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THE_STANDARD_MODEL

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Lecture notes by Dr T Teubner of the University of Liverpool, presented at the School for Experimental High Energy Physics Students, Somerville College, Oxford, September 2009. Chapters cover QED and non-abelian gauge theories, QCD, spontaneous symmetry breaking and the Higgs mechanism, the one-family Standard Model, additional generations and CP violation, neutrinos, and supersymmetry. It is a downloaded book in Phil's particle theory collection, not his own work.

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THE STANDARD MODEL By Dr T Teubner University of Liverpool Lecture presented at the School for Experimental High Energy Physics Students Somerville College, Oxford, September 2009 - 115 - /g1 - 116 - Contents Introduction .......................................................................................................... 119 1 QED as an Abelian Gauge Theory ..................................................... 121 1.1 Preliminaries............................................................................................ 121 1.2 Gauge Transformations ......................................................................... 121 1.3 Covariant Derivatives ............................................................................ 124 1.4 Gauge Fixing............................................................................................ 125 1.5 Summary .................................................................................................. 127 2 Non-Abelian Gauge Theories ............................................................. 129 2.1 Global Non-Abelian Transformations ................................................. 129 2.2 Non-Abelian Gauge Fields .................................................................... 131 2.3 Gauge Fixing............................................................................................ 133 2.4 The Lagrangian for a General Non-Abelian Gauge Theory ............. 134 2.5 Feynman Rules........................................................................................ 135 2.6 An Example ............................................................................................. 136 2.7 Summary .................................................................................................. 138 3 Quantum Chromodynamics ................................................................ 139 3.1 Running Coupling .................................................................................. 139 3.2 Quark (and Gluon) Confinement ......................................................... 142 3.3 /.notdef.g0001-Parameter of QCD ............................................................................... 144 3.4 Summary .................................................................................................. 145 4 Spontaneous Symmetry Breaking ...................................................... 146 4.1 Massive Gauge Bosons and Renormalizability .................................. 146 4.2 Spontaneous Symmetry Breaking ........................................................ 148 4.3 The Abelian Higgs Model...................................................................... 149 4.4 Goldstone Bosons.................................................................................... 151 4.5 The Unitary Gauge ................................................................................. 153 4.6 R /.notdef.g0002 Gauges (Feynman Gauge) ................................................................ 154 4.7 Summary .................................................................................................. 155 5 The Standard Model with one Family............................................... 157 5.1 Left- and Right- Handed Fermions ...................................................... 157 5.2 Symmetries and Particle Content ......................................................... 159 5.3 Kinetic Terms for the Gauge Bosons.................................................... 160 5.4 Fermion Masses and Yukawa Couplings............................................ 161 5.5 Kinetic Terms for Fermions................................................................... 163 5.6 The Higgs Part and Gauge Boson Masses........................................... 166 5.7 Classifying the Free Parameters ........................................................... 168 5.8 Summary .................................................................................................. 169 6 Additional Generations........................................................................ 177 6.1 A Second Quark Generation ................................................................. 177 6.2 Flavour Changing Neutral Currents.................................................... 179 6.3 Adding Another Lepton Generation.................................................... 180 6.4 Adding a Third Generation (of Quarks).............................................. 182 6.5 CP Violation............................................................................................. 184 6.6 Summary .................................................................................................. 187 - 117 - 7 Neutrinos................................................................................................. 189 7.1 Neutrino Oscillations ............................................................................. 189 7.2 Oscillations in Quantum Mechanics (in Vaccum and Matter) ......... 192 7.3 The See-Saw Mechanism ....................................................................... 196 7.4 Summary .................................................................................................. 198 8 Supersymmetry ...................................................................................... 199 8.1 Why Supersymmetry?............................................................................ 199 8.2 A New Symmetry: Boson /.notdef.g0001 Fermion ................................................ 200 8.3 The Supersymmetric Harmonic Oscillator.......................................... 202 8.4 Supercharges ........................................................................................... 204 8.5 Superfields ............................................................................................... 205 8.6 The MSSM Particle Content (Partially)................................................ 207 8.7 Summary .................................................................................................. 208 Acknowledgements ............................................................................................. 209 - 118 - Introduction An important feature of the Standard Model (SM ) is that “it works”: it is consistent with, or veriﬁed by, all available data, with no compelling evidence for physics beyond.1Secondly, it is a uniﬁed description, in terms of “gauge theories” of all the interactions of known particles (except gravity). A gauge theory is one that possesses invariance under a set of “local transformations”, i.e. transformations whose parameters are space-time dependent. Electromagnetism is a well-known example of a gauge theory. In this case the gauge trans- formations are local complex phase transformations of the ﬁelds of charged particles, and gauge invariance necessitates the introduction of a massless vector (spin-1) particle, called the photon, whose exchange mediates the electromagnetic interactions. In the 1950’s Yang and Mills considered (as a pur ely mathematical exercise) extending gauge invariance to include local non-abelian (i.e . non-commuting) transformations such as SU(2). In this case one needs a set of massless vector ﬁelds (three in the case of SU(2)), which were formally called “Yang-Mills” ﬁelds, but are now known as “gauge ﬁelds”. In order to apply such a gauge theory to weak interactions, one considers particles which transform into each other under the weak interaction, such as a u-quark and a d-quark, or an electron and a neutrino, to be arranged in doublets of weak isospin. The three gauge bosons are interpreted as the W±andZbosons, that mediate weak interactions in the same way that the photon mediates electromagnetic interactions. The diﬃculty in the case of weak interactions was that they are known to be short range, me- diated by very massive vector bosons, whereas Yang-Mills ﬁelds are required to be massless in order to preserve gauge invariance. The apparent paradox was solved by the applica- tion of the “Higgs mechanism”. This is a prescription for breaking the gauge symmetryspontaneously. In this scenario one starts with a theory that possesses the required gauge invariance, but where the ground state of the theory is notinvariant under the gauge transfor- mations. The breaking of the invariance arises in the quantization of the theory, whereas theLagrangian only contains terms which areinvariant. One of the consequences of this is that the gauge bosons acquire a mass and the theory can thus be applied to weak interactions. Spontaneous symmetry breaking and the Higgs mechanism have another extremely impor- tant consequence. It leads to a renormalizable theory with massive vector bosons. This means that one can carry out a programme of re normalization in which the inﬁnities that 1In saying so we have taken the liberty to allow for neutrino masses (see chapter 7) and discarded some deviations in electroweak precision measurements which are far from conclusive; however, note that there is a 3−4σdeviation between measurement and SM prediction of g−2 of the muon, see the remarks in chapter 8. - 119 - arise in higher-order calculations can be reabsorbed into the parameters of the Lagrangian (as in the case of QED). Had one simply broken the gauge invariance explicitly by addingmass terms for the gauge bosons, the resulting theory would not have been renormalizable and therefore could not have been used to carry out perturbative calculations. A consequence of the Higgs mechanism is the existence of a scalar (spin-0) particle, the Higgs boson. The remaining step was to apply the ideas of gauge theories to the strong interaction. The gauge theory of the strong interaction is calle d “Quantum Chromo Dynamics” (QCD). In this theory the quarks possess an internal property called “colour” and the gauge transformationsare local transformations between quarks of diﬀerent colours. The gauge bosons of QCD are called “gluons” and they medi ate the strong interaction. The union of QCD and the electroweak gauge theory, which describes the weak and elec-tromagnetic interactions, is known as the Standard Model. It has a very simple structure and the diﬀerent forces of nature are treated in the same fashion, i.e. as gauge theories.It has eighteen fundamental parameters, mos t of which are associated with the masses of the gauge bosons, the quarks and leptons, and the Higgs. Nevertheless these are not allindependent and, for example, the ratio of the WandZboson masses are (correctly) pre- dicted by the model. Since the theory is renormalizable, perturbative calculations can beperformed at higher order that predict cross sections and decay rates for both strongly and weakly interacting processes. These predictions, when confronted with experimental data, have been conﬁrmed very successfully. As both predictions and data are becoming more and more precise, the tests of the Standard Mode l are becoming increasingly stringent. - 120 - 1 QED as an Abelian Gauge Theory The aim of this lecture is to start from a symmetry of the fermion Lagrangian and show that “gauging” this symmetry (= making it well behaved) implies classical electromagnetismwith its gauge invariance, the e¯eγinteraction, and that the photon must be massless. 1.1 Preliminaries In the Field Theory lectures at this school, the quantum theory of an interacting scalar ﬁeld was introduced, and the voyage from the Lagrangian to the Feynman rules was made. Fermions can be quantised in a similar way, and the propagators one obtains are the Green functions for the Dirac wave equation (the inverse of the Dirac operator) of the QED/QCD course. In this course, I will start from the Lagrangian (as opposed to the wave equation) of a free Dirac fermion, and add interactions, to construct the Standard Model Lagrangian in classical ﬁeld theory. That is, the ﬁelds are treated as functions, and I will not discuss creation and annihiliation operators. However, to extra ct Feynman rules from the Lagrangian, I will implicitly rely on the rules developed for scalar ﬁelds in the Field Theory course. 1.2 Gauge Transformations Consider the Lagrangian density for a free Dirac ﬁeld ψ: L= ψ(iγμ∂μ−m)ψ (1.1) This Lagrangian density is invariant under a phase transformation of the fermion ﬁeld ψ→eiQωψ, ψ→eiQω ψ, (1.2) where Qis the charge operator ( Qψ=+ψ,Q¯ψ=−¯ψ),ωis a real constant (i.e. independent ofx)a n d ψis the conjugate ﬁeld. The set of all numbers e−iωform a group2. This particular group is “abelian” which is to say that any two elements of the group commute. This just means that e−iω1e−iω2=e−iω2e−iω1. (1.3) 2A group is a mathematical term for a set, where multiplication of elements is deﬁned and results in another element of the set. Furthermore, there has to be a 1 element (s.t. 1 ×a=a)a n da ni n v e r s e( s . t . a×a−1= 1) for each element aof the set. - 121 - This particular group is called U(1) which means the group of all unitary 1 ×1 matrices. A unitary matrix satisﬁes U+=U−1withU+being the adjoint matrix. We can now state the invariance of the Lagrangian eq. (1.1) under phase transformations in a more fancy way by saying that the Lagrangian is invariant under global U(1) transformations. By global we mean that ωdoes not depend on x. For the purposes of these lectures it will usua lly be suﬃcient to consider inﬁnitesimal group transformations, i.e. we assume that the parameter ωis suﬃciently small that we can expand inωand neglect all but the linear term. Thus we write e−iω=1−iω+O(ω2). (1.4) Under such inﬁnitesimal phase transformations the ﬁeld ψchanges according to ψ→ψ+δψ=ψ+iQ ωψ, (1.5) and the conjugate ﬁeld ψby ψ→ ψ+δ ψ= ψ+iQ ω ψ= ψ−iω ψ, (1.6) such that the Lagrangian density remains unchanged (to order ω). At this point we should note that global tra nsformations are not very attractive from a theoretical point of view. The reason is that making the same transformation at everyspace-time point requires that all these points ’know’ about the transformation. But if I were to make a certain transformation at the top of Mont Blanc, how can a point somewhere in England know about it? It would take some time for a signal to travel from the Alps to England. Thus, we have two options at this point. Either, we simply note the invariance of eq. (1.1) under global U(1) transformations and put this aside as a curiosity, or we insist that in- variance under gauge transformations is a funda mental property of nature. If we take the latter option we have to require invariance unde r local transformations. Local means that the parameter of the transformation, ω, now depends on the space-time point x. Such local (i.e. space-time dependent) transformatio ns are called “gauge transformations”. If the parameter ωdepends on the space-time point then the ﬁeld ψtransforms as follows under inﬁnitesimal transformations δψ(x)=iω(x)ψ(x); δ ψ(x)=−iω(x) ψ(x). (1.7) Note that the Lagrangian density eq. (1.1) now is no longer invariant under these trans- formations, because of the partial derivative between ψandψ. This derivative will act on - 122 - the space-time dependent parameter ω(x) such that the Lagrangian density changes by an amount δL,w h e r e δL=− ψ(x)γμ[∂μQω(x)]ψ(x). (1.8) The square brackets in [ ∂μQω(x)] are introduced to indicate that the derivative ∂μacts only inside the brackets. It turns out that we can restore gauge invariance if we assume that the fermion ﬁeld interacts with a vector ﬁeld Aμ, called a “gauge ﬁeld”, with an interaction term −e ψγμAμQψ (1.9) added to the Lagrangian density which now becomes L= ψ(iγμ(∂μ+ieQ A μ)−m)ψ. (1.10) In order for this to work we must also assume that apart from the fermion ﬁeld transform- ing under a gauge transformation according to eq. (1.7) the gauge ﬁeld, Aμ, also changes according to −eQA μ→−eQ(Aμ+δAμ(x)) =−eQ A μ+Q∂μω(x). (1.11) SoδAμ(x)=−Q∂μω(x)/e. Exercise 1.1 Using eqs. (1.7) and (1.11) show that under a gauge transformation δ(−e ψγμAμψ)= ψ(x)γμ[∂μQω(x)]ψ(x). This change exactly cancels with eq. (1.8), so that once this interaction term has been added the gauge invariance is restored. We recognize eq. (1.10) as being the fermionic part of the Lagrangian density for QED, where eis the electric charge of the fermion and Aμis the photon ﬁeld. In order to have a proper quantum ﬁeld theory, in which we can expand the photon ﬁeld Aμ in terms of creation and annihilation operators for photons, we need a kinetic term for the photon, i.e. a term which is quadratic in the derivative of the ﬁeld Aμ. Without such a term the Euler-Lagrange equation for the gauge ﬁeld would be an algebraic equation and we coulduse it to eliminate the gauge ﬁeld altogether from the Lagrangian. We need to ensure thatin introducing a kinetic term we do not spoil the invariance under gauge transformations. This is achieved by deﬁning the ﬁeld strength tensor, F μν,a s Fμν≡∂μAν−∂νAμ, (1.12) where the derivative is understood to act on the A-ﬁeld only.3It is easy to see that under the gauge transformation eq. (1.11) each of th e two terms on the right hand side of eq. (1.12) 3Strictly speaking we should therefore write Fμν=[∂μAν]−[∂νAμ]; you will ﬁnd that the brackets are often omitted. - 123 - change, but the changes cancel out. Thus we may add to the Lagrangian any term which depends on Fμν(and which is Lorentz invariant, thus, with all Lorentz indices contracted). Such a term is aFμνFμν. This gives the desired term which is quadratic in the derivative of the ﬁeld Aμ. If we choose the constant ato be−1/4 then the Lagrange equations of motion match exactly (the relativistic formulation of) Maxwell’s equations.4 We have thus arrived at the Lagrangian density for QED, but from the viewpoint of de-manding invariance under U(1) gauge transformations rather than starting with Maxwell’s equations and formulating the equivalent quantum ﬁeld theory. The Lagrangian density for QED is: L=−1 4FμνFμν+ ψ(iγμ(∂μ+ieQ A μ)−m)ψ. (1.13) Exercise 1.2 Starting with the Lagrangian density for QED write down the Euler-Lagrange equations for the gauge ﬁeld Aμand show that this results in Maxwell’s equa- tions. In the Field Theory lectures, we have seen that a term λφ4in the Lagrangian gave 4! λas the coupling of four φs in perturbation theory. Neglecting the combinatoric factors, it is plausible that eq. (1.13) gives the γ¯eeF e y n m a nR u l eu s e di nt h eQ E Dc o u r s e , −ieγμ,f o r negatively charged particles. Note that we are notallowed to add a mass term for the photon. A term such as M2AμAμ added to the Lagrangian density is not invari ant under gauge transformations as it would lead to δL=2M2 eAμ(x)∂μω(x)/negationslash=0. (1.14) Thus the masslessness of the photon can be unde rstood in terms of the requirement that the Lagrangian be gauge invariant. 1.3 Covariant Derivatives Before leaving the abelian case, it is useful to introduce the concept of a “covariant deriva- tive”. This is not essential for abelian gauge theories, but will be an invaluable tool when we extend these ideas to non-abelian gauge theories. 4The determination of this constant ais the onlyplace that a match to QED has been used. The rest of the Lagrangian density is obtained purely from the requirement of local U(1) invariance. A diﬀerent constant would simply mean a diﬀerent normalization of the photon ﬁeld. - 124 - The covariant derivative Dμis deﬁned to be Dμ≡∂μ+ieA μ. (1.15) It has the property that given the transformations of the fermion ﬁeld eq. (1.7) and the gauge ﬁeld eq. (1.11) the quantity Dμψtransforms in the same way under gauge transformations asψ. Exercise 1.3 Show that under an inﬁnitesimal gauge transformation Dμψtransforms as Dμψ→Dμψ+δ(Dμψ)w i t h δ(Dμψ)=iω(x)Dμψ. We may thus rewrite the QED Lagrangian density as L=−1 4FμνFμν+ ψ(iγμDμ−m)ψ. (1.16) Furthermore the ﬁeld strength Fμνcan be expressed in terms of the commutator of two covariant derivatives, i.e. Fμν=−i e[Dμ,Dν]=−i e[∂μ,∂ν]+[∂μ,Aν]+[Aμ,∂ν]+ie[Aμ,Aν] =∂μAν−∂νAμ, (1.17) where in the last line we have adopted the conventional notation again and left out the square brackets. Notice that when using eq. (1.17) the derivatives act only on the A-ﬁeld. 1.4 Gauge Fixing The guiding principle of this chapter has been to hold onto the U(1) symmetry. This forced us to introduce a new massless ﬁeld Aμwhich we could interpret as the photon. In this subsection we will try to quantise the photon ﬁeld (e.g. calculate its propagator) by naivelyfollowing the prescription used for scalars and fermions, which will not work. This should not be surprising, because A μhas four real components, introduced to maintain gauge symmetry. However the physical photon has two polarisation states. This diﬃculty can be resolved by“ﬁxing the gauge” (breaking our precious gauge symmetry) in the Lagrangian in such a way as to maintain the gauge symmetry in observables. 5 5The gauge symmetry is also preserved in the Path Integral, which is a sum over all ﬁeld conﬁgurations weighted by exp {i/integraltext Ld4x}. In path integral quantisation, which is an alternative to the canonical approach used in the Field Theory lectures, Green functions are calculated from the path integral and it is unimportantthat the gauge symmetry seems broken in the Lagrangian. - 125 - In general, if the part of the action that is quadratic in some ﬁeld φ(x) is given in terms of the Fourier transform ˜φ(p)b y Sφ=/integraldisplay d4p˜φ(−p)O(p)˜φ(p), (1.18) then the propagator for the ﬁeld φmay be written as iO−1(p). (1.19) In the case of QED the part of the Lagrangian that is quadratic in the photon ﬁeld is given by−1/4FμνFμν=−1/2Aμ(−gμν∂σ∂σ+∂μ∂ν)Aν, where we have used partial integration to obtain the second expression. In momentum space, the quadratic part of the action is then given by SA=/integraldisplay d4p1 2˜Aμ(−p)/parenleftBig −gμνp2+pμpν/parenrightBig˜Aν(p). (1.20) Unfortunately the operator ( −gμνp2+pμpν) does not have an inverse. This can be most easily seen by noting ( −gμνp2+pμpν)pν= 0. This means that the operator ( −gμνp2+pμpν) has an eigenvector ( pν) with eigenvalue 0 and is therefore not invertible. Thus it seems we are not able to write down the propagator of the photon. We solve this problem by adding to the Lagrangian density a gauge ﬁxing term −1 2(1−ξ)(∂μAμ)2. (1.21) With this term included (again in momentum space), SAbecomes SA=/integraldisplay d4p1 2˜Aμ(−p)/parenleftBigg −gμνp2−ξ 1−ξpμpν/parenrightBigg ˜Aν(p), (1.22) and, noting the relation /parenleftBigg gμνp2+ξ 1−ξpμpν/parenrightBigg/parenleftBigg gνρ−ξpνpρ p2/parenrightBigg =p2gρ μ, (1.23) we see that the propagator for the photon may now be written as −i/parenleftBigg gμν−ξpμpν p2/parenrightBigg1 p2. (1.24) The special choice ξ= 0 is known as the Feynman gauge. In this gauge the propagator eq. (1.24) is particularly simple a nd we will use it most of the time. This procedure of gauge ﬁxing seems strange: ﬁrst we worked hard to get a gauge invariantLagrangian, and then we spoil gauge invariance by introducing a gauge ﬁxing term. The point is that we have to ﬁx the gauge in order to be able to perform a calculation. Once we have computed a physical quantity, th e dependence on the gauge cancels. In other - 126 - words, it does not matter how we ﬁx the gauge, and in particular, what value for ξwe take. The choice ξ= 0 is simply a matter of convenience. A more careful procedure would be to leaveξarbitrary and check that all ξ-dependence in the ﬁnal result cancels. This gives us a strong check on the calculation, however, at the price of making the computation much more tedious. The procedure of ﬁxing the gauge in order to be able to perform a calculation, even though the ﬁnal result does not depend on how we have ﬁxed the gauge, can be understood by the following analogy. Assume we wanted to calculate some scalar quantity (say the time it takes for a point mass to get from one point t o another) in our ordinary 3-dimensional Euclidean space. To do so, we choose a coordinate system, perform the calculation andget our ﬁnal result. Of course, the result does not depend on how we choose the coordinate system, but in order to be able to perform the c alculation we have to ﬁx it somehow. Picking a coordinate system corresponds to ﬁxing a gau ge, and the independence of the result on the coordinate system chosen corresponds to the gauge invariance of physical quantities. To takethis one step further we remark that not all quantities are independent of the coordinate system. For example, the x-coordinate of the position of the point mass at a certain time depends on our choice. Similarly, there are important quantities that are gauge dependent.One example is the gauge boson propagator given in eq. (1.24). However, all measurable quantities (observables) are gauge invariant. This is where our analogy breaks down: in our Euclidean example there are measurable qu antities that do depend on the choice of the coordinate system. Finally we should mention that eq. (1.21) is by far not the only way to ﬁx the gauge but it will be suﬃcient for these lectures to consider gauges deﬁned through eq. (1.21). These gauges are called covariant gauges. 1.5 Summary •It is possible for the Lagrangian for a (com plex) Dirac ﬁeld to be invariant under localU(1) transformations (phase rotations) , in which the phase parameter depends on space-time. In order to accomplish this we include an interaction with a vector gauge boson which transforms under the loca l (gauge) transformation according to eq. (1.11). •This interaction is encoded by replacing the derivative ∂μby the covariant derivative Dμdeﬁned by eq. (1.15). Dμψtransforms under gauge transformations as e−iωDμψ. •The kinetic term for the gauge boson is −1 4FμνFμν,w h e r e Fμνis proportional to the - 127 - commutator [ Dμ,Dν] and is invariant under gauge transformations. •The gauge boson must be massless, since a term proportional to AμAμisnotinvariant under gauge transformations and hence not included in the Lagrangian. •The resulting Lagrangian is identical to that of QED. •In order to deﬁne the propagator we have to specify a certain gauge; the resulting gauge dependence cancels in physical observables. - 128 - 2 Non-Abelian Gauge Theories In this lecture, the “gauge” concept will be constructed so that the gauge bosons have self- interactions — as are observed among the gluons of QCD, and the W±,Zandγof the electroweak sector. However, the gauge bosons will still be massless. (We will see how togive the W ±andZtheir observed masses in the Higgs chapter.) 2.1 Global Non-Abelian Transformations We apply the ideas of the previous lecture t o the case where the tra nsformations do not commute with each other, i.e. the group is “non-abelian”. Consider nfree fermion ﬁelds {ψi}, arranged in a multiplet ψ: ψ=⎛ ⎜⎜⎜⎜⎜⎜⎜⎜⎝ψ1 ψ2 . . ψn⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎠ (2.1) for which the Lagrangian density is L= ψ(iγμ∂μ−m)ψ, ≡ ψi(iγμ∂μ−m)ψi, (2.2) where the index iis summed from 1 to n. Eq. (2.2) is therefore a shorthand for L= ψ1(iγμ∂μ−m)ψ1+ ψ2(iγμ∂μ−m)ψ2+.... (2.3) The Lagangian density (2.2) is invariant under (space-time independent ) complex rotations inψispace: ψ→Uψ, ¯ψ→¯ψU†, (2.4) where Uis ann×nmatrix such that UU†=1, det[U]=1. (2.5) The transformation (2.4) is called an internal symmetry, which rotates the ﬁelds (e.g. quarks of diﬀerent colour) among themselves. The group of matrices satisfying the conditions (2.5) is called SU(n). This is the group of special, unitary n×nmatrices. Special in this context means that the determinant is - 129 - equal to 1. In order to specify an SU(n) matrix completely we need n2−1 real parameters. Indeed, we need 2 n2real parameters to determine an arbitrary complex n×nmatrix. But there are n2constraints due to the unitary requirements and one additional constraint due to the requirement det = 1. An arbitrary SU(n) matrix can be written as U=e−i/summationtextn2−1 a=1ωaTa≡e−iωaTa(2.6) where we again have adopted Einstein’s summation convention. The ωa,a∈{1...n2−1}, are real parameters, and the Taare called the generators of the group. Exercise 2.1 Show that the unitarity of the SU(n) matrices entails hermiticity of the gen- erators and that the requirement of det = 1 means that the generators have to be traceless. I nt h ec a s eo f U(1) there was just one generator. Here we have n2−1 generators Ta. There is still some freedom left of how to normalize the generators. We will adopt the usualnormalization convention tr(T aTb)=1 2δab. (2.7) The reason we can always enforce eq. (2.7) is that tr( TaTb) is a real matrix symmetric in a↔b. Thus it can be diagonalized. If you have problems getting on friendly terms with the concept of generators, for the moment you can think of them as traceless, hermitian n×n matrices. (This is, however, not the complete picture.) The crucial new feature of the group SU(n) is that two elements of SU(n) generally do not commute, i.e. e−iωa 1Tae−iωb 2Tb/negationslash=e−iωb 2Tbe−iωa 1Ta(2.8) (compare to eq. (1.3)). To put this in a diﬀerent way, the group algebra is not trivial. For the commutator of two generators we have [Ta,Tb]≡ifabcTc/negationslash=0 ( 2 . 9 ) where we deﬁned the structure constants of the group, fabc, and used the summation conven- tion again. The structure constants are totally antisymmetric. This can be seen as follows: from eq. (2.9) it is obvious that fabc=−fbac. To convince us of the antisymmetry in the other indices as well, we note that multiplying eq. (2.9) by Tdand taking the trace, using eq. (2.7), we get 1 /2ifabd= tr( TaTbTd)−tr(TbTaTd) = tr( TaTbTd)−tr(TaTdTb). - 130 - 2.2 Non-Abelian Gauge Fields Now suppose we allow the transformation Uto depend on space-time. Then the Lagrangian density changes by δLunder this “non-abelian gau ge transformation”, where δL= ψU†γμ(∂μU)ψ. (2.10) The local gauge symmetry can be restored by introducing a covariant derivative Dμ, giving interactions with gauge bosons, such that DμU(x)ψ(x)=U(x)Dμψ(x). (2.11) This is like the electromagnetic case, except that Dμis now a matrix, iDμ=iI∂μ−gAμ (2.12) where Aμ=TaAa μ.I t c o n t a i n s n2−1 vector (spin one) gauge bosons, Aaμ,o n ef o re a c h generator of SU(n). Under a gauge transformation U,Aμshould transform as Aμ→UA μU†+i g(∂μU)U†. (2.13) This ensures that the Lagrangian density L= ψ(iγμDμ−m)ψ (2.14) is invariant under local SU(n) gauge transformations. It can be checked that eq. (2.13) reduces to the gauge transformation of el ectromagnetism in the abelian limit. Exercise 2.2 (For algebraically ambitious people): p erform an inﬁnitesimal gauge transfor- mation on ψ,¯ψandD, using (2.6), and show that to linear order in the ωa, ¯ψγμDμψis invariant. Exercise 2.3 Show that in the SU(2) case, the covariant derivative is iDμ=⎛ ⎝i∂μ−g 2W3 μ−g 2(W1 μ−iW2 μ) −g 2(W1 μ+iW2 μ)i∂μ+g 2W3 μ⎞⎠ , and ﬁnd the usual charged current interactions for the lepton doublet ψ=⎛ ⎝ν e⎞⎠ by deﬁning W±=(W1∓iW2)/√ 2. - 131 - Exercise 2.4 Include the U(1) hypercharge interaction in t he previous question; show that the covariant derivative acting on the lepton doublet (of hypercharge Y= −1/2) is iDμ=⎛ ⎝i∂μ−g 2W3 μ−g/primeYBμ−g 2(W1 μ−iW2 μ) −g 2(W1 μ+iW2 μ)i∂μ+g 2W3 μ−g/primeYBμ⎞⎠ . Deﬁne⎛ ⎝Zμ Aμ⎞⎠ =⎛⎝ cosθW−sinθW sinθWcosθW⎞⎠⎛⎝ W3 μ Bμ⎞⎠ and write the diagonal (neutral) interactions in terms of ZμandAμ. Extract sinθWin terms of gandg/prime. (Recall that the photon does not interact with the neutrino.) The kinetic term for the gauge bosons is again constructed from the ﬁeld strengths Fa μνwhich are deﬁned from the commutator of two covariant derivatives, Fμν=−i g[Dμ,Dν], (2.15) where the matrix Fμνis given by Fμν=TaFa μν, (2.16) with Fa μν=∂μAa ν−∂νAaμ−gfabcAbμAcν. (2.17) Notice that Fμνis gauge variant , unlike the U(1) case. We know the transformation of D from (2.13), so [Dμ,Dν]→U[Dμ,Dν]U†. (2.18) The gauge invariant kinetic term for the gauge bosons is therefore −1 2Tr F μνFμν=−1 4Fa μνFaμ ν, (2.19) where the trace is in SU(n) space, and summation over the index ais implied. In sharp contrast with the abelian case, this term does not only contain terms which are quadratic in the derivatives of the gauge boson ﬁelds, but also the terms gfabc(∂μAa ν)AbμAcν−1 4g2fabcfadeAbμAcνAdμAeν. (2.20) This means that there is a very important diﬀe rence between abelian and non-abelian gauge theories. For non-abelian gauge theories the gauge bosons interact with each other via both - 132 - three-point and four-point interaction terms. The three point interaction term contains a derivative, which means that the Feynman rule for the three-point vertex involves themomenta of the particles going into the vertex. We shall write down the Feynman rules in detail later. Once again, a mass term for the gauge bosons is forbidden, since a term proportional to A a μAaμisnotinvariant under gauge transformations. 2.3 Gauge Fixing As in the case of QED, we need to add a gauge-ﬁxing term in order to be able to derive a propagator for the gauge bosons. In Feynman gauge this means adding the term −1 2(∂μAa μ)2 to the Lagrangian density, and the propagator (in momentum space) becomes −iδabgμν p2. There is one unfortunate complic ation, which is mentioned brieﬂy here for the sake of com- pleteness, although one only needs to know abo ut it for the purpose of performing higher loop calculations with non-abelian gauge theories: If one goes through the formalism of gauge-ﬁxing carefully, it turns out that at higher orders extra loop diagrams emerge. These diagrams involve additional particles that are mathematically equivalent to interacting scalar particles and are known as a “Faddeev-Popov ghosts”. For each gauge ﬁeld there is such a ghost ﬁeld. These are notto be interpreted as physical scalar particles which could in principle be observed experimentally, but merelyas part of the gauge-ﬁxing programme. For this reason they are referred to as “ghosts”. Furthermore they have two peculiarities: 1. They only occur inside loops. This is beca use they are not really particles and cannot occur in initial or ﬁnal states, but are introduced to clean up a diﬃculty that arises in the gauge-ﬁxing mechanism. 2. They behave like fermions even though they are scalars (spin zero). This means that we need to count a minus sign for each loop of Faddeev-Popov ghosts in any Feynman diagram. We shall display the Feynman rules for these ghosts later. Thus, for example, the Feynman diagrams which contribute to the one-loop corrections to the gauge boson propagator are - 133 - +- - (a) (b) (c) (d) Diagram (a) involves the three-point interaction between the gauge bosons, diagram (b) involves the four-point interaction between the gauge bosons, diagram (c) involves a loop of fermions, and diagram (d) is the extra diagram involving the Faddeev-Popov ghosts. Note that both diagrams (c) and (d) have a minus sign in front of them because both fermionsand Faddeev-Popov ghosts obey Fermi statistics. 2.4 The Lagrangian for a General Non-Abelian Gauge Theory Let us summarize what we have found so far: Consider a gauge group Gof “dimension” N(forSU(n):N≡n2−1), whose Ngenerators, Ta, obey the commutation relations/bracketleftBig Ta,Tb/bracketrightBig =ifabcTc,w h e r e fabcare called the “structure constants” of the group. The Lagrangian density for a gauge theory with this group, with a fermion multiplet ψi,i s given (in Feynman gauge) by L=−1 4Fa μνFaμ ν+i ψ(γμDμ−mI)ψ−1 2(∂μAa μ)2+LFP (2.21) where Fa μν=∂μAaν−∂νAaμ−gfabcAbμAcν, (2.22) Dμ=∂μI+igTaAaμ(2.23) and LFP=−ξa∂μ∂μηa+gfacbξaAcμ(∂μηb). (2.24) Under an inﬁnitesimal gauge transformation the Ngauge bosons Aa μchange by an amount that contains a term which is not linear in Aa μ: δAa μ(x)=−fabcAbμ(x)ωc(x)+1 g∂μωa(x), (2.25) whereas the ﬁeld strengths Fa μνtransform by a change δFa μν(x)=−fabcFb μν(x)ωc. (2.26) - 134 - In other words, they transform as the “adjoint” representation of the group (which has as many components as there are generators). This means that the quantity Fa μνFaμ ν(summa- tion over a, μ, ν implied) is invariant unde r gauge transformations. 2.5 Feynman Rules The Feynman rules for such a gauge theory can be read oﬀ directly from the Lagrangian. As mentioned previously, the propagators are ob tained by taking all terms bilinear in the ﬁeld and inverting the corresponding operator (and multiplying by i). The rules for the vertices are obtained by simply taking ( itimes) the factor which multiplies the corresponding term in the Lagrangian. The explicit rules are given in the following. Vertices: (Note that all momenta are deﬁned as ﬂowing into the vertex!) μa p1 ρcp3 νbp2−gfabc/parenleftBig gμν(p1−p2)ρ+gνρ(p2−p3)μ+gρμ(p3−p1)ν/parenrightBig −ig2feabfecd(gμρgνσ−gμσgνρ) −ig2feacfebd(gμνgρσ−gμσgνρ) −ig2feadfebc(gμνgρσ−gμρgνσ)μa νb σd ρc μa j i−igγμ(Ta)ij μa c bqgfabcqμ - 135 - Propagators: Gluon: −iδabgμν/p2pa μbν Fermion: iδij(γμpμ+m)/(p2−m2)pij Faddeev-Popov ghost: iδab/p2 pab 2.6 An Example As an example of the application of these Feynman rules, we consider the process of Compton scattering, but this time for the scattering of non-abelian gauge bosons and fermions, ratherthan photons. We need to calculate the amplitude for a gauge boson of momentum p 2and gauge label ato scatter oﬀ a fermion of momentum p1and gauge label iproducing a fermion of momentum p3and gauge label jand a gauge boson of momentum p4and gauge label b.N o t e t h a t i, j∈{1...n}whereas a, b∈{1...n2−1}. In addition to the two Feynman diagrams one gets in the QED case there is a third diagram involving the self-interaction of the gauge bosons. ik jp1 p3(p1+p2)p2 p4 μνab ik jp1 p3p2 p4 μνab ijp1 p3(p4−p2)ρc σp2 p4 ab μν (a) (b) (c) We will assume that the fermions are massless (i.e. that we are at suﬃciently high energies so that we may neglect their masses), and work in terms of the Mandelstam variables s=(p1+p2)2=(p3+p4)2, t=(p1−p3)2=(p2−p4)2, u=(p1−p4)2=(p2−p3)2. The polarizations are accounted for by contracting the amplitude obtained for the abovediagrams with the polarization vectors /epsilon1 μ(λ2)a n d /epsilon1ν(λ4). Each diagram consists of two vertices and a propagator and so their contributions can be read oﬀ from the Feynman rules. - 136 - For diagram (a) we get /epsilon1μ(λ2)/epsilon1ν(λ4) uj(p3)/parenleftBig −igγν(Tb)k j/parenrightBig/parenleftBigg iγ·(p1+p2) s/parenrightBigg/parenleftBig −igγμ(Ta)ik/parenrightBig ui(p1) =−ig2 s/epsilon1μ(λ2)/epsilon1ν(λ4) u(p3)(γνγ·(p1+p2)γμ)/parenleftBig TbTa/parenrightBig u(p1). For diagram (b) we get /epsilon1μ(λ2)/epsilon1ν(λ4) uj(p3)/parenleftBig −igγμ(Ta)k j/parenrightBig/parenleftBigg iγ·(p1−p4) u/parenrightBigg/parenleftBig −igγν(Tb)ik/parenrightBig ui(p1) =−ig2 u/epsilon1μ(λ2)/epsilon1ν(λ4) u(p3)(γνγ·(p1−p4)γμ)/parenleftBig TaTb/parenrightBig u(p1). Note that here the order of the Tmatrices is the other way around compared to diagram (a). Diagram (c) involves the three-point gauge-boson self-coupling. Since the Feynman rule for this vertex is given with incoming momenta, it is useful to replace the outgoing gauge-boson momentum p4by−p4and understand this to be an incoming momentum. Note that the internal gauge-boson line carries momentum p4−p2coming into the vertex. The three incoming momenta that are to be substituted into the Feynman rule for the vertex aretherefore p 2,−p4,p4−p2. The vertex thus becomes −gfabc(gμν(p2+p4)ρ+gρν(p2−2p4)μ+gμρ(p4−2p2)ν), and the diagram gives /epsilon1μ(λ2)/epsilon1ν(λ4) uj(p3)/parenleftBig −igγ σ(Tc)i j/parenrightBig ui(p1)/parenleftbigg −igρσ t/parenrightbigg ×(−gfabc)(gμν(p2+p4)ρ+gρν(p2−2p4)μ+gμρ(p4−2p2)ν) =−ig2 t/epsilon1μ(λ2)/epsilon1ν(λ4) u(p3)/bracketleftBig Ta,Tb/bracketrightBig γρu(p1)/parenleftBig gμν(p2+p4)ρ−2(p4)μgνρ−2(p2)νgμρ/parenrightBig , where in the last step we have used the commutation relation eq. (2.9) and the fact that the polarization vectors are transverse so that p2·/epsilon1(λ2)=0a n d p4·/epsilon1(λ4)=0 . Exercise 2.4Draw all the Feynman diagrams for the tree level amplitude for two gauge bosons with momenta p 1andp2to scatter into two gauge bosons with momenta q1andq2. Label the momenta of the external gauge boson lines. - 137 - 2.7 Summary •A non-abelian gauge theory is one in which the Lagrangian is invariant under local transformations of a non-abelian group. •This invariance is achieved by introducing a gauge boson for each generator of thegroup. The partial derivative in the Lagrangian for the fermion ﬁeld is replaced by a covariant derivative as deﬁned in eq. (2.23). •The gauge bosons transform under inﬁnitesima l gauge transformations in a non-linear way given by eq. (2.25). •The ﬁeld strengths, F a μν, are obtained from the commutator of two covariant derivatives and are given by eq. (2.22). They transform as the adjoint representation under gauge transformations such that the quantity Fa μνFaμ νis invariant. •Fa μνFaμ νcontains terms which are cubic and quartic in the gauge bosons, indicating that these gauge bosons interact with each other. •The gauge-ﬁxing mechanism leads to the introduction of Faddeev-Popov ghosts whichare scalar particles that occur only inside loops and obey Fermi statistics. - 138 - 3 Quantum Chromodynamics Quantum Chromodynamics (QCD) is the theory o f the strong interaction. It is nothing but a non-abelian gauge theory with the group SU(3). Thus, the quarks are described by a ﬁeldψiwhere iruns from 1 to 3. The quantum number associated with the label iis called colour. The eight gauge bosons which have to be introduced in order to preserve local gauge invariance are the eight ‘gluons’. These are taken to be the carriers which mediate the strong interaction in the same way that photons are the carriers which mediate the electromagnetic interactions. The Feynman rules for QCD are therefore sim ply the Feynman rules listed in the previous lecture, with the gauge coupling constant, g, taken to be the strong coupling, gs,( m o r e about this later), the generators Tataken to be the eight generators of SU(3) in the triplet representation, and fabc,a ,b ,c , =1...8 are the structure constants of SU(3) (you can look them up in a book but normally you will not need their explicit form). Thus we now have a quantum ﬁeld theory which can be used to describe the strong interac- tion. 3.1 Running Coupling The coupling for the strong interaction is the QCD gauge coupling, gs. We usually work in terms of αsdeﬁned as αs=g2 s 4π. (3.1) Since the interactions are strong, we would expect αsto be too large to perform reliable calculations in perturbation theory. On the o ther hand the Feynman rules are only useful within the context of perturbation theory. This diﬃculty is resolved when we understand that ‘coupling constants’ are not constant at all. The electromagnetic ﬁne structure constant, α, has the value 1 /137 only at energies which are not large compared to the electron mass. At higher energies it is larger than this.For example, at LEP energies it takes a value close to 1 /129. In contrast to QED, it turns out that in the non-abelian gauge theories of the Standard Model the weak and the strong coupling decrease as the energy increases. To see how this works within the context of QCD we note that when we perform higherorder perturbative calculations there are loop diagrams which have the eﬀect of ‘dressing’the couplings. For example, the one-loop diagrams which dress the coupling between a quark and a gluon are: - 139 - where = + - - are the diagrams needed to calculate the one-loop corrections to the gluon propagator. These diagrams contain UV divergences and need to be renormalized, e.g. by subtracting at some renormalization scale μ. This scale then appears inside a logarithm for the renormalized quantities. This means that if the squared momenta of all the external particles coming intothe vertex are of order Q 2,w h e r e Q/greatermuchμ, then the above diagrams give rise to a correction which contains a logarithm of the ratio Q2/μ2: −α2 sβ0ln/parenleftBig Q2/μ2/parenrightBig . (3.2) This correction is interpreted as the correction to the eﬀective QCD coupling, αs(Q2), at momentum scale Q, i.e. αs(Q2)=αs(μ2)−αs(μ2)2β0ln/parenleftBig Q2/μ2/parenrightBig +.... (3.3) The coeﬃcient β0is calculated to be β0=11Nc−2nf 12π, (3.4) where Ncis the number of colours (=3), nfis the number of active ﬂavours, i.e. the number of ﬂavours whose mass threshold is below the momentum scale Q.N o t et h a t β0ispositive , which means that the coeﬃcient in front of the logarithm in eq. (3.3) is negative , so that the eﬀective coupling decreases as the momentum scale is increased. A more precise analysis shows that the eﬀective coupling obeys the diﬀerential equation ∂αs(Q2) ∂ln(Q2)=β/parenleftBig αs(Q2)/parenrightBig , (3.5) where βhas the perturbative expansion β(αs)=−β0α2 s−β1α3 s+O(α4 s)+.... (3.6) - 140 - 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 1 10 100 αs(Q2) √ Q2(GeV) Figure 3.1: The running of αs(Q2)w i t h βtaken to two loops. In order to solve this diﬀerential equation w e need a boundary value. Nowadays this is usually taken to be the measured value of the coupling at scale of the Zboson mass, MZ=9 1.19 GeV, which is measured to be αs(M2 Z)=0 .118±0.002. (3.7) This is one of the free parameters of the Standard Model.6 The running of αs(Q2) is shown in ﬁgure 3.1. We can see that for momentum scales above about 2 GeV the coupling is less than 0.3 so that one can hope to carry out reliable pertur- bative calculations for Q CD processes with energy s cales larger than this. Gauge invariance requires that the gauge coupling for the interaction between gluons must be exactly the same as the gauge coupling for the interaction between quarks and gluons. Theβ-function could therefore have been calculated from the higher order corrections to the three-gluon (or four-gluon) vertex and must yi eld the same result, despite the fact that it is calculated from a completely diﬀerent set of diagrams. 6Previously the solution to eq. (3.5) (to leading order) was written as αs(Q2)=4π/β 0ln(Q2/Λ2 QCD)a n d the scale Λ QCDwas used as the standard parameter which sets the scale for the magnitude of the strong coupling. This turns out to be rather inconvenient since it needs to be adjusted every time higher order corrections are taken into consideration and the number of active ﬂavours has to be speciﬁed. The detour via Λ QCDalso introduces additional truncation errors and can complicate the error analysis. - 141 - Exercise 3.1 Draw the Feynman diagrams needed for the calculation of the one-loop cor- rection to the triple gluon coupling (don’t forget the Faddeev-Popov ghost loops). Exercise 3.2Solve equation (3.5) using βto leading order only, and calculate the value of α sat a momentum scale of 10 GeV. Use the value at MZgiven by eq. (3.7). Calculate also the error in αsat 10 GeV. 3.2 Quark (and Gluon) Conﬁnement This argument can be inverted to provide an answer to the question of why we have never seen quarks or gluons in a laboratory. Asymptotic Freedom tells us that the eﬀective coupling be- tween quarks becomes weaker at shorter distances (equivalent to higher energies/momentum scales). Conversely it implies that the eﬀectiv e coupling grows as we go to larger distances. Therefore, the complicated system of gluon exchanges which leads to the binding of quarks(and antiquarks) inside hadrons leads to a s tronger and stronger binding as we attempt to pull the quarks apart. This means that we can never isolate a quark (or a gluon) at large distances since we require more and more ene rgy to overcome the binding as the distance between the quarks grows. Instead, when the energy contained in the ‘string’ of bound glu-ons and quarks becomes large enough, the colour-string breaks and more quarks are created, leaving more colourless hadrons, but no isolated, coloured quarks. The upshot of this is that the only free parti cles which can be observed at macroscopic distances from each other are colour singlets . This mechanism is known as “quark conﬁne- ment”. The details of how it works are not fu lly understood. Nevertheless the argument presented here is suggestive of such conﬁnement and at the level of non-perturbative ﬁeld theory, lattice calculations have conﬁrmed that for non-abelian gauge theories the binding energy does indeed grow as the distance between quarks increases. 7 Thus we have two diﬀerent pictures of the worl d of strong interactions: On one hand, at suf- ﬁciently short distances, which can be probed at suﬃciently large energies, we can considerquarks and gluons (partons) interacting with each other. In this regime we can perform calculations of the scattering cross sections between quarks and gluons (called the “par- tonic hard cross section”) in p erturbation theory because the running coupling is suﬃciently 7Lattice QCD simulations have also succeeded in calculating the spectrum of many observed hadrons and also hadronic matrix elements for certain processes from ‘ﬁrst principles’, i.e. without using perturbative expansions or phenomenological models. - 142 - small. On the other hand, before we can make a direct comparison with what is observed in accelerator experiments, we need to take into account the fact that the quarks and glu-ons bind (hadronize) into colour singlet hadro ns, and it is only these colour singlet states that are observed directly. The mechanism for this hadronization is beyond the scope ofperturbation theory and not understood in deta il. Nevertheless Monte Carlo programs have been developed which simulate the hadronization in such a way that the results of the short-distance perturbative calculations at the level of quarks and gluons can be confronted with experiments measuring hadrons in a successful way. Thus, for example, if we wish to calculate the cross section for an electron-positron annihila- tion into three jets (at high energies), we ﬁrst calculate, in perturbation theory, the process for electron plus positron to annihilate into a virtual photon (or Zboson) which then de- cays into a quark and antiquark, and an emi tted gluon. At leading order the two Feynman diagrams for this process are: 8 e+ e−q ¯qγ∗ge+ e−q ¯qγ∗ g However, before we can compare the results of this perturbative calculation with experi- mental data on three jets of observed hadrons, we need to perform a convolution of thiscalculated cross section with a Monte Carlo simulation that accounts for the way in which the ﬁnal state partons (quarks and gluons) b ind with other quarks and gluons to produce observed hadrons. It is only after such a convolution has been performed that one can get a reliable comparison of the calculated observ ables (like cross sections or event shapes) with data. Likewise, if we want to calculate scattering pr ocesses including initial state hadrons we need to account for the probability of ﬁnding a parti cular quark or gluon inside an initial hadron with a given fraction of the initial hadron’s momentum (these are called “parton distribution functions”). Exercise 3.3 Draw the (tree level) Feynman diagrams for the process e+e−→4jets. Con- sider only one photon exchange plus t he QCD contributions (do not include Z boson exchange or WW production). 8The contraction of the one loop diagram (where a gluon connects the quark and antiquark) with the e+e−→q¯qamplitude is of the same order αsand has to be taken into account to get an infra-red ﬁnite result. However, it does not lead to a three-jet event (on the partonic level). - 143 - 3.3 θ-Parameter of QCD There is one more gauge invariant term that can be written down in the QCD Lagrangian: Lθ=θg2 s 64π2/epsilon1μνρσFa μνFa ρσ. (3.8) Here/epsilon1μνρσis the totally antisymmetric tensor (in fo ur dimensions). Since we should work with the most general gauge invariant Lagrangian there is no reason to omit this term. However, adding this term to the Lagrangian leads to a problem, called the “strong CP problem”. To understand the nature of the problem, we ﬁrst convince ourselves that this term violates CP. In QED we would have /epsilon1μνρσFμνFρσ=E·B, (3.9) and for QCD we have a similar expression except that EaandBacarry a colour index — they are known as the chromoelectric and chromomagnetic ﬁelds. Under charge conjugationboth the electric and magnetic ﬁeld change sign. But under parity the electric ﬁeld, whichis a proper vector, changes sign, whereas the magnetic ﬁeld, which is a polar vector, does not change sign. Thus we see that the term E·Bis odd under CP. For this reason, the parameter θin front of this term must be exceedingly small in order not to give rise to strong interaction contributions to CPviolating quantities such as the electric dipole moment of the neutron. The current experimental limits on this dipole moment tellus that θ<10 −10. Thus we are tempted to think that θis zero. Nevertheless, strictly speaking θis a free parameter of QCD, and is sometimes considered to be the nineteenth free parameter of the Standard Model. Of course we simply could set θto zero (or a very small number) and be happy with it.9 However, whenever a free parameter is zero or extremely small, we would like to understand the reason. The fact that we do not know why this term is absent (or so small) is the strong CPproblem. There are several possible solutions to the strong CPproblem that oﬀer explanations as to why this term is absent (or small). One possible solution is through imposing an ad-ditional symmetry, leading to the postulation of a new, hypothetical, weakly interacting particle, called the “(Peccei-Quinn) axion”. Unfortunately none of these solutions have been conﬁrmed yet and the problem is still unresolved. Another question is why is this not a proble m in QED? In fact a term like eq. (3.8) can also 9To be precise, setting θ→0 in the Lagrangian would not be enough, as θ/negationslash= 0 can also be generated through higher order electroweak radiative corrections, requiring a ﬁne-tuning beyond θ→0. - 144 - be written down in QED. A thorough discussion of this point is beyond the scope of this lecture. Suﬃce to say that this term can be written (in QED and QCD) as a total divergence,so it seems that it can be eliminated from the Lagrangian altogether. However, in QCD (but not in QED) there are non-perturbative eﬀects from the non-trivial topological structure of the vacuum (somewhat related to so called “instantons” you probably have heard about) which prevent us from neglecting the θ-term. 3.4 Summary •Quarks transform as a triplet representation of colour SU(3) (each quark can have one of three colours). •The eight gauge bosons of QCD are the gluons which are the carriers that mediate the strong interaction. •The coupling of quarks to gluons (and gluons to each other) decreases as the energyscale increases. Therefore, at high energies one can perform reliable perturbative cal- culations for strongly interacting processes. •As the distance between quarks increases t he binding increases, such that it is impos- sible to isolate individual quarks or gluons. The only observable particles are colour singlet hadrons. Perturbative calculations performed at the quark and gluon level must be supplemented by accounting for the recomb ination of ﬁnal state quarks and gluons into observed hadrons as well as the probability of ﬁnding these quarks and gluonsinside the initial state hadrons (if applicable). •QCD admits a gauge invariant strong CPviolating term with a coeﬃcient θ.T h i s parameter is known to be very small from limits on CPviolating phenomena such as the electric dipole moment of the neutron. - 145 - 4 Spontaneous Symmetry Breaking We have seen that in an unbroken gauge theory the gauge bosons must be massless. This is exactly what we want for QED (massless photon) and QCD (massless gluons). However, if we wish to extend the ideas of describing interactions by a gauge theory to the weak interactions,the symmetry must somehow be broken since the carriers of the weak interactions ( Wand Zbosons) are massive (weak interactions are very short range). We could simply break the symmetry by hand by adding a mass term for the gauge bosons, which we know violates the gauge symmetry. However, this would destroy renormalizability of our theory. Renormalizable theories are preferred because they are more predictive. As discussed in the Field Theory and QED lectures, there ar e divergent results (inﬁnities) in QED and QCD, and these are said to be renormalizable theories. So what could be worse abouta non-renormalizable theory? The critical issue is the number of divergences: few in a renormalizable theory, and inﬁnite in the non -renormalizable case. Associated to every divergence is a parameter that must be extracted from data, so renormalizable theories canmake testable predictions once a few parameters are measured. For instance, in QCD, thecoupling g shas a divergence. But once αsis measured in one process, the theory can be tested in other processes.10 In this chapter we will discuss a way to give masses to the WandZ, called “spontaneous symmetry breaking”, which maintains the renormalizability of the theory. In this scenariothe Lagrangian maintains its symmetry under a se t of local gauge transformations. On the other hand, the lowest energy state, which we interpret as the vacuum (or ground state),isnota singlet of the gauge symmetry. There is an inﬁnite number of states each with the same ground-state energy and nature choose s one of these states as the ‘true’ vacuum. 4.1 Massive Gauge Bosons and Renormalizability In this subsection we will convince ourselve s that simply adding by hand a mass term for the gauge bosons will destroy t he renormalizability of the theory. It will not be a rigorous argument, but will illustrate the diﬀerence between introducing mass terms for the gauge bosons in a brute force way and introducing them via spontaneous symmetry breaking. Higher order (loop) corrections generate ultraviolet divergences. In a renormalizable theory, 10It should be noted that eﬀective ﬁeld theories, though formally not renormalizable, can nevertheless be very valuable as they often allow for a simpliﬁed description of a more ‘complete’ or fundamental theory in a resticted energy range. Popular examples are Chiral Perturbation Theory, Heavy Quark Eﬀective Theory and Non-Relativistic QCD. - 146 - these divergences can be absorbed into the parameters of the theory we started with, and in this way can be ‘hidden’. As we go to higher orders we need to absorb more and moreterms into these parameters, but there are only as many divergent quantities as there are parameters. So, for instance, in QED the Lagrangian we start with contains the fermion ﬁeld, the gauge boson ﬁeld, and interactions whose strength is controlled by eandm.B e i n g a renormalizable theory, all divergences of diagrams can be absorbed into these quantities(irrespective of the number of loops or legs), and once eandmare measured, all other observables (cross sections, g−2, etc.) can be predicted. In order to ensure that this programme can be c arried out there have to be restrictions on the allowed interaction terms. Furthermore all the propagators have to decrease like 1 /p 2 as the momentum p→∞ . Note that this is how the massless gauge-boson propagator eq. (1.24) behaves. If these conditions are not fulﬁlled, then the theory generates more andmore divergent terms as one calculates to higher orders, and it is not possible to absorbthese divergences into the parameters of the theory. Such theories are said to be “non- renormalizable”. Now we can convince ourselves that simply adding a mass term M 2AμAμto the Lagrangian given in eq. (2.21) will lead to a non-renormalizable theory. To start with we note that such a term will modify the propagator. Collecting all terms bilinear in the gauge ﬁelds in momentum space we get (in Feynman gauge) 1 2Aμ/parenleftBig −gμν(p2−M2)+pμpν/parenrightBig Aν. (4.1) We have to invert this operator to get the propagator which now takes the form i p2−M2/parenleftbigg −gμν+pμpν M2/parenrightbigg . (4.2) Note that this propagator, eq. (4.2), has a much worse ultraviolet behavior in that it goesto a constant for p→∞. Thus, it is clear that the ultraviolet properties of a theory with a propagator as given in eq. (4.2) are worse than for a theory with a propagator as givenin eq. (1.24). According to our discussion at the beginning of this subsection we conclude that without the explicit mass term M 2AμAμthe theory is renormalizable, whereas with this term it is not. In fact, it is precisely the gauge symmetry that ensur es renormalizability. Breaking this symmetry results in t he loss of renormalizability. The aim of spontaneous symmetry breaking is to break the gauge symmetry in a more subtle way, such that we can still give the gauge bo sons a mass but retain renormalizability. - 147 - 4.2 Spontaneous Symmetry Breaking Spontaneous symmetry breaking is a pheno menon that is by far not restricted to gauge symmetries. It is a subtle way to break a symmetry by still requiring that the Lagrangian remains invariant under the symmetry transfo rmation. However, the ground state of the symmetry is notinvariant, i.e. nota singlet under a symmetry transformation. In order to illustrate the idea of spontaneo us symmetry breaking, consider a pen that is completely symmetric with respect to rotations around its axis. If we balance this pen onits tip on a table, and start to press on it with a force precisely along the axis we have a perfectly symmetric situation. This corresp onds to a Lagrangian which is symmetric (under rotations around the axis of the pen in this case). However, if we increase the force, at some point the pen will bend (and eventually break) . The question then is in which direction will it bend. Of course we do not know, since all di rections are equal. But the pen will pick one and by doing so it will break the rotati onal symmetry. This is spontaneous symmetry breaking. A better example can be given by looking at a point mass in a potential V(/vector r)=μ 2/vector r·/vector r+λ(/vector r·/vector r)2. (4.3) This potential is symmetric under rotations and we assume λ>0 (otherwise there would be no stable ground state). For μ2>0 the potential has a minimum at /vector r=0 ,t h u st h e point mass will simply fall to this point. The situation is more interesting if μ2<0. For two dimensions the potential is shown in Fig. 4.1. If the point mass sits at /vector r=0t h e system is not in the ground state but the sit uation is completely symmetric. In order to reach the ground state, the symmetry has to be broken, i.e. if the point mass wants to roll down, it has to decide in which direction. An y direction is equally good, but one has to be picked. This is exactly what spontaneous symmetry breaking means. The Lagrangian (herethe potential) is symmetric (he re under rotations around the z-axis), but the ground state (here the position of the point mass once it rolled down) is not. Let us formulate this ina slightly more mathematical way for gauge symmetries. We denote the ground state by |0/angbracketright. A spontaneously broken gauge theory is a theory whose Lagrangian is invariant under gauge transformations, which is exactly wha t we have done in chapters 1 and 2. The new feature in a spontaneously broken theory is that the ground state is not invariant under gauge transformations. This means e −iωaTa|0/angbracketright /negationslash=|0/angbracketright (4.4) which entails Ta|0/angbracketright /negationslash=0 f o rs o m e a. (4.5) - 148 - yV(r) x Figure 4.1: A potential that leads to spontaneous symmetry breaking. Eq. (4.5) follows from eq. (4.4) upon expansion in ωa. Thus, the theory is spontaneously broken if there exists at least one generator that does not annihilate the vacuum. In the next section we will explore the concept of spontaneous symmetry breaking in the context of gauge symmetries in more detail, and we will see that, indeed, this way of breaking the gauge symmetry has all the desired features. 4.3 The Abelian Higgs Model For simplicity, we will start by spontaneously breaking the U(1) gauge symmetry in a theory of one complex scalar ﬁeld. In the Standard Model, it will be a non-abelian gauge theorythat is spontaneously broken, but all the important ideas can simply be translated from the U(1) case considered here. The Lagrangian density for a gauged complex scalar ﬁeld, with a mass term and a quarticself-interaction, may be written as L=(D μΦ)∗DμΦ−1 4FμνFμν−V(Φ), (4.6) where the potential V(Φ), is given by V(Φ) = μ2Φ∗Φ+λ|Φ∗Φ|2, (4.7) - 149 - and the covariant derivative Dμand the ﬁeld-strength tensor Fμνare given in eqs. (1.15) and (1.12) respectively. This Lagrangian is invariant under U(1) gauge transformations Φ→e−iω(x)Φ. (4.8) Provided μ2is positive this potential has a minimum at Φ = 0. We call the Φ = 0 state the vacuum and expand Φ in terms of creation and annihilation operators that populate the higher energy states. In terms of a quantum ﬁeld theory, where Φ is an operator, the precise statement is that the operator Φ has zero vacuum expectation value, i.e. /angbracketleft0|Φ|0/angbracketright=0 . Now suppose we reverse the sign of μ2, so that the potential becomes V(Φ) = −μ2Φ∗Φ+λ|Φ∗Φ|2, (4.9) withμ2>0. We see that this potential no longer has a minimum at Φ = 0, but a (local) maximum . The minimum occurs at Φ=eiθ/radicalBigg μ2 2λ≡eiθv √ 2, (4.10) where θcan take any value from 0 to 2 π. There is an inﬁnite number of states each with the same lowest energy, i.e. we have a degene rate vacuum. The symmetry breaking occurs in the choice made for the value of θwhich represents the true vacuum. For convenience we shall choose θ= 0 to be our vacuum. Such a choice constitutes a spontaneous breaking of theU(1) invariance, since a U(1) transformation takes us to a diﬀerent lowest energy state. In other words the vacuum breaks U(1) invariance. In quantum ﬁeld theory we say that the ﬁeld Φ has a non-zero vacuum expectation value /angbracketleftΦ/angbracketright=v √ 2. (4.11) But this means that there are ‘excitations’ with zero energy, that take us from the vacuum to one of the other states with the same energy. T he only particles which can have zero energy are massless particles (with zero momentum). We therefore expect a massless particle insuch a theory. To see that we do indeed get a massless particle, let us expand Φ around its vacuum expec- tation value, Φ=e iφ/v √ 2/parenleftBiggμ √ λ+H/parenrightBigg /similarequal1 √ 2/parenleftBiggμ √ λ+H+iφ/parenrightBigg . (4.12) The ﬁelds Handφhave zero vacuum expectation values and it is these ﬁelds that are expanded in terms of creation and annihilation operators of the particles that populate the excited states. Of course, it is the H-ﬁeld that corresponds to the Higgs ﬁeld. - 150 - We now want to write the Lagrangian in terms of the Handφﬁelds. In order to get the potential we insert eq. (4.12) into eq. (4.9) and ﬁnd V=μ2H2+μ√ λ/parenleftBig H3+φ2H/parenrightBig +λ 4/parenleftBig H4+φ4+2H2φ2/parenrightBig +μ4 4λ. (4.13) Note that in eq. (4.13) there is a mass term for the H-ﬁeld, μ2H2≡MH/2H2,w h e r ew e have deﬁned11 MH=√ 2μ. (4.14) However, there is nomass term for the ﬁeld φ.T h u s φis a ﬁeld for a massless particle called the “Goldstone boson”. We will look at this i ssue in a more general way in section 4.4. Next let us consider the kinetic term. We plug eq. (4.12) into ( DμΦ)∗DμΦa n dg e t (DμΦ)∗DμΦ=1 2∂μH∂μH+1 2∂μφ∂μφ+1 2g2v2AμAμ+1 2g2AμAμ(H2+φ2) −gAμ(φ∂μH−H∂μφ)+gvA μ∂μφ+g2vAμAμH. (4.15) There are several important features in eq. (4.15). Firstly, the gauge boson has acquired a mass term 1 /2g2v2AμAμ≡1/2M2 AAμAμ, where we have deﬁned MA=gv. (4.16) Secondly, there is a coupling of the gauge ﬁeld to the H-ﬁeld, g2vAμAμH=gMAAμAμH. (4.17) It is important to remember that this coupling is proportional to the mass of the gaugeboson. Finally, there is also the bilinear term gvA μ∂μφ, which after integrating by parts (for the action S) may be written as −MAφ∂μAμ. This mixes the Goldstone boson, φ,w i t h the longitudinal component of the gauge boson, with strength MA(when the gauge-boson ﬁeldAμis separated into its transvers e and longitudinal components, Aμ=AL μ+ATμ, where ∂μAT μ= 0). Later on, we will use the gauge freedom to get rid of this mixing term. 4.4 Goldstone Bosons In the previous subsection we have seen that there is a massless boson, called the Goldstone boson, associated with the ﬂat direction in the potential. Goldstone’s theorem describes the appearance of massless bosons when a global (n ot gauge) symmetry is spontaneously broken. 11Note that for a real ﬁeld φrepresenting a particle of mass mthe mass term is1 2m2φ2,w h e r e a sf o ra complex ﬁeld the mass term is m2φ†φ. - 151 - Suppose we have a theory whose Lagrangi an is invariant under a symmetry group Gwith Ngenerators Taand the symmetry group of the vacuum forms a subgroup HofG,w i t h mgenerators. This means that the vacuum stat e is still invariant under transformations generated by the mgenerators of H, but not the remaining N−mgenerators of the original symmetry group G.T h u sw eh a v e Ta|0/angbracketright=0 a=1...m , Ta|0/angbracketright /negationslash=0 a=m+1...N. (4.18) Goldstone’s theorem states that there will be N−mmassless particles (one for each broken generator of the group). The case considered in this section is special in that there is only one generator of the symmetry group (i.e. N= 1) which is broken by the vacuum. Thus, there is no generator that leaves the vacuum invariant (i.e. m=0 )a n dw eg e t N−m=1 Goldstone boson. Like all good general theorems, Goldstone’s theorem has a loophole, which arises when one considers a gauge theory, i.e. when one allows the original symmetry transformations to be local. In a spontaneously broken gauge theory, the choice of which vacuum is the true vacuum is equivalent to choosing a gauge, which is necessary in order to be able to quantize the theory. What this means is that the Goldsto ne bosons, which can, in principle, transform the vacuum into any of the states degenerate with the vacuum, now aﬀect transitions intostates which are not consistent with the original gauge choice. This means that the Goldstonebosons are “unphysical” and are often called “Goldstone ghosts”. On the other hand the quantum degrees of freed om associated with the Goldstone bosons are certainly there ab initio (before a choice of gauge is made). What happens to them? A massless vector boson has only two degrees of freedom (the two directions of polarization of a photon), whereas a massive vector (spin- one) particle has three possible values for the helicity of the particle. In a spontaneously broken gauge theory, the Goldstone bosonassociated with each broken generator provides the third degree of freedom for the gauge bosons. This means that the gauge bosons become massive. The Goldstone boson is said to be “eaten” by the gauge boson. This is related to the mixing term between A μ Landφof the previous subsection. Thus, in our abelian model, the two degrees of freedom of the complex ﬁeld Φ turn out to be the Higgs ﬁeld and the longitudinal component of the (now massive)gauge boson. There is no physical, massless particle associated with the degree of freedom φpresent in Φ. - 152 - 4.5 The Unitary Gauge As mentioned above, we want to use the gauge freedom to choose a gauge such that there are no mixing terms between the longitudinal component of the gauge ﬁeld and the Goldstone boson. Recall Φ=1 √ 2(v+H)eiφ/v=1 √ 2/parenleftBiggμ √ λ+H+iφ+.../parenrightBigg , (4.19) where the dots stand for nonlinear terms in φ. Next we make a gauge transformation (see eq. (1.2)) Φ→Φ/prime=e−iφ/vΦ. (4.20) In other words, we ﬁx the gauge such that the imaginary part of Φ vanishes. Under thegauge transformation eq. (4.20) the gauge ﬁe ld transforms according to (see eq. (1.11)) A μ→A/prime μ=Aμ+1 gv[∂μφ]. (4.21) It is in fact the superposition of Aμandφwhich make up the physical ﬁeld. Note that the change from AμtoA/prime μmade in eq. (4.21) aﬀects only the longitudinal component. If we now express the Lagrangian in terms of Φ/primeandA/prime μthere will be no mixing term. Even better, the φﬁeld vanishes altogether! This can ea sily be seen by noting that under a gauge transformation the covariant derivative DμΦ transforms in the same way as Φ, thus DμΦ→(DμΦ)/prime=e−iφ/vDμΦ=e−iφ/v1 √ 2/parenleftBig ∂μH+igA/prime μ(v+H)/parenrightBig , (4.22) and (DμΦ)/prime∗(DμΦ)/primeis independent of φ. Performing the algebra (and dropping the/primefor the A-ﬁeld) we get the Lagrangian in the unitary gauge L=1 2∂μH∂μH+M2 A 2AμAμ−1 4FμνFμν−M2 H 2H2 +gMAAμAμH+g2 2AμAμH2−λ 4H4−/radicalBigg λ 2MHH3, (4.23) withMAandMHas deﬁned in eqs. (4.16) and (4.14), respectively. All the terms quadratic inAμmay be written (in momentum space) as Aμ(−p)/parenleftBig −gμνp2+pμpν+gμνM2 A/parenrightBig Aν(p). (4.24) The gauge boson propagator is the inverse of the coeﬃcient of Aμ(−p)Aν(p), which is −i/parenleftBigg gμν−pμpν M2 A/parenrightBigg1 (p2−M2 A). (4.25) This is the usual expression for the propagator of a massive spin-one particle, eq. (4.2). The only other remaining particle is the scalar, H,w i t hm a s s mH=√ 2μ,w h i c hi st h e - 153 - Higgs boson. This is a physical particle, which interacts with the gauge boson and also has cubic and quartic self-interactions. The Lagrangian given in eq. (4.23) leads to the followingvertices and Feynman rules: μ ν μ ν2ie2gμν 2ie M Agμν 6iλ 6imH√ 2λ The advantage of the unitary gauge is that no unphysical particles appear, i.e. the φ-ﬁeld has completely disappeared. The disadvant age is that the propagator of the gauge ﬁeld, eq. (4.25), behaves as p0forp→∞ . As discussed in section 4.1 this seems to indicate that the theory is non-renormalizable. It seems that we have not gained anything at all by breaking the theory spontaneously rath er than by simply adding a mass term by hand. Fortunately this is not true. In order to see that the theory is still renormalizable, in spiteof eq. (4.25), it is very useful to consider a diﬀerent type of gauges, namely the R ξgauges discussed in the next subsection. 4.6 RξGauges (Feynman Gauge) The class of Rξgauges is a more conventional way to ﬁx the gauge. Recall that in QED we ﬁxed the gauge by adding a term, eq. (1.21), in the Lagrangian. This is exactly what we dohere. The gauge ﬁxing term we are adding to the Lagrangian density eq. (4.6) is L R≡−1 2(1−ξ)(∂μAμ−(1−ξ)MAφ)2 =−1 2(1−ξ)∂μAμ∂νAν+MAφ∂μAμ−1−ξ 2M2 Aφ2. (4.26) - 154 - Again, the special value ξ= 0 corresponds to the Feynm an gauge. The second term in eq. (4.26) cancels precisely the mixing term in eq. (4.15). Thus, we have achieved our goal. Note however, that in this case, contra ry to the unitary gauge, the unphysical φ-ﬁeld does not disappear. The ﬁrst term in eq. (4.26) is b ilinear in the gauge ﬁeld, thus it contributes to the gauge-boson propagat or. The terms bilinear in the A-ﬁeld are −1 2Aμ(−p)/parenleftBigg −gμν(p2−M2 a)+p μpν−pμpν 1−ξ/parenrightBigg Aν(p) (4.27) which leads to the gauge boson propagator −i (p2−M2 A)/parenleftBigg gμν−ξpμpν p2−(1−ξ)M2 A/parenrightBigg . (4.28) In the Feynman gauge, the propagator becomes particularly simple. The crucial feature of eq. (4.28), however, is that this propagator behaves as p−2forp→∞ .T h u s , t h i s c l a s s of gauges is manifestly renormalizable. There is, however, a price to pay: The Goldstoneboson is still present. It has acquired a mass, M A, from the gauge ﬁxing term, and it has interactions with the gauge boson, with the Higgs scalar and with itself. Furthermore, for thepurposes of higher order correct ions in non-Abelain theories, we need to introduce Faddeev- Popov ghosts which interact with the gauge bosons, the Higgs scalar and the Goldstonebosons. Let us stress that there is no contradiction at a ll between the apparent non-renormalizability of the theory in the unitary gauge and t he manifest renormalizability in the R ξgauge. Since physical quantities are gauge invariant, any physical quantity can be calculated in a gauge where renormalizability is manifest. As mentio ned above, the price we pay for this is that there are more particles and many more interactions, leading to a plethora of Feynmandiagrams. We therefore only work in such gauges if we want to compute higher order corrections. For the rest of these lectures we shal l conﬁne ourselves to tree-level calculations and work solely in the unitary gauge. Nevertheless, one cannot over-stress the f act that it is only when the gauge bosons ac- quire masses through the Higgs mechanism that we have a renormalizable theory. It is this mechanism that makes it possible to write down a consistent Quantum Field Theory which describes the weak interactions. 4.7 Summary •In the case of a gauge theory the Goldstone bo sons provide the longitudinal component of the gauge bosons, which therefore acquire a mass. The mass is proportional to the - 155 - magnitude of the vacuum expectation value and the gauge coupling constant. The Goldstone bosons themselves are unphysical. •It is possible to work in the unitary gauge where the Goldstone boson ﬁelds are set to zero. •When gauge bosons acquire masses by this (Higgs) mechanism, renormalizability ismaintained. This can be seen explicitly if one works in a R ξgauge, in which the gauge boson propagator decreases like 1 /p2asp→∞ . This is a necessary condition for renormalizability. If one does work in such a gauge, however, one needs to work with Goldstone boson ﬁelds, even though the Gol dstone bosons are unphysical. The number of interactions and the number of Feynman g raphs required for the calculation of some processes is then greatly increased. - 156 - 5 The Standard Model with one Family To write down the Lagrangian of a theory, one ﬁrst needs to choose the symmetries (gauge and global) and the particle content, and th en write down every allowed renormalizable interaction. In this section we shall use this recipe to construct the Standard Model withone family. The Lagrangian should contain pieces L (SM,1)=Lgauge bosons +Lfermion masses +LfermionKT +LHiggs. (5.1) The terms are written out in eqns. (5.15), (5.29), (5.30) and (5.55). 5.1 Left- and Right- Handed Fermions The weak interactions are known to violate parity. Parity non-invariant interactions forfermions can be constructed by giving diﬀerent interactions to the “left-handed” and “right- handed” components deﬁned in eq. (5.4). Th us, in writing down the Standard Model, we will treat the left-handed and right-handed parts separately. A Dirac ﬁeld, ψ, representing a fermion, can be expressed as the sum of a left-handed part, ψ L, and a right-handed part, ψR, ψ=ψL+ψR, (5.2) where ψL=PLψwith PL=(1−γ5) 2, (5.3) ψR=PRψwith PR=(1 +γ5) 2. (5.4) PLandPRare projection operators, i.e. PLPL=PL,P RPR=PRandPLPR=0= PRPL. (5.5) They project out the left-handed (negative) and right-handed (positive) chirality states of the fermion, respectively. This is the deﬁnition of chirality, which is a property of fermion ﬁelds, but not a physical observable. The kinetic term of the Dirac Lagrangian and the interaction term of a fermion with a vector ﬁeld can also be written as a sum of two terms, each involving only one chirality ψγμ∂μψ= ψLγμ∂μψL+ ψRγμ∂μψR, (5.6) ψγμAμψ= ψLγμAμψL+ ψRγμAμψR. (5.7) - 157 - On the other hand, a mass term mixes the two chiralities: m ψψ=m ψLψR+m ψRψL. (5.8) Exercise 5.1 Use (γ5)2= 1 to verify eq. (5.5) and ψ=ψ†γ0,γ5†=γ5as well as γ5γμ= −γμγ5to verify eq. (5.7). In the limit where the fermions are massless (or suﬃciently relativistic), chirality becomes helicity , which is the projection of the spin on the direction of motion and which is a physical observable. Thus, if the fermions are massless, we can treat the left-handed and right-handedchiralities as separate particles of conserved h elicity. We can understand this physically from the following simple consideration. If a fermion is massive and is moving in the positive z direction, along which its spin is having a positive component so that the helicity is positive in this frame, one can always boost into a frame in which the fermion is moving in thenegative zdirection, but with this spin component unchanged. In the new frame the helicity will hence be negative . On the other hand, if the particle is massless and travels with the speed of light, no such boost is possible, and in tha t case helicity/chirality is a good quantum number. Exercise 5.2For a massless spinor u(p)=1 √ E⎛ ⎝Eχ /vector σ·/vectorpχ⎞⎠ , where χis a two-component spinor, show that (1±γ5)u(p) are eigenstates of /vector σ·/vectorp/Ewith eigenvalues ±1, respectively. Take γ5=⎛⎝ 01 10⎞ ⎠, a n di n4 ×4 matrix notation /vector σ·/vector pmeans ⎛ ⎝/vector σ·/vector p0 0/vector σ·/vector p⎞⎠ . - 158 - 5.2 Symmetries and Particle Content We have made all the preparations to write down a gauge invariant Lagrangian. We now only have to pick the gauge group and the matter content of the theory. It should be noticed that there are no theoretical reasons to pick a certain group or certain matter content. To match experimental observations we pick the gauge group for the Standard Model to be U(1)Y×SU(2)×SU(3). (5.9) To indicate that the abelian U(1) group is notthe gauge group of QED but of hypercharge a subscript Yhas been added. The corresponding co upling and gauge boson is denoted by g/primeandBμrespectively. TheSU(2) group has three generators ( Ta=σa/2), the coupling is denoted by gand the three gauge bosons are denoted by W1 μ,W2 μ,W3 μ. None of these gauge bosons (and neither Bμ) are physical particles. As we will see, linea r combinations of these gauge bosons will make up the photon as well as the W±and the Zbosons. Finally, the SU(3) is the group of the strong interaction. The corresponding eight gauge bosons are the gluons. In this section we will concentrate on the other two groups, with one generation of fermions. The strong interaction is dealt with in section 3, and extragenerations are introduced in the next chapter. As matter content for the ﬁrst family, we have q L≡⎛ ⎝uL dL⎞⎠ ;uR;dR;/lscriptL≡⎛⎝ νL eL⎞⎠ ;eR;{νR!!}. (5.10) Note that a right-handed neutrino νRhas appeared. It is a gauge singlet (no strong interac- tion, no weak interactions, no electric charg e), so is unneccessary in a model with massless neutrinos. However, neutrinos are now known to have small masses, which can be described by adding the right-handed ﬁeld νR. Neutrino masses will be discussed further in chapter 7. Note also that the left- and right-handed fermion components have been given diﬀerent weak interactions. The Standard Model is construct ed this way, because the weak interactions are known to violate parity. The left-handed components form doublets under SU(2) whereas the right-handed components are si nglets. This means that under SU(2) gauge transformations we have eR→e/prime R=eR, (5.11) /lscriptL→/lscript/primeL=e−iωaTa/lscriptL. (5.12) - 159 - Thus, the SU(2) singlets eR,νR,uRanddRare invariant under SU(2) transformations and do not couple to the corresponding gauge bosons W1 μ,W2 μ,W3 μ. Since this separation of the electron into its le ft- and right-handed helicity only makes sense for a massless electron we also need to assume that the electron ismassless in the exact SU(2) limit and that the mass for the electron arises as a result of spontaneous symmetry breaking in a similar way as the masses for the gauge bosons arise. We will come back to this later. Under U(1)Ygauge transformations the matter ﬁelds transform as ψ→ψ/prime=e−iωY(ψ)ψ (5.13) where Yis the hypercharge of the particle under consideration. It is chosen to give the observed electric charge of the particles. The explicit values for the hypercharges of the particles listed in eq. (5.10) are as follows: Y(/lscriptL)=−1 2,Y(eR)=−1,Y(νR)=0,Y(qL)=1 6,Y(uR)=2 3,Y(dR)=−1 3.(5.14) Under SU(3) the lepton ﬁelds /lscriptL,eR,νRare singlets, i.e. they do not transform at all. This means that they do not couple to the gluons. T he quarks on the other hand form triplets under SU(3). The strong interaction does not distinguish between left- and right-handed particles. We have now listed all fermions that belong to the ﬁrst family, together with their transfor- mation properties under the various gauge trans formations. However, since we ultimately want massive weak gauge bosons, we will have to break the U(1)Y×SU(2) gauge group spontaneously, by introducing some type of Higgs scalar. The transformation properties of this scalar will be deduced in the discussion of fermion masses. 5.3 Kinetic Terms for the Gauge Bosons The gauge kinetic terms for abelian and non-ab elian theories were presented in the ﬁrst two lectures. From the general expression of eq. (2.21), we extract for the SM gauge bosons: L=−1 4BμνBμν−1 4Fa μνFaμ ν−1 4FA μνFAμ ν+Lgauge− ﬁxing+LFP ghosts . (5.15) HereBμν=∂μBν−∂νBμis the hypercharge ﬁeld strength, the second term contains the SU(2) ﬁeld strength, so aruns from one to three (over the three vector bosons of SU(2)), and the third term is the gluon kinetic term, so A=1...8. To do an explicit perturbative calculation, additional gauge ﬁxing terms, and Fadeev-Popov ghosts, must be included. The form of these terms depends on the choice of gauge. - 160 - 5.4 Fermion Masses and Yukawa Couplings We cannot have an explicit mass term for the quarks or electrons, since a mass term mixes left-handed and right-handed fermions and we ha ve assigned these to diﬀerent multiplets of weak SU(2). However, if an SU(2) doublet Higgs is introduced, there is a gauge invariantinteraction that will look like a mass when the Higgs gets a vacuum expectation value (“vev”). Such an interaction is called a ‘Yukawa interaction’ and is written as L Yukawa =−Ye lLiΦieR+h . c . , (5.16) where h.c. means ‘hermitian conjugat e’. Note that the Higgs doublet must have Y=1/2t o ensure that this term has zero weak hypercharge. Recalling eq. (5.19) we introduce a scalar “Higgs” ﬁeld, which is a doublet under SU(2), singlet under SU(3) (no colour), and has a scalar potential as given in eq. (4.9), i.e. V(Φ) = −μ2Φ∗Φ+λ|Φ∗Φ|2. (5.17) This potential has a minimum at Φ∗Φ=1 2μ2/λ, so some component of the Higgs doublet should get a vev. In the unitary gauge, this vev can be written as /angbracketleftΦ/angbracketright=1 √ 2⎛ ⎝0 v⎞⎠ (5.18) withv=μ/√ λ. Recall from the previous chapter that Φ can be written as its “radial” degree of freedom times an exponential containing the broken generators of the gauge symmetry: Φ=ei(ωaTa−ω3Y) √ 2⎛ ⎝0 v+H⎞⎠ . (5.19) The unitary gauge choice consists of absorbing t his exponential with a gauge transformation, so that in the unitary gauge eq. (5.16) is LYukawa =−Ye √ 2/parenleftBig νL eL/parenrightBig⎛ ⎝0 v+H⎞⎠ eR+h . c . . (5.20) The part proportional to the vev is simply −Yev √ 2( eLeR+ eReL)=Yev √ 2 ee, (5.21) - 161 - and we see that the electron has acquired a m ass which is proportional to the vev of the scalar ﬁeld. This immediately gives us a relation for the Yukawa coupling in terms of the electron mass, me,a n dt h e Wmass, MW: Ye=gme √ 2MW. (5.22) Thus, as for the gauge bosons, the strength of the coupling of the Higgs to fermions is proportional to the mass of the fermions. The quarks also acquire a mass through the s pontaneous symmetry breaking mechanism, via their Yukawa coupling with the scalars. The interaction term −Yd qLiΦidR+ h.c. (5.23) gives a mass to the dquark when we replace Φ iby its vev. This mass mdis given by md=Yd √ 2v=√ 2YdMW g. (5.24) Since the vev is in the lower component of the Higgs doublet, we must do a little more work to obtain a mass for the upper element uof the quark doublet. In the case of SU(2) there is a second way in which we can construct an invariant for the Yukawa interaction: −Yu/epsilon1ij qLiΦ∗juR+h . c . ( i, j=1,2), (5.25) where /epsilon1ijis the two-dimensional antisymmetric tensor. Note that Φc=/epsilon1ijΦj∗=⎛ ⎝01 −10⎞⎠⎛⎝ Φ∗ + Φ∗ 0⎞ ⎠ (5.26) hasY=−1/2, as required by the U(1) symmetry. This term does indeed give a mass mu to the uquark, where mu=Yu √ 2v=√ 2YuMW g. (5.27) So the SM Higgs scalar couples to both the uanddquark, with interaction terms −gmu 2MW uHu −gmd 2MW dHd. (5.28) The terms in the Lagrangian that give masses to the ﬁrst generation quarks and charged leptons are Lfermion masses =−Ye lLiΦieR−Yd qLiΦidR−Yu/epsilon1ij qLiΦ∗juR+h . c . . (5.29) We could also have included a Yukawa mass term for the neutrinos: −Yν/epsilon1ij /lscriptLiΦ∗jνR+h . c . However, neutrino masses do not neccessarily arise from a Yukawa interaction (this will bediscussed in chapter 7). - 162 - 5.5 Kinetic Terms for Fermions The fermionic kinetic terms sh ould be familiar from chapter 2: LfermionKT =i /lscriptLTγμDμ/lscriptL+i eRγμDμeR+i νRγμ∂μνR +i qLTγμDμqL+i dRγμDμdR+i uRγμ∂μuR (5.30) where the covariant derivatives include the hypercharge, SU(2) and SU(3) gauge bosons as required. For instance: Dμ=∂μ+igTaWa μ+ig/primeY(/lscriptL)Bμ for/lscriptL, (5.31) Dμ=∂μ+ig/primeY(eR)Bμ foreR, (5.32) Dμ=∂μ+igsTsaGa μ+ig/primeY(dR)BμfordR, (5.33) where the strong coupling (g s), the eight generators of SU(3) ( Tsa) and the corresponding gluon ﬁelds ( Ga μ) have been introduced, and Y(f) is the hypercharge of fermion f. This gives the following interaction terms between the leptons and the gauge bosons: −g 2⎛ ⎝ νL eL⎞⎠T γμ⎛⎝⎛⎝ W3 μ√ 2W+ μ√ 2W− μ−W3 μ⎞⎠ −tanθWBμ⎞⎠⎛⎝ νL eL⎞⎠ −igtanθW eRγμBμeR, (5.34) w h e r ew eh a v eu s e d W±=(W1∓iW2)/√ 2. The ﬁelds BμandW3 μare replaced by the physical particles ZμandAμthrough the ‘rotation’ Zμ≡cosθWW3 μ−sinθWBμ, (5.35) Aμ≡cosθWBμ+s i nθWW3 μ. (5.36) (In the exercises of chapter 2 these deﬁnitions f ollowed from requiring that the photon does not interact with the neutrino. In section 5.6 we will see that the photon is also massless.) Writing out the projection operators for left- an d right-handed fermion s, eqs. (5.3) and (5.4), we obtain the following interactions: 1. A coupling of the charged vector bosons W±which mediate transitions between neu- trinos and electrons (or uanddquarks) with an interaction term −g 2√ 2 νγμ/parenleftBig 1−γ5/parenrightBig eW− μ−g 2√ 2 uγμ/parenleftBig 1−γ5/parenrightBig dW− μ+ h.c. (5.37) (h.c. means ‘hermitian conjugate’ and gives the interaction involving an emitted W+ μ where the incoming particle is a neutrino (or u) and the outgoing particle is an electron (ord).) - 163 - 2. The usual coupling of the photon with the char ged fermions is (using, for instance, the relation eq. (5.54)): gsinθW eγμeAμ−2 3gsinθW uγμuAμ+1 3gsinθW dγμdAμ. (5.38) Note that the left- and right-handed fermions have exactly the same coupling to the photon so that the electromagnetic couplin g turns out to be purely vector (i.e. no γ5 term). 3. The coupling of neutrinos to the neutral weak gauge boson Zμ: −g 4c o sθW νγμ/parenleftBig 1−γ5/parenrightBig νZμ. (5.39) 4. The coupling of both the left- and right-handed electron to the Z: g 4c o sθW e/parenleftBig γμ/parenleftBig 1−γ5/parenrightBig −4s i n2θWγμ/parenrightBig eZμ. (5.40) 5. The coupling of the quarks to the Zcan be written in the general form −g 2c o sθW qi/parenleftBig T3 iγμ/parenleftBig 1−γ5/parenrightBig −2Qisin2θWγμ/parenrightBig qiZμ, (5.41) where quark ihas the third component of weak isospin T3 iand electric charge Qi. From these terms in the Lagrangian we can directly read oﬀ the Feynman rules for the three-point vertices with two fermions and one weak gauge boson. Then we can use these vertices to calculate weak interactions of the quarks and leptons. This allows us, for example, to calculate the total decay width of the ZorWboson, by calculating the decay width into all possible quarks and leptons. However, quarks are not free particles, so for exclusiveprocesses, in which we trigger on known initia l or ﬁnal state hadrons, information is needed about the probability to ﬁnd a quark with given properties inside an initial hadron or the probability that a quark with given properti es will decay (“fragment”) into a ﬁnal state hadron. - 164 - Exercise 5.3 The decay rate for the Zinto a fermion-antifermion pair, Z→f¯f,i s Γ=1 2MZ/integraldisplay dLIPS|M|2=1 64π2MZ/integraldisplay dΩ|M|2, where dLIPSstands for the Lorentz invariant phase space measure for the two ﬁnal-state fermions, and/integraltextdΩ is the integral over the solid angle (of one ﬁnal- state particle).Write the general interaction term for the coupling of the Zb o s o nt oaf e r m i o n as −g 2c o sθWγμ/parenleftBig vf−afγ5/parenrightBig . Show that the squared matrix element, s ummed over the spins of the (outgoing) fermions and averaged over the spin of the (incoming) Zboson is |M|2=−1 12gμνg2 cos2θW/parenleftBig (vf)2+(af)2/parenrightBig Tr (γμγ·k1γνγ·k2), where k1andk2are the momenta of the outgoing fermions and the gauge polarization sum is/summationdisplay λ/epsilon1(λ)∗ μ/epsilon1(λ) ν=−gμν+qμqν M2 Z (q=k1+k2is the initial momentum of the Zboson). Hence show that Γ=1 48πg2 cos2θW/parenleftBig (vf)2+(af)2/parenrightBig MZ. Neglect the masses of the fermions in comparison to the Zmass. Exercise 5.4TheZboson can decay leptonically into a pair of neutrinos or charged leptons of all three generations and hadronically into uquarks, dquarks, cquarks, s quarks, or bquarks ( cquarks couple like uquarks, whereas squarks and b quarks couple like dquarks). Deduce the values of v fandaffor each of these cases and consequently estimate the decay width of the Zboson. (The current experimental value is 2 .4952±0.0023 GeV.) [Take MZ=9 1.19 GeV, sin2θW=0.23, and the ﬁne-structure constant α= 1/129 (why this value?)]. - 165 - 5.6 The Higgs Part and Gauge Boson Masses The Higgs doublet Lagrangian should contain a “spontaneous symmetry breaking” potential which will give the Higgs a vev and self-intera ctions, and kinetic terms which will generate the gauge boson masses and interactions between the Higgs and the gauge bosons. We ﬁrstconsider the potential: V(Φ) = −μ 2Φ∗ iΦi+λ/parenleftBig Φ∗iΦi/parenrightBig2. (5.42) This potential has a minimum at Φ+ iΦi=1 2μ2/λ. Writing Φ in the form of eq. (5.19) and replacing this in the potential eq. (5.42), we ﬁnd that we get a mass term for the real Higgs ﬁeldH,w i t hv a l u e mH=√ 2μ. As expected, the ωado not appear in the potential. In an ungauged theory, they would be the massless goldstone bosons. In a gauge theory like theStandard Model, they will reappear as the longitudinal degrees of freedom of the massive gauge bosons. The remaining term of the Φ Lagrangian is the kinetic term ( D μΦ)†(DμΦ). Looking at this term more carefully will help us to understand where the “physical” gauge bosons (i.e. the W±,Zand photon) come from, and how they are related to the W1 μ,W2 μ,W3 μ,Bμ.T os e et h e eﬀect of the Higgs vev on the gauge boson masses, it is most simple to work in the unitary gauge, that is, we absorb the exponential of eq. (5.19) with a gauge transformation. In this gauge, the covariant derivative acting on the Higgs doublet is DμΦ=1 √ 2⎛ ⎝∂μ+ig 2⎛⎝ W3 μ√ 2W+ μ√ 2W− μ−W3 μ⎞⎠ +ig/prime 2Bμ⎞⎠⎛⎝ 0 v+H⎞⎠ , (5.43) so that |DμΦ|2=1 2(∂μH)2+g2v2 4W+μW− μ+v2 8/parenleftBig gW3 μ−g/primeBμ/parenrightBig2+ interaction terms ,(5.44) where the ‘interaction terms’ are terms involving three ﬁelds (two gauge ﬁelds and the H- ﬁeld). Eq. (5.44) tells us that the W3 μandBμﬁelds mix (as do W1 μandW2 μ)a n dt h ep h y s i c a l gauge bosons must be superpositions of these ﬁ elds, such that there are no mixing terms. Thus we deﬁne Zμ≡cosθWW3 μ−sinθWBμ, (5.45) Aμ≡cosθWBμ+s i nθWW3 μ, (5.46) with the weak mixing angle θW(“Weinberg angle”) deﬁned by tanθW≡g/prime g. (5.47) - 166 - With this eq. (5.44) is rewritten as |DμΦ|2=1 2(∂μH)2+g2v2 4W+ μW−μ+v2g2 8c o s2θWZμZμ+0AμAμ. (5.48) Here we see how SU(2) and U(1) are uniﬁed (or at least ‘en tangled’) in the sense that the neutral gauge boson that acquires a mass through the Higgs mechanism is the linear superposition of a gauge boson from the SU(2) and the U(1)Ygauge boson. From eq. (5.48) we can read oﬀ the masses of the gauge bosons. The last term tells us thatthe linear combination eq. (5.46) remains ma ssless. This ﬁeld is identiﬁed with the photon. For the other ﬁelds we have M W=1 2gv, M Z=1 2gv cosθW. (5.49) TheZboson mediates the neutral current weak interactions. These were not observed until after the development of the theory. From the magnitude of amplitudes involving weak neutral currents (exchange of a Zboson), one can infer the (tree level) magnitude of the weak mixing angle, θW. The ratio of the masses of the ZandWbosons is a prediction of the Standard Model. More precisely, we deﬁne a quantity known as the ρ-parameter by M2 W=ρM2 Zcos2θW. (5.50) In the Standard Model ρ= 1 at tree level. In higher orders there is a small correction, which depends on the deﬁnition used for sin θW(that is, which loop corrections are included in sin θW). Note that the ρ-parameter would be very diﬀerent from one if the symmetry breaking were due to a scalar multiplet whic h was not a doublet of weak isospin. Accurate measurements of the ρ-parameter and other so-called electro-weak precision observables, together with their prediction at higher order within the SM, serve as very powerful tests ofthe SM. The Higgs enters in virtual loops, allowing for an indirect determination of its mass through ﬁts of the predictions to the data (see the homepage of the Electroweak Working Group, http://lepewwg.web.cern.ch/LEPEWWG for more information). The spontaneous symmetry breaking mechanism breaks SU(2)×U(1) Ydown to U(1). It is this surviving U( 1 )t h a ti si d e n t i ﬁ e da st h e U(1) of electromagnetism. It is not the U(1)Y of the original gauge group but a set of transformations generated by a particular linearcombination of the original U(1) and rotations about the third axis of weak isospin. To see this we note that the explicit representation of the generator Yas a 2×2 matrix, which can be combined with the explicit representation of T 1,T2andT3,i sg i v e nb y Y=1 2⎛ ⎝10 01⎞ ⎠. (5.51) - 167 - The factor 1 /2 ensures the normalization12condition eq. (2.7). Using eq. (5.51) it can easily be seen that the symmetry associated with the generator Q≡Y+T3=⎛ ⎝10 00⎞ ⎠ (5.52) is not broken, i.e. Q|0/angbracketright= 0 (see eq. (4.5)). Thus, starting with four generators, we get only three Goldstone bosons. These will become the longitudinal components of three gauge bosons, thereby giving them a mass, whereas the fourth is left massless. The coupling of any particle to the photon is always proportional to gsinθW(Y+T3)=gsinθWQ. (5.53) Thus we can identify gsinθWwith one unit of electric charge, and we have the relationship between the weak coupling gand the electron charge e, e=gsinθW. (5.54) We end this subsection by giving the remaining pieces of the SM Lagrangian from eqs. (5.44) and (5.42), LHiggs =|DμΦ|2−μ2Φ∗ iΦi+λ/parenleftBig Φ∗iΦi/parenrightBig2 =1 2(∂μH)2+μ2H2+g2v2 4W+μW− μ+v2g2 8c o s2θWZμZμ + interaction terms . (5.55) 5.7 Classifying the Free Parameters The free parameters in the Standard Model for one generation are: •The two gauge couplings for the SU(2) and U(1) gauge groups, gandg/prime. •The two parameters μandλin the scalar potential V(Φ). •The Yukawa coupling constants Yu,Yd,YeandYν. It is convenient to replace these parameters by others, which are more directly measurable in experiments, namely e,sinθW,meandmW,a n dmH,mu,mdandmν. (Note that the gauge 12We warn the reader that in the literature sometimes a diﬀerent normalization is used such that eq. (5.52) reads Q=Y/2+T3. - 168 - sector is well measured, but the quark masses are not directly observable; we have yet to ﬁnd the Higgs, and although we see neutrino mass diﬀerences, measuring the absolute mass scale is diﬃcult — and the neutrino masses might not be directly proportional to Yukawa couplings anyway.) The relation between these physical parameters and the parameters of the initial Lagrangian are tanθW=g/prime g, (5.56) e=gsinθW, (5.57) mH=√ 2μ, (5.58) MW=gμ 2√ λ, (5.59) me=Yeμ √ λ. (5.60) Note that when we add more generations of fermions, we will acquire more parameters:additional masses (or yukawa couplings, i.e. 4 parameters per generation), and also mixing angles, as we will see in the next chapter. In terms of these measured quantities, the Zmass, M Z, and the Fermi-coupling, GF,a r e pre- dictions of the SM (although historically GFwas known for many years before the discovery of the Wboson, and its value was used to predict the Wmass). 5.8 Summary •Weak interactions are mediated by the SU(2) gauge bosons, which act only on the left-handed components of fermions. •The (left-handed) neutrino and left-ha nded component of the electron form an SU(2) doublet, whereas the right-handed com ponents of the electron and neutino are SU(2) singlets. Similarly for the quarks. •There is also a weak hypercharge U(1)Ygauge symmetry. Both left- and right-handed quarks transform under this U(1)Ywith a hypercharge which is related to the elec- tric charge by the relation eq. (5.54). The left-handed leptons and the eRalso carry hypercharge, but the νRhas no SM gauge interactions. •In the symmetry limit (before spontaneous symmetry breaking) the fermions with SU(2) gauge interactions are massless.13The spontaneous symmetry breaking mech- anism which gives a vev to the scalar ﬁeld also generates the fermion masses. 13This does not apply to νR,w h i c h canhave an explicit mass term - 169 - •The scalar multiplet that is responsible for the spontaneous symmetry breaking also carries weak hypercharge. As a result, one neutral gauge boson (the Z)a c q u i r e sa mass, whereas its orthogonal superposition is the massless photon. The magnitude of the electron charge, e,i st h e ng i v e nb y e=gsinθW. •The weak interactions proceed via the exchange of massive charged or neutral gaugebosons. The old four-fermi weak Hamiltonia n is an eﬀective Hamiltonian which is valid for low energy processes in which all momenta are small compared with the Wmass. The Fermi coupling is obtained in terms of e, M Wand sin θWby eq. (6.16). For completeness, a full set of Feynman rules for the case of a single family of leptons isgiven as an appendix to this lecture. - 170 - Feynman Rules in the Unitary Gauge (for one Lepton Generation) Propagators: (All propagators carry momentum p.) W μν −i(gμν−pμpν/M2 W)/(p2−M2 W) Z μν −i(gμν−pμpν/M2 Z)/(p2−M2 Z) A μν −igμν/p2 e i(γ·p+me)/(p2−m2 e) ν iγ·p/p2 H i/(p2−m2 H) - 171 - Three-point gauge-boson couplings: (All momenta are deﬁned as incoming.) Aρ W− μ W+ νp3 p1 p2igsinθW((p1−p2)ρgμν+(p2−p3)μgνρ+(p3−p1)νgρμ) Zρ W− μ W+ νp3 p1 p2igcosθW((p1−p2)ρgμν+(p2−p3)μgνρ+(p3−p1)νgρμ) - 172 - Four-point gauge-boson couplings: W− ρ W+ σ W− μ W+ νig2(2gμρgνσ−gμνgρσ−gμσgνρ) Zρ Zσ W− μ W+ νig2cos2θW(2gμνgρσ−gμρgνσ−gμσgνρ) Aρ Aσ W− μ W+ νig2sin2θW(2gμνgρσ−gμρgνσ−gμσgνρ) Zρ Aσ W− μ W+ νig2cosθWsinθW(2gμνgρσ−gμρgνσ−gμσgνρ) - 173 - Three-point couplings with Higgs scalars: HH−3 2igm2 H/MW H ee−1 2igm e/MW H W− μ W+ νigM Wgμν H Zμ Zνi(g/cos2θW)MWgμν - 174 - Four-point couplings with Higgs scalars: HH HH−3 4ig2(m2 H/M2 W) HH W− μ W+ ν1 2ig2gμν HHZ μ Zν1 2i(g2/cos2θW)gμν - 175 - Fermion interactions with gauge bosons: W− μ eν−i/parenleftBig g/2√ 2/parenrightBig γμ(1−γ5) Aμ eeigsinθWγμ Zμ ee1 4i(g/cosθW)γμ/parenleftBig 1−4s i n2θW−γ5/parenrightBig Zμ νν−1 4i(g/cosθW)γμ(1−γ5) - 176 - 6 Additional Generations In the previous section, the Lagrangian of the Standard Model with one family was given. Here we include additional “families” (or “gene rations”) and brieﬂy outline the phenomeno- logical consequences in the quark sector. Family-changing processes among the leptons willbe discussed in the neutrino chapter. 6.1 A Second Quark Generation The second generation of quarks consists of a c(“charm”) quark, which has electric charge +2 3and an s(“strange”) quark, with electric charge −1 3. We can just add a copy of the left- handed isodoublet and copies of the right-handed s inglets in order to include this generation. The only diﬀerence would be in the Yukawa interaction terms where the coupling constantsare chosen to reproduce the correct masses for the new quarks. But in this case there is a further complication. It is possible to write down Yukawa terms which mix quarks of diﬀerent generations, e.g. the Yukawa couplings of the previous section become matrices in ﬂavour space, −[Y d]ij qLiΦdRj−[Yu]ij qLiΦcuRj+ h.c. (6.1) where i, jare generation indices. The oﬀ-diagonal element [ Yd]12seems to gives rise to a mass mixing between dandsquarks. The Yukawa matrices are nf×nfmatrices, where nfis the number of ﬂavours, and can be diagonalised by independant unitary transformations on the left and right (because YY† andY†Yare hermitian). The physical particles are those that diagonalize the mass matrix. So it is convenient to rotate to the eigenbasis of the mass matrix, where there is noYukawa mixing between quarks of diﬀerent generations. Notice that when we add a second generation, i t has the same gauge interactions as the ﬁrst. So if we make a unitary transformation in generation space, the fermion kinetic terms remain unchanged. Taking advantage of this freedom, we can rotate uR,dRandqLrespectively to the mass eigenstate bases of the uR,dRanduL. This means, however, that the quark doublets which couple to the gauge bosons are,i n general, superpositions of physical quarks, because we have written the dLiin the uLimass eigenstate basis:⎛ ⎝u ˜d⎞⎠ L, (6.2) - 177 - and ⎛ ⎝c ˜s⎞⎠ L, (6.3) where ˜dand ˜sare related to the physical dandsquarks by ⎛⎝˜d ˜s⎞⎠ =VC⎛⎝ d s⎞⎠ , (6.4) where VCis a unitary 2 ×2 matrix. Terms which are diagonal in the quarks are unaﬀected by this unitary transformation of the quarks. Thus the coupling to photons or Zbosons is the same whether written in terms of ˜d,˜sor simply s,d. We will return to this later. On the other hand the coupling to the charged gauge bosons is −g 2√ 2 uγμ(1−γ5)˜dW− μ−g 2√ 2 cγμ(1−γ5)˜sW− μ+ h.c. (6.5) which we may write as −g 2√ 2⎛ ⎝ u c⎞⎠T γμ(1−γ5)VC⎛⎝ d s⎞⎠ W− μ+h . c . . (6.6) The most general 2 ×2 unitary matrix may be written as ⎛⎝ e−iγ 1⎞⎠⎛⎝ cosθCsinθC −sinθCcosθC⎞⎠⎛⎝ eiα eiβ⎞⎠ , (6.7) w h e r ew eh a v es e to n eo ft h ep h a s e st o1s i n c ew ec a na l w a y sa b s o r ba no v e r a l lp h a s eb y adjusting the remaining phases, α,βandγ. The phases, α,β,γcan be absorbed by performing a global phase transformation on the d, sanduquarks respectively. This again has no eﬀect on the neutral terms. Thus the most general observable unitary matrix is given by VC=⎛ ⎝cosθCsinθC −sinθCcosθC⎞⎠ , (6.8) where θCis the Cabibbo angle. In terms of the physical quarks, the charged gauge boson interaction terms are −g 2√ 2/parenleftBig cosθC uγμ(1−γ5)d+s i n θC uγμ(1−γ5)s +c o s θC cγμ(1−γ5)s−sinθC cγμ(1−γ5)d/parenrightBig W− μ+h . c . .(6.9) - 178 - This means that the uquark can undergo weak interactions in which it is converted into ansquark, with an amplitude that is proportional to sin θC. I ti st h i st h a tg i v e sr i s et o strangeness violating weak interaction processes, such as the leptonic decay of K−into a muon and antineutrino. The Feynman diagram for this process is K−s ¯u ¯νμ W− 6.2 Flavour Changing Neutral Currents Although there are charged weak interactions that violate strangeness conservation, there are no known neutral weak interactions tha t violate strangeness. For example, the K0does not decay into a muon pair or two neutrinos (branching ratio <10−5). This means that theZboson only interacts with quarks of the same ﬂavour. We can see this by noting that theZboson interaction terms are unaﬀected by a unitary transformation. This absence of ﬂavour changing neutral currents (FCNC) in e xperimental data is rather important. As we will see, in the Standard Model there are no FCNC at tree level, and the absence of FCNCis an important constraint for many extensions of the Standard Model. TheZboson interactions with dandsquarks are proportional to ˜d˜d+ ˜s˜s (6.10) (we have suppressed the γ-matrices which act between the fermion ﬁelds). Writing this out in terms of the physical quarks we get cos2θC dd+s i n θCcosθC sd+c o s θCsinθC ds+s i n2θC ss +c o s2θC ss−sinθCcosθC ds−cosθCsinθC sd+s i n2θC dd. (6.11) We see that the cross terms cancel out and we are left with simply dd+ ss. (6.12) This cancellation is the reason for the absen ce of FCNC and is simply a consequence of the unitarity of the mixing matrix eq. (6.7). This eﬀect is also known as the “GIM” (Glashow- Iliopoulous-Maiani) mechanism. It w as used to predict the existence of the cquark. - 179 - There can be a small contribution to strangeness changing neutral processes from higher order corrections in which we do not exchange a Zboson, but two charged Wbosons. The Feynman diagrams for such a contribution to the leptonic decay of a K0(which consists of adquark and an santiquark) are: d ¯sW− W+μ ¯μν uc +d ¯sW− W+μ ¯μν These diagrams diﬀer in the ﬂavour of the internal quark which is exchanged, being a u quark in the ﬁrst diagram and a cquark in the second. Both of these diagrams are allowed because of the Cabibbo mixing. The ﬁrst of the se diagrams gives a contribution proportional to +s i nθCcosθC, which arises from the product of the two couplings involving the emission of the Wbosons. The second diagram gives a term proportional to −cosθCsinθC. If thecanduquarks had identical masses then these tw o contributions would cancel precisely. However, because the cquark is much more massive than the uquark, there is some residual contribution. This was used to limit the mass of the cquark to <5 GeV, before it was discovered. 6.3 Adding Another Lepton Generation We ﬁrst neglect the νRand neutrino masses. In this approximation, there will be no gen- eration mixing in the lepton sector, so we c an include other lepton families, the muon and its neutrino, and the tau-lepton with its neutrino, simply as copies of what we have for theelectron and its neutrino. For each family we have a weak isodoublet of left-handed leptons and a right-handed isosinglet for the charged lepton. Thus, the mechanism which determines the decay of the muon ( μ) is one in which the muon converts into its neutrino and emits a charged W −, which then decays into an electron and (electron-) antineutrino. The Feynman diagram is - 180 - μνμ ¯νeeW− The amplitude for this process is given by the product of the vertex rules for the emission (or absorption) of a W−with a propagator for the Wboson between them. Up to corrections of order m2 μ/M2 W, we may neglect the eﬀect of the term qμqν/M2 Win the W-boson propagator, so that we have /parenleftBigg −ig 2√ 2 νμγρ(1−γ5)μ/parenrightBigg/parenleftBigg−igρσ q2−M2 W/parenrightBigg/parenleftBigg −ig 2√ 2 eγσ(1−γ5)νe/parenrightBigg , (6.13) where qis the momentum transferred from the muon to its neutrino. Since this is negligible in comparison with MWwe may neglect it and the expression for the amplitude simpliﬁes to ig2 8M2 W νμγρ(1−γ5)μ eγρ(1−γ5)νe. (6.14) Before the development of this model, weak interactions were described by the “four-fermi model” with a weak interaction Hamiltonian given by Hijkl=GF √ 2 ψiγμ(1−γ5)ψj ψkγμ(1−γ5)ψl. (6.15) We now recognize this as an eﬀective low-energy Hamiltonian which may be used when theenergy scales involved in the weak process are negligible compared with the mass of the W boson. The Fermi coupling constant, G F, is related to the electric charge, e,t h eWmass and the weak mixing angle by GF=e2 4√ 2M2 Wsin2θW. (6.16) This gives us a value for GF, GF=1.12×10−5GeV−2, (6.17) which is very close to the value of 1 .17×10−5GeV−2as measured from the lifetime of the muon. We see that the weak interactions are ‘weak’, not because the coupling i s particularly small (theSU(2) gauge coupling is about twice as large as the electromagnetic coupling), but - 181 - because the exchanged boson is very massive, so that the Fermi coupling constant of the four-fermi theory is very small. The large mass of the Wboson is also responsible for the fact that the weak interactions are short range (of order 10−18m). In the Standard Model, however, we also have neutral weak currents. Thus, for example, wecan have elastic scattering of muon-type neutrinos against electrons via the exchange of the Zboson. The Feynman diagram for such a process is: νμνμ eeZ Exercise 6.1 Let us write the four-fermi interaction for this process as H=GF √ 2 νeγρ(1−γ5)νe μγρ(v−aγ5)μ, where vandagive us the vector and axial-vector coupling of the muon to the Zboson (the muon couples in an identical way to the electron). Determine v andain terms of θW. 6.4 Adding a Third Generation (of Quarks) Adding a third generation is achieved in a similar way. In this case the three weak isodoubletsof left-handed fermions are ⎛ ⎝u ˜d⎞⎠ ,⎛⎝ c ˜s⎞⎠ ,⎛⎝ t ˜b⎞⎠ , (6.18) where ˜d,˜sand˜bare related to the physical d,sandbquarks by ⎛ ⎜⎜⎝˜d ˜s ˜b⎞ ⎟⎟⎠=VCKM⎛ ⎜⎜⎝d s b⎞ ⎟⎟⎠. (6.19) The 3×3 unitary matrix VCKMis called Cabibbo-Kobayashi-Maskawa (CKM) matrix. Once again it only aﬀects the charged weak processes in which a Wboson is exchanged. For this - 182 - reason the elements are written as ⎛ ⎜⎜⎝VudVusVub VcdVcsVcb VtdVtsVtb⎞ ⎟⎟⎠. (6.20) A3×3 unitary matrix can have nine independent parameters (counting the real and imag- inary parts of a complex element as two parameters). In this case there are six possible fermions involved in the charged weak pro cesses and so we can have ﬁve relative phase transformations, thereby absorbing ﬁve of the nine parameters. This means that whereas the Cabibbo matrix o nly has one parameter (the Cabibbo angle, θC), the CKM matrix has four independent parameters. If the CKM matrix were real it would only have three independent paramete rs. This means that in the case of the CKM matrix some of the elements may be complex. The four independent parameters can be thought of as three mixing angles between the three pairs of generations and a complex phase. The requirement of unitarity puts various constraints on the elements of the CKM matrix. For example we have VudV∗ ub+VcdV∗ cb+VtdV∗ tb=0. This can be represented as a triangle in the complex plane known as the “unitarity triangle”: VudV∗ ub VtdV∗ tb VcdV∗ cbα β γ The angles of the triangle are related to ratios of elements of the CKM matrix α=−arg/braceleftBiggVtdV∗ tb VudV∗ ub/bracerightBigg , (6.21) β=−arg/braceleftBiggVtdV∗ tb VcdV∗ cb/bracerightBigg , (6.22) γ=−arg/braceleftBiggVudV∗ ub VcdV∗ cb/bracerightBigg . (6.23) A popular representation of the CKM matrix is the Wolfenstein parameterisation which uses the parameters A, which is assumed to be of order unity, a complex number ( ρ+iη)a n d as m a l ln u m b e r λ, which is approximately equal to sin θC. In terms of these parameters the - 183 - CKM matrix is written as VCKM =⎛ ⎜⎜⎝1−λ2/2 λA λ3(ρ−iη) −λ 1−λ2/2 Aλ2 Aλ3(1−ρ−iη)−Aλ21⎞ ⎟⎟⎠+O(λ4). (6.24) We see that whereas the Wbosons can mediate a transition between a uquark and a b quark ( Vub) or between a tquark and a dquark ( Vtd), the amplitude for such transitions are suppressed by the cube of the small quantity wh ich determines the amplitude for transitions between the ﬁrst and second generations, λ.T h eO(λ4) corrections are needed to ensure the unitarity of the CKM matrix and these corrections have several matrix elements which are complex. 6.5 CP Violation The possibility that some of the elements of the CKM matrix may be complex provides amechanism for the violation of CPconservation. Violation of CPconservation has been observed in the K 0− K0system, and is currently being investigated for Bmesons. Higher-order corrections to the masses of B0and B0give rise to mixing between the two states. Thus the mass matrix can be written as ⎛ ⎝MB0ΔM (ΔM)∗MB0⎞⎠ . (6.25) The mass eigenstates are therefore |BL/angbracketright=p|B0/angbracketright+q| B0/angbracketright, (6.26) whose mass is M−1 2Δm,a n d |BH/angbracketright=p|B0/angbracketright−q| B0/angbracketright, (6.27) whose mass is M+1 2Δm, where we have introduced the mass diﬀerence between the two mass eigenstates, Δm ≡2/radicalBig ΔM(ΔM)∗. If ΔM were real then we would have p=q=1/√ 2 and these mass eigenstates would be CPeigenstates, using the fact that CP|B0/angbracketright=−| B0/angbracketright. However, the non-zero phases in the CKM matrix give rise to a complex phase for Δ M, so that the ratio of pandqis a complex phase, indicating that BLandBHare notCP eigenstates. - 184 - A typical weak interaction contribution to the mass-mixing term, Δ M,i sg i v e nb yt h e Feynman diagram b dW+ W−d bqj qi Note that on the left we have a B0, consisting of a bquark and a dantiquark, whereas on the r i g h tw eh a v ea B0consisting of a dquark and a bantiquark. The internal quarks marked qiandqjcan each be u,cortquarks, and each of the vertices carries some element of the CKM matrix. The total contribution, therefore, may be written as /summationdisplay i=u,c,t/summationdisplay j=u,c,tVibV∗ idV∗ jbVjdaij. Once again, if all the masses of the quarks were equal then the amplitudes aijwould all be equal, and the sum would vanish by the unitarity constraints imposed on the elements Vik. Since the quarks do not all have the same mass, there is some residual contribution. Indeed, the above diagram is dominated by the term in which a tquark is exchanged on both sides, since this quark is much more massive than the rest. Restricting ourselves to the tquark exchange contribution, we can read oﬀ the phase of this contribution, without calculating the diagram itself. It is given by the phase of the products of the CKM matrix elements entering in the diagram, namely (V∗ tdVtb)2. The phase of this quantity is the square of the ratio of pandq,s ow eh a v e p q=V∗ tdVtb VtdV∗ tb. Now suppose that at time t= 0 we prepare a state which is purely B0. Accounting for the fact that the B0meson has a decay rate Γ, we can use eqs. (6.26, 6.27) to write the state at timetas |B(t)/angbracketright=e−iMte−Γt/2/parenleftBigg cos/parenleftbiggΔm 2t/parenrightbigg |B0/angbracketright+iq psin/parenleftbiggΔm 2t/parenrightbigg | B0/angbracketright/parenrightBigg . (6.28) Now suppose that the amplitude for a state |B0/angbracketrightto decay into some CPeigenstate |f/angbracketrightis Af, whereas the amplitude for a state | B0/angbracketrightto decay into the state |f/angbracketrightis Af. Once again, if - 185 - CPwere conserved, we would have Af=± Af, but the CPviolating phases give rise to a more general complex phase for the ratio of these two amplitudes. This means that the amplitude to ﬁnd the state |f/angbracketrightafter time tis given by /angbracketleftf|Hwk|B(t)/angbracketright=e−iMte−Γt/2/parenleftBigg cos/parenleftbiggΔm 2t/parenrightbigg Af+iq psin/parenleftbiggΔm 2t/parenrightbigg Af/parenrightBigg . (6.29) Similarly, if we had prepared a B0att= 0 the amplitude to ﬁnd the state |f/angbracketrightwould be /angbracketleftf|Hwk| B(t)/angbracketright=e−iMte−Γt/2/parenleftBigg cos/parenleftbiggΔm 2t/parenrightbigg Af−ip qsin/parenleftbiggΔm 2t/parenrightbigg Af/parenrightBigg . (6.30) Taking the moduli squared for the decay rates we derive the result Γ(B(t)→f)−Γ( B(t)→f) Γ(B(t)→f)+Γ ( B(t)→f)=−sin (Δmt )/Ifracturm/parenleftBiggq p Af Af/parenrightBigg . (6.31) For example, if the state |f/angbracketrightis the CPeven two-pion state |π0π0/angbracketright, the Feynman diagram at the quark level for A2πis W−b ¯du ¯u d¯d To fully calculate the decay amplitudes we would need to know the wave functions for the mesons in terms of the constituent quark-antiquark pairs, but for the ratio A2π/A2πwe just need the ratios of the CKM matrix elements occurring in this diagram, namely A2π A2π=VubV∗ ud V∗ ubVud, so that (using eq. (6.21)) /Ifracturm/parenleftBiggq p A2π A2π/parenrightBigg =VtdV∗ tb V∗ tdVtbVubV∗ ud V∗ ubVud=−sin (2α). (6.32) As a further example we consider the so-called “golden channel” where |f/angbracketrightis the state |J/ψK S/angbracketright. In this case the quark level Feynman diagram is - 186 - W−b ¯dc ¯c s¯d Here there is a further complication since the outgoing state ( s¯d) is actually a K0(and likewise for the B0decay it would be a K0). As in the B0system, the mass eigenstates are given by |KS/angbracketright=pK|K0/angbracketright+qK| K0/angbracketright, |KL/angbracketright=pK|K0/angbracketright−qK| K0/angbracketright. (6.33) Once again, if CPwere conserved we would have pK=qK=1/√ 2, and these mass eigen- states would be eigenstates of CP. The phases in the CKM matrix introduce a phase in the ratio of pKandqK, calculated from diagrams similar to the ones for the B0system (but with thebquark replaced by an squark). In this case it is the diagram with an internal cquark exchange that dominates (although the mass of the cis much smaller than the tquark mass, the CKM matrix elements are much larger for cquark exchange than for tquark exchange and this eﬀect dominates), so we have a factor qK pK=V∗ cdVcs VcdV∗cs which enters in the ratio of the decay amplitudes, giving AJ/ψ K S AJ/ψ K S=−VcbV∗ cs V∗ cbVcsV∗ cdVcs VcdV∗cs=−VcbV∗ cd V∗ cbVcd (a minus occurs because the J/ψK Sstate is CPodd), so that (using eq. (6.22)) /Ifracturm/parenleftBiggq p AJ/ψ K S AJ/ψ K S/parenrightBigg =−VtdV∗ tb V∗ tdVtbVcbV∗ cd V∗ cbVcd=s i n ( 2 β). (6.34) 6.6 Summary •Additional generations may be added, with gauge interactions copied from the ﬁrst, but in this case one can have mass-mixing between quarks of diﬀerent generations. In terms of the mass eigenstates, the charged Wbosons mediate transitions between - 187 - aT3=+1 2quark ( u,cort) and a superposition of T3=−1 2quarks ( d,sandb). In two generations, this mechanism allows weak interactions that violate strangeness conservation, and the mixing matrix has only one independent parameter, the Cabibbo angle. •The unitarity of the mixing matrix guarantees that there are no strangeness changingneutral processes. Weak interact ions involving the exchange of a Zboson do not change ﬂavour. There is a small violation of this in higher orders owing to the mass splittingbetween the quarks. •Including a third generation, the mixing matrix for the T 3=−1/2q u a r k s( d,sand b) is the CKM matrix. This matrix has four i ndependent parameters, so that some of the matrix elements may be complex. •The possibility that some of the elements of the CKM matrix may be complex leads to a weak interaction contribution to the mass mixing of B0and B0which can be complex. This gives rise to CPviolation, since the eigenstates of the B0mass matrix are then no longer eigenstates of CP. The CKM matrix also introduces phases in the ratios of the decay amplitudes for B0and B0to a given CPeigenstate. Products of the phase of the mass mixing and the ratio of the decay amplitudes can be observed directly intagged Bmeson experiments, and the angles αandβof the unitarity triangle can be directly measured. - 188 - 7 Neutrinos In its original formulation, the Standard Mo del had massless neutrinos — neutrino masses were not measured at the time. We now know that neutrinos have a (very small) mass, which can be accommodated in the SM in a straightforward way. We will discuss this in thesecond part of this chapter. There are two possible types of neutrino mass terms, “Dirac” and “Majorana”, because the neutrino has zer o electric charge. This makes neutrino mass terms a bit diﬀerent from those of the other fermions and may explain why neutrinos aremuch lighter than SM fermions. In the ﬁrst part of this chapter we focus on t he currently observed consequence of small neutrino masses, neutrino oscillations. 7.1 Neutrino Oscillations Recall that in the quark sector, there were ﬂavour changing charged current processes, that is, the Wcould interact with an up-type quark of one generation, and a down-type quark of another. If the neutrinos have mass, we should get exactly the same eﬀect in the leptonsector, except that the mixing matrix U fmis called the PMNS matrix (for Pontecorvo, Maki, Nakagawa and Sakata), rather than CKM. The index order “ﬂavour-mass” in Ufmindicates thatUrotates a vector from the neutrino mass basis to the neutrino “ﬂavour” basis, which is the charged lepton mass basis. The physical consequences of mixing angles are quite diﬀerent between the lepton sector and the quarks. This is because neutrinos are very light and have only weak interactions. In the quark sector one can diﬀerentiate D→K¯μνfromD→π¯μν, because the πandKhave strong and electromagnetic interactions, which allows us to track them in the detector, and they have suﬃciently diﬀerent masses that the tracks are distinguishable. This is not the case in trying to distinguish μ→eν3¯ν2fromμ→eν3¯ν1. The small masses and weak interactions of neutrinos imply that the wave packets corre- sponding to diﬀerent neutrino mass eigenstate s remain superposed over long distances. The eﬀects of ﬂavour mixing can therefore be seen via oscillations. For simplicity we will consider the case of two ge nerations which in the charged lepton sector we will take to be the electron and muon.14We label the neutrino mass eigenstates as ν1 14Of course, in the Standard Model we have three families, but the important concepts can be understood in the simpler case. - 189 - andν2. They are related by an equation very similar to eq. (6.4), ⎛ ⎝νe νμ⎞⎠ =⎛⎝ cosθsinθ −sinθcosθ⎞⎠⎛⎝ ν1 ν2⎞⎠ . (7.1) Now we would like to compute the amplitude for an oscillation process. Suppose that we have an initial beam of muons which decays to relativisitic neutrinos of energy Eand momentum k. The neutrinos travel a distance L=τto a detector where they produce an eor aμby charged current (CC) scattering. The amplitude will be Aμα∼/summationdisplay jUμj×e−i(Ejτ−kjL)×U∗ αj, (7.2) where the three pieces arise from production, p ropagation and detection. (From your ﬁeld theory notes, you can check that the Feynman propagator in position space, G(0,(τ,L)), is the exponential, where the momentum integ ral in the propagator was taken care of in the production process of the neutrinos with 4-momentum ( E,k).) First, suppose that we can neglect the neutrino masses, so ( Ej,kj)=(En,kn) for any j, n. The propagation exponential can then be factored out, and (7.2) is the unitarity condition forU, UμjU∗ αj=δμα. (7.3) Recall that for quarks, with three generations, this relation gives the unitarity triangle. Now we allow the neutrinos to have small masses, m/lessmuchE,k,s ot h a t L/similarequalτremains. Then the exponent can be written as −i(Ejτ−kjL)/similarequal−i(Ej−kj)L=−iE2 j−k2 j E+kL/similarequal−im2 j 2EL, (7.4) such that Pμα=|Aμα|2=|/summationdisplay jUμje−iΔm2 jL/(2E)U∗ αj|2. (7.5) Using the explicit form of Ugiven in eq. (7.1) one obtains the muon survival probability Pμμ=1−sin22θsin2(m2 2−m21)L 4E. (7.6) In reality, there are three generations of leptons in the SM, so the MNS matrix Uis 3×3, and there are three mass eigenstates in the sum of eq. (7.5). As in the case of CKM, MNS can be written in terms of three angles and one phase: ˆU=⎡ ⎢⎢⎣c13c12 c13s12 s13e−iδ −c23s12−s23s13c12eiδc23c12−s23s13s12eiδs23c13 s23s12−c23s13c12eiδ−s23c12−c23s13s12eiδc23c13⎤ ⎥⎥⎦ (7.7) - 190 - /similarequal⎡ ⎢⎢⎣c12 s12 s13e−iδ −s12/√ 2c12/√ 21/√ 2 s12/√ 2−c12/√ 21/√ 2⎤ ⎥⎥⎦, (7.8) where the “solar” angle θ12/similarequalπ/6, and we have used the approximate measured value of the atmospheric angle θ23/similarequalπ/4.s13=s i nθ13≤0.2 is known from experimental bounds and θ13is signiﬁcantly smaller than the other two angles. Note that, unlike the quark sector, some mixing angles are large. Combined with the small neutrino masses, this is puzzling and provoking to theorists, who expend much eﬀort into building models of this. It is often said that MNS has three phases, so let us recall the phase choices that allow us to write eq. (7.7), so as to understand where the other two phases could be: •A3×3 complex matrix has 18 real parameters. •The unitarity condition UU†= 1 reduces this to 9, which can be taken as 3 angles and 6 phases. •Five of those phases are relative phases between the ﬁelds e,μ,τ,ν1,ν2andν3, •... so if we are free to choose the phases of a ll the left-handed fermions, we are left with one phase in the mixing matrix. This was the case with the quarks, where any potential phase in the quark masses could be absorbed by the right-handed fermion ﬁelds. •If the right-handed ﬁelds do not appear in our physical process (which means the masses appear as mm∗), then we are free to make the above phase choice, and our process is independant of any possible phase of the masses. This is the case for neutrino oscillations. •We will see in a later section that the νLcan have so-called “Majorana” masses, between themselves and their antiparticle. This me ans that it is the left-handed neutrino ﬁeld which must absorb the phase of the Majorana m ass. So in physical processes where the Majorana mass appears linearly (and not as mm∗; this is the case e.g. in neutrinoless double-beta decay), one can choose the phase such that the mass is real — in whichcase one can remove one less phase from MNS, or one can keep MNS with one phase, and allow complex masses. •It is always possible to remove the phase from one majorana mass by using the global overall phase of all the leptons. (This ove rall phase corresponds to the global symmetry of lepton number conservation in a theor y without majorana masses and is the sixth - 191 - phase of e,μ,τ,ν1,ν2andν3, which we could not use to remove phases from the lepton number conserving MNS matrix.) So, in three generations, there are possibly two complex majorana neutrino masses, so two “Majorana” phases in addition to the “Dirac” phase δof MNS. Although there are three generations, it is we ll known that for the oscillation probabilities we observe, with the mixing angles that are me asured, two neutrino probabilities are a very good approximation. Why is this? Let us return to the oscillation amplitude Aαβ(L), and imagine it as the sum of three vectors in the complex plane. If α=β, the unitarity condition at L= 0 says they should sum to a vector of length one. If α/negationslash=β, then they should sum to zero and this is the unitarity triangle. At non-zero L, two of the vectors rotate in the complex plane, with frequencies ( m2 j−m21)/2E — so neutrino oscillations correspond, in some sense, to time-dependent non-unitarity. Consider the oscillation probabilities Pμα, measured for atmospheric neutrinos, on length scales corresponding to m2 3−m21. The solar mass diﬀerence can be neglected, because m22−m21/lessmuchm23−m21, so there is only one relevant mass diﬀerence, and the survival probability behaves as for two generations. This is easy t o visualise in the complex plane, where only the vector Uμ3U∗ α3rotates with L. The stationary sum Uμ2U∗ α2+Uμ1U∗ α1can be treated as a single vector, so this looks like a two generat ion system. So “atmospheric” oscillations can be approximated as two-neutrino oscillati ons because the atmospheric mass diﬀerence is very large compared to the solar one. In the case of the solar mass diﬀerence, measur ed for instance at KamLAND, the two neutrino approximation is good because θ13is small. The observed survival probability is Peeand sinceUe3/lessmuchUej,j=1,2, the last term can be dropped in Aee=/summationdisplay jUeje−iΔm2 jL/(2E)U∗ ej. (7.9) 7.2 Oscillations in Quantum Mechanics (in Vaccum and Matter) This subsection reviews a more conventional der ivation of neutrino oscillations in two gener- ations, and includes neutrino oscillations in m atter. Electron neutrinos acquire an eﬀective mass term from their interactions with dense matter — this is the MSW eﬀect — which can have signiﬁcant eﬀects in the sun and in supernovae, and over long baselines in the earth. In the mass eigenbasis we have the Schr¨ odinger equation id dtΨ=H·Ψ (7.10) - 192 - with a diagonal Hamiltonian H=⎛ ⎝E10 0E2⎞⎠ . (7.11) This Schr¨ odinger equation can easily be solved. Deﬁning our initial states at t=0a s |1/angbracketright≡|1(t=0 )/angbracketright,|2/angbracketright≡|2(t=0 )/angbracketrightwe get the time dependent states |1(t)/angbracketright=e−iE1t|1/angbracketright, |2(t)/angbracketright=e−iE1t|2/angbracketright. (7.12) Let us repeat the last few steps in the interaction eigenbasis. Multiplying eq. (7.10) by V from the left we get the corresponding Schr¨ odinger equation as id dt˜Ψ=˜H·˜Ψ (7.13) with ˜H≡V·H·V−1=⎛⎝ a+bc ca −b⎞⎠ , (7.14) where a=1 2(E1+E2), (7.15) b=1 2(E1−E2)c o s ( 2 θ), (7.16) c=−1 2(E1−E2)s i n ( 2 θ). (7.17) The crucial feature of the new Hamiltonian is that it is no longer diagonal. As a result, i fw es t a r ta tt i m e t= 0 with an interaction eigenstate |α/angbracketright, then at a later time we get a superposition of |α/angbracketrightand|β/angbracketrightinteraction eigenstates. Indeed, using eq. (7.1) for the time dependent states we get |α(t)/angbracketright=e−iE1tcosθ|1/angbracketright+e−iE2tsinθ|2/angbracketright, (7.18) |β(t)/angbracketright=−e−iE1tsinθ|1/angbracketright+e−iE2tcosθ|2/angbracketright. (7.19) Let us now use this relation to compute the oscillation probability Pα→β(t). What we mean by this is the following: assume that at t= 0 we know that our state is a pure interaction eigenstate |α/angbracketright. To be concrete we can assume this is an electron neutrino νecreated in the sun.Pα→β(t) then gives us the probability that at a later time tthis state has evolved into an interaction eigenstate |β/angbracketright. Of course, this probability is simply the absolute value of the amplitude squared Pα→β(t)= |/angbracketleftβ|α(t)/angbracketright|2 - 193 - =/vextendsingle/vextendsingle/vextendsingle−sinθcosθ/parenleftBig e−iE1t−e−iE2t/parenrightBig/vextendsingle/vextendsingle/vextendsingle2 =1 2sin2(2θ)( 1−cos(E2−E1)t) =s i n2(2θ)s i n2/parenleftbiggE2−E1 2t/parenrightbigg . (7.20) In the ﬁrst step we have used eq. (7.18) and the orthogonality of the mass eigenstates /angbracketlefti|j/angbracketright=δij. The expression for Pα→β(t) can be brought into a more useful form by noting that Ei=/radicalBig p2+m2 i=p+m2 i 2p+... (7.21) and, therefore, 1 2(E2−E1)/similarequalm2 2−m21 4E≡Δm2 4E(7.22) where Eis the energy of the beam.15Furthermore, since the neutrinos travel at the speed of light, we have L=vt=ct=t,w h e r e Lis the distance travelled by the neutrino. Thus, we arrive at the ﬁnal expression fo r the oscillation probability, Pα→β(t)=s i n2(2θ)s i n2/parenleftBigg LΔm2 4E/parenrightBigg . (7.23) Eq. (7.23) has the expected properties in that the probability vanishes for L→0,θ→0 and most notably for Δ m2→0. This last limit tells us that there is no mixing if the two neutrino species have the same mass and, in particular, if they are massless. So far we have considered oscillations in vac uum, i.e. we have assumed that the neutrinos were travelling through the vacuum. While th is is true most of the time, the neutrinos produced in the sun ﬁrst have to travel through the sun before they can reach us. The matter surrounding the neutrinos can have a crucial eﬀect on the oscillation probability for the neutrinos. This eﬀect is called the matter eﬀect or the Mikheyev-Smirnov-Wolfenstein (MSW) eﬀect. The question at the heart of the prob lem is: how does the Hamiltonian ˜H, eq. (7.14), change through interactions of the neutrinos with surrounding matter? There are basically neutral and charged current interactions. As we have learnt, neutral current interactions are mediated by the exchange of a Zboson. Taking into accoun t that the surrounding matter is basically made of protons, neutrons and e lectrons, a typical F eynman diagram is: 15This argument can be made more rigorously using wave packets. - 194 - νx νx p ,n ,e p ,n ,eZ The important point is that these interactions are independent of the ﬂavour xof the neu- trino. Thus they aﬀect the two diagonal entries of the Hamiltonian in the same way. This means they change a, eq. (7.15), i.e. the Hamiltonian is modiﬁed by a→¯a. As we will see later, this change is irrelevant. The charged current interactions are mediated by a W±. A typical Feynman diagram is: νe e− e−νeW These interactions take place only for electron neutrinos since there are no μ’s (or τ’s) in the surrounding matter. In our convention where we identify the |α/angbracketrightstate with an electron neutrino, this means that only the top-left en try of the Hamiltonian, eq. (7.14), is modiﬁed. Thus, including the matter eﬀects we arrive at the following Hamiltonian, ˜HMSW=⎛ ⎝¯a+b+wc c ¯a−b⎞⎠ , (7.24) where wcomes from the charged current interactions. The explicit form of wis not important for us. What we want to know is how the w-term modiﬁes the mixing angle. To ﬁnd the modiﬁed mixing angle θMSWwe have to diagonalize ˜HMSW,i . e .w eh a v et oﬁ n d VMSW=⎛⎝ cosθMSW sinθMSW −sinθMSW cosθMSW⎞⎠ (7.25) such that HMSW≡V−1 MSW·˜HMSW·VMSW (7.26) - 195 - is diagonal. If we plug the explicit forms for VMSW, eq. (7.25), and ˜HMSW, eq. (7.24), into eq. (7.26) we ﬁnd the oﬀ-diagonal terms of HMSWto be ccos(2θMSW)+2b+w 2sin(2θMSW). (7.27) This vanishes for tan(2θMSW)=−2c 2b+w=−Δm2sin(2θ) 4Ew−Δm2cos(2θ)(7.28) where we have used eqs. (7.16) and (7.17). We note that θMSWdoes not depend on a, thus as mentioned above, the change a→¯ainduced by the neutral current interactions does not matter at all. The important point is that for 4Ew∼Δm2cos(2θ) there can be a dramatic eﬀect and t he oscillation probability can increase substantially. In fact, this eﬀect is ve ry important in the explanation of experimental results. 7.3 The See-Saw Mechanism In this section we are concerned with neutrino masses and oﬀer a possible explanation as to why they might be so small compared to other fermion masses. We will restrict ourselves tot h ec a s eo fo n ef a m i l y . As mentioned previously, introducing a right-handed neutrino allows us to write down the same kind of Yukawa coupling as for the u-type quarks, eq. (5.25). This will result in a ‘usual’ Dirac mass term for the neutrinos of the form m D νν=mD( νLνR+ νRνL) (7.29) (compare to eq. (5.27)). There is no doubt that such a term can be introduced in theLagrangian, but it leads immediately to the question of why the νmass is so much smaller than the other fermion masses. In fact, we would expect that the Yukawa couplings of allfermions are roughly of the same order. This would lead to neutrino masses roughly of the same size as the masses of the other leptons, obviously in sharp contrast to observations. However, the very special properties of the right-handed neutrinos allow us to write down yet another term in the Lagrangian. Recall that we want to write down the most generalgauge invariant Lagrangian, given the gauge gr oup and the matter content. In fact, since ν R is a singlet under all gauge transformations , we can (or even have to) add a term like MνRνR+h . c . . (7.30) - 196 - Note that for this term to be gauge invariant it is mandatory that Y(νR)=0a n dt h a t νR neither couples to SU(2) nor SU(3) gauge bosons. This is a Majorana mass term, but its fermion index contraction is perhaps unclear, so let us consider this with some care: •The Dirac mass for a four-component spinor is of the form mD¯ψψ=mDψ†γ0(P2 L+P2 R)ψ=mD( ψRψL+ ψLψR). (7.31) So to get a Lorentz scalar, a left-handed two-component fermion must be contracted with a rigth-handed two-component fermion. •Recall that the antiparticle of a chiral fermion has opposite chirality from the particle: 1) The negative energy solutions of momentum /vector pbecame the positive energy solutions of−/vector p. 2) For a massless (= chiral) particle, helicity = chirality, and helicity is /vector s·/vector p,s ot h e antiparticle has opposite chirality from the particle. In analogy with the Dirac mass term, one could try to write a mass term between the chiral ψLand its antiparticle as m (ψL)cψL+h . c . . (7.32) O n es h o u l dt a k ec a r ew i t hs u c he x p r e s s i o n si n the literature, beca use the operations−,C andPLdo not commute, and diﬀerent authors perform them in diﬀerent order. Eq. (7.32) is a Lorentz scalar and can also be expressed as mψT Liσ2ψLand is often written as mψLψL, with the index contraction understood . This is the notation of eq. (7.30). Whereas mDis expected to be of the same size as charged-lepton masses, the most natural value for Mis much larger. Ultimately we expect that at a high energy scale (maybe the GUT scale M∼1015GeV) there is a theory that explains all of the fermion masses. Then, the natural value for the fermion masses is of the order M. However, all fermion masses except for the νRare ‘protected’ by chiral symmetry. This explains why mD/lessmuchM.T o understand the consequences of M/greatermuchmDconsider the neutrino mass matrix /parenleftBig (νL)c νR/parenrightBig⎡ ⎣0mD mDM⎤⎦⎛⎝ νL (νR)c⎞⎠ . (7.33) In order to get the masses of the physical parti cles, i.e. the eigenstates of the mass matrix, we have to diagonalize this matrix. The eigenvalues are approximately given by m2 D Mand M, (7.34) - 197 - w h e r ew eu s e d mD/lessmuchM. Thus we can see that the physic al neutrinos are a (nearly) left-handed neutrino with mass m2 D/Mand a (nearly) right-handed neutrino with mass M. Taking mD∼mtandM∼1015GeV, we get m∼0.03 eV, which is not too far from the measured atmospheric mass diﬀerence. This may serve as an explanation as to why the mass of the left-handed neutrino is so much sma ller than the mass of the other leptons. If this explanation is correct, then there shou ld also be very heavy (nearly sterile) right- handed neutrinos. If they have GUT-scale masses, they may not be interesting for collider experiments, but they can be relevant in cosmology. If the νRare produced in the universe after inﬂation, they could produce a lepton asymmetry in their decay. The Standard Model has non-perturbative B+L violating interactions, which are rapid at temperatures T>m W, which would partially transform this lepton asymmetry into a baryon asymmetry. Thisscenario, called leptogenesis, appears to work (it may require CPviolation beyond the SM) and adds to the attraction of the seesaw model. 7.4 Summary •When neutrino masses are included in the Lagrangian, mixing angles appear at thecharged current vertex, as in the quark case. •The experimental signature of (small) neutrino masses is oscillations: a neutrino pro- duced from one ﬂavour of charged lepton, can be detected by the appearance of a diﬀerent charged lepton. Thus, an electron neutrino produced in the sun can arrive as a neutrino of a diﬀerent ﬂavour on earth. •If the neutrinos travel through matter rat her than the vacuum the oscillation pattern can change dramatically. •The see-saw mechanism provides us with an explanation of why the neutrino masses are so much smaller than t he other lepton masses. - 198 - 8 Supersymmetry This is the only section truly beyond the Standard Model. However, supersymmetry (SUSY) plays an important role in particle physics phe nomenology, so in this section we will outline the basic ideas of this new symmetry, why so many theorists like it and sketch how to‘supersymmetrise’ the Standard Model. Supersymmetry is a big topic, and this is a short lecture. There are books and review articles for readers of all tastes. In preparing this lecture, I have used, among others, a phenomenological introduction by S. Martin, hep-ph/9709356 (∼100 pages) — which uses the space-time metric ( −,+,+,+), and also a review of physics beyond the Standard Model (BSM) by M. Peskin, hep-ph/9705479 . 8.1 Why Supersymmetry? We have learned from LEP and other experiments that loop calculations work. This is ashining success for the Standard Model: we calculate, as a function of a few input parameters, quantum corrections to many (precision) observables, and what is measured agrees very well with the calculations. Nevertheless, there are several arguments as to why the Standard Model is probably not valid for energies up to the GUT scale. First of all, the Standard Model req uires a ‘light’ Higgs boson of mass ∼100 GeV. However, if one calculates loop corrections to the Higgs boson mass, they are “quadratically divergent”, that is proportional to Λ 2 NPwhere Λ NPis the scale of New (BSM) Physics. There are various conclusions that one can draw: there is new physics close to the electroweak scale that does not contribute visibly in the precision observ ables of LEP, or the loop contributions cancel against each other, or the Higgs mass in the Lagrangian has just the right value to cancel thequadratic divergences (this is called “ﬁne tuning”, and unpopular not only among theorists). We will see that supersymmetry is a combination of the ﬁrst and the second solution. Secondly, the running of the gauge couplings indi cates that, at a very high energy scale, the strong, electromagnetic and weak interactions may combine into one uniﬁed force, with one unique coupling strength. Within the SM, this G UT scenario does not quite work, but it can be achieved within supersymmetric extensions due to the additional particle content whichcontributes to the running of the couplings. Thirdly, even though the SM works amazingly well in the sector of electroweak precision observables, one of the most precise tests of all fails by about 3 −4σ: the measurement of the anomalous magnetic moment of the muon, g−2, from BNL, is larger than the SM - 199 - prediction. This discrepancy could well be solved within supersymmetry, but less easily (or not at all) in other extensions of the SM. In addition to the above arguments, SUSY coul d also supply a much sought after dark matter candidate, e.g. with the neutralino as the lightest stable neutral SUSY particle. In the following we will give a brief introduction into the formalism and consequences of SUSY. Supersymmetry is a transformation which turns bosons into fermions, and fermions into bosons. If it is a symmetry of the Lagrangian, then every fermion must have a bosonic partner and vice versa, and the interactions are restricted by the symmetry. When we supersymmetrise (exactly) the Standard Model, we will therefore (more than) double the number of particles — but the number of coupling constants stays (almost) the same. Exercise 8.1Consider the interaction Lagrangian L=y fH( tLtR+ tRtL)+y2 s 2H2(T1T∗ 1+T2T∗ 2) where tL,tRare chiral fermions (the top?), His a real scalar and T1andT2 are complex scalars.•Draw the Feynman diagrams for the one-loop contributions to the Higgs mass from t, T 1andT2. •Using Feynam rules from the lectures, calculate the leading (= most diver- gent) part of the diagrams at zero external momentum. •Find a desirable relation between yfandyS, such that the divergences can- cel.•Now include soft scalar masses δL=m 2 T(T1T∗ 1+T2T∗ 2), take the supersymmetric relation that you have found between yfandys,a n d estimate the same one-loop diagrams. 8.2 A New Symmetry: Boson ↔Fermion Recall that a symmetry, be it local gauge, or global like Poincar´ e, is deﬁned by operators which generate the transformations under which the Lagrangian transforms to itself, plus a total divergence. These operators are called generators. - 200 - We are looking for an operator Q, acting on bosons |b/angbracketrightand fermions |f/angbracketrightsuch that Q|b/angbracketright=|f/angbracketright,Q |f/angbracketright=|b/angbracketright. (8.1) Bosons have even spin and mass dimension (where I am counting the mass dimension of a ﬁeld in four dimensions), fermions have odd spin and mass dimension, so we conclude that the operator Qshould have spin 1/2 and mass dimension 1/2. And since it transforms bosons into fermions, and fermions into boso ns, our supersymmetric Lagrangian should have exactly the same number of fermionic and bosonic degrees of freedom. So there is a complex scalar for every chiral fermion, a chiral fer mion for each massless vector, and fundamental real scalars are not allowed. Since Qis a fermion, it should have a spinor index. By statistics and dimensional analysis, we can imagine it acting on ﬁelds (operators) as [Qα,φ]∼ψα, {Qα,ψ}∼∂μφ+mφ+gφ2,...A μ.(8.2) It is clear that Qαchanges spin, so mixes into the Poincar´ e group of translations and rotations. It can be shown that there is one way, and only one way, of extending the commutation relations of the Poincar´ e group (Haag-Lopuszanski-Sohnius extension of the Coleman-Mandula theorem). And this extension is supersymmetry, with the properties wewere looking for above. More precisely, one may introduce fermionic generators Q α,i na d - dition to the bosonic symmetry generators ( Pμfor translations and Mμνfor proper Lorentz transformations), which satisfy the following algebra: {Qα, Qβ}=2σμ αβPμ, (8.3) {Qα,Qβ}={ Qα, Qβ}=0, (8.4) [Qα,Pμ]=0 , (8.5) [Qα,Mμν]= i(σμν)β αQβ. (8.6) The labels αandβare spinor indices taking the values 1 and 2, the bar denotes conjugation and the algebra involves anticommutators an d commutators. Another important point to note is that in eqs. (8.3), (8.5) and (8.6) the new generators mix with the other Poincar´ e generators. A theory is supersymmetric if it is invariant unde r the group of transformations generated byPμ,MμνandQα. In such a theory, for every bosonic state there is a fermionic state with the same energy, and vice-versa. This follows directly from the fact that the Hamiltonian ( P0)c o m m u t e sw i t h Q. - 201 - Another interesting feature is that the cosmological constant vanishes: the Hamiltonian is bounded from below and the ground state has zero energy (if SUSY is not spontaneously broken). To understand this we simply have to note that since σ0is equal to the unit matrix andP0is the Hamiltonian, eq. (8.3) entails {Qα, Qβ}=2δαβH. (8.7) From this we conclude for an arbitrary state |ψ/angbracketright /angbracketleftψ|H|ψ/angbracketright=/angbracketleftψ|Q Q|ψ/angbracketright=/bardbl Q|ψ/angbracketright/bardbl ≥0. (8.8) A tt h es a m et i m ew es e et h a t /angbracketleftψ|H|ψ/angbracketright=0 ⇔ Q|ψ/angbracketright=0, (8.9) which is precisely the condition for SUSY not to be spontaneously broken (see eq. (4.18)). 8.3 The Supersymmetric Harmonic Oscillator In this subsection we will consider the simplest supersymmetric model and convince ourselves that this model indeed has all the nice properties we expect. Let us start with the usual (bosonic) harmo nic oscillator. The Hamiltonian is given by HB=1 2/parenleftBig p2+ω2 Bx2/parenrightBig . (8.10) If we deﬁne creation and annihilation operators a≡1 √ 2ωB(p−iωBx),a+≡1 √ 2ωB(p+iωBx), (8.11) then the canonical commutation relation [ p, x]=−ientails the usual commutation relations for the creation and annihilation operators [a, a+]=1,[a, a]=[ a+,a+]=0 . (8.12) If we write the Hamiltonian eq. (8.10) in term s of the creation and annihilation operators, we get HB=ωB 2/parenleftBig a+a+aa+/parenrightBig =ωB/parenleftbigg NB+1 2/parenrightbigg , (8.13) where we have deﬁned the counting operator NB≡a+a. The energy spectrum of this Hamiltonian (i.e. its eigenvalues) is given by EnB=ωB/parenleftbigg nB+1 2/parenrightbigg with nB=0,1,2,3,.... (8.14) - 202 - A point to note is that the ground state energy E0is 1/2 and not 0. In a quantum ﬁeld theory this leads to the problem with inﬁnite gr ound state energy. This problem is solved by normal ordering. Let us now repeat these steps for a fermionic harmonic oscillator. We introduce fermionic creation and annihilation operators bandb+.T h e ys a t i s f y {b, b+}=1,{b, b}={b+,b+}=0. (8.15) These relations correspond to eq. (8.12). However, since we are dealing with fermionic operators now, the commutators are replaced by anticommutators. In analogy to eq. (8.13) we write the Hamiltonian of the fe rmionic harmonic oscillator as HF=ωF 2/parenleftBig b+b−bb+/parenrightBig =ωF/parenleftbigg NF−1 2/parenrightbigg , (8.16) where we have introduced another counting operator, NF≡b+b. Note that there is a relative minus sign between the b+bandbb+term. This sign is due to the fermionic nature of the creation and annihilation operators. The energy spectrum of this Hamiltonian is given by EnF=ωF/parenleftbigg nF−1 2/parenrightbigg with nF=0,1. (8.17) Note that contrary to eq. (8.14), nFcan only take the values 0 or 1. This is a reﬂection of Pauli’s exclusion principle in that there c annot be two fermions in the same state. If we wish we can ﬁnd an explicit representation of the creation and annihilation operators in terms of Pauli matrices, b=σ1−iσ2=⎛ ⎝00 10⎞ ⎠, b+=σ1+iσ2=⎛⎝ 01 00⎞ ⎠. (8.18) In this representation the Hamiltonian eq. (8.16) is given by HF=ωF 2σ3=⎛⎝ 10 0−1⎞ ⎠, (8.19) and we see that the eigenvalues of HFare indeed ±ωF/2 as given in eq. (8.17). Now we are ready to combine the fermionic and the bosonic harmonic oscillator. If we just add the two, we do not increase the symmetry of the theory. In order to do this we also have - 203 - to require ωB=ωF≡ω. Only in this case do we end up with a supersymmetric model. The Hamiltonian then is H≡HB+HF/vextendsingle/vextendsingle/vextendsingle ωB=ωF≡ω=ω 2/parenleftBig a+a+aa++b+b−bb+/parenrightBig =ω/parenleftBig a+a+b+b/parenrightBig .(8.20) First of all we naively see that Hhas an additional symmetry a↔b.A s t a t e i s n o w determined by two quantum numbers nBandnF, and the energy spectrum is EnB,nF=ω(nB+nF)w i t h nB=0,1,2,3,..., n F=0,1. (8.21) Note that the ground state energy is E0,0= 0. Thus as advertised above, the ground state has zero energy. This is simply becaus e the bosonic ground state energy +1 /2a n dt h e fermionic ground state energy −1/2 cancel. The other feature mentioned before, namely that the states appear in pairs (a fermionic and a bosonic state) with the same energy can be seen from eq. (8.21). Indeed, the states |nB,nF=0/angbracketrightand|nB−1,nF=1/angbracketrighthave the same energy. Furthermore |nB,nF=0/angbracketrightis a bosonic state (integer spin), whereas |nB−1,nF=1/angbracketrightis a fermionic state (half-integer spin). 8.4 Supercharges In this subsection we want to look at the sy mmetry found in subsection 8.2 in a somewhat more formal way. Through the Noether theorem, a symmetry is rela ted to a conserved current and a conserved charge. Thus, in a supersymmetric theory there is a conserved supercurrent and a conserved supercharge. It is the latter that generates the transformations and we denote it by Q.S i n c e it is conserved it has to commute with the Hamiltonian. For the supersymmetric harmonic oscillator the supercharge is given by Q1=√ ω/parenleftBig a+b+ab+/parenrightBig , Q2=i√ ω/parenleftBig a+b−ab+/parenrightBig , (8.22) where, as mentioned after eq. (8.3), Qhas a spinor index. We now show that the supercharges as deﬁned in eq. (8.22) have the desired properti es. Using the (anti-) commutation rules for the creation and annihilation operators, eqs. (8.12) and (8.15), we can compute {Q1,Q1}=ω{a+b+ab+,a+b+ab+} =ω{a+b, a b+}+ω{ab+,a+b} =2ω/parenleftBig a+a(1−b+b)+( 1+ a+a)b+b/parenrightBig =2ω(a+a+b+b)=2 H. (8.23) - 204 - In a similar way we can compute the remaining anticommutators of Q1andQ2,a n dw eg e t {Q1,Q1}={Q2,Q2}=2H,{Q1,Q2}=0. (8.24) Note that this is in agreement with eq. (8.7). Now we can see that, as promised, the supercharge is conserved: [Q1,H]=/bracketleftBig Q1,(Q1)2/bracketrightBig =0, (8.25) [Q2,H]=/bracketleftBig Q2,(Q2)2/bracketrightBig =0. (8.26) Eqs. (8.25) and (8.26) allow us to see that the states in this theory come in pairs. In fact,let|Ψ/angbracketrightbe an eigenstate of H, i.e. H|Ψ/angbracketright=E Ψ|Ψ/angbracketright.T h e n Q1|Ψ/angbracketrightis an eigenstate of Hwith t h es a m ee n e r g y , HQ1|Ψ/angbracketright=Q1H|Ψ/angbracketright=Q1EΨ|Ψ/angbracketright=EΨQ1|Ψ/angbracketright. (8.27) If|Ψ/angbracketrightis a bosonic state containing nBbosons and no fermions, then Q1|Ψ/angbracketright=Q1|nB,0/angbracketright=√ ω/parenleftBig a+b+ab+/parenrightBig |nB,0/angbracketright=|nB−1,1/angbracketright (8.28) is a fermionic state with the same energy. Similarly, if |Ψ/angbracketright=|nB,1/angbracketrightis a fermionic state, then Q1|Ψ/angbracketright=|nB+1,0/angbracketrightis a bosonic state with the same energy. Thus the states come indeed in pairs with the same energy, one fermionic and one bosonic. Of course, the same argument could have been made with Q2rather than with Q1. However, Q2acting on a state |Ψ/angbracketrightproduces the same state as Q1acting on a state |Ψ/angbracketright.T h u s ,t h e r e are not four but only two states with the same energy.What we have seen is that if we start with the us ual bosonic harmonic oscillator and want to make this theory supersymmetric, then we are led to introduce for every bosonic (fermionic) state a fermionic (bosonic) state with the sa me energy. This is exactly what happens if we want to make the Standard Model supersymmetric: For each boson (fermion) we have tointroduce a fermionic (bosonic) partner, t hereby doubling the particle spectrum. 8.5 Superﬁelds The superﬁeld is a very convenient piece of SUSY notation, which rests on the abstract idea of supersymmetrising space-time. Suppos e that for the four (bosonic) dimensions we know, that is x, y, z andt, we add a pair of fermionic dimensions ηand ¯η. The SUSY transformations Qand¯Qare translations in the fermionic d irections of this “superspace”. Being ηand ¯ηfermions, they anticommute with themse lves, so the Taylor expansion in these fermionic dimensions ends quickly! - 205 - The superﬁeld associated with, say, t he Higgs, is a function of superspace: H(xμ,η)=H(xμ)+ηh (xμ)+ηηF (xμ). (8.29) His an example of a (left-handed) “chiral sup erﬁeld”, a simple sort of superﬁeld that is independent of ¯ η, suitable for decribing a matter multiplet made of a left-handed fermion and complex scalar. By a standard abuse of not ation, the superﬁeld has the same symbol as its scalar component. So on the RHS of the equality, His the scalar Higgs, his the higgsino, andFis a bosonic ﬁeld of mass dimension two, which therefore cannot have kinetic terms and can be removed from the Lagrangian by using its equations of motion (something like a Lagrange multiplier). We make no more mention of F, other than to note that it is the origin of calling part of the SUSY Lagrangian “F-terms”. The reason that superﬁelds are convenient , is that one can compactly write all the SM Yukawa interactions, and their supersymmetric relatives (of which there are very many), as the “superpotential”: W=YeHdLEc+YνHuLNc+YdHdLDc+YuHuLUc. (8.30) For simplicity, let us consider only one generation. Yfis the Yukawa coupling for fermion f, and the right-handed fermions ( e.g. eR) have been written as left-handed anti-particles (ec). Notice that there are two physically distinct Higgs doublets HuandHd, where in the SM we have used one doublet and its charge conjugate. We will return later to the reasonfor this extra ﬁeld. To obtain supersymmetric interactions of component ﬁelds, in ordinary four-dimensional space, one should extract the F-term of W. That is, expand each ﬁeld as in eq. (8.29) and pick out all the terms ∝η 2. It is clear that this will include the SM Yukawa couplings, because each fermion comes with an η. It also gives scalar four point interactions. The full expression is LSSM = kinetic terms +/summationdisplay ij∂2W ∂Φi∂Φjψiψj−/summationdisplay k|∂W ∂Φi|2 =...+YeHd/lscriptec+...−|Ye|2(HdL)(HdL)∗−..., (8.31) where i, j, k run over all the superﬁelds in W, and on the second line are the parts coming from derivatives with respect to Ec. The fermion index contraction is in the same shorthand as eq. (7.30). The kinetic terms and gauge interactions come from another function of thesuperﬁelds. It is possible to draw diagrams and do calculations in superspace; this can be useful for obtaining exact supersymmetric cancellations. - 206 - 8.6 The MSSM Particle Content (Partially) The Lagrangian for the Minimal Supersymmetric SM (MSSM) can be motivated as follows: 1. Add a boson for all SM fermions, and a fermion for all SM bosons. 2. “Supersymmetrise” the SM Feynman diagrams. 3. Observe that step 2 gave superpartners with the same masses as their SM relatives. As we have not observed any superpartn ers, add “SUSY breaking” mass terms to make them heavier than current experiment al sensitivities. (T hese masses are called “soft” because the quadratic divergences still cancel — as you have discovered in the problem.) This heuristic recipe will give a Lagrangian with ∼125 free parameters, compared to 19 in the SM. The vast majority of the additional parameters come in the SUSY breaking sector and make the theory unwieldy to study. It is t herefore common to work within simpliﬁed SUSY breaking scenarios with f ewer parameters, like e.g. mSUGRA, the minimal version of supergravity grand uniﬁcation. In this mo del universality of the soft SUSY breaking parameters is assumed (there are only four additional new parameters plus one sign), leading to a suppression of ﬂavour changing neutral currents. In this subsection we restrict ourselves to the ﬁrst step outlined above, describing the particle content of the MSSM. Feynman rules can be found elsewhere. Superpartners are often written as capitalised, or “tilded” SM particles. The partners of one generation of SM leptons are a slepton do ublet, a singlet selectron and a “right-handed” sneutrino: /lscript=⎛ ⎝eL νL⎞⎠ →˜/lscript=⎛⎝ ˜eL ˜νL⎞⎠ orL=⎛⎝ EL NL⎞⎠ , ec→/tildewideec orEc, (νR)c→/tildewidest(νc R)o r Nc,(8.32) and sometimes, abusively, thecis dropped from the singlets, although they remain “left- handed”. Similarly, one introduces squark partners, of all colours and ﬂavours, for the quarks. The spartners of the SM bosons are the “-inos”, who can be names according to whether they are added before (Bino and three Wino s) or after (Photino, Zino and two Winos) - 207 - spontaneous symmetry breaking: γ →˜γ, Z →˜zorz, W±→˜w±orw±, H=⎛ ⎝H+ H0⎞⎠ →˜hu=⎛⎝˜h+ u ˜h0 u⎞ ⎠.(8.33) In supersymmetry, we need a second Higgs doublet. One can see this from the formal structure of the theory, or from considerations of anomaly cancellation, or by counting fermionic degrees of freedom. Let us do the l ast: Suppose we break th e electroweak gauge symmetry in an exactly supersymmetric SM. The spartners must therefore have the same masses as the SM particles, and notice in the S M after spontaneous symmetry breaking, there are no massless charged bosons. However , among the inos in eq. (8.33), there are three chiral charged fermions, and it takes two chiral fermions to make a massive charged “Dirac” fermion (a Majorana mass would break charge conservation). The solution to this problemis to add a second Higgs, H d=⎛ ⎝H0 H−⎞⎠ →˜hd=⎛⎝˜h0 d ˜h− d⎞ ⎠ (8.34) which gives mass to the dquarks and charged leptons. Recall that we must add soft masses for all these new fermions, to ensure that they should not have been discovered yet, so the physical mass eigenstates will be four neutralinos and two (four component fermion) charginos, respectively linear combinations of ˜ γ,˜z,huandhd, and ˜w±,h+ u,h− d. 8.7 Summary •Supersymmetry transforms bosons ↔fermions. It is an (the only possible) extension of the Poincar´ ea l g e b r a . •Since fermion loops come with a relative minus sign, the Higgs mass would have no quadratic divergence in an exactly supersymmetric theory. •To supersymmetrise the SM, one has to add a boson (sfermion) for every fermion, and a fermion (-ino) for every boson. Then one adds a second Higgs doublet and its SUSY partners. •No spartners have been observed so far, so one gives them masses in excess of currentexperimental bounds. This breaks the supe rsymmetry, and allows ﬁnite corrections to the Higgs mass. - 208 - •At the time of writing this sentence, supersymmetry has not been found. Acknowledgements I can by no means claim authorship for these lectures which I have inherited from Sacha Davidson, who in turn had taken over from Adrian Signer. Large parts of these notes (in particular chapters 3 and 6) go back to an even e arlier version by Douglas Ross, modiﬁed by Adrian, whereas chapters 1, 2, 4 and 5 were rew ritten by Sacha, following Guido Altarelli’s Standard Model course at the Les Houches summer school 1990. I thank all of them for their invaluable work, Sacha for letting me us e the LaTeX-ﬁles and lecture notes, and Cathy Nockles for carefully reading the manuscript . However I take full responsibility for all the minor editing, typos and mistakes introduced or overlooked by me. It is my pleasure to thank all lecturers, tutors and students for their questions and comments. Finally, big thanks go to Bill and Margaret for their tireless eﬀorts and for creating such a friendly and productive atmosphere at the school. - 209 -