Lectures on the Symmetries and Interactions of Particle Physics

Transcription

1 Lectures on the Symmetries and Interactions of Particle Physics James D. Wells CERN Theory, Geneva, Switzerland, and Physics Department, University of Michigan, Ann Arbor, MI USA July 4-10, 013, CERN Summer Student Lecture Programme Lectures discuss: - Spacetime symmetries of the Standard Model - Internal (gauge) symmetries of the Standard Model - Observables and their precision tests of the theory - Higgs boson theory and its discovery 1

2 Special Relativity We begin with a statement of our most cherished symmetries. Laws of physics should be invariant under special relativity transformations: rotations (3 of them) and velocity boosts (3 of them). This implies that the length ds should be invariant under transformations ds = c dt dx i dx i = c dt dx idx i (1) where c is the same in all reference frames. Define it to be c = 1. Construct Lorentz four-vector dx µ = (dt, dx 1, dx, dx 3 ) where µ = 0, 1,, 3. Define metric tensor Then ds = g µν dx µ dx ν. g µν = ()

3 Any position 4-vector x µ contracted with itself is its length and should be invariant. x µ x µ = x µ g µν x ν = x µ x µ (3) What are the transformations on x µ that leave its length invariant? x µ = Λ µ νx ν, suppressing indices : x = Λx (4) Substituting this into the invariance-requirement equation above gives x T gx = x T gx = (Λx) T g(λx) = x T (Λ T gλ)x (5) Thus we have to find matrices Λ that satisfy g = Λ T g Λ (6) If g = E where E = diag(1, 1, 1, 1) is the identity matrix, it would be much more familiar to you. In that case E = Λ T Λ = Λ T = Λ 1 (7) This last condition is the definition of special orthogonal matrices, which must have det Λ = ±1. The set of matrices is then SO(4), 4 4 orthogonal matrices. 3

4 Definition of a group SO(4) is a group, which has a very precise mathematical meaning. A group G is a collection of elements g G endowed with a multiplication operator that satisfies four axioms: 1. Closure: For every g 1, g G, g 1 g G. Associativity: For all g 1, g, g 3 G, (g 1 g )g 3 = g 1 (g g 3 ) 3. Identity: There exists an e G such that for all g G, eg = ge = g 4. Inverse: For every g G there is a g 1 G such that gg 1 = g 1 g = e The mathematics of group theory plays a significant role in the description of symmetries, which includes the symmetries of the Standard Model. Note, there is no requirement that g 1 g = g g 1. If this equality is satisfied it is an Abelian group (e.g., U(1), SO()); otherwise, it is called a Non-Abelian group (e.g., SU(), SO(3), etc.). 4

5 Z discrete group One of the simplest groups of all is the Z group. It has two elements {1, 1} and group multiplication is normal multiplication. Multiplication table: This is sometimes called the even/odd group. Forms a group because the four axioms are respected: 1. Closure: check. Associativity: check 3. Identity: check 4. Inverse: check This is an example of a discrete abelian group. 5

6 Example of a Group, SO() These are simply two-dimensional rotations that you are used to. Every group element is parameterized by a rotation angle θ: g(θ) = cos θ sin θ sin θ cos θ Multiply two elements together and we get g(θ 1 )g(θ ) = = = (8) cos θ 1 sin θ 1 cos θ sin θ (9) sin θ 1 cos θ 1 sin θ cos θ cos θ 1 cos θ sin θ 1 sin θ cos θ 1 sin θ + cos θ 1 sin θ cos θ 1 sin θ cos θ 1 sin θ cos θ 1 cos θ sin θ 1 sin θ cos θ 3 sin θ 3 = g(θ sin θ 3 cos θ 3 ) (10) 3 Thus, closure is satisfied by g(θ 1 )g(θ ) = g(θ 3 ) where θ 3 = θ 1 + θ. Associativity obviously works; the identity element is when θ = 0; and, the inverse of g(θ) is g( θ), which is in SO(). Thus, all the group axioms are satisfied. 6

7 Back to the Lorentz Group, SO(3, 1) However, in our case our metric tensor g is not the identity matrix but rather has a mixed metric of three 1 entries and one +1 entry. Nevertheless the elements Λ that satisfy g = Λ T g Λ form a group, called SO(3, 1). Here are a few examples of elements in SO(3, 1): Λ R = R, where R are the 3 3 rotation matrices SO(3). (11) where Λ Bx = cosh η sinh η 0 0 sinh η cosh η cosh η = γ and sinh η = βγ, with γ = velocity boost in the x direction (1) 1 1 v /c and β = v/c. (13) 7

8 With a little algebra you can see that x = Λ Bx x is equivalent to what you are used to seeing ct = x = 1 v 1 v /c ct c x (14) 1 (x vt) (15) 1 v /c y = y (16) z = z (17) Summary of this: SO(3, 1) is a group and the matrices Λ are its elements, and their transformations on Lorentz vectors are rotations and boosts that we are familiar with. What does all of this have to do with particle physics? The Symmetry Invariance Principle 8

9 Symmetry Invariance Principle When we say that nature is invariant under some symmetry, it means All objects in the theory have well defined transformation properties (i.e., well defined representation of the symmetry group) under the symmetry, and Every interaction is invariant (i.e., a singlet ) under the symmetry transformations The objects of particle physics are particle fields. The interactions in particle physics are the operators in the lagrangian. 9

10 Singlet and Triplet representations of SO(3) Representations of groups can be intuitively understood from tensor analysis from rotations, the group SO(3) with elements R ij satisfying the condition R T = R 1 and det R = 1. Let s start with a vector. If we rotate the vector v we get v v = R v, or equivalently v i = R ij v j. (18) The vector v is definite transformations properties under SO(3) and it has three independent elements (v x, v y, v z ) and so it defines a three-dimensional representation or triplet representation of SO(3). Or, for short, 3. There is always the trivial or singlet representation: c c = c singlet representation. (19) This is the 1 representation, or sometimes called the scalar representation. We have just defined rather precisely the 1 and 3 representations of SO(3) from the scalar and vector. What about tensors? Does a tensor form a separate representation of SO(3). Yes, but it s slightly more complicated! 10

11 Tensor representations of SO(3) Let us look at the tensor formed from two vectors: T ij = a i b j. This tensor has 9 elements. However, there are subspaces of these 9 elements that have definite and closed transformation properties under SO(3). The most obvious is the trace: τ = Tr(T ) = a i b i. Under rotation it is preserved. τ R il a l R ik b k = R T kir il a l b k = δ kl a l b k = a l b l = τ (0) The trace of the tensor therefore is a singlet 1 representation of SO(3). Now let us look at the anti-symmetric tensor A T = A, A ij = a i b j a j b i = 0 A 1 A 13 A 1 0 A 3 A 13 A 3 0 (3 independent elements) (1) The anti-symmetric tensor does not change its character under transformations A ml = R mi R lj A ij = R mi R lj ( A ji ) = R lj R mi A ji = R li R mj A ij = A lm Thus the anti-symmetric tensor forms a 3 A representation of SO(3). 11

12 Let s now take the symmetric tensor S = S T, S 11 S 1 S 13 S ij = a i b j + a j b i = S 1 S S 3 S 13 S 3 S 33 (6 distinct entries) () However, we have already used the trace to form a representation, so we need to subtract out the trace. What we really need is the traceless symmetric tensor Ŝ T = Ŝ with Tr(Ŝ) = 0: S 11 S 1 S 13 Ŝ = S 1 S S 3 (5 independent elements) (3) S 13 S 3 S 11 S This traceless symmetric tensor preserves its character under transformations Ŝ ml = R mi R lj Ŝ ij = R mi R lj Ŝ ji = R li R mj Ŝ ij = Ŝlm (4) Thus, the traceless symmetric tensor forms a 5 S representation of SO(3). So, the 9 elements of the tensor form a reducible rep of SO(3), which can be decomposed into three irreps of dimension 1, 3 and 5. In group theory language we we did was show that 3 3 = 3 A (1 5) S (5) 1

13 There are an infinite number of representations of SO(3) All representations can found by taking tensor products of the vectors T ij = a i b j (6) T ijk... = a i b j c k... (7) There are techniques to do this, and tables exist that classify all representations. Representation of dimension d: Group elements g G are mapped to d d matrices M(g) that preserve all the group multiplications. I.e., if g 1 g = g 3 then M(g 1 )M(g ) = M(g 3 ). Warning: Often it is said that X is a representation r of the symmetry group G, whereas what is really meant is X is an object such that when a symmetry transformation of G is applied, it transforms under the r representation 13

14 What does this have to do with the Lorentz group? The relevance is because 1. SO(3) and SU() are closely related, and. The representations of SO(3, 1) can be classified in terms of representations of SU() SU() Next, I will remind you why point 1 is correct. In order to show you that point is correct, I will need to tell you about generators for group elements and the Lie algebras that they form. 14

15 The connection between SO(3) and SU() SU() is the set of all complex matrices of det = +1 that satisfies A A = I, A = α β β α, where α + β = 1. (8) There are three independent parameters required to specify an element here of SU(). Here is one A = cos(θ/) exp{ i (ψ + φ)} sin(θ/) exp{ i (ψ φ)} sin(θ/) exp{ i (ψ φ)} cos(θ/) exp{ i (ψ + φ)} (9) where 0 θ π, 0 ψ 4π, 0 φ π. (30) The objects that transform under SU() elements like this are called spinors χ α, where α = 1,. They are the analogy of vectors in three dimensional rotation group (SO(3)): χ α = A αβ χ β, α, β = 1,. (31) 15

16 -to-1 homomorphic mapping of SU() onto SO(3) Theorem (Cornwell 1984): There exists a two-to-one homomorphic mapping of the group SU() onto the group SO(3). If A SU() maps onto R(A) SO(3), then R(A) = R( A) and the mapping may be chosen so that R(A) jk = 1 Tr ( σ j Aσ k A 1), j, k = 1,, 3 (3) where σ 1 = are the Pauli spin matrices., σ = 0 i i 0, σ 3 = (33) This implies that the representations of SO(3) are also representations of SU(). In addition there are spinor representations of SU() that have no analog in SO(3). The defining representation of SU() with elements A αβ is the lowest dimension spinor representation. 16

17 Representations of SU() You already know the SU() representations well from quantum mechanics! In QM you analyzed spin carefully and found basis functions of the form lm where l = 0, 1/, 1, 3/,,..., and m = l, l + 1,..., l 1, l Each l labels a distinct irreducible representation of SU(), and the number of m s (= l + 1) is the dimensionality of the representation. The representations of integer l are equivalent to SO(3) reps (bosonic reps), and the half-integer representations are the spinor representations (fermionic reps). 17

18 Ok, now we understand that SU() and SO(3) have lots of different representations of different dimensions. Before we show how this relates to the Lorentz group and the labels we give to elementary particles, we must introduce two more concepts: generators and Lie algebras. We will do this first through the more intuitive SO(3) and then show them for SU() and then for the Lorentz group SO(3, 1). 18

19 Generators of SO(3) group elements Let s consider the matrix And now compute e iφs x : s x = i 0 i 0 e iφs x = 1 + iφs x + 1! i φ s x ! i3 φ 3 s 3 x + = cos φ sin φ 0 sin φ cos φ (34) (35) which is a rotation about the x axis. In this sense s x is a generator of the rotations. The word generator comes from the fact that an infinitesimal rotation about the x-axis is φ, and so one can build up or generate the full rotation by adding an infinite number of infinitesimal rotations: R x (φ) = lim 1 + i φ N N s x N e iφs x. (36) 19

20 Similarly, rotations about the y and z axis are generated by s y = 0 0 i i 0 0, and s z = 0 i 0 i (37) The set of three generators s x, s y and s z of SO(3) forms a real Lie algebra. What is a Lie algebra? 0

21 Reminder: Algebra (Szekeres 004) An algebra consists of a vector space A over a field K together with a law of composition or product of vectors, A A A denoted which satisfies a pair of distributive laws: (A, B) AB A (A, B A), (38) A(aB + bc) = aab + bac, (aa + bb)c = aac + bbc (39) for all scalars a, b K and vectors A, B and C. 1

22 Definition of a Lie algebra Real Lie algebra: Real lie algebra L of dimension n ( 1) is a real vector space of dimension n equipped with a Lie product or commutator [a, b] defined for every a and b of L such that (Cornwell 1984) 1. [a, b] L for all a, b L. [αa + βb, c] = α[a, c] + β[b, c] for all a, b, c L and all real numbers α and β 3. [a, b] = [b, a] for all a, b L 4. for all a, b, c L the Jacobi identity holds [a, [b, c]] + [b, [c, a]] + [c, [a, b]] = 0

23 The commutator algebra of the SO(3) generators is [s x, s y ] = is z et cyclic (40) and s x, s y and s z are the basis vectors of the vector space. These commutator relationships are very familiar! They are exactly the same as SU(), whose generators are σ i /: σ x, σ y = i σ z et cyclic (41) Thus, we see that although the SU() group is different than the SO(3) group, the Lie algebras are isomorphic. This is another way in which SO(3) and SU() are very similar. Now, all this will help in understanding why SO(3, 1) SU() SU(). 3

24 Generator algebra of SO(3, 1) The Lorentz group has three rotations and three boosts, and thus there are six total generators acting on the four-vectors. The three generators of rotation are J x = i 0 0 i 0, J y = The three generators of boosts are K x = 0 i 0 0 i , K y = i i i i , J z =, K z = i 0 0 i i i (4). (43) 4

25 The algebra of Lorentz group generators The generators form an algebra with commutation relations [J x, J y ] = ij z et cyclic; [K x, K y ] = ij z et cyclic; (44) [J x, K y ] = ik z et cyclic; [J x, K x ] = 0 et cetera. (45) The algebra can be recast more simply be redefining which leads to A = 1 (J + ik) and B = 1 (J ik) (46) [A x, A y ] = ia z et cyclic; [B x, B y ] = ib z et cyclic; (47) and [A i, B j ] = 0, for all i, j = x, y, z. (48) This appears to be the same algebra as that of two independent SU() algebras. This is why SO(3, 1) representations can be classified as SU() SU() representations. This is also why SU() spin shows up everywhere in quantum mechanics. It s because of Lorentz symmetry! 5

26 SU() SU() Representations of the Lorentz Group Lorentz Rep Total Spin Elementary Particle Quantum Field (0, 0) 0 scalars (Higgs boson) ( 1, 0) 1 left spinor (leptons, quarks, neutrinos) ( ) 0, 1 1 right spinor (leptons, quarks, neutrinos) ( 1, 1 ) 1 vector gauge field (γ, Z, W ±, gluons) ( ) 1, 1 3 Rarita-Schwinger field (no SM particle) (1, 1) spin- field (graviton) The Lorentz group has an operation that allows a spinor ξ that transforms under SU() L to transform as a spinor under SU() R : χ = iσ ξ. Related by parity transformation K K (i.e., A B generators.) * The fermions (leptons, quarks, and neutrinos) are often treated as a four-dimensional representation ( 1, 0) ( 0, 1 ). The gauge fields A µ take (1/, 1/) representation form by σ α α µ A µ, where α = 1, is spinor index for SU() L and α = 1, is spinor index for SU() R. We ll call this σa for short. 6

27 Interactions preserving Lorentz invariance From mathematical point of view particle: an object that has a well-defined transformation property under Lorentz symmetry. If nature is to be invariant under the Lorentz symmetry then the only interactions allowed among particles are those that are singlets under both left and right spin groups. You know how to do this! There are two facts you learned from early days that help: 1. Tensor product of spin s 1 and s give spin s 1 s and s 1 + s results, and. If all tensor indices are contracted, the result is a scalar invariant! We can manipulate and understand invariants using these facts, and build up all Lorentz invariants of the theory. 7

28 Products of spin Majorana fermion mass Just as in QM we realized that 1/ 1/ = 0 1 (49) we can do the same thing with particles, with the slight complication that we need to keep track of the left and right SU() s separately. Consider the spinor f R = (0, 1/), and let s ask if f R f R interaction is ok: left : 0 0 = 0 contains singlet (50) right : 1/ 1/ = 0 1 contains singlet (51) so this is an invariant. It is the mass operator: mf R f R. To be more precise, there is a spinor-metric on the contraction which is iσ = ε. The mass operator is m f T Riσ f R (Majorana mass) (5) If f R has charge (e.g., electric charge) this term is not invariant, and not allowed. In the Standard Model only right-handed neutrinos qualify for this type of mass: M R ν T Riσ ν R. (53) 8

29 Dirac fermion mass What if we have f L = (1/, 0) and f R = (0, 1/). We learned early that a Lorentz invariant is χ T Riσ f R, and we also learned that iσ f L transforms like a RH-fermion. Thus, identifying χ R = iσ f L, we have χ T R iσ f R = (iσ f L) T iσ f R = f Li(σ ) T iσ f R = f Lf R (54) which used the facts that (σ ) T = σ and σ σ = 1. Likewise f Rf L is an invariant, which is just the conjugate of f Lf R. Therefore, we have identified a new fermion bilinear invariant (i.e., mass term): This is often called Dirac mass. m f (f Lf R + f Rf L ). (55) 9

30 Four component spinor representation We have been talking about f L and f R, and making mass terms that connect the two. But you have more commonly heard only labels like electron, muon, quarks. We can put the two-component f L and f R spinors, Weyl spinors, into a four-component spinor, the Dirac spinor as this: Ψ = ψ L ψ R (56) We can construct the projection operators P L Ψ = (ψ L 0) and P R Ψ = (0 ψ R ) from P L,R = 1 (1 γ5 ) where γ 5 = (57) The four dimensional analogy to the σ µ matrices are the γ µ matrices, where in the Weyl representation they are γ 0 = , γ i = 0 σ i σ i 0 The γ 0 acting on Ψ interchanges ψ L ψ R (parity operation).. (58) 30

31 Four-component spinor invariants In this four component notation, we can write the Majorana and Dirac mass terms. For the Majorana mass, let us define the built-up four-component spinor of χ L (which transforms under (1/, 0) representation) to be Ψ M = χ L iσ χ L and Ψ D = ψ L ψ R (59) The Majorana and Dirac mass terms are then m Ψ T M( iγ 0 γ )Ψ M (Majorana mass) (60) m D ΨD Ψ D where Ψ D = Ψ Dγ 0 (Dirac mass) (61) 31

32 Vector particle invariants A vector particle A µ (e.g., photon) has many invariants. Easy: just contract all the Lorentz indices. The invariants up to dimension four are A µ A µ, µ A µ, µ A µ ν A ν, µ A ν µ A ν, A µ A ν A µ A ν. (6) If A µ is the gauge field of a U(1) invariant theory, such as QED, interactions must be invariant also under gauge transformations A µ A µ + 1 e µλ (63) The only interaction that is invariant under both Lorentz symmetry and gauge symmetry is F µν F µν, where F µν µ A ν ν A µ. (64) This is the well-known kinetic energy term in the QED lagrangian L KE QED = 1 4 F µνf µν. (65) 3

33 Vector particles interacting with fermions The vector A µ interaction with fermions requires us to consider its (1/, 1/) representation: σ µ A µ. A general interaction (1/, 0) (1/, 1/) (0, 1/) = contains (0, 0) contains singlet (66) If χ R = (0, 1/) and f L = (1/, 0) then we can have interaction χ T R σ µ A µ f L (67) where the first (second) refers to SU() R (SU() L ) contraction. Consider χ R = iσ f L. Interaction becomes (iσ f L) T σ µ A µ f L = if L(σ ) T σ σ µ A µ f L = if Lσ µ A µ f L (68) In four component language we see this interaction as i Ψγ µ A µ Ψ. (69) 33

34 Scalar interactions with itself, fermions and vector bosons The invariant interactions of vector bosons with scalars is also easy. If we assume real scalar φ, we have µ φ µ φ, φ, φ, µ µ φ, φ 3, φ 4, etc. (70) For charged complex scalar Φ (like the Higgs boson doublet) invariance under Lorentz and charge symmetry allow Interactions with the vector bosons are Interactions with fermions include ( µ Φ )( µ Φ), Φ Φ, (Φ Φ), etc. (71) A µ A ν Φ Φ, A µ Φ µ Φ, etc. (7) φν T Riσ ν R, Φν T Riσ ν R, φf Rf L, Φf Rf L, etc. (73) 34

35 Lorentz invariance is too general for what is witnessed in nature Lorentz invariance alone allows us to classify particles and gives strong constraints on what particles are allowed to interact. For example, one cannot have the interactions A µ f R, f Lf L f R, Φf L A µ, etc. (Lorentz forbidden) (74) But there are many other interactions forbidden that Lorentz invariance alone does not preclude. These include A µ A µ, e Lσ µ A µ u L, τ T Rσ τ R, µ LΦ t R, etc. (75) These are forbidden by internal gauge symmetries. The Standard Model particles are charged not only under SU() L SU() R Lorentz symmetry, but also under SU(3) c SU() W U(1) Y gauge symmetries. Interactions must be invariant under the transformations of every symmetry. We discuss next the gauge symmetries of the Standard Model. 35

36 Strong, weak and hypercharge forces The Standard Model particles also transform as representations of the strong, weak and hypercharge forces, which in group theory language is SU(3) c SU() W U(1) Y (Standard Model gauge groups). (76) If a particle ϕ transforms as d dimensional representation R of group G, then ϕ ϕ = e iθ kt R k ϕ (77) where T R are d d dimensional generator matrices associated with the representation R, and θ k are the parameters of the group, analogous to the angle of rotation in SO(). Global symmetries mean θ k do not depend on spacetime, whereas with local symmetries they do, θ k (x). Gauge symmetries are local internal symmetries. 36

37 Hypercharge gauge symmetry Hypercharge is a U(1) gauge symmetry, and its generator is the hypercharge operator Y, and the parameter we can define as α. Under gauge transformation ψ ψ = e iα(x)y ψ (78) Let s look at the transformation of the kinetic operator ψ Lσ µ µ ψ L (e iαy ψ L ) σ µ µ e iαy ψ L (79) = ψ Le iαy σ µ e iαy (iy µ αψ L + µ ψ L ) (80) = ψ Lσ µ µ ψ L + ψ Lσ µ (iy µ α)ψ L (81) The kinetic term would be invariant if it weren t for µ α 0 contribution. Introduce covariant derivative D µ = µ iy A µ (introducing gauge field A µ ), and one finds ψ Lσ µ D µ ψ L, is invariant when (8) A µ A µ = A µ + i µ α. (83) 37

38 Field SU(3) SU() L T 3 Y Q = T 3 + Y gµ a (gluons) (W µ ±, Wµ) (±1, 0) 0 (±1, 0) Bµ ) ) ) Q L = ( ul d L 3 ( ( u R d R ( ) 3 1 ( 0 ) ( 3 ) 1 νl E L = 1 0 e L e R ( ) ( ) ( ) φ + 1 Φ = φ ( ) ( ) ( ) φ Φ c 0 1 = φ Invariants under SU(3) c and SU() are found when tensor products yield a singlet 1. Under SU(3) conjugate representations are distinct (i.e., QL is 3). Conjugate reps for SU() are not distinct. SU(3) c : 3 3 = (84) 3 3 = (85) 8 8 = 1 + (86) SU() W : = (87) 3 3 = 1 + (88) 38

39 The problem with elementary particle masses The fermions of the Standard Model and some of the gauge bosons have mass. This is a troublesome statement since gauge invariance appears to allow neither. First, we illustrate the concepts with a massive U(1) theory spontaneously broken QED. The lagrangian of QED is where L QED = 1 4 F µνf µν + ψ(iγ µ D µ m)ψ (89) D µ = µ + iea µ (90) and Q = 1 is the charge of the electron. This lagrangian respects the U(1) gauge symmetry ψ e iα(x) ψ (91) A µ A µ + 1 e µα(x). (9) 39

40 Vector gauge boson mass If we wish to give the photon a mass we may add to the lagrangian the mass term L mass = m A A µa µ. (93) However, this term is not gauge invariant since under a transformation A µ A µ becomes A µ A µ A µ A µ + e Aµ µ α + 1 e µα µ α (94) Simply adding a mass term does not respect the gauge symmetry. Way to give photon mass while retaining the gauge symmetry: Higgs boson. Higgs boson is a condensing complex scalar field Φ = 0, charged under the symmetry in question. 40

41 The Higgs mechanism Let us suppose that the photon in QED has a mass. Invoke the Higgs mechanism in a gauge invariant manner by introducing the field Φ with charge q to the lagrangian: where L = L QED + (D µ Φ) (D µ Φ) V (Φ) (95) and it is assumed that λ > 0 and µ < 0. V (Φ) = µ Φ + λ Φ 4 (96) With µ < 0 the minimum of this potential is not at Φ = 0 (algebra on next slide). The field Φ is then said to get a vacuum expectation value or a vev, which freezes some charge (i.e., the charge of Φ) into the vacuum. 41

42 Since Φ is a complex field we have the freedom to parametrize it as Φ = 1 φ(x)e iξ(x), (97) where φ(x) and ξ(x) are real scalar fields. The scalar potential with this choice simplifies to V (Φ) V (φ) = µ φ + λ 4 φ4. (98) Minimizing the scalar potential one finds dv dφ φ=φ 0 = µ φ 0 + λφ 3 0 = 0 = φ 0 = µ λ. (99) 4

43 In the unitary gauge, where α(x) = ξ(x)/φ 0, one finds (D µ Φ) (D µ Φ) 1 ( µφ) + 1 e q φ A µ A µ (100) At the minimum of the potential φ = φ 0, so one can expand the field φ about its vev, φ = φ 0 + h, and identify the fluctuating degree of freedom h with a propagating real scalar boson. The Higgs boson mass and self-interactions are obtained by expanding the lagrangian about φ 0. The result is where L Higgs = m h h + µ 3! h3 + η 4! h4 (101) m h = λφ 0, µ = 3m h φ 0, η = 6λ = 3 m h. (10) φ 0 The mass of the Higgs boson is not dictated by gauge couplings here, but rather by its self-interaction coupling λ and the vev. 43

44 The complex Higgs boson kinetic terms can be expanded to yield L = 1 e q φ 0A µ A µ + e q ha µ A µ + 1 e q h A µ A µ. (103) The first term is the mass of the photon, m A = e q φ 0. A massive vector boson has a longitudinal degree of freedom, in addition to its two transverse degrees of freedom, which accounts for the degree of freedom lost by virtue of gauging away ξ(x). The second and third terms of eq. 103 set the strength of interaction of a single Higgs boson and two Higgs bosons to a pair of photons: ha µ A ν Feynman rule : ie q φ 0 g µν = i m A φ 0 (104) hha µ A ν Feynman rule : ie q g µν = i m A φ 0 after appropriate symmetry factors are included. (105) The general principles that we have learned here 1/ Higgs mechanism gives gauge bosons mass in gauge-invariant way / Higgs boson couples with strength of particle s mass 44

45 Chiral fermion masses In quantum field theory a four-component fermion can be written in its chiral basis as ψ = ψ L ψ R (106) where ψ L,R are two-component chiral projection fermions. A mass term in quantum field theory is equivalent to an interaction between the ψ L and ψ R components m ψψ = mψ Lψ R + mψ Rψ L. (107) In vectorlike QED, the ψ L and ψ R components have the same charge and a mass term can simply be written down. However, let us now suppose that in our toy U(1) model, there exists a set of chiral fermions where the P L ψ = ψ L chiral projection carries a different gauge charge than the P R ψ = ψ R chiral projection. In that case, we cannot write down a simple mass term without explicitly breaking the gauge symmetry. 45

46 The resolution to this conundrum of masses for chiral fermions resides in the Higgs sector. If the Higgs boson has just the right charge, it can be utilized to give mass to the chiral fermions. For example, if the charges are Q[ψ L ] = 1, Q[ψ R ] = 1 q and Q[Φ] = q we can form the gauge invariant combination L f = y ψ ψ LΦψ R + c.c. (108) where y f is a dimensionless Yukawa coupling. Now expand the Higgs boson about its vev, φ = (φ 0 + h)/, and we find L f = m ψ ψ Lψ R + m ψ φ 0 where m ψ = y ψ φ 0 / is mass of fermion! hψ Lψ R + c.c. (109) This same Yukawa interaction gives rise to an Higgs-fermion interaction: h ψψ (Feynman rule) : i m ψ φ 0. (110) Just as was the case with the gauge bosons, the generation of fermion masses by the Higgs boson leads to an interaction of the physical Higgs bosons with the fermion proportional to the fermion mass. 46

47 Standard Model Electroweak Theory The bosonic electroweak lagrangian is an SU() L U(1) Y gauge invariant theory L bos = D µ Φ µ Φ λ Φ B µνb µν 1 4 W a µνw a,µν (111) where Φ is an electroweak doublet with Standard Model charges of (, 1/) under SU() L U(1) Y (Y = +1/). In our normalization electric charge is Q = T 3 + Y, and the doublet field Φ can be written as two complex scalar component fields φ + and φ 0 : Φ = φ + φ 0. (11) The covariant derivative and field strength tensors are D µ Φ = µ + ig τ a W µ a + ig Y B µ Φ (113) B µν = µ B ν ν B µ (114) Wµν a = µ Wν a ν Wµ a gf abc WµW b ν c (115) The minimum of the potential does not occur at Φ = 0 if µ < 0. Instead, one finds that the minimum occurs at a non-zero value of Φ its vacuum expectation value (vev). 47

48 With gauge transformation the vev of Φ is Φ = 1 0 v where v µ λ. (116) This vev carries hypercharge and weak charge into the vacuum, and what is left unbroken is electric charge. Our symmetry breaking pattern is then simply SU() L U(1) Y U(1) EM. The original group, SU() L U(1) Y, has a total of four generators and U(1) EM has one generator. Thus, three generators are broken. Goldstone s theorem tells us that for every broken generator of a symmetry there must correspond a massless field. These three massless Goldstone bosons we can call φ 1,,3. We now can rewrite the full Higgs field Φ as Φ = 1 0 v + 1 φ 1 + iφ h + iφ 3 (117) The fourth degree of freedom of Φ is the Standard Model Higgs boson h. It is a propagating degree of freedom. The other three states φ 1,,3 can all be absorbed as longitudinal components of three massive vector gauge bosons Z, W ±. 48

49 The four electroweak gauge boson mass eigenstates are W µ ± = 1 ( W (1) µ iw µ () ), M W = 1 4 g v (118) Z µ = gw (3) µ g A µ g + g, M Z = 1 4 (g + g )v (119) A µ = gb µ + g W (3) g + g, M A = 0 (10) It is convenient to define tan θ W = g /g. By measuring interactions of the gauge bosons with fermions it has been determined experimentally that g = 0.65 and g = 0.35, and therefore sin θ W = 0.3. Important Exercise: From eqs. 111, 113 and 117 derive the mass eigenvector and eigenvalue equations for A µ, Z µ and W µ ± given above. 49

50 Higgs boson interactions to bosons After performing the redefinitions of the fields above, the kinetic terms for the W ± µ, Z µ, A µ will all be canonical. Expanding the Higgs field about the vacuum, the contributions to the lagrangian involving Higgs boson interaction terms are L h int = m WW µ + W,µ + m Z Z µz µ 1 + h v (11) where m h h ξ 3! h3 η 4! h4 (1) m W = 1 4 g v, m Z = 1 4 (g + g )v = m W m Z = 1 sin θ W (13) m h = λv, ξ = 3m h v, η = 6λ = 3m h v. (14) From our knowledge of the gauge couplings, the value of the vev v can be determined from the masses of the gauge bosons: v 46 GeV. 50

51 The Feynman rules for Higgs boson interactions among the bosons are hhh : i3m h v (15) hhhh : i 3m h v (16) hw + µ W ν hz µ Z ν hhw + µ W ν hhz µ Z ν : i m W v gµν (17) : i m Z v g µν (18) : i m W v g µν (19) : i m Z v g µν (130) 51

52 Fermion masses and their couplings to the Higgs boson Fermion masses are also generated in the Standard Model through the Higgs boson vev, which in turn induces an interaction between the physical Higgs boson and the fermions. Let us start by looking at b quark interactions. The relevant lagrangian for couplings with the Higgs boson is L = y b Q LΦb R + c.c. where Q L = (t L b L) (131) where y b is the Yukawa coupling. The Higgs boson, after a suitable gauge transformation, can be written simply as Φ = 1 0 v + h and the interaction lagrangian can be expanded to L = y b Q LΦb R + c.c. = y b (t L b L) (13) 0 v + h = m b (b Rb L + b Lb R ) 1 + h = m b bb 1 + h v v where m b = y b v/ is the mass of the b quark. b R + h.c. (133) (134) 5

53 The quantum numbers are good for b Lb R mass term. Under SU() the interaction Q LΦb R is invariant because 1 1 contains a singlet. And under U(1) Y hypercharge the interaction is invariant because Y Q L Y br = sums to zero. + Y Φ + Thus, the interaction is invariant under all gauge groups, and we have found a suitable way to give mass to the bottom quark. Field SU(3) SU() L T 3 Y Q = T 3 + Y gµ a (gluons) (W µ ±, Wµ) (±1, 0) 0 (±1, 0) Bµ ) ) ) Q L = ( ul d L 3 ( ( u R d R ( ) 3 1 ( 0 ) ( 3 ) 1 νl E L = 1 0 e L e R ( ) ( ) ( ) φ + 1 Φ = φ ( ) ( ) ( ) φ Φ c 0 1 = φ

54 How does this work for giving mass to the top quark? Obviously, Q LΦt R is not invariant. However, we have the freedom to create the conjugate representation of Φ which still transforms as a under SU() but switches sign under hypercharge: Φ c = iσ Φ. This implies that Y Φ c = 1 and Φ c = 1 v + h (135) 0 when restricted to just the real physical Higgs field expansion about the vev. Therefore, it becomes clear that y t Q LΦ c t R + c.c. is now invariant since the SU() invariance remains 1 1 and U(1) Y invariance follows from Y Q + Y Φ c + Y tr = = 0. Similar to the b quark one obtains an L expression for the mass and Higgs boson interaction: L = y t Q LΦ c t R + c.c. = y t (t L b v + h L) t 0 R + c.c. (136) = m t (t Rt L + t Lt R ) 1 + h = m t tt 1 + h (137) v v where m t = y t v/ is the mass of the t quark. The mass of the charged leptons follows in the same manner, y e E LΦe R + c.c., and interactions with the Higgs boson result. 54

55 In all cased the Feynman diagram for Higgs boson interactions with the fermions at leading order is We see from this discussion several important points. h ff : i m f v. (138) Higgs boson can give mass to all massive vector bosons (W ±, Z) Higgs boson can give mass to all fermions Higgs boson phenomenology (production, decays) entirely determined once its mass is known 55

56 Standard Model Lagrangian We now put the elements together to write the SM lagrangian: L SM = D µ Φ µ Φ λ Φ 4 (139) 1 4 B µνb µν 1 4 W a µνw a,µν 1 4 Gk,µν G k µν if Liγ µ D µ f Li if Rγ µ D µ f Ri y d ijq LiΦd Rj y u ijq LiΦ c u Rj y e ije LiΦe Rj + h.c. where D µ = µ + ig τ a W µ a + ig Y B µ + ig s λ k G k µ (140) B µν = µ B ν ν B µ (141) Wµν a = µ Wν a ν Wµ a gf abc WµW b ν c (14) G k µν = µ G k ν ν G k µ g s f klm G l µg m ν (143) Lagrangian is Lorentz invariant and gauge invariant and allows mass for all particles. 56

57 Observables in the Standard Model The input parameters of the Standard Model are g, g, g s, y t, y b, y e,... (144) From these input parameters, and knowledge of the elementary particle identities, we can calculate all the observables, which include σ(e + e µ + µ ), σ(p p W + Z), (145) M Z, M W, m t, m e, (146) Γ(µ e ν e ν µ ), Γ(Z τ τ), Γ(h γγ), (147) A F B (Z b b), A LR (Z e + e ) (148) There are an infinite number of observables that can be defined. 57

58 Calculating Observables To learn how to compute observables is the subject of other lectures. We shall just give the answers to the calculation of the observables in terms of the lagrangian parameters. To do that, it is convenient to define three derived parameters, e, s W and c W, where gg e = g + g, s g W = g + g, c g W = (149) g + g These are defined merely for convenience, and not strictly necessary. Note also that g = e s W, g = e c W, and c W = 1 s W (150) 58

59 Leading-order results for some key observables ˆM Z = 1 4 (g + g )v, ˆM W = 1 4 g v, m f = y f v (151) i4π ˆα q q 0 = e q q 0 (γ e + e Compton scattering) (15) Ĝ F = g 8 ˆM W ˆΓ(Z f f) = N c 48π g c W = 1 v (µ decay) (153) M Z (g V + g A), where g V = T (3) f Q f s W, g A = T (3) f, and T (3) f = ± 1 for up/down fermions (154) Â e LR = Γ(Z e+ Le L) Γ(Z e + Re R) Γ(Z e + Le L) + Γ(Z e + Re R) (1/ ŝ eff) (ŝ eff) (1/ ŝ eff) (ŝ eff) (155) ŝ eff is often called sin θ eff l, and at leading order it is ŝ = s W! 59

60 Six key observables and their measured values Let s simplify and exchange {g, g, v} for {e, s, v} and calculate ˆα = e 4π = 1/ (46) (156) Ĝ F = 1 v = (1) 10 5 GeV (157) ˆM Z ˆM W = ev = ± GeV sc (158) = ev = ± GeV s (159) ŝ eff = s W = ± (SLAC A LR only) (160) ˆΓ l + l = v e 3 96π s 3 c s W + 1 = ± MeV (161) 4 Left column is name of observable, center column is calculation in terms of lagrangian parameters, and right column is the experimental measurement. 60

61 Over-constrained system = ability to test the theory In our analysis we have six observables {ˆα, Ĝ F, ˆMZ, ˆMW, ŝ eff, ˆΓl + l } (16) that are written in terms of just three parameters {e, s, v}. Test by solving for {e, s, v} using the first three very well measured observables: e = 4π ˆα = (163) v = 1 Ĝ F 1/ = 46. GeV (164) s = π ˆα Ĝ F ˆM Z 1/ = (165) 61

62 Calculation of M W prediction in terms of ˆα, Ĝ F and ˆM Z Having expressed e, v and s in terms of observables ˆα, Ĝ F and ˆM Z, we can now compute the remaining observables M W, ŝ eff and ˆΓ ll in terms of ˆα, Ĝ F and ˆM Z. Examples: ˆM W (prediction) = ev s = 4π ˆα/ ĜF 1 1 4π ˆα Ĝ F ˆM Z 1/ = GeV (166) ŝ eff(prediction) = s = π ˆα = (167) Ĝ F ˆM Z 6

63 Tree-level predictions for ˆM W, ŝ eff and Γ l Using the calculations of {e, s, v} we find the following predictions for ˆM W, ŝ eff and ˆΓ l, their corresponding experimental measurement, and the number of σ the prediction is off from the measurement Observable Prediction Measurement Error ˆM W GeV ± GeV 37σ ŝ eff ± σ ˆΓ l MeV ± MeV 8.5σ Even if we put ˆα α(m Z ) = 1/19.0, we still find problems Observable Prediction Measurement Error ˆM W GeV ± GeV 7σ ŝ eff ± σ ˆΓ l MeV ± MeV 5.6σ 63

64 Does this mean the Standard Model is a very bad theory? No, the problem is that we have pretended that the theory computation was perfect, whereas they were only tree-level results. There are radiative corrections to all the observables: quantum loops that are higher-order in the couplings. The measurements are so good that we need these higher-order calculations to test the theories compatibility. The next slide shows a set of observables and their compatibilities with the Standard Model after doing state-of-the-art calculations. From LEP Electroweak Working Group. 64

65 Measurement Fit O meas O fit /σ meas α (5) had (m Z ) ± m Z [GeV] ± Γ Z [GeV].495 ± σ 0 had [nb] ± R l ± A 0,l fb ± A l (P τ ) ± R b ± R c ± A 0,b fb ± A 0,c fb ± A b 0.93 ± A c ± A l (SLD) ± sin θ lept eff (Q fb ) 0.34 ± m W [GeV] ± Γ W [GeV].085 ± m t [GeV] ± March This figure is from the LEP Electroweak Working Group. 65

66 Discovery of the Higgs Boson It was on announced on July 4, 01 that a boson of mass 16 GeV was discovered. Overtime it is looking more and more like the Standard Model Higgs boson. Discovery takes place mainly from two channels: σ(gg H) Br(H γγ) and, (168) σ(gg H) Br(H ZZ l + l l + l ) (169) The gg H cross-section comes about from a loop of top quarks gg (t t) loop H. (170) The H γγ decay comes about from loop of W bosons and top quarks: H (W + W ) loop + (t t) loop γγ. (171) 66

67 ATLAS (Peters 013) 67

68 ATLAS (Peters 013) 68

69 ATLAS (Peters 013) 69

70 ATLAS (Peters 013) 70

71 The discovered scalar is acting like a Higgs boson to within about 30% or so. Many new physics ideas suggest that corrections on the order of 10% or less are to be expected. This could take LHC many years to get to that level, and may require a future e + e linear collider. 71

72 Conclusion Our discussion of the Standard Model has been focused on understanding - Particle identities with respect to spacetime symmetries (Lorentz Group) - Particle identities with respect to internal symmetries (gauge symmetries) - Particle interactions - Observables in which particles manifest their behavior and properties - Higgs boson solution to the elementary particle mass problem In other lectures you will learn more about how particles interact among the generations (flavor physics) and how they violate CP symmetry, and other special features derivable from the Standard Model lagrangian you have learned here. And finally, you will learn that the Standard Model is incomplete! 7