Week 3: Renormalizable lagrangians and the Standard model lagrangian 1 Reading material from the books

Week 3: Renormalizable lagrangians and the Standard model lagrangian 1 Reading material from the books Burgess-Moore, Chapter Weiberg, Chapter 5 Donoghue, Golowich, Holstein Chapter 1, 1 Free field Lagrangians We will be mostly interested in Lagrangian descriptions for fields of spins 0, 1, 1. We are interested in finding a Lagrangian that reproduces the free field equations that we obtained from representation theory of the Lorentz group. 1.1 Spin 0 For a real scalar field, these equations are µ µ φ m φ = 0 (1) It is easy to see that these follow from variation of the action given by S = d 4 x [ 1 ] µφ µ φ m φ () We can then use the apparatus of canonical quantization to verify that the quantum scalar field we constructed using the representation theory of the Lorentz group gives identical results to the lagrangian free field described by φ. In this notation, the Feynman propagator is the inverse of the partial differential operator that appears in the quadratic action above. For complex scalar fields, we usually use the following Lagrangian S = d 4 x [ µ φ µ φ m φ φ ] (3) 1 c David Berenstein 009 1

The sign in the quadratic kinetic term is the convention that we have used for η µν = diag( 1, 1, 1, 1). This is the sign that assigns positive action to time variations and positive potential to space gradients (remember that Lagrangians are of the structure T V, where T is kinetic energy and V is potential). For many real scalar fields φ i, the most general lagrangian is given by S = d 4 x 1 A ij φ i φ j 1 B ijφ i φ j (4) Positivity of the energy and reality of the action requires that both A, B be positive definite real matrices. This is, that v i A ij v j > 0 v 0 (5) It is well known that a positive definite real matrix can be diagonalized by an orthogonal transformation. Consider such transformation O that diagonalizes A, so that à = OAO T (6) and à is diagonal. We can use this field redefinition φ = Oφ to make the kinetic term couplings diagonal. This is, we have that S = d 4 x 1 Ãij φ i φ j 1 B ij φi φj (7) where B = OBO T and where Ãij = κ iδ ij, and κ i 0. We can further redefine the field φ i = κ i s i, so that without loss of generality we get the following action S = d 4 x 1 si s i 1 b ijs i s j (8) In this notation, b ij is also positive. Now, the kinetic term is said to be normalized. A further orthogonal transformation can be used to diagonalize b, so that in the end, we have that the initial system is equivalent to a collection of free scalar fields S i, with masses M i. 1. Spin 1/ There are two canonical representations of spin 1/ in four dimensions. These are the Weyl spinors ψ α, ψ α with dotted and un-dotted spinor indices. They

are complex conjugate to each other. The first is called a left-handed spinor and the other a right handed spinor. These are conventions. The Dirac equation mixes both of these. In particular, we should find that ( µ µ m )ψ α = 0 (9) just like for the scalar fields. The field ψ α is complex, so in principle it should have four different degrees of freedom. However, a spin one half particle in the Lorentz group only needs two different degrees of freedom: up spin and down spin in the center of mass frame. This means that the complex conjugate field should be related linearly to the field itself (and via the little group invariants because we are using the rest frame. This invariant is the momentum). The combination µ σ µ αα ψ α (10) transforms as a right handed spinor, so this is our candidate to relate the right handed spinor to a left handed spinor. In the equation above spinor index is raised and lowered with ǫ α β and ǫ α β. These are antisymmetric matrices, so the order of the spinor indices is important. Also, the Clebsch Gordon coefficients relating a vector to a left plus a right handed spinor are notated by σ µ αα. The other ones, with opposite spinor index orientation are called σ µ α α (11) These are related by σ µ αα = ǫ α βǫ αβ σ µ β β The relation to cut on the number of components is then (1) i µ σ µ α α ψ α mψ α (13) We include here m as a unit of mass, but it is in general a complex number with mass dimension. Via a field redefinition of the phase of ψ α we can get rid of this phase. This,, together with it s complex conjugate is the Dirac equation. The action that reproduces this equation is given by S = i ψ α µ σ µ αα ψ α m ψα ψ α m ψ α ψ α 3 (14)

These equations put together also predict the Klein Gordon equation for the field ψ and ψ. Again, it can be shown that the Feynman propagtor is the inverse of the differential operator appearing in the quadratic terms in the action. Since fermions are complex, we can solve the theory of many free fermion fields similarly to the case of spin zero bosons. Instead of orthogonal transformations we use unitary transformations instead. Again, the only physical parameters are the mass. For fields that have additional symmetries, it is usually better to choose a charge basis rather than a real basis of spinors as above. In this case, a left handed spinor of charge one has an associated right handed spinor of the same charge, this implies another set of left fields with the opposite charge. The corresponding action is S = i ψ σ µ µ ψ + i χ σ µ µ χ mψχ m χ ψ (15) in the equation above all spinor contractions have been suppressed. This is quite common. The Dirac field is then ( ) ψα Ψ = χ α (16) and the Diract equation can be written as iγ µ µ Ψ mψ = 0 (17) by collecting terms and defining the gamma matrices appropriately. Doing many fermions is just as straightforward as with bosons. The difference is that to diagonalize the mass matrix we need to use Unitary transformations. 1.3 Spin 1 Here, it is rather similar to the problem for spin zero scalar fields. We need to reproduce the two equations µ µ A ν m A ν = 0 (18) and µ A µ = 0 (19) 4

The first equation is again the mass-shell condition. The second equation is the transversality constraint. We should look for a Lagrangian with two derivatives. The most general combination is given by α( µ A µ ) + β ν A µ ν A µ m A µ A µ (0) Remember that a spin one particle has only three polarizations, and that in the rest frame the time component should vanish. The easiest way to guarantee that is if the time component is not dynamical: there are no time derivatives on it. As such, in the action it would be a Lagrange multiplier and enforce some type of constraint and there would be no canonical conjugate variable to it. This can be accomplished if we choose α = β. After a bit of massaging the indices, we end up with the following 1 S = 4 F µνf µν m Aµ A µ (1) where we have defined the field strength F µν = µ A ν ν A µ. This is antisymmetric in two indices. This is called the Procca Lagrangian. The equations of motion are given by µ F µν m A ν = 0 () Taking the divergence, we find that ν µ F µν = 0 = m ν A ν. This vanishes because derivatives commute, and the F is antisymmetric. This Lagrangian (the Procca Lagrangian) satisfies our requirements. Notice that the quadratic operator in front of A µ is not invertible. One of the polarizations of A is after all not propagating. The propagator is then designed so that the Klein-Gordon wave equation acting on the propagator vanishes everywhere expect if we are evaluating it at zero distance. This is the property of locality on the Green s function. The propagator in momentum space given by Massless particles δ µν (p) = η µν + pµpν m (3) p + m We can think of massless particles as the limit of massive particle when we send the mass to zero. For spin zero and spin one half particles, this is not a 5

problem. Here we find that the antiparticle of a helicity one half particle is a helicity minus one half particle. Both of these can be accommodated in a single (chiral) Weyl spinor, so long as there are no extra conserved charges. For spin one particles this is a little bit more delicate. This is because a massive spin one particle has three polarizations, while a massless spin one particle has two possible helicities only. If we try to couple a vector particle to some other fields in a linear way, we would get a term in the action of the form A µ J µ. The Procca equation gets a source, and we find that the constraint µ A µ = 0 gets modified to M µ A µ µ J µ = 0 (4) In the massless limit we find that A µ has to couple to a conserved current µ J µ = 0. Under these conditions, we can ask what happened to the third polarization of the vector boson. The conditions to be transverse require that for the particle momentum and polarization we have p µ A µ = 0. If the mass vanishes, the polarization can be taken proportional to p µ, this is, we can take ǫ µ p µ. This just says that the field is a gradient of a scalar. Putting that polarization in the action, we can integrate by parts on the coupling to the current, and the amplitude for producing or absorbing such a particle vanishes because the current is conserved! In the end, the amplitudes are invariant if we make the substitution ǫ µ ǫ µ + αp µ for external particles. Indeed, these are generated by Lorentz transformations that belong to the little group of a lightlike particle. By contrast, the field strength F µν (p) is invariant and proportional to p µ ǫ ν p ν ǫ µ, which is invariant if we add a term proportional to p µ to ǫ. Notice that the stress tensor (how we measure energy) depends only on F in this case, so the lightlike polarization would not seem to carry any energy. If we want to make this true also off-shell, for any value of the gradient, then we need to make sure that another contribution to the action compensates and eliminates this extra degree of freedom completely. The principle that lets us do that is called gauge invariance. The way we write such actions is that we begin with a symmetry, with an associated conserved charge. The coupling of the vectors is done by what is called minimal coupling: all derivatives are replaced by covariant derivatives. It is such that on variation of the field, to linearized order, we recover the current as before. 6

If the vector particles are charged under the symmetry as well, then we get a non-abelian gauge theory. The m 0 limit of the Feynman propagator for a massive particle as written above is problematic. However, we see that the powers of momentum can be shifted because a gradient polarization decouples in the end. This is the gauge freedom. There are a lot of ways to make choices here. We can take our propagator to be given by µν (p ) = η µν p (5) This is called Feynman gauge. One can take other gauge conditions as well. The end results of calculations should be independent of these choices. The different gauges are most useful when trying to compute loops in perturbation theory. 3 The Standard Model Lagrangian The standard model of particle physics is a special case of a gauge theory. It contains scalar fields, massive and massless vector fields and spin one half particles. The simplest way to describe it is by using the gauge principle to build the interactions. The gauge group is SU W () U(1) Y SU c (3). The first two groups together are called the electroweak theory. The SU() is called the weak interactions group, while the U(1) is called hypercharge. The electromagnetic field is derived from the hypercharge and the weak gauge groups. The SU(3) is called the color group and it is responsible for the strong interactions. It is believed that it confines and that it is unbroken. We will study the SU c (3) much later in the course. In the mean time, you should know that at high energies it becomes weakly coupled (this happens roughly at about 1GeV energy transfers). In the meantime we will study the electroweak theory in detail. We will assume that for calculations color is free (we are at high enough energies) and use the color degree of freedom just to count states. This is justified by using asymptotic freedom. However, we should always question the validity of these calculations by comparing to experiments and trying to understand the corrections to this naive reasoning in a systematic way. 7

The notation for the SU() gauge bosons is to call them Wµ a, where a = 1,, 3. The U(1) Y gauge boson is called B µ, and for the SU(3) we use the letter G a µ, where a = 1,...8. The spin one half particles that are charged under color are called quarks. The spin one half particles that are not charged under color are called leptons. The standard model is built of three copies of a general pattern of fermions. These copies are called families. A single family contains the following left handed fermion fields (spinor indices are suppressed) Field name Field nomenclature ( ) Representation NL Lepton doublet (1,, E 1) L Right handed lepton ( Ē R ) (1, 1, 1) The Dirac UL Quark doublet (3,, 1) 6 Right handed Up quark Ū R ( 3, 1, ) 3 Right handed Down quark DR ( 3, 1, 1) 3 particles are usually split by adding a left handed and a right handed spinor. For example, the electron would be a combination of the form ( ) el (6) e R where the lack of a bar and an R subindex indicates a right handed spinor index. The same would be done for the quarks. You can verify the anomaly cancellation conditions D L tr(y ) = 0, tr(y 3 ) = 0, tr(y W ) = 0, tr(y G ) = 0, tr(g 3 ) = 0 (7) The field content uniquely specifies the coupling of the fermions to the vector particles and by the vector particles to themselves. Written as above, all gauge bosons are massless (we only see one such in nature) and all quarks and leptons are massless as well: there is no gauge invariant quadratic mass term in the action that one can write down. In the standard model, we add a single spin zero scalar field doublet (dubbed the Higgs doublet) that will be responsible for giving masses to to the fermions (by Yukawa couplings) and to the vector bosons by giving the Higgs field a vacuum expectation value. There can be more than one Higgs field. The Higgs field is given by ( ) h + ) h = (8) 1 (v + H + i h 8

and it has the same electro-weak quantum numbers as the electron doublet. The vacuum expectation value of the Higgs field is ( ) 0 h = v (9) We can always choose a gauge where h = h = 0. This is called unitary gauge. We are left with one degree of freedom H, the so called Higgs particle. If we add a standard kinetic term for the Higgs field, when we impose the physical gauge we get a mass matrix for some of the electro-weak gauge bosons. Also, if we study the most general Yukawa coupling of the higgs doublet to the quarks and leptons we obtain that all particles except neutrinos can become massive. We will now study the massive spin one particles that appear in the lagrangian, as well as their couplings to matter. It s common to write the W field as The kinetic term of h is given by ( W 3 W µ = µ W + µ W µ Wµ 3 The covariant derivatives are given by ) (30) D µ φ D µ φ (31) D µ φ = µ φ i g W a µ σa φ + i g 1 B µφ (3) where the σ a are the Pauli matrices. here we have a convention where the covariant derivative carries the powers of the coupling constant. There is another standard convention where A has absorbed the g above and the coupling shows up instead in the kinetic term for the gauge bosons as follows: L = 1 4g F µνf µν (33) This second convention is more convenient for non-perturbative calculations and this is the convention where the gauge connection transforms in the more covariant way (without any powers of g in the gauge transformation). However, this is not the canonical normalization for the kinetic term. Keeping 9

track of the powers of g for gauge invariance is more confusing in the canonical normalization. However, this is a simpler system to extract quantities like masses and mixings. In this convention we have that F µν i g [D µ, D ν ] = ( µ A a ν ν A a µ)τ a + gf abc A a µa b ντ c (34) The τ are the infinitesimal generators of the group in any representation (they are taken to be hermitian with the structure constants having an extra factor of i). The combination in the kinetic term of h (this is a complex field) that depends on v is g /W g 1 /Bh = 1 4 (g v) W µ W + µ (35) + 1 1 4 (g vw 3 g 1 vb) (36) The one half in the expression above is the usual one in the Pauli matrices as generator of rotations. The W 1,,3 are real, but W ± are complex. Hence the extra. We find that the W mass is given by M W = 1 gv, while the mass for W 3, B is nondiagonal and given by the matrix M = v 4 ( ) g g 1 g g 1 g g 1 (37) There is one massless vector particle (called the photon), and another massive one called the Z 0 boson. The matrix that diagonalizes M is a rotation with one angle called θ W, the weak mixing angle (or the Weinberg angle). We write Z 0 = g W 3 g 1 B g + g 1 = cos(θ W )W 3 sin(θ W )B (38) The mass is given by M Z = g + g1v/. It is easy to see that M W = M Z cos(θ W ) (39) The photon is the orthogonal combination A µ = s w W 3 + c w B (40) 10

and this can be used to determine the electric charge coupling constant from g 1, g by demanding that it couples with the right strength to the electron. This relation is e = g 1 c w = g s w (41) or equivalently 1 e = 1 + 1 (4) g1 g Notation: we use c W and s W to indicate the sine and cosine of the weak mixing angle. Remember that coupling constants run under the renormalization group.thus, the mixing angle suffers renormalization group effects from high energies to low energies. These are usually logarithmic corrections. We can express W 3 = c W Z + s W A, B = c W A s w Z. The observed values of s W (M Z) MS 0.31 at the Z-pole mass in the modified minimal subtraction scheme.. 11