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Week 3: Renormalizable Lagrangians and the Standard Model Lagrangian 1 Reading Material from the Books

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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 2 • Weiberg, Chapter 5 • Donoghue, Golowich, Holstein Chapter 1,2 • 1 Free field Lagrangians We will be mostly interested in Lagrangian descriptions for fields of spins 1 0, 2 , 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 ∂ ∂µφ m2φ =0 (1) µ − It is easy to see that these follow from variation of the action given by 1 m2 S = d4x ∂ φ∂µφ φ2 (2) −2 µ − 2 Z 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 4 µ 2 S = d x ∂ φ∗∂ φ m φ∗φ (3) − µ − Z 1 c David Berenstein 2009 1 The sign in the quadratic kinetic term is the convention that we have used for η = diag( 1, 1, 1, 1). This is the sign that assigns positive action to µν − 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 φi, the most general lagrangian is given by 1 1 S = d4x A ∂φi∂φj B φiφj (4) − 2 ij − 2 ij Z Positivity of the energy and reality of the action requires that both A, B be positive definite real matrices. This is, that viA vj > 0 v =0 (5) ij ∀ 6 It is well known that a positive definite real matrix can be diagonalized by an orthogonal transformation. Consider such transformation O that di- agonalizes A, so that A˜ = OAOT (6) and A˜ is diagonal. We can use this field redefinition φ˜ = Oφ to make the kinetic term couplings diagonal. This is, we have that 1 1 S = d4x A˜ ∂φ˜i∂φ˜j B˜ φ˜iφ˜j (7) − 2 ij − 2 ij Z ˜ T ˜ 2 where B = OBO and where Aij = κi δij, and κi = 0. We can further i i 6 redefine the field φ˜ = κis , so that without loss of generality we get the following action 1 1 S = d4x ∂si∂si b sisj (8) − 2 − 2 ij Z In this notation, bij 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 i 2 collection of free scalar fields S , with masses Mi . 1.2 Spin 1/2 There are two canonical representations of spin 1/2 in four dimensions. These are the Weyl spinors ψα, ψ¯α˙ with dotted and un-dotted spinor indices. They 2 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 m2)ψ =0 (9) µ − α just like for the scalar fields. The field ψα 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 µαα˙ ∂µσ¯ ψα (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 α˙ β˙ is raised and lowered with ǫ and ǫα˙ 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 byσ ¯µαα˙ . The other ones, with opposite spinor index orientation are called µ σαα˙ (11) These are related by ˙ σ¯µαα˙ = ǫα˙ βǫαβσµ (12) ββ˙ The relation to cut on the number of components is then i∂ σµ ψ¯α˙ mψ (13) µ αα˙ ∼ α 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 ψα 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 µαα˙ m α m∗ α˙ S = iψ¯ ˙ ∂ σ¯ ψ ψ ψ ψ¯ ˙ ψ¯ (14) α µ α − 2 α − 2 α Z 3 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 trans- formations 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 µ µ S = iψ¯σ¯ ∂ ψ + iχ¯σ¯ ∂ χ mψχ m∗χ¯ψ¯ (15) µ µ − − Z in the equation above all spinor contractions have been suppressed. This is quite common. The Dirac field is then ψ Ψ= α (16) χ¯α˙ and the Diract equation can be written as iγµ∂ Ψ mΨ=0 (17) µ − 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 ∂ ∂µA m2A =0 (18) µ ν − ν and µ ∂µA =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(∂ Aµ)2 + β∂ Aµ∂ν A m2AµA (20) µ ν µ − µ 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 m2 S = − F F µν AµA (21) 4 µν − 2 µ Z where we have defined the field strength Fµν = ∂µAν ∂ν Aµ. This is anti- symmetric in two indices. This is called the Procca Lagrangi− an. The equations of motion are given by ∂µF m2A =0 (22) µν − ν ν µ 2 ν Taking the divergence, we find that ∂ ∂ Fµν =0= m ∂ Aν. 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 Aµ 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 pµpν ηµν + 2 δ (p)= m (23) µν − p2 + m2 2 Massless particles 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, µ we would get a term in the action of the form AµJ . The Procca equation − µ gets a source, and we find that the constraint ∂µA = 0 gets modified to M 2∂ Aµ ∂ J µ =0 (24) µ − µ In the massless limit we find that Aµ has to couple to a conserved current µ ∂µJ = 0. Under these conditions, we can ask what happened to the third polariza- tion of the vector boson.
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