupled Majorana fields Υ and Υ, related via ( ) ( ) ( ) ( ) ΥL = U nl ΥR n c Υ L n c, = U L. (0) R Υ R n R for each of which one finds the lagrangians L = 4 Υ i i Υ i M i Υ i Υ i () for i =, with real masses M i. For the situation M L = 0 and M R M D (taking M D real) one finds M M D /M R and M M R. 7

21 Exercises Exercise Consider the case of the Weyl mode for symmetries. Prove that if the generators Q a generate a symmetry, i.e. [Q a, H] = 0, and a and a belong to the same multiplet (there is a Q a such that a = Q a a ) then H a = E a a implies that H a = E a a, i.e. a and a are degenerate states. Exercise Derive for the vector and axial vector currents, V µ = ψγ µ T ψ and A µ = ψγ µ γ 5 T ψ µ V µ µ A µ = i ψ [M, T ] ψ, = i ψ {M, T } γ 5 ψ. Exercise 3 (a) The coupling of the Z 0 particle to fermions is described by the vertex g ( ) i c f V cos θ γµ c f A γµ γ 5, W with C V = I 3 W Q sin θ W, C A = I 3 W. Write down the matrix element squared (averaged over initial spins and summed over final spins) for the decay of the Z 0. Neglect the masses of fermions and use the fact that the sum over polarizations is 3 ɛ (λ) µ (p)ɛ(λ) ν (p) = g µν + p µp ν M. λ= to calculate the width Γ(Z 0 ff), Γ(Z 0 ff) = M Z 48π g ( ) cos C f V θ + Cf A. W (b) Calculate the width to electron-positron pair, Γ(Z 0 e + e ), and the width to a pair of neutrino s, Γ(Z 0 ν e ν e ). The mass of the Z 0 is M Z = 9 GeV, the weak mixing angle is given by sin θ W = 0.3. Exercise 4 Calculate the lifetime τ = /Γ for the top quark (t) given that the dominant decay mode is t b + W +. In the standard model this coupling is described by the vertex i g (γµ γ µ γ 5 ). The masses are m t 75 GeV, m b 5 GeV and M W 80 GeV. 8

22 Exercise 5 Show that the coupling to the Higgs (W + W h, ZZh, hhh and e + e h) are proportional to the mass squared (bosons) or mass (fermions) of the particles. Note that you can find the answer without explicit construction of the interaction terms lagrangian. Exercise 6 Check that the Wolfenstein parametrization of the CKM matrix respects unitarity up to the required (which?) order in λ. Exercise 7 In this exercise two limits are investigated for the two-majorana case. (a) Calculate for the special choice M L = M R = 0 and M D real, the mass eigenvalues and show that the mixing matrix is U = ( ) i i which enables one to rewrite the Dirac field in terms of Majorana spinors. Give the explicit expressions that relate ψ and ψ c with Υ and Υ. (solution) One finds M = M = M D. For both left- and righthanded fields the relations between ψ, ψ c and Υ and Υ are the same, ψ = (Υ + i Υ ), ψ c = (Υ i Υ ). (b) A more interesting situation is 0 = M L < M D M R, which leads to the socalled seesaw mechanism. Calculate the eigenvalues M L = 0 and M R = M X. Given that neutrino masses are of the order of /0 ev, what is the mass M X if we take for M D the electroweak symmetry breaking scale v (about 50 GeV). (solution) The eigenvalues are M MD /M X and M M. For a neutrino mass of the order of /0 ev, and a fermion mass of the order of the electroweak breaking scaling 50 GeV, this leads to M X 0 5 GeV. The recoupling matrix in this case is ( ) i cos θs i sin θ U = S, sin θ S cos θ S with sin θ S M D /M X. The weak current couples to n L = sin θ S Υ i cos θ S Υ, where Υ is the light neutrino (mass) eigenstate. 9

23 A kinematics in scattering processes Phase space The -particle state is denoted p. It is determined by the energy-momentum four vector p = (E, p) which satisfies p = E p = m. A physical state has positive energy. The phase space is determined by the weight factors assigned to each state in the summation or integration over states, i.e. the - particle phase space is d 3 p (π) 3 E = This is generalized to the multi-particle phase space and the reduced phase space element by dr(p,..., p n ) = d 4 p (π) 4 θ(p0 ) (π)δ(p m ). () n i= d 3 p i (π) 3 E i, (3) dr(s, p,..., p n ) = (π) 4 δ 4 (P i p i ) dr(p,..., p n ), (4) which is useful because the total 4-momentum of the final state usually is fixed by overall momentum conservation. Here s is the invariant mass of the n-particle system, s = (p p n ). It is a useful quantity, for instance for determining the threshold energy for the production of a final state n. In the CM frame the threshold value for s obviously is ( n ) s threshold = m i. (5) For two particle states p a, p b we start with the four vectors p a = (E a, p a ) and p b = (E b, p b ) satisfying p a = m a and p b = m b, and the total momentum four-vector P = p a + p b. For two particles, the quantity s = P = (p a + p b ), (6) is referred to as the invariant mass squared. Its square root, s is for obvious reasons known as the center of mass (CM) energy. To be specific let us consider two frequently used frames. The first is the CM system. In that case i= p a = (Ea cm, q), (7) p b = (Eb cm, q), (8) It is straightforward to prove that the unknowns in the particular system can be expressed in the invariants (m a, m b and s). Prove that q = (s m a m b ) 4 m a m b 4s E cm a E cm b λ(s, m = a, m b ), (9) 4s = s + m a m b, (0) s = s m a + m b. () s The function λ(s, m a, m b ) is a function symmetric in its three arguments, which in the specific case also can be expressed as λ(s, m a, m b ) = 4(p a p b ) 4p a p b. 0

24 The second frame considered explicitly is the socalled target rest frame in which one of the particles (called the target) is at rest. In that case p a = (Ea trf a ), () p b = (m b, 0), (3) Also in this case one can express the energy and momentum in the invariants. Prove that E trf a p trf a = = s m a m b, (4) m b λ(s, m a, m b ). (5) m b One can, for instance, use the first relation and the abovementioned threshold value for s to calculate the threshold for a specific n-particle final state in the target rest frame, ( Ea trf (threshold) = ( ) m i ) m a m b. (6) m b i Explicit calculation of the reduced two-body phase space element gives dr(s, p, p ) = (π) d 3 p E d 3 p E δ 4 (P p p ) CM = (π) which using q d q = E de = E de gives = d 3 q 4 E E δ( s E E ) (π) dω(ˆq) q d q 4 E E δ( s E E ) dr(s, p, p ) = = q (π) dω(ˆq)d(e + E ) q 4π dω(ˆq) = s 4π 4(E + E ) δ( s E E ) λ 8π s dω(ˆq) 4π, (7) where λ denotes λ(s, m, m ). Kinematics of scattering processes The simplest scattering process is particles in and particles out. Examples appear in π + p π + p (8) π 0 + n (9) π + + π + n (30).... (3) The various possibilities are referred to as different reaction channels, where the first is referred to as elastic channel and the set of all other channels as the inelastic channels. Of course there are not only -particle channels. The initial state, however, usually is a -particle state, while the final state often arises from a series of -particle processes combined with the decay of an intermediate particle (resonance).

25 Consider the process a + b c + d. An often used set of invariants are the Mandelstam variables, s = (p a + p b ) = (p c + p d ) (3) t = (p a p c ) = (p b p d ) (33) u = (p a p d ) = (p b p c ) (34) which are not independent as s + t + u = m a + m b + m c + m d. The variable s is always larger than the minimal value (m a + m b ). A specific reaction channel starts contributing at the threshold value (Eq. 5). Instead of the scattering angle, which for the above process in the case of azimuthal symmetry is defined as ˆp a ˆp c = cos θ one can use in the CM the invariant t (p a p c ) CM = m a + m c E ae c + qq cos θ cm, with q = λ ab / s and q = λ cd / s. The minimum and maximum values for t correspond to θ cm being 0 or 80 degrees, t max min = m a + m c E a E c ± qq = m a + m c (s + m a m b )(s + m c m d ) s ± λλ s. (35) Using the relation between t and cos θ cm it is straightforward to express dω cm in dt, dt = qq d cos θ cm and obtain for the two-body phase space element dr(s, p c, p d ) = = q 4π s dt 8π = λ ab dω cm 4π = λcd 8π s dω cm 4π (36) dt 6π q s. (37) B Cross sections and lifetimes Scattering process For a scattering process a + b c +... (consider for convenience the rest frame for the target, say b) the cross section σ(a + b c +...) is defined as the proportionality factor in N c T = σ(a + b c +...) N b flux(a), where V and T indicate the volume and the time in which the experiment is performed, N c /T indicates the number of particles c detected in the scattering process, N b indicates the number of (target) particles b, which for a density ρ b is given by N b = ρ b V, while the flux of the beam particles a is flux(a) = ρ a va trf. The proportionality factor has the dimension of area and is called the cross section, i.e. σ = N T V ρ a ρ b va trf. (38) Although this at first sight does not look covariant, it is. N and T V are covariant. Using ρ trf ρ (0) a γ a = ρ (0) a Ea lab/m a (where ρ (0) a is the rest frame density) and va lab = p lab a /Elab a we have ρ a ρ b v trf a = ρ(0) a ρ (0) b λ ab 4 m a m b a = or with ρ (0) a = m a, σ = λ ab N T V. (39)

26 Decay of particles For the decay of particle a one has macroscopically dn dt = Γ N, (40) i.e. the amount of decaying particles is proportional to the number of particles with proportionality factor the em decay width Γ. From the solution one knows that the decay time τ = /Γ. Microscopically one has or N(t) = N(0) e Γ t (4) N decay T Γ = = N a Γ N T V ρ a. (4) This quantity is not covariant, as expected. The decay time for moving particles τ is related to the decay time in the rest frame of that particle (the proper decay time τ 0 ) by τ = γ τ 0. For the (proper) decay width one thus has Γ 0 = N m a T V. (43) Fermi s Golden Rule In both the scattering cross section and the decay constant the quantity N/T V appears. For this we employ in essence Fermi s Golden rule stating that when the S-matrix element is written as S fi = δ fi (π) 4 δ 4 (P i P f ) im fi (44) (in which we can calculate im fi using Feynman diagrams), the number of scattered or decayed particles is given by N = (π) 4 δ 4 (P i P f ) im fi dr(p,..., p n ). (45) One of the δ functions can be rewritten as T V (remember the normalization of plane waves), (π) 4 δ 4 (P i P f ) = (π) 4 δ 4 (P i P f ) = (π) 4 δ 4 (P i P f ) V,T V,T d 4 x e i(pi P f ) x d 4 x = V T (π) 4 δ 4 (P i P f ). (Using normalized wave packets these somewhat ill-defined manipulations can be made more rigorous). The result is N T V = M fi dr(s, p,..., p n ). (46) Combining this with the expressions for the width or the cross section one obtains for the decay width Γ = m body decay = dr(m, p,..., p n ) M (47) q 3π m dω M. (48) 3

27 The differential cross section (final state not integrated over) is given by and for instance for two particles dσ = q q dσ = λ ab M fi dr(s, p,..., p n ), (49) M(s, θ cm ) 8π s dω cm = π M(s, t) 4π λ ab dt. (50) This can be used to get the full expression for dσ/dt for eµ and e + e scattering, for which the amplitudes squared have been calculated in the previous chapter. The amplitude M/8π s is the one to be compared with the quantum mechanical scattering amplitude f(e, θ), for which one has dσ/dω = f(e, θ). The sign difference comes from the (conventional) sign in relation between S and quantummechanical and relativistic scattering amplitude, respectively. C Unitarity condition The unitarity of the S-matrix, i.e. (S ) fn S ni = δ fi implies for the scattering matrix M, [ δfn + i(π) 4 δ 4 (P f P n ) (M ) fn ] [ δni i(π) 4 δ 4 (P i P n ) M ni ] = δfi, or i [ M fi (M ] ) fi = (M ) fn (π) 4 δ 4 (P i P n ) M ni. (5) n Since the amplitudes also depend on all momenta the full result for two-particle intermediate states is (in CM, see 36) i [ M fi (M ] ) fi = dω(ˆq n )M nf (q q n f, q n ) 6π s M ni(q i, q n ). (5) Partial wave expansion n Often it is useful to make a partial wave expansion for the amplitude M(s, θ) or M(q i, q f ), M(s, θ) = 8π s l (l + )M l (s) P l (cos θ), (53) (in analogy with the expansion for f(e, θ) in quantum mechanics; note the sign and cos θ = ˆq i ˆq f ). Inserted in the unitarity condition for M, [ ] M i 8π s M 8π = M nf dω n s fi 8π q n M ni s π 8π s, n we obtain LHS = i l (l + ) P l (ˆq i ˆq f ) ( ) (M l ) fi (M l ) fi, while for the RHS use is made of P l (ˆq ˆq ) = m 4π l + Y m (l) (l) (ˆq) Y m (ˆq ) 4