Supersymmetric Field Theories in Four and Two Dimensions

Transcript

1 Bachelor Thesis Lukas Gnam Supersymmetric Field Theories in Four and Two Dimensions Institute for Theoretical Physics Advisors/Supervisors: Privatdoz. Dipl.-Ing. Dr.techn. Herbert Balasin Univ.Ass. Dipl.-Ing. Dr.rer.nat. Johanna Knapp Date of hand over:

2 Contents 1 Introduction Motivation Outline of the Thesis Field Theory Classical Mechanics Field Theory Natural Units Supersymmetry A Four-Dimensional Model: Wess-Zumino Model Invariance of the Action The Supersymmetry Transformations Ansatz Determing the Constants A Two-Dimensional Model: Ramond-Neveu-Schwarz Superstring Invariance of the Action Comparing the two Models A Appendix 20 A.1 Gamma Matrices A.2 Charge Conjugation A.3 Weyl Spinors A.4 Majorana condition A.5 Majorana flips i

3 1 Introduction 1.1 Motivation In the last 40 years scientists developed supersymmetric field theories to answer open questions in astro- and particle physics. In 1974 Julius Wess and Bruno Zumino introduced a supersymmetric model. It was the first one which transformed fermions into bosons and vice versa. Though it was unrealistic, this model became popular for showing important properties of supersymmetric field theories due to its simplicity. Later on, string theories were developed in order to solve problems that arise when combining quantum theory and general relativity. The Ramond-Neveu-Schwarz string is such a supersymmetric string theory in two dimensions, a so called superstring theory. These two models are the basis of this work whose goal is to highlight some important properties of supersymmetric field theories. 1.2 Outline of the Thesis The work is organized as follows: Chapter 2 gives a short summary of classical mechanics and field theory. We will review the elements of those theories needed in the later work. Additionally, we want to point out some problems of the Standard Model of particle physics that supersymmetric theories, and in particular superstring theory, could solve. The central chapter of this bachelor thesis, chapter 3, deals with the four-dimensional Wess-Zumino model. In section 3.1 the invariance of this model under certain supersymmetric transformations is shown, following the steps in the lecture notes from the supersymmetry course [7]. The next section, 3.2, is about pursuing the question if the transformations given in the previous section are of the most general form possible. This will be done by motivating an ansatz and deriving the most general solutions possible. Chapter 4 deals with a different model, the two-dimensional Ramond-Neveu-Schwarz superstring. Again we will show its invariance under supersymmetry transformations, now following the book by Becker et al. [4]. At last a short comparison of the two models is given. In the appendix we explain our conventions and derive some usefule formulae. 1

4 2 Field Theory 2.1 Classical Mechanics From classical mechanics we know that we can derive the equations of motion using the principle of stationary action, known as Hamilton s principle. An action S for every path q(t, parametrized by the time t [t 0,t 1 ], is defined as S = t 1 t 0 L dt, (2.1.1 where L is a Lagrange function L( q, q = T V, with q denoting the space-time coordinates, q being the momenta, T being the kinetic energy and V being the potential energy. Hamilton s principle states that the development in time for a mechanical system is such that the variaton of the action is stationary. That is equivalent to t 1 δs = δ (T V dt = δ t 0 t 1 t 0 L dt = 0. (2.1.2 Based on that it is possible to get the Lagrange-equations (see for instance [1]. 2.2 Field Theory In contrast to classical mechanics, in field theory we will now associate every point of a spatial region with a continous field variable q i (t φ( x,t. (2.2.1 Additionally, the Lagrange function is now a functional of the field, denoted by square brackets. It depends on φ and φ [ L(t = L φ( x,t, φ( x,t ]. (2.2.2 In a local field theory the Lagrange function L is the volume integral over the Lagrange density function L, which depends on its fields and their derivatives L(t = d 3 x L(φ, µ φ. (

5 Thus we have for the action S S = L dt = L d 4 x, (2.2.4 where we used the common notation of a 4-vector known from the theory of relativity d 4 x = d 3 x c dt. Similar to (2.1.2 Hamilton s principle yields δs = δ L[φ, φ] dt = δ L d 4 x = 0, (2.2.5 where it is again possible to get the Lagrange-equations [11]. 2.3 Natural Units Additionally, we want to think about units and dimensions, which will help us to construct a Lagrange density. Using natural units we set h = c = e = 1. The rest energy can be calculated by E = m}{{} c 2. (2.3.1 =1 Now we easily see that the dimensions, denoted by square brackets, of energy E and mass m are equal, leading to [E] = [M]. (2.3.2 Energy can be expressed in natural units using mass and vice versa. Considering the quantum mechanical energy E = }{{} h ω, (2.3.3 =1 we find a relation between energy and time Finally looking at a four-vector we recognize the following relations [E] = [T] 1. (2.3.4 x µ = ( ct x (2.3.5 [T] = [L] = 1 [E] = 1 [M]. (2.3.6 Thus we have shown that in natural units everything can be expressed in units of mass. The integral in (2.2.4 has a d 4 x with x being a length. Therefore we can relate the dimension of d 4 x with the mass leading to [d 4 x] = [M] 4 = 4. (2.3.7 By definition the action has the dimension 0, thus we get for the Lagrangian [L] = +4. (

6 2.4 Supersymmetry One of the problems of the Standard Model of particle physics is the asymmetric number of degrees of freedom connected with bosons and fermions. For the fermions we have in the first family 4 degrees of freedom for the electron, 2 for the neutrino and 4 for each of the two quarks, which makes 14 for every family and hence 42 in sum. Bosons, however, have 1 degree of freedom for the Higgs-boson, 2 for the photon, 3 for each of the three bosons mediating the weak force and 2 for every one of the 8 gluons. This sums up to 28 degrees of freedom. In supersymmetric theories a symmetry between bosons and fermions is achieved by introducing additional fields. This leads to the possibility for fermions to transform into bosons and vice versa. Supersymmetry doubles the amount of particles, by assigning every particle a supersymmetric partner. The difference of the spin of a particle and its partner is 1/2. If supersymmetry really exists in nature the supersymmetric partners of the particles would have the same rest mass. In the case of light particles we would have discovered their partners by now. Therefore it is a widely held belief that supersymmetry is a theory with broken symmetry. This means the rest masses of the partner particles could be much higher. So why do people still believe in supersymmetry? First, because it can solve the so called hierarchy problem. It is about the difference of the rest mass between the W and Z bosons which mediate the weak force and the rest mass of X-bosons which occur in Grand Unified Theories (GUTs. These theories combine the electromagnetic, weak and strong interaction, and the X-bosons are hypothetical elementary particles analogous to the W and Z bosons, occuring in some of the GUTs. The second point is, if supersymmetry is introduced as a local gauge symmetry, Einstein s field equations can be derived. This is called supergravity [3]. As third point I want to mention the mysterious dark matter. It was hypothesized when physicists calculated gravitational effects caused by visible matter and found differences between the theoretical and experimental data. Supersymmetric theories could provide an explanation [2]. As last argument I want to bring up the problem of renormalization in the attempts of combining quantum field theory and general relativity. String theory purports to overcome these difficulties and to provide a consistent quantum theory of relativity [5]. 4

7 3 A Four-Dimensional Model: Wess-Zumino Model The four-dimensional Wess-Zumino model introduced in 1974, which we will work with in this chapter, contains only one bosonic and one fermionic field and no mass or interaction terms. Given the possibility to transform fermions into bosons and vice versa makes it the simplest supersymmetric model possible, describing freely moving particles. 3.1 Invariance of the Action We want to show the invarianve of the action S = L d 4 x, (3.1.1 where the Lagrangian of the Wess-Zumino model is given by L = µ φ µ φ + i 4 ψγ µ µ ψ. (3.1.2 With ψ being a Majorana spinor as defined in sections A.3 and A.4, and ψ being its Dirac adjoint, as defined in (A.4.4. The term containing the spinors denote the fermion. γ µ represents a Dirac matrix (A.1.1. The symbol means complex conjugation and µ stands for µ µ µ, (3.1.3 where µ acts from the left and µ from the right. The complex scalar field φ, representing the bosons, can be split up into two real scalar fields A and B φ = A+iB 2. (3.1.4 Multiplying with yields µ φ = 1 2 ( µ A+i µ B (3.1.5 µ φ = 1 2 ( µ A i µ B (3.1.6 µ φ µ φ = 1 ( µ A( µ A+( µ B( µ B i( µ A( µ B+i( µ A( µ B. ( }{{} =0 5

8 We can rewrite our Lagrangian using the real scalar fields L = 1 2 µa µ A+ 1 2 µb µ B + i 4 ψγ µ µ ψ. (3.1.8 The supersymmetry transformations transforming bosons into fermions and vice versa are given by δa = εψ, δb = i εγ 5 ψ, δψ = iγ µ ε µ A+γ µ γ 5 ε µ B, δ ψ = i εγ µ µ A εγ 5 γ µ µ B, (3.1.9 where ε is a constant Majorana spinor with ε being its Dirac adjoint, as defined in (A.4.4. In order to show the invariance we have to compute the variation δ(a µ b, where a and b are arbitrary fields. Assuming that the variation δ and the derivative µ commute we find δ(a µ b = δ(a( µ b ( µ ab = (δa µ b+a µ (δb µ (δab ( µ aδb = (δa µ b+ µ (a(δb ( µ aδb µ ((δab+δa( µ b ( µ aδb = µ (a(δb (δab +2[(δa µ b ( µ aδb], ( where we have used the following relations a µ (δb = µ (aδb ( µ aδb, µ (δab = µ ((δab (δa µ b. ( Now we have everything prepared to verify the invariance of the action (3.1.1 under the supersymmetry transformations ( The first step is to derive the variation of the scalar part of the Lagrangian taking into account that the spinors ε and ε are constant and commute with A and B: δ[ 1 2 ( µa( µ A+ 1 2 ( µb( µ B] = = 1 2 [( µδa( µ A+( µ A( µ δa]+ 1 2 [( µδb( µ B+( µ B( µ δb] = ( µ A( µ δa+( µ B( µ δb = ( µ A( µ εψ+( µ B( µ i εγ 5 ψ = ( µ A( ε µ ψ+( µ B(i εγ 5 µ ψ = ε[( µ A+i( µ Bγ 5 ] µ ψ. ( As second step, we are computing the variation of the spinor part, using the Majorana 6

9 flip (A.5.15 and relation ( δ[ i 4 ψγ µ µ ψ] = i 2 [(δ ψγ µ µ ψ ( µ ψγ µ δψ]+ i 4 µ[ ψγ µ (δψ (δ ψγ µ ψ] = i[(δ ψγ µ µ ψ] i 2 µ[(δ ψγ µ ψ] = i[(δ ψγ µ µ ψ] i 2 [ µ(δ ψγ µ ψ] i 2 [(δ ψγ µ µ ψ] ( = i 2 [(δ ψγ µ µ ψ] i 2 [ µ(δ ψγ µ ψ] Inserting the supersymmetry transformations (3.1.9 we get δ[ i 4 ψγ µ µ ψ] = i 2 [(i εγρ ρ A εγ 5 γ ρ ρ Bγ µ µ ψ] i 2 [ µ(i εγ ρ ρ A εγ 5 γ ρ ρ Bγ µ ψ] = 1 2 [ εγρ γ µ ρ A µ ψ i εγ 5 γ ρ γ µ ρ B µ ψ] [ εγρ γ µ ( µ ρ Aψ +i εγ 5 γ ρ γ µ ( µ ρ Bψ] ( = ε 2 [ γρ γ µ ρ A µ ψ iγ 5 γ ρ γ µ ρ B µ ψ] + ε 2 [γρ γ µ ( µ ρ Aψ +iγ 5 γ ρ γ µ ( µ ρ Bψ]. The term ρ A µ ψ can be written as ρ A µ ψ = ( ρ µ Aψ + µ ( ρ Aψ ( and the same applies to ρ B µ ψ = ( ρ µ Bψ + µ ( ρ Bψ. ( This leads us to δ[ i 4 ψγ µ µ ψ] = ε 2 [γρ γ µ ( ρ µ Aψ γ ρ γ µ µ ( ρ Aψ +iγ 5 γ ρ γ µ ( ρ µ Bψ iγ 5 γ ρ γ µ µ ( ρ Bψ] + ε 2 [γρ γ µ ( µ ρ Aψ +iγ 5 γ ρ γ µ ( µ ρ Bψ] ( = ε[γ ρ γ µ ( ρ µ Aψ +iγ 5 γ ρ γ µ ( ρ µ Bψ] ε 2 [γρ γ µ µ ( ρ Aψ+iγ 5 γ ρ γ µ µ ( ρ Bψ]. Since ρ µ is symmetric in the indices ρ and µ, only the symmetric part of the matrix γ ρ γ µ produces a nonvanishing contribution (γ ρ γ µ sym = 1 2 (γρ γ µ +γ µ γ ρ = 1 2 {γρ,γ µ } ( = η ρµ. 7

10 Knowing this, it follows from equation ( δ[ i 4 ψγ µ µ ψ] = ε[( µ µ Aψ +iγ 5 ( µ µ Bψ] ε 2 [ µ( µ Aψ+iγ 5 µ ( µ Bψ], ( Now we can rewrite the terms ( µ µ Aψ and ( µ µ Bψ, obeying the following relation ( µ µ Aψ = µ ( µ Aψ µ A µ ψ, ( yielding δ[ i 4 ψγ µ µ ψ] = ε[ µ ( µ Aψ µ A µ ψ +iγ 5 µ ( µ Bψ iγ 5 µ B µ ψ] ε 2 [ µ( µ Aψ+iγ 5 µ ( µ Bψ] ( = ε[ µ A+iγ 5 µ B] µ ψ + ε 2 [ µ( µ Aψ+iγ 5 µ ( µ Bψ]. Finally we simply have to combine the variation of the scalar ( and the spinor part ( and get δl = δ[ 1 2 ( µa( µ A+ 1 2 ( µb( µ B+ i 4 ψγ µ µ ψ] = ε[ µ A+ µ Bγ 5 ] µ ψ ε[ µ A+iγ 5 µ B] µ ψ ( ε 2 [ µ( µ Aψ+iγ 5 µ ( µ Bψ]. With the third term being a total derivative, which can be dropped because we assume no boundaries and that the fields go to 0 as x goes to infinity, this result implies the invariance of the action (3.1.1 under the supersymmetric transformations ( Note that the Lagrangian itself is not invariant. 3.2 The Supersymmetry Transformations The next section is about the previously used supersymmetry transformations ( We want to findout if they are of the simplest and most general form possible. Therefore we will motivate a proper ansatz and derive its possible solutions Ansatz Before we try to find an ansatz we have to think about which demands it has to meet: 1. Lorentz invariance 2. Bosons need to transform into fermions and vice versa. 3. The infinitesimal transformations have to be linear in its parameters and fields. and which mathematical elements we have at our disposal: 8

11 1. transformation parameters (spinors: ε, ε 2. bosons: A, B 3. fermions: ψ, ψ 4. gamma-matrices: γ µ,γ 5 5. derivatives: µ The next problem we have to solve is about the dimension of the Lagrangian and its constituents. From (2.3.8 we know that all terms of our Lagrangian are of dimension +4. Looking at the first two terms in the Wess-Zumino-Lagrangian (3.1.8 and knowing that derivatives have dimension +1, we find that A and B are of dimension +1 too. For the last term things are a bit more complicated. Gamma matrices are of dimension 0, the spinors ψ and ψ have the same dimension, namely [ ψγ µ µ ψ ]! = +4 [ ψ]+[ψ] = +3 [ ψ] = [ψ] = (3.2.1 We start with the scalar part (bosons δa. Since it has dimension +1, our possibilities for the ansatz are limited. Every single term on the right has to be of dimension +1 (note that the spinors ε and ε have dimension 1/2. Knowing this, we construct the following δa = α εψ +β ψε+γ εγ 5 ψ +ν ψγ 5 ε, (3.2.2 where α,β,γ and ν are constants. Due to the fact that A is a scalar, it is obvious that on the right side we have to multiply a row vector with a colum vector (e.g. εψ. We cannot add any further terms, because they would not have the correct dimension. For the second scalar field B we get a similar ansatz δb = α εψ + β ψε+ γ εγ 5 ψ + ν ψγ 5 ε, (3.2.3 where α, β, γ and ν are constants. Next we are making an ansatz for δψ, considering its form as a column vector and its dimension of 3/2 δψ = c µ ε µ A+d µ ε µ B +k µ γ 5 ε µ A+l µ γ 5 ε µ B, (3.2.4 where c µ,d µ,k µ and l µ have constant entries and therefore dimension 0. Same goes now for δ ψ, where we have a row vector in difference to δψ, leading to δ ψ = ε c µ µ A+ ε d µ µ B + εγ 5 kµ µ A+ εγ 5 lµ µ B, (3.2.5 wheresimilarly c µ, d µ, l µ and k µ have constant entries and dimension 0. Onemight claim that the gamma matrices are missing in these two ansätze, but we assume that they are hidden in c µ,d µ,k µ etc. 9

12 3.2.2 Determing the Constants Starting with the scalar part of our Lagrangian 1 2 µa µ A+ 1 2 µb µ B (3.2.6 and using ( we get [ 1 δ 2 µa µ A+ 1 ] 2 µb µ B = µ A µ (δa+ µ B µ (δb. (3.2.7 Inserting our ansätze (3.2.2 and (3.2.3 yields µ A µ (α εψ +β ψε+γ εγ 5 ψ +ν ψγ 5 ε + µ B µ ( α εψ + β ψε+ γ εγ 5 ψ + ν ψγ 5 ε. (3.2.8 Using ( we obtain α ε( µ µ Aψ β( µ µ A ψε γ εγ 5 ( µ µ Aψ ν( µ µ A ψγ 5 ε α ε( µ µ Bψ β( µ µ B ψε γ εγ 5 ( µ µ Bψ ν( µ µ B ψγ 5 ε +tot.der.. (3.2.9 With the metric η µρ we finally find the scalar part in the following form α ε(η µρ ρ µ Aψ β ψ(η µρ ρ µ Aε γ εγ 5 (η µρ ρ µ Aψ ν ψ(η µρ ρ µ Aγ 5 ε α ε(η µρ ρ µ Bψ β ψ(η µρ ρ µ Bε γ εγ 5 (η µρ ρ µ Bψ ν ψ(η µρ ρ µ Bγ 5 ε = α ε(γ µ γ ρ ρ µ Aψ β ψ(γ µ γ ρ ρ µ Aε γ εγ 5 (γ µ γ ρ ρ µ Aψ ν ψ(γ µ γ ρ ρ µ Aγ 5 ε α ε(γ µ γ ρ ρ µ Bψ β ψ(γ µ γ ρ ρ µ Bε γ εγ 5 (γ µ γ ρ ρ µ Bψ ν ψ(γ µ γ ρ ρ µ Bγ 5 ε, ( where the total derivative has been dropped using the same arguments as before at the end of section (3.1. We switched ψ with the scalar fields δa and δb before inserting the metric in order to simplify later comparisons. Now we take the spinor part of our Lagrangian with its variation i 4 ψγ µ µ ψ ( [ ] i δ 4 ψγ µ µ ψ = i(δ ψγ µ µ ψ i 2 [ µ (δ ψγ µ ψ ]. ( The second term, which is a total derivative, can be dropped. Inserting our ansatz 10

13 (3.2.5 yields [ ] i δ 4 ψγ µ µ ψ i( ε c µ µ A+ ε d µ µ B + εγ 5 kµ µ A+ εγ 5 lµ µ Bγ ρ ρ ψ = i ε c µ µ Aγ ρ ρ ψ +i ε d µ µ Bγ ρ ρ ψ +i εγ 5 kµ µ Aγ ρ ρ ψ +i εγ 5 lµ µ Bγ ρ ρ ψ i ε c µ γ ρ ( µ ρ Aψ i ε d µ γ ρ ( µ ρ Bψ i εγ 5 kµ γ ρ ( µ ρ Aψ i εγ 5 lµ γ ρ ( µ ρ Bψ, ( where we used ( again. The symbol means up to total derivatives. Comparing now ( with ( yields the following relations for our constants i c µ = αγ µ c µ = iαγ µ, i d µ = αγ µ d µ = i αγ µ, i k µ = γγ µ k µ = iγγ µ, i l µ = γγ µ l µ = i γγ µ, β = 0, β = 0, ν = 0, ν = 0. ( Inserting into (3.2.5 gives us δ ψ = iα εγ µ µ A+i α εγ µ µ B +iγ εγ 5 γ µ µ A+i γ εγ 5 γ µ µ B. ( A second way to determine the constants is to use the Majorana flip (A This will determine δψ instead of δ ψ. For the variation of the scalar part we get [ ] i δ 4 ψγ µ µ ψ i(δ ψγ µ µ ψ = i µ ψγ µ δψ. ( Together with (3.2.4 and ( we find [ i δ 4 ψγ µ ] ρ µ ψ i ρ ψγ [ c µ ε µ A+d µ ε µ B +k µ γ 5 ε µ A+l µ γ 5 ε µ B ] = i[ ρ ψ µ Aγ ρ c µ + ρ ψ µ Bγ ρ d µ + ρ ψ µ Aγ ρ k µ γ 5 + ρ ψ µ Bγ ρ l µ γ 5 ]ε i[( µ ρ A ψγ ρ c µ +( µ ρ B ψγ ρ d µ +( µ ρ A ψγ ρ k µ γ 5 +( µ ρ B ψγ ρ l µ γ 5 ]ε, ( where we dropped all total derivatives. Again we compare with the scalar part (

14 and find ic µ = γ µ β c µ = iγ µ β, id µ = γ µ β d µ = iγ µ β, ik µ = γ µ ν k µ = iγ µ ν, il µ = γ µ ν l µ = iγ µ ν, α = 0, α = 0, γ = 0, γ = 0. ( We expected the same results for our constants, since we have used a simple and obviously allowed Majorana flip, but this is in stark contrast to ( So how can we solve the problem concerning our constants? Let us start with a look at our ansatz for the bosonic fields and δa = α εψ +β ψε+γ εγ 5 ψ +ν ψγ 5 ε δb = α εψ + β ψε+ γ εγ 5 ψ + ν ψγ 5 ε. Since we wanted to find the simplest and most general supersymmetric transformations, we should use the simplest ansatz for our bosonic field. Applying the Majorana flips (A.5.15 and (A.5.18 on the second and the last term of the equations above we get δa = α εψ +β εψ +γ εγ 5 ψ +ν εγ 5 ψ, ( δb = α εψ + β εψ + γ εγ 5 ψ + ν εγ 5 ψ. ( We easily see that the first and last two terms are the same for Majorana spinors. Therefore we can choose β, β,ν and ν to be 0, which is the result we get in ( Would we choose α, α,γ and γ to be 0 we would have the results from ( So if we simplify our ansätze, which was our original goal, the paradox vanishes. However, we still have not got the transformations ( Having eliminated half of the constants we get δa = α εψ +γ εγ 5 ψ, ( In the Wess-Zumino model the transformations are δb = α εψ + γ εγ 5 ψ. ( δa = εψ, ( δb = i εγ 5 ψ. ( Comparing with our equations, we find for the remaining constants α = 1, γ = 0, α = 0 and γ = i. Inserting all into ( yields δ ψ = i εγ µ µ A εγ 5 γ µ µ B, (

15 which is equal to δ ψ in ( Thus we have found with our ansätze the equations in ( After having derived the transformations used in the Wess-Zumino model, we want to investigate the question if those are the simplest and most general transformations possible. To do this, we create linear combinations of δa and δb using their form we got in ( and ( δc = δa+δb = (α+ α εψ +(γ + γ εγ 5 ψ }{{}}{{} =v =w = v εψ +w εγ 5 ψ, ( So now we have δd = δa δb = (α α εψ +(γ γ εγ 5 ψ }{{}}{{} =x =y = x εψ +y εγ 5 ψ. δc = v εψ +w εγ 5 ψ, δd = x εψ +y εγ 5 ψ. ( ( Finally we have to determine the constants v,w,x and y. A trivial solution would be if we choose all of them to be 0. Since every transformation consists of two terms of the same form, the simpelst transformations, avoiding them to be equal, would have v = 1, w = 0, x = 0, and y = 1. This leads to Rearranging ( and ( gives us which yields for ( and ( v = α+ α = 1, ( w = γ + γ = 0, ( x = α α = 0, ( y = γ γ = 1. ( γ = γ, ( α = α, ( α = 1, ( γ = 1. (

16 We insert this into our transformations δc and δd δc = (α+ α εψ +(γ + γ εγ 5 ψ = (2α εψ = εψ, δd = (α α εψ +(γ γ εγ 5 ψ = (γ γ ν + ν εγ 5 ψ = εγ 5 ψ. ( ( Comparing with (3.1.9 we find a difference in the choice of the constants. In the Wess- Zumino model we have δb = i εγ 5 ψ in contrast to what we have found. The reason for this is how we choose v,w,x and y. For the most general form we wanted the transformations to have only one term, avoiding them to be equal. And since choosing y to be i or 1 is a matter of taste we have shown that it is possible to get the transformations of the Wess-Zumino-model (3.1.9 by a change of basis from {A,B} {C,D}. Finally we want to check the fermionic transformations δψ and δ ψ too. Since they are connected via Dirac conjugation, we only derive the latter. Therefore we determine the coefficients in our ansätze (3.2.2 and (3.2.3 by comparing with ( and ( This yields for the constants in δa and δb α = 1, α = 0, γ = 1, α = 0. ( Inserting into ( gives us δ ψ = i εγ µ µ A+i εγ 5 γ µ µ B. ( Again the difference of a factor i in the second term does not change the form of our transformation. So the transformations for the fermions are also of the most general form possible. 14

17 4 A Two-Dimensional Model: Ramond-Neveu-Schwarz Superstring The Ramond-Neveu-Schwarz formalism is a superstring theory which is supersymmetric on the string world sheet. This is one approach developed to include supersymmetry into string theory [4]. 4.1 Invariance of the Action The action containing bosons and fermions in the RNS formalism is given by S = 1 d 2 σ( α X µ α X µ + 2π ψ µ ρ α α ψ µ, (4.1.1 where X µ (σ,τ represent the bosonic fields of the two-dimensional world-sheet of a string with µ being space time indices. σ parametrizes X µ at a given time τ on the world sheet [6]. The fermionic partners in this action are ψ µ (σ,τ. ρ α, with α = 0,1, represent the two-dimensional Dirac matrices, which obey To follow the steps in [4] the metric in this chapter is {ρ α,ρ β } = 2η αβ. (4.1.2 η αβ = diag( 1,1. (4.1.3 In order further to simplify our work we choose a basis in which these matrices take the following form ( ( ρ =,ρ =. ( The spinor ψ µ, which is a Majorana spinor, has two components ( ψ µ ψ µ =. (4.1.5 Since we use Majorana spinors in two dimensions it follows from the Majorana condition that the spinor components are real. ψ µ + The Dirac conjugate of this spinor ψ is defined as ψ + = ψ +, ψ = ψ (4.1.6 ψ = ψ iρ 0 = i(ψ +, ψ (

18 which for a Majorana spinor is simply ψ T iρ 0. We now want to seperate our action(4.1.1 introduced at the beginning into a fermionic andabosonicpart. Beginningwiththefermionicpart,thatisS f = 1 2π d 2 σ( ψ µ ρ α α ψ µ, we use light-cone coordinates, where the derivatives take the form ± = 1 2 ( 0± 1, leading to ( ( ρ α 0 1 α = 0 0 = 2. ( Then we simply insert into the fermionic part of the action and get S f = 1 [ ( ( ] d 2 0 ψ σ i(ψ +, ψ 2 2π + 0 ψ + = 1 [ ( ] 2π 2i d 2 ψ σ (ψ +, ψ + + ψ = i d 2 σ[ ψ + ψ + ψ + ψ ] π = i d 2 σ[ψ + ψ +ψ + ψ + ]. π (4.1.9 For the bosonic part we can use the metric (4.1.3 and get S b = 1 d 2 σ( α X µ α X µ 2π = 1 d 2 σ( α X µ η αβ β X µ 2π = 1 d 2 σ( 0 X µ 0 X µ + 1 X µ 1 X µ 2π = 1 d 2 σ( ( + + X µ ( + + X µ +( + X µ ( + X µ 2π = 1 d 2 σ( + X µ + X µ X µ X µ + + X µ + X µ + X µ X µ 2π + X µ X µ X µ + X µ + X µ X µ X µ + X µ = 1 d 2 σ( 2 + X µ X µ 2 X µ + X µ 2π = 1 d 2 σ( 4 + X X 2π = 1 d 2 σ(2 + X X, π ( where we suppressed the space-time indices in the last two steps. The action S is now the sum of the bosonic and fermionic part S = S b +S f = 1 d 2 σ(2 + X X +iψ + ψ +iψ + ψ +, ( π where we summed over the µ-indices. 16

19 Similar to the Wess-Zumino model our goal is to show the invariance of the action (4.1.11, respectively (4.1.1, under the following supersymmetry transformations δx µ = εψ µ, ( δψ µ = ρ α α X µ ε, ( where ε is a constant Majorana spinor. The spinor written in components is ( ε ε =. ( ε + Inserting this into the transformation for the bosonic part yields ( δx µ = εψ µ ψ µ = i(ε +, ε = i(ε + ψ µ ε ψ µ +. ( And for the fermionic part δψ µ = ρ α α X µ ε, ( ( ψ µ where we use (4.1.8 and ψ =, which leads to ψ µ + ( ψ µ δ = 2 ψ µ + A simple matrix multiplication then gives us ψ µ + ( ( 0 X µ ε + 0 ε +. ( δψ µ = 2 X µ ε +, ( δψ µ + = 2 +X µ ε. ( In order to verify the invariance under the supersymmetry transformations (4.1.18, (4.1.19, (4.1.15, we vary the action ( and calculate the terms proportional to ε + and ε separately. Starting with δ + we get δ + S = 1 d 2 σ(δ + (2 + X X+iδ + (ψ + ψ +iδ + (ψ + ψ +. ( π The first term leads to δ + (2 + X X = 2[ + (δ + X X + + X (δ + X] = 2[ + (iε + ψ X + + X (iε + ψ ] = 2iε + [ + ψ X + + X ψ ], ( and the second one gives us iδ + (ψ + ψ = i(δ + ψ + ψ +iψ + (δ + ψ = i( 2 Xε + + ψ +iψ + ( 2 Xε + = 2iε + ( X + ψ +2iε + (ψ + X = 2iε + ( X + ψ +ψ + X, (

20 where we used the fact that the spinors ψ and ε anticommute (ψ ε + = ε + ψ. The third term does not need to be taken into account due to the fact that in ( no δ + -parts occur, and therefore we have δ + ψ + = 0. Finally, we have the following: δ + S = 2iε + π = 2iε + π = 2iε + π d 2 σ( + ψ X + + X ψ X + ψ }{{ +ψ } + X = + ψ X d 2 σ( + X ψ +ψ + X d 2 σ ( + Xψ. ( Since the integrand is a total derivative, we find the equivalence to 0 by assuming a world-sheet without boundaries and 2π-periodicity on the string world-sheet. Next, we need to calculate the variation proportional to ε δ S = 1 d 2 σ(δ (2 + X X+iδ (ψ + ψ +iδ (ψ + ψ +. ( π Again we compute the terms separately. The first one yields δ (2 + X X = 2[ + (δ X X + + X (δ X] = 2[ + ( iε ψ + X + + X ( iε ψ + ] = 2iε ( + ψ + X + X ψ +. ( The second term can be ignored because in ( we have no δ -parts, which leads to δ ψ = 0. The third term gives us iδ (ψ + ψ + = i[(δ ψ + ψ + +ψ + (δ ψ + ] = i[2 + Xε ψ + +ψ Xε ] = 2iε ( + X ψ + ψ + + X, ( where we again need to use ψ + ε = ε ψ +. Combining the results for the single terms we find δ S = 2iε d 2 σ( + X ψ + ψ + + X + ψ + X + X ψ + π = 2iε d 2 σ( ψ + + X + ψ + X π = 2iε ( d 2 σ + (ψ + X π = 0. With the same argument that has been used for δ + S, we find the equivalence to 0. 18

21 4.2 Comparing the two Models As we could see in the above section, it is much easier to show the invariance of the action in the Ramon-Neveu-Schwarz formalism than in the four-dimensional Wess-Zumino model. Listed in the following table below are some differences and common properties of the two models. 4-dim 2-dim matrices four 4 4-matrices two 2 2-matrices spinors two 2-dim. complex components 2 real components 1 bosons 2 ( µa µ A+ µ B µ B α X α X i fermions 4 µ ψ ψρ α α ψ 1 Lagrangian 2 ( µa µ A+ µ B µ B+ i ψγ µ 4 µ ψ α X α X + ψρ α α ψ degrees of A,B : 2 X : 1 freedom ψ, ψ : 2 ψ, ψ : 1 Majorana condition OK OK Table 4.1: Comparison between the four dimensional and the two dimensional model In the two-dimensional case we have half the amount of Dirac matrices and degrees of freedom for our bosonic and fermionic fields. Our equations describing the bosons and fermions are of a very similar structure. The terms describing the bosons have derivatives acting on their fields. The closely related structure is more obvious when we take a look at the fermions. There we have a Dirac conjugated spinor, a Dirac matrix and then a derivative acting on a spinor. Looking at the spinors we have on the one hand 2 two-dimensional complex entities and on the other hand only 2 real components. Additionally, in both cases it is possible to impose a Majorana condition, which simplifies our work in four dimensions as well as in two. 19

22 A Appendix A.1 Gamma Matrices The Gamma matrices in 4 dimensions are given by ( γ µ 0 σ µ = σ µ 0, (A.1.1 with σ µ being a Pauli matrix. The pauli matrices are defined as ( ( ( σ :=, σ 2 0 i :=, σ := 1 0 i 0 0 1, (A.1.2 in addition with the matrix σ 0 σ 0 := The matrix σ µ is defined as ( ( σ µ := (σ 0, σ i.. (A.1.3 Therefore the Gamma matrices are γ 0 = , γ 1 = i γ 2 = 0 0 i 0 0 i 0 0, γ 3 = i They fulfill the Clifford algebra (A.1.4 (A.1.5 {γ µ,γ ν } = 2η µν ½ 4, {γ µ,γ 5 } = 0, {γ 5,γ 5 } = 2½ 4, (A.1.6 with the Minkowski metric η µν := diag(1, 1, 1, 1. (A

23 The matrix γ 5 is defined as γ 5 := iγ 0 γ 1 γ 2 γ 3 = (A.1.8 Additionally, we just want to list the tranposed gamma matrices (γ 0 T =γ 0, (γ 1 T = γ 1, (γ 2 T =γ 2, (γ 3 T = γ 3, (γ 5 T =γ 5. (A.1.9 A.2 Charge Conjugation Wearetakingacloser lookat thediracequation whichdescribesafreemassivefermionic particle with spin 1 2 and mass m, (iγ µ µ mψ(x = 0, (A.2.1 with ψ(x being a 4-spinor (i.e. Dirac-spinor. If we want to couple a spin- 1 2 particle with electric charge q and the electromagnetic field A µ, which can be assumed to be real, we get [γ µ (i µ qa µ (x m]ψ(x = 0. (A.2.2 For antiparticles we have to take into account that they have the opposite electrical charge but the same mass and spin as the corresponding particle. This yields [γ µ (i µ +qa µ (x m]ψ c (x = 0, (A.2.3 where ψ c (x is the charge conjugated spinor describing the antiparticle. Our goal now is to relate the spinors ψ(x and ψ c (x. Therefore we take the complex conjugate of the Dirac equation (A.2.2 [(γ µ ( (i µ +qa µ (x m]ψ (x = 0 (A.2.4 and compare it with the charge conjugated equation (A.2.3. We see that that complex conjugation changes the sign of the derivative and the charge, which is in some way similar to charge conjugation. This similarity is the reason for our ansatz for the charge conjugated spinor ψ c = Uψ, (A

24 where the matrix U represents a linear operator. Inserting U 1 Uψ = U 1 ψ c in the complex conjugated Dirac equation gives us [ (γ µ (i µ +qa µ m]u 1 ψ c = 0. (A.2.6 Now we multiply with U from the left and get [ U(γ µ U 1 (i µ +qa µ m]ψ c = 0. (A.2.7 Comparing this equation with the one for the antiparticle (A.2.3 leads to the relation U(γ µ U 1 = γ µ. With ψ = (ψ T the charge conjugated spinor becomes ψ c = Uψ = U(ψ T (A.2.8 (A.2.9 (A.2.10 and since the Dirac conjugate is ψ = ψ γ 0, it makes sense to use the following ansatz U = Cγ 0. (A.2.11 Inserting into (A.2.8 yields Cγ 0 (γ µ (γ 0 1 C 1 = γ µ. (A.2.12 Using the relations (γ 0 1 = γ 0 and γ 0 (γ µ γ 0 = (γ µ T, which can be easily verified, we get C(γ µ T C 1 = γ µ. (A.2.13 Multiplying with C from the right and rearranging gives us C(γ µ T +γ µ C = 0. (A.2.14 With the properties of the transposed gamma matrices (A.1.9 and the above equation, we find the following conditions for the charge conjugation matrix C {C,γ 0 } = {C,γ 2 } = 0, (A.2.15 [C,γ 1 ] = [C,γ 3 ] = 0. A matrix satisfying these relations has the form ( ( C = iγ 2 γ 0 0 σ 2 0 ½ = i σ 2 0 ½ 0 ( σ 2 0 = i 0 σ 2 (A (A.2.17 So we have for the matrix U: ( U = Cγ 0 = iγ 2 0 σ 2 = i σ 2 0. (A

25 A.3 Weyl Spinors The 4-spinor ψ(x may be decomposed as ψ(x = ( χ(x η(x, (A.3.1 where χ and η are 2-spinors, in our case Weyl spinors. We now make the Lorentz transformation ( ( ( χ η = ψ s 0 χ = S(Λψ = 0 (s 1, (A.3.2 η with S(Λ being the matrix used for the Lorentz transformation of a Dirac spinor [9]. A simple matrix multiplication then gives us χ χ = sχ, (A.3.3 η η = (s 1 η. (A.3.4 In our attempt of finding Lorentz-invariant quantities, we take the Lorentz transformation of the 2-spinor (iσ 2 χ into consideration, using (s 1 T = σ 2 sσ 2 [9], and its transpose (iσ 2 χ T, (iσ 2 χ (iσ 2 χ = iσ 2 χ = iσ 2 sχ = (s 1 T (iσ 2 χ, (A.3.5 (iσ 2 χ T (iσ 2 χ T = (iσ 2 χ T s 1. (A.3.6 This vields the Lorentz invariant expression (iσ 2 χ T χ = (iσ 2 χ T s 1 sχ = (iσ 2 χ T χ. (A.3.7 Introducing the covariant components of the 2-spinor χ, ( χ1 χ := (χ α =, (A.3.8 we define the contravariant components of the Weyl spinor iσ 2 χ, ( iσ 2 χ =: (χ α χ 1 =, (A.3.9 χ 2 χ 2 leading to ( χ 1 = χ 2 ( ( χ1 χ 2 = ( χ2 χ 1 (A

26 and thereby obtaining χ 1 = χ 2, χ 2 = χ 1. Now we can rewrite (A.3.7 as (χ 1,χ ( 2 χ 1 χ = (χ 1,χ 2 2 or ( χ1 χ 2 χ α χ α = χ α χ α = Lorentz invariant. (A.3.11, (A.3.12 (A.3.13 The order of the indices is very important when writing Lorentz-invariant expressions without indices. The left χ has an upper index, the right one a lower index. Using the relation (A.3.11, the Lorentz invariant expression χ α χ α turns into χ α χ α = χ 1 χ 1 +χ 2 χ 2 = χ 2 χ 1 χ 1 χ 2. (A.3.14 The previously used relation (A.3.11 implies the metric of the two-dimensional spinor space ( (ε αβ 0 1 = = iσ 2 (A and the relation becomes Similarly we have with (ε αβ = The spinor metric is antisymmetric χ α = ε αβ χ β. χ α = ε αβ χ β, (A.3.16 (A.3.17 ( 0 1 = iσ 2. (A ε αβ = ε βα, ε αβ = ε βα, (A.3.19 and it fulfills ε αβ ε βγ = ε γβ ε βα = δ γ α. (A.3.20 This is just the two-dimensional ε-tensor. The fact that the spinor metric is antisymmetric has some unusual consequences. For example, the Lorentz invariant expression χ α χ α may be written as χ α χ α = χ 1 χ 1 +χ 2 χ 2 = χ 2 χ 1 χ 1 χ 2 = χ 2 χ 2 χ 1 χ 1 = χ α χ α. (A

27 Since the spinors χ and η transform differently under Lorentz transformations, the contravariant components of η carry a dotted index. η := ( η α = ( η 1 η 2 (A.3.22 The covariant components are defined as iσ 2 η =: ( η α = ( η 1 η 2. (A.3.23 We now have ( η 1 η 2 ( = iσ 2 η 1 η 2 ( ( η 1 ( 0 1 η 2 = = 1 0 η 2 η 1 (A.3.24 and thereby obtain η 1 = η 2, η 2 = η 1. Like (A (A.3.7 for the spinor χ, we now get for η η α η α = η α η α. (A.3.25 (A.3.26 Again it is very important to maintain the order of the spinor indices. The relation between dotted covariant and contravariant indices, η α = ε α β η β, η α = ε α β η β, (A.3.27 uses the metric tensor β ε α = ε αβ = iσ 2, ε α β = ε αβ = iσ 2, (A.3.28 which is the same as used with χ At the end of this section about Weyl spinors we take a quick look at how the matrices σ µ and σ µ transform using the metric tensors (A.3.28 We define [9] ( σ µ αβ := ε α γ ε βδ ( σ µ γδ. (A.3.29 Using we get (σ µ T = σ 2 σ µ σ 2 (A.3.30 ((σ µ T αβ = ( iσ 2 ( σ µ ( iσ 2 αβ = ε α γ ( σ µ γδ ε δβ (A.3.31 = ε α γ ε βδ ( σ µ γδ 25

28 and since we find that which is nothing else than (( σ µ T αβ = (σ µ β α (A.3.32 (σ µ β α = ε α γ ε βδ ( σ µ γδ, (A.3.33 (σ µ β α = ( σ µ αβ. (A.3.34 A.4 Majorana condition The charge conjugated spinor is given by where ψ c = C ψ T, ( σ 2 0 C = i 0 σ 2 (A.4.1 (A.4.2 is the charge conjugation matrix (A ψ is a Dirac spinor written in terms of Weyl spinors ( ( χ (χα ψ = = η ( η α. (A.4.3 Using (A.2.9 the Dirac adjoint is given by ψ = ψ γ 0 = ( ( χ α (η α ( 0 ½ 2 ½ 2 0 and therefore the spinor ψ c is ( ( ψ c = C ψ T σ 2 0 η α = i 0 σ 2 = χ α = ( (η α ( χ α (A.4.4 ( iσ 2 (η α iσ 2 ( χ α. (A.4.5 Now we use the definitions for the metric (A.3.15 and (A.3.18 and get ( ψ c (εαβ η β ( (ηα = = ( χ α. (A.4.6 (ε α β χ β Since ψ c has the same structure as ψ we can impose the Majorana Condition ψ c = ψ, leading us to ( (ηα ( χ α = A Majorana spinor is then given by ψ = ψ c = ( (χα ( η α ( (χα ( χ α (A.4.7. (A.4.8. (A.4.9 This is a special kind of a four-dimensional spinor. A particle fulfilling this condition is its own antiparticle, reducing the number of real degrees of freedom from eight to four [10]. 26

29 A.5 Majorana flips In this work we often need to deal with a terms of the form ε 2 ε 1, ε 2 γ µ ε 1, or ε 2 γ 5 ε 1. To simplify our work, we will now derive so called Majorana flip properties. The Majorana spinors ε 1 and ε 2 written in components are ( χα ε 1 = χ α, ( (A.5.1 φα ε 2 =, φ α with their Dirac adjoints ε 1 = (χ α, χ α, ε 2 = (φ α, φ α. Starting with ε 2 ε 1 we find with use of the metric tensor ε 2 ε 1 = φ α χ α + φ α χ α = ε αβ φ β ε αγ χ γ +ε α β φ βε α γ χ γ. (A.5.2 (A.5.3 (A.3.19 together with (A.3.28 leads us to ε αβ φ β ε αγ χ γ +ε α β φ βε α γ χ γ = ε βα ε αγ φ β χ γ ε β α ε α γ φ β χ γ. (A.5.4 Using (A.3.20 we arrive at ε βα ε αγ φ β χ γ ε β α ε α γ φ β χ γ = δ β γ φ βχ γ δ γ β φ β χ γ = φ β χ β φ β χ β. (A.5.5 For the last step we use (A.3.21 φ β χ β φ β χ β = χ β φ β + χ β φ β = ε 1 ε 2. (A.5.6 So we found the following important relation ε 2 ε 1 = ε 1 ε 2. For the second flip property for ε 2 γ µ ε 1 we get ( ε 2 γ µ ε 1 = (φ α 0 (σ, φ α µ ( α α χα ( σ µ αα 0 χ α ( = (φ α (σ, φ α µ α α χ α ( σ µ αα χ α = φ α (σ µ α α χ α + φ α ( σ µ αα χ α. (A.5.7 (A

30 We start with calculating the first term Then we use (A.3.34 and find φ α (σ µ α α χ α = ε αγ φ γ (σ µ α α ε α γ χ γ. (A.5.9 ε αγ φ γ (σ µ α α ε α γ χ γ = ε αγ φ γ ( σ µ αα ε α γ χ γ Considering (A.3.19 we get = ε αγ ε α γ φ γ ε α λε αν ( σ µ λν χ γ = ε αγ ε αν ε α γ ε α λφ γ ( σ µ λν χ γ. (A.5.10 ε αγ ε ε α γ αν ε α λφ γ ( σ µ λν χ γ = ( ε γα ε ( ε γ α αν ε α λφ γ ( σ µ λν χ γ. Applying (A.3.20 yields ( ε γα ε ( ε γ α αν ε α λφ γ ( σ µ λν χ γ = ( δ ν( δ γ λφ γ γ ( σ µ λν χ γ = φ γ ( σ µ γγ χ γ. And since the spinors anti-commute we get φ γ ( σ µ γγ χ γ = χ γ ( σ µ γγ φ γ. For the second term we follow the same steps: φ α ( σ µ αα β χ α = ε α β φ ( σ µ αα ε αβ χ β (A.5.11 (A.5.12 (A.5.13 = ε α βε αβ φ βε α γ ε αδ ( σ µ γδ χ β = ε α βε α γ ε αβ ε αδ φ β(σ µ δ γ χ β = ε β α ε α γ ε βα ε αδ φ β(σ µ δ γ χ β = δ γ βδ δ β φ β (σ µ γδ χ β = φ γ (σ µ δ γ χ δ = χ δ (σ µ δ γ φ γ = χ α (σ µ α α φ α. Combining the results for the two terms leads us to ε 2 γ µ ε 1 = χ α (σ µ α α φ α χ γ ( σ µ γγ φ γ = ε 1 γ µ ε 2. The last flip we are looking at is of the form ε 2 γ 5 ε 1, ( ε 2 γ 5 ε 1 = (φ α, φ α ½2 0 0 ½ 2 ( = (φ α, φ α χα χ α = φ α χ α + φ α χ α. ( χα χ α (A.5.14 (A.5.15 (A

31 Using the anti-commutation property for spinors and the same steps with the metric tensor as before, we get φ α χ α + φ α χ α = χ α φ α χ α φ α = ε αβ χ β ε αδ φ δ ε α β χ βε α δ φ δ = ε αβ ε αδ χ β φ δ ε α βε α δ χ β φ δ = ε βα ε αδ χ β φ δ +ε β α ε α δ χ β φ δ = δ δ β χβ φ δ +δ β δ χ β φ δ = χ β φ β + χ β φ β (A.5.17 Finally, we find = χ α φ β + χ α φ α = ε 1 γ 5 ε 2. ε 2 γ 5 ε 1 = ε 1 γ 5 ε 2. (A

32 Bibliography [1] Steindl A. Vorlesungsunterlagen Mechanik für Technische Physiker, pages TU Wien, [2] Greene B. Das elegante Universum, page 263. Berliner Taschenbuch Verlag, Berlin, [3] Rebhan E. Theoretische Physik II, pages Elsevier Gmbh, München, [4] Becker K., Becker M., and Schwarz J. String Theory and M-Theory, pages Cambridge University Press, New York, [5] Becker K., Becker M., and Schwarz J. String Theory and M-Theory, page 2. Cambridge University Press, New York, [6] Becker K., Becker M., and Schwarz J. String Theory and M-Theory, pages 7 8. Cambridge University Press, New York, [7] Schweda M. et al. Supersymmetry, pages TU Wien. [8] Schweda M. et al. Supersymmetry. TU Wien. [9] Schweda M. et al. Supersymmetry, pages TU Wien. [10] Schweda M. et al. Supersymmetry, page 46. TU Wien. [11] Greiner W. and Reinhardt J. Field Quantization, pages Springer-Verlag, Berlin,