1 Poincare group and spinors

Transcript

1 Introduction on Supersymmetry Di Xu 1 Poincare group and spinors A Poincare transformation P is a proper Lorentz transformation Λ followed by a translation a. The generators of the Poincare group are the six generator M µν of the Lorentz Group plus the four generators P λ of the translation group. Then the Poincare algebra reads [P λ, P µ ] = 0 [M µν, P λ ] = i(η νλ P µ η µλ P ν ) (1) [M µν, M ρσ ] = i(ηνρm µσ + η µσ M νρ η µρ M νσ η νσ M µρ ) When discussing massless solutions of Dirac equation, it is particularly useful to use the Weyl representation for the gamma matrices. In the Weyl representation ( γ µ = ) 0 σ µ (2) σ µ 0 From now on, I will use Ψ as 4-component Dirac spinor and ψ as two component Weyl spinors ( ) Ψ L = ( Ψ R = ψ α 0 0 χ α ) (3) (4) It is to easy to verify that σ 2 ψ transforms in the same way as a righthand Weyl spinor does under Lorentz boosts, and that σ 2 χ transform like left-hand. We therefor introduce the following notation: ψ α ψα χ α ( χ α ) (5)

2 and define two matrices ( (iσ 2 ) αβ = ɛ αβ = ɛ α β = ( (iσ 2 ) α β = ɛ α β = ɛ αβ = ) ) (6) (7) (8) Then we could use these matrices to raise indices, thus χ α ɛ αβ χ β (9) χ α = ɛ αβ χ β (10) ψ α ɛ α β ψ β (11) β ψ α = ɛ α β ψ (12) is defined as one that is equal to its charge- A Majorana spinor Ψ M conjugate spinor Ψ c M = Ψ M (13) And given a Weyl spinor ψ α, we can always construct a Majorana spinor from it: Ψ M = 2 Supersymmetry algebra and multiplets ( ψ α ψ α 2.1 Simple supersymmetry algebra ) (14) Supersymmetry involves the introduction of a spinor generator to supplement the usual (bosonic) generators of the Poincare group. The simplest way to do this, is to introduce a (two-componenet) Weyl spinor generator Q α. Of course, we can then construct a (four componenet) Majorana spinor from it.

3 After introducing this generator, we need to express the various commutation and anti-commutation relations satisfied by Q α. The algebra involves Q α reads [P µ, Q α ] = 0 (15) [P µ, Q β] = 0 (16) [M µν, Q α ] = i(σ µν ) β αq β (17) [M µν, Q α ] = i( σ µν β ) α βq (18) {Q a, Q β } = 0 = { Q α, Q β} (19) {Q α, Q β} = 2σ µ α β P µ (20) where Q β is defined as Q β (Q β ) (21) We can write these algebra in terms of the Majorana spinor Q M constructed from Q α and Q α, [P µ, Q M ] = 0 (22) [M µν, Q M ] = 1 2 Σµν Q M (23) {Q M, Q M } = 2γ µ P µ (24) 2.2 Supersymmetry multiplets Now we want to work out the representation of the supersymmetry algebra that can be realized by one-particle states, and let us start with massless case. First, the mass-squared operator P 2 P µ P µ, which is a Casimir operator of the Poincare algebra, and is also a Casimir operator of the supersymmetry algebra. Then, introduce the Pauli-Lubanski spin vevtor W µ = 1 2 ɛµνρσ P ν M ρσ (25)

4 which also gives a Poincare group Casimir W 2 = m 2 J 2 (26) where m 2 is the mass-squared eigenvalue, and J 2 = j(j + 1) is the angular momentum eigenvalue. For massless particles W 2 = P 2 = 0, and in fact the spin vector W µ and the energy-momentum vector P µ are parallel: It is easy to see that λ is just the helicity where is the total angular momentum. state p, λ with momentum p. Then where and λ is the helicity We may choose p, λ in such a way that We can do this because Q 2 α = 0 W µ = λp µ (27) λ = (J P )P 1 0 (28) J i = 1 2 ɛijk M jk (29) Now consider the (normalized) massless P µ p, λ = p µ p, λ (30) p µ = (E, 0, 0, E) (31) W µ p, λ = λp µ p, λ (32) Q α p, λ = 0 (33) Other possible states in the same supersymmetric representation as p, λ are Q α p, λ ( α = 1, 2). However, Q 1 p, λ is a state of zero norm,which

5 means Q 1 p, λ = 0, this could be seen by using the anticommutation relation between Q α and Q β Thus, the only other state is ψ (4E) 1/2 Q 2 p, λ = (4E) 1/2 Q 1 p, λ (34) Obviously ψ has momentum P µ and is also a massless state. Now, using the definition of spin vector and supersymmetry algebra, it follows that Applying this to the state p, λ, gives [W µ, Q α ] = i 2 ɛ µνρσp ν ( σ ρσ ) α β Q β (35) Thus gives [W 0, Q α ] p, λ = 1 2 p 0(σ 3 Q) α p, λ (36) W 0 ( Q 1 p, λ ) = (λ 1 2 )p 0( Q 1 p, λ ) (37) With some further work, one will find there are no other states in this representation. There are just two states in the supersymmetry representation. The most common of these representations that we shall encounter are those with λ = 1, 1, 2 These examples are summarized in Table 1. 2

6 Particle λ Helicity Degeneracy T CP conjugate helicity Super-multiplet Quark, lepton, Higgsino 1 Squark, slepton, Chiral 2 2 Higgs Chiral Gauge boson Vector 1 1 Gaugino 1 1 Vector 2 2 Graviton Gravity 2 3 Gravitino 1 3 Gravity 2 2 Table 1: N = 1 supermultiplet examples 2.3 Supersymmetry free-field theory Under a general supersymmetry transformation, the field operator transforms according to φ (x) = U φ(x)u (38) For an infinitesimal supersymmetry transformation, it is characterized by (constant) anti-commuting Grassmann parameter ξ α and ξ α. U(ξ) = 1 i(ξ α Q α + ξ α Q α ) (39) Combine these two equations, using commutation and anticommutation relations of operators, it turns out δ ξ φ(x) = [i(ξq + ξ Q, φ(x)] (40) We have already seen that supersymmetry has its simplest realization in the massless(chiral) supermultiplet that has just scalar and spinor particles. Thus we might anticipate a field theory realization involving just scalar and spinor fields. Suppose we start with a complex scalar field φ(x). The mass dimension of φ is 1. And that of Q is [Q] = [ Q] = 1 2 (41)

7 and [ξ] = [ ξ] = 1 2 (42) since So the simplest possibility linear in ξ and having the right dimension is δ ξ φ(x) = aξψ(x) + b ξ ψ(x) (43) [ψ] = [ ψ] = 3 2 (44) Do the same dimension analysis for δ ξ ψ, we will find δ ξ ψ(x) = cσ µ ξ µ φ (45) The transformations are representations of the supersymmetry algebra provided ψ is a massless non-interacting Weyl field. In fact it is straightforward verify that the action can be invariant under supersymmetry transformation. Taking L = ( µ φ )( µ φ) + i ψ σ µ µ ψ (46) which decribes a free massless complex scalar field, and a free massless Weyl spinor field, and is supersymmetric with supersymmetry transformation δ ξ φ = 2ξψ (47) δ ξ ψ = i 2σ µ ξ µ φ (48) 3 Chiral superfields 3.1 Superfield representations of the supersymmetry algebra Although it is possible to construct supersymmetric Lagrangians directly from the fields belonging to a supermultiplet, there is an efficient way to do

8 this through introduction of superfields, which is a function of anticommuting Grassmann variables θ α and θ α transforming as two-component Weyl spinors. A finite element of the corresponding group is It s not difficult to show that G(x µ, θ, θ) = exp(i(θq + θ Q x µ P µ )) (49) G(x µ, θ, θ)g(a µ, ξ, ξ) = G(x µ + a µ iξσ µ θ + iθσ µ ξ, θ + ξ, θ + ξ) (50) For a function S(x µ, θ, θ), we have S(x µ + a µ iξσ µ θ + iθσ µ ξ, θ + ξ, θ + ξ) (51) = S(x µ, θ, θ) + (a µ iξσ µ θ + iθσ µ S S ξ) + ξα x µ θ + ξ S α α θ α (52) it follows that the action of the supersymmetry algebra on superfields is generated by P µ = i µ (53) iq α = θ α iσµ α α θ α µ (54) iq α = θ + α iθα σ µ α α µ (55) This is the linear representation of the supersymmetry algebra. It is easy to check that they satisfy the commutation relations. The superfiled S(x µ, θ, θ) may be expanded as a power series in θ and θ involving not more than two powers. The coefficient of the various powers of θ and θ in this expansion are ordinary fields. The general superfiled S turns out to contain more than one supermultiplet. To get an irreducible representations of supersymmetry algebra, one need to impose constraints that are covariant under the supersymmetry algebra. For chiral supermultiplet, we use supersymmetric covariant derivatives to construct chiral superfield, whose component fields form the chiral supermultiplet. These supersymmetric covariant derivatives are defined by id α = θ α + iσµ α α θ α µ (56) i D α = θ α iθα σ µ α α µ (57)

9 Then we donate Φ as a chiral superfield with covariant condition The general expansion of a chiral superfield reads D α Φ = 0 (58) Φ(x µ, θ, θ) = φ + 2θψ + θθf + i µ φθσ µ θ (59) i θθ µ ψσ µ θ µ µ φθθ θ θ (60) It follows immediately that the conjugate superfield Φ has the component field expansion Φ = φ + 2 θ ψ + θ θf i µ φ θσ µ θ (61) + i θ θθσ µ 1 µ ψ 2 4 µ µ φ θθ θ θ (62) We can refer to Φ as a left chiral superfield nad Φ superfield because they involve left- and right-hand Weyl spinors. as a right chiral The behaviour of the superfiled Φ under an infinitesimal supersymmetry transformation is δφ = i(ξq + ξ Q)Φ (63) = ξ α ( θ α iσµ α α θ α µ ) + ξ α ( θ α iθα σ µ α α µ)φ (64) Plug it into the expansion of Φ, we find δφ = 2ξψ (65) δψ = 2ξF 2 µ φσ µ ξ (66) δf = i 2 µ ψσ µ ξ (67) To construct Lagarangians for supersymmetry, it s necessary to study the product of chiral superfields. It turns out that any product of two chiral superfields is still a chiral superfield. By direct calculation Φ i (y, θ)φ j (y, θ) have the expansions in the form (where y µ = x µ + iθσ µ θ) Φ i (y, θ)φ j (y, θ) = φ i φ j + 2θ(ψ i φ j + φ i ψ j ) (68) +θθ(φ i F j + φ j F i ψ i ψ j )) (69)

10 Also Φ i Φ j =φ i φ j + 2θψ j φ i (70) + 2 θ ψ i φ j + 2 θ ψ i θψ j + F j φ iθθ (71) + F i φ θ θ j + 2θθ θ ψ i F j (72) + 2 θ θθψ j F i + θ θθθf i F j (73) It turns out Φ Φ is a vector superfield with constrain S = S. Of particular importance for the construction of supersymmertic Lagrangians is the coefficient of θθ in the expansion ΦΦ and ΦΦΦ, referred to as F-term and the coefficient of θ θθθ in the expansion of Φ Φ, referred to as D-term. 3.2 Renormalizable supersymmetric Lagrangians for chiral superfields When constructing a suppersymmetric Lagrangian using products of chiral superfields Φ i, we need to use terms that are invariant under a supersymmetry transformation up to a total divergence. It turns out that the viration of F-terms of chiral superfields and D-terms in vector superfieldss are total divergence, thus we can use these terms to construct supersymmetric Lagrangians. In general, L has the form L = i [Φ i Φ i] D + ([W (Φ)] F + HC) (74) where W (Φ), which is referred to as the superpotential, must involve only up to the third power of superfields Φ i to obtain a renormalizable Lagrangian. We may write W (Φ) = 1 2 m ijφ i Φ j λ ijkφ i Φ j Φ k (75) Now we can write the Lagrangian in component fields L = µ φ i µ φ i + i ψ i σ µ µ ψ i + F i F i (76) +(m ij φ i F j 1 2 m ijψ i ψ j + λ ijk φ i φ j F k λ ijk ψ i ψ j φ k + HC) (77)

11 Since there is no derivatives of F, the fields F are merely auxiliary fields which may now be eliminated by equation of motion. In particular, the tree level effective potential V is given by with V = F i F i F i 2 (78) F i = W φ i (79) Now consider a simple model with three superfields, Φ x, Φ y and Φ z, with the first two of these superfields having no mass, and the third having a large mass. The superpotential for the model is and the Lagrangian reads W = λφ x Φ y Φ z + m z Φ z Φ z (80) L = µ φ x µ φ x + µ φ y µ φ y + µ φ z µ φ z (81) + i 2 Ψ x γ µ µ Ψ x + i 2 Ψ y γ µ µ Ψ y + i 2 Ψ z γ µ µ Ψ z (82) λ 2 (φ yφ y φ zφ z + φ xφ x φ zφ z + φ xφ x φ yφ y ) (83) λ 2 ( Ψ x Ψ y Ψ x γ 5 Ψ y )φ z λ 2 ( Ψ x Ψ y + Ψ x γ 5 Ψ y )φ z (84) λ 2 ( Ψ x Ψ z Ψ x γ 5 Ψ z )φ y λ 2 ( Ψ x Ψ z + Ψ x γ 5 Ψ z )φ y (85) λ 2 ( Ψ y Ψ z Ψ y γ 5 Ψ z )φ x λ 2 ( Ψ y Ψ z + Ψ y γ 5 Ψ z )φ x (86) m 2 zφ zφ z λm z (φ xφ yφ z + φ x φ y φ z) m z 2 Ψ z Ψ z (87) The corresponding Feynman rules for vertices are then = iλ 2 (88)

12 y d z e = iλ 2 (I γ 5) (89) x y dh e = iλ z 2 (I + γ 5) (90) x y JH z x I = iλm z (91) y z = iλm z (92) x Now calculate the one-loop diagrams involving m z renormalization of the φ x mass are = λ 2 d 2ω q 1 (2π) 2ω (q 2 m 2 z) contributing to the (93) (94) = λ2 4 d 2ω q (2π) 2ω T r((i + γ 5 ) 1 /q (I γ 5) 1 (/q m z) ) = 2λ 2 d 2ω q (2π) 2ω 1 (q 2 m 2 z ) (95) = λ 2 m 2 z d 2ω q 1 (2π) 2ω q 2 (q 2 m 2 z) (96) (97)

13 Thus, sum of digrams reads λ 2 d 2ω q (2π) 2ω 1 q 2 (98) So there is no contribution to the mass renormalization of m x proportional m z. Actually in supersymmetric theories of chiral superfields there is also no infinite renormalization of coupling constants other than by wave function renormalization. 3.3 F-term supersymmetry breaking The simplest example of a model without any supersymmetric minima is the O Raifeartaigh model which has three chiral superfiels, with superpotential For this model, W = λ 1 Φ 1 (Φ 2 3 M 2 ) + µφ 2 Φ 3 (99) F 1 = W φ 1 = λ 1 (φ 2 3 m 2 ) (100) F 2 = W = µφ 3 φ 2 (101) F 3 = W = 2λ 1 φ 1 φ 3 + µφ 2 φ 3 (102) there is no solution with F 1, F 2 and F 3 all zero. Thus, supersymmetry is spontaneously broken. The effective potential reads V = 3 F i 2 = λ 2 1 φ 2 3 M µ 2 φ µφ 2 + 2λ 1 φ 1 φ 3 2 (103) i=1 For M 2 < µ 2 /2λ 2 1, the φ 1 will have non-zero vacuum expectation value. At this minium, F 1 = λ 1 M 2, F 2 = F 3 = 0

14 Because F 1 is non-zero we expect that ψ 1 is Goldstone fermion, we can verify this by looking at fermion matrix as L F m = 1 2 (2µψ 2ψ 3 + 2λ 1 < ψ 1 > ψ3 2 + HC) (104) And through some algebra we can combine two Weyl spinors into one Dirac spinor with mass µ For the bosonic section, the mass term reads L B m = λ 2 1M 2 (φ (φ 3) 2 ) µ 2 (φ 2φ 2 + φ 3φ 3 ) (105) we can write φ 3 in two real fields a 3 and b 3 then we can work out the mass for bosons are m φ2 = µ (106) m 2 a 3 = µ 2 2λ 2 1M 2 (107) m 2 b 3 = µ 2 + 2λ 2 1M 2 (108) Now we find φ 1 and φ 2 are still degenerate in mass with their superpartner fermions ψ 1 and ψ 2. But symmetry breaking splits the mass of two degrees of freedom of φ 3 which differ from the mass µ of their superpartner ψ. The reason for this outcome is that only superfield Φ 3 couples to the superfield Φ 1 containing the Goldstone fermion. References [1] D Balin, A Love,Supersymmetric gauge field theory and string theory