MA4447: The Standard Model of Elementary Particle Physics

Transcript

1 MA4447: The Standard Model of Elementary Particle Physics David Whyte Updated November 2, Natural units We will use natural units: ħ = c = 1. This means that [p] = [m] = [T ] 1 = [L] 1 = [E] = MeV. To convert from standard units to natural units, remember that ħc 2 MeV fm. For example, to convert from seconds to natural units: τ = 1 23 s = 1 23 c s c = 3 fm c 3ħc 1 c 2 MeV τ = MeV 1 If τ is the lifetime of a strongly decaying particle, the decay width Γ = 1 τ 7 MeV. 2 Klein Gordon equation 2.1 Derivation In non-relativistic quantum mechanics, the Schrödinger equation is based on an energy: E = p2 2m, E i t, p i We keep this recipe, but use the relativistic energy: We rewrite this to avoid the square root: E = ± p 2 + m 2 E 2 = p 2 + m 2. 1

2 and note that: E 2 2 t 2, p2 2. So we have: 2 t 2 ϕx, t = 2 + m 2 ϕx, t or, noting that the d Alembertian operator = µ µ = 2 t 2 2, the Klein Gordon equation. + m 2 ϕx, t =, 2.2 Example We will test the equation on the plane wave solution: since px p µ x µ = p x p x = Et p x. We calculate derivatives: ϕx, t = Ne iet+ip x = Ne i px, µ e i px = µ i pxe i px = i p µ e i px. µ µ e i px = e i px = i p µ i p µ e i px = p 2 i px e Note that p 2 p µ p µ = E 2 p 2, so: + m 2 Ne i px = N p 2 + m 2 i px e = E 2 + p 2 + m 2 }{{} =, as expected. i px Ne 3 Probability current 3.1 Derivation Recall that we used E = ± p 2 + m 2, meaning that for any p there are two solutions, E < and E >. We use the K G equation for ϕ, and its complex conjugate ϕ : iϕ µ µ + m 2 ϕ iϕ µ µ + m 2 ϕ =, }{{}}{{} 2

3 and rewrite the LHS: = m 2 i ϕ ϕ ϕϕ +iϕ µ µ ϕ iϕ µ µ ϕ }{{} = i µ ϕ µ ϕ ϕ µ ϕ i µ ϕ µ ϕ + i µ ϕ µ ϕ }{{} ] = µ [iϕ µ ϕ ϕ µ ϕ µ j µ. j µ can be expressed as ρ,j, where ρ is probability density and j is probability current: ρ = iϕ ϕ t ϕ ϕ t j = iϕ ϕ ϕ ϕ µ j µ = can be written in 3D as the continuity equation: ρ t + j =. 3.2 Example We evaluate ρ for the plane wave ϕ = Ne i px : ρ = i N 2 e i px iee i px e i px iee i px = N 2 i 2iE = 2E N 2 If E < this gives a negative probability density. This is a problem! 4 Dirac equation 4.1 Derivation The K G equation is second-order in space and time because of. Dirac tried to obtain an equation which was first-order. The equation he proposed was: i t ψ = iα + βmψ, where α 1, α 2, α 3 and β are to be determined. This is the Dirac equation. In order to find α and β we require: the relativistic energy-momentum equation 3

4 Lorentz covariance of the equation discussed later We also want ϕ to satisfy the K G equation. We square both sides of the Dirac equation: i 2 ψ = iα + βm iα + βmψ. t and carefully expand, bearing in mind that α i and β are not necessarily scalar: 2 t 2 ψ = i α 2 i 2 ψ x i 2 2 ψ α i α j + α j α i i<j x i x j + β2 m 2 ψ im i α i β + βα i ψ x i, i, j = 1,2,3. In order for this to match the K G equation, the RHS must equal 2 ψ + m 2 ψ. And for this, we have the requirements: 1. α 2 = 1, i = 1,2,3. i 2. β 2 = α i α j + α j α i =, i j. 4. α i β + βα i =, i = 1,2, Dirac spinors Requirements 3 and 4 mean that α i and β cannot be scalars, so they must be matrices. Furthermore, we require that they are Hermitian in order for the Hamiltonian to be Hermitian. A tempting choice for α i is the Pauli spin matrices σ i, since we know that they satisfy requirements 1 and 3. However no nonzero β exists which anticommutes with σ i for requirement 4. Take requirement 4: α i β + βα i = α i = βα i β trα i = trβα i β = trβ 2 α i = trα i, so trα i =. By a similar argument, trβ =. So α i and β are Hermitian matrices with square 1 and trace. This tells us that they have: real eigenvalues eigenvalues = ±1 4

5 equal number of +1 and 1 eigenvalues even dimension. There is no set of 2 2 matrices which satisfy these conditions, so we try 4 4. One possible choice is: σi 12 α i =, β =, σ i 1 2 where σ i are the Pauli spin matrices and 1 2 is the 2 2 identity matrix. This choice is not unique. We can construct α i = Uα iu 1 and β = UβU 1 such that: α i 2 = UαU 1 UαU 1 = Uα 2 U 1 = UU 1 = 1 2 and similarly for β. We want α i = α i, so: U 1 α i U = α i U 1 α i U = α i, so U = U 1, i.e. U is unitary. Hence ψ is a column vector with four components, called a Dirac spinor: ψ = ψ 1 ψ 2 ψ 3 ψ 4, ψ = ψ 1 ψ 2 ψ 3 ψ Probability current Consider again the Dirac equation: i t ψ = iα + βmψ 1 Pre-multiply 1 by ψ : ψ i t ψ = ψ iα + βmψ. 2 Take the complex conjugate of 1: i t ψ = ψ iα i t ψ ψ = ψ iα + βm + βmψ,3 where the means that the operates to the left, i.e. on ψ here. Subtracting 3 from 2 gives: i t ψ ψ = iψ α + α ψ4 5

6 Define: ρ = ψ ψ = 4 ψ σ ψ σ = ψ σ 2. σ=1 This means no more negative probability densities! Defining j k = ψ α k ψ allows 4 to be written: ρ t + j =, or in four-vector form, µ j µ =, where j µ = ρ,j as before. 4.4 Non-relativistic correspondence Take an electron at rest: p =, so: i ψ t = βmψ. This has four solutions: 1 ψ 1 = e imt, ψ 3 = e imt 1, ψ2 = e imt 1 ψ4 = e imt 1 positive energy negative energy We will focus on the positive energy solutions, and work out the connections to Pauli theory in non-relativistic QM by introducing an EM potential: p µ p µ e A µ, A µ = ϕ,a Substitute this into the Dirac equation: i t eϕψ = α i ea + βmψ. 5 Introduce the kinetic momentum operator π = i ea, and let: ψ = φ where φ and χ are 2-component objects. Now rewrite 5 using α = σ σ, β = 1 π = i ea: σ π χ φ φ i t = + eϕ + m φ χ σ π φ χ χ χ 1 and 6 6

7 Remember we are looking at the nonrelativistic limit, so all energies m. We are considering only the positive energy solutions, so we can write: φ = e imt φ χ χ By assumption, φ and χ are slowly oscillating functions, so 6 gives: φ χ φ i t = σ π + eϕ 2m. 7 χ φ χ χ Since t χ, eϕχ 2mχ, σ π 2mχ χ = σ π φ } 2m {{} Ov/c so χ are the small components of the 4-spinor relative to φ. Insert this expression for χ into 7 and the top line gives: We want to simplify σ π 2. σ π 2 i t φ = 2m + eϕ φ. 8 σ aσ b = σ i a i σ k b k = a i b k σ i σ k = a i b k δ ik iϵ ikl σ l = a b1 2 + iσ l ϵ lik a i b k = a b1 2 + iσ a b So σ π 2 = σ π σ π = π 2 +iσ π π. Note that π π, since π is an operator. π π i = [ i ea i ea ] i = ϵ i j k i j ea j i k ea k = ϵ i j kl j k + iea j k + ie j A k + iea k j + e 2 A j A k Since j k, e 2 A j A k and iea j k + iea k j are symmetric in j k they vanish when summed over j and k due to the antisymmetric ϵ i j kl, leaving: π π i = ieϵ i j k j A k = ie A i = ieb i σ π 2 = π 2 eσ B. Using this expression in 8 gives the non-relativistic Pauli equation: i ea 2 i t φ = e 2m 2m σ B + eϕ φ 7

8 The equation contains a term which reflects the electron magnetic moment µ: µ = e 2m σ = g e 2m S. The electron g -factor has been found experimentally to be This deviation from 2 can be explained by QFT. 5 Covariance of Dirac equation 5.1 Lorentz group Lorentz transformations are denoted by x µ x µ = Λ µ νx ν. The Lorentz group consists of those Λ under which the 4D scalar product, x y = x µ y µ = x y x y = x µ g µν y ν, is invariant. We will use g = diag1, 1, 1, 1. What is the implication of invariance? x µ y µ = Λµ νx ν g µρ y ρ = Λ µ νx ν g µρ Λ ρ σy σ = x ν y σ[ Λ µ νg µρ Λ ρ ] σ In order for this to equal x µ y µ, we require [ Λ µ νg µρ Λ ρ σ] = gνσ. This can be written: Λ T ν µ g µρ Λ ρ σ = g νσ, where Λ T is the transpose of Λ, or, in matrix form, Λ T g Λ = g. 9 The Lorentz group is hence made up of all Λ which satisfy Observations Look at detλ: detλ T g Λ = det g detλ T detλ = 1 detλ = ±1, since Λ has only real entries. 8

9 Look at g : Λ T g Λ = g = 1 Λ µ g µρ Λ ρ = 1 Expand the sum, noting that g αβ = for α β: Λ g Λ + Λ 1 g 11 Λ 1 + Λ 2 g 22 Λ 2 + Λ 3 g 33 Λ 3 = 1 = Λ 2 Λ 1 2 Λ 2 Λ 3 = 1. This means that Λ 2 1, i.e. either Λ 1 or Λ 1. The two properties detλ = ±1 and Λ 1 or 1 define four disconnected components of the Lorentz group: detλ signλ contains L Λ = 1 L 1 +1 Λ = I s L Λ = I st L 1 1 Λ = I t I s is the parity transformation x,x x, x, I t is the time reversal transformation x,x x,x and I st = I s I t. L ± are called orthochronous transformations since they do not change the sign of x, i.e. the direction of time. Usually only L + is discussed, since the others can be obtained using I s and I t. 5.3 Transforming wavefunctions We want to find the connection between a wavefunction as seen by observers O and O, where x µ = Λ µ νx ν. How is an equation for ψ related to an equation for ψ? Consider the Klein Gordon equation: O : µ µ + m 2 ϕx =. O : µ µ + m 2 ϕ x =. Note that m 2 is Lorentz invariant. We use the Ansatz: ϕ x = f Λϕx = f ΛϕΛ 1 x It turns out that µ µ = µ µ, so we have: µ µ + m 2 ϕ x = µ µ + m 2 f ΛϕΛ 1 x. f Λ = 1 is a solution, so ϕ is a scalar field. 9

10 6 Gamma matrices 6.1 Definition Define γ = β and γ k = βα k, k = 1,2,3. This gives the property: Verify: γ µ γ ν + γ ν γ µ {γ µ,γ ν } = 2g µν. 1 µ = ν = : γ γ + γ γ = 2β 2 = 2 = 2g µ =,ν = k: γ γ k + γ k γ = ββα k + βα k β = α k + ββα k = α k α k = = 2g k µ = ν = k: γ k γ k + γ k γ k = 2βα k βα k = 2 α k ββα k = 2α k β 2 α k = 2α 2 k = 2 = 2g kk µ = k, ν = k l: γ k γ l + γ l γ k = βα k βα l + βα l βα k = β 2 α k α l β 2 α l α k = α k α l + α l α k = = 2g kl 6.2 Properties α k and β are Hermitian. γ = β so obviously γ = γ, and γ 2 = β 2 = 1. Since γ k = βα k, γ k = α k β = α β k = βα k = γ k, i.e. γ k are anti-hermitian. γ k 2 = βα k βα k = β 2 α 2 k = 1. A useful form is: γ µ = { γ µ if µ = γ µ if µ = k } = γ γ µ γ. 11 1

11 6.3 Explicit representation Using our earlier choice of α k and β gives: 1 σ k γ =, γ k = 1 σ k 7 Back to the Dirac equation 7.1 Alternate expression We want to rewrite the Dirac equation in terms of γ matrices. i t + iα βm ψ = iβ t + iβα m ψ = iγ + iγ m ψ = iγ µ µ mψ =. This can be written in Feynman slash notation as i / mψ = where B / γ µ B µ. 7.2 Adjoint Dirac equation i ψ = ψ iα t ψ i t + iα ψ β i 2 t + iα + βm + βm + βm = = Now define ψ = ψ β = ψ γ. So: the adjoint Dirac equation. ψ βi t + iβα ψ γ i + iγ ψiγ µ + β 2 m + m = = µ + m = ψi / + m =, 11

12 8 Lorentz transformations 8.1 Formulation Again we have observers O and O where x µ = Λ µ νx ν. The Dirac equations for the observers read: iγ µ µ mψ x = and iγ µ µ mψx =. Use the Ansatz: ψ x = SΛψx = SΛψΛ 1 x SΛ must have an inverse to get from O back to O, so: ψx = SΛ 1 ψ x = SΛ 1 ψ Λx ψx = SΛ 1 ψ x. This means that SΛ 1 = SΛ 1. Also SΛ 1 Λ 2 = SΛ 1 SΛ 2, so SΛ defines a representation of the Lorentz group. i / mψx = SΛi / msλ 1 SΛψx = }{{} ψ x isλγ µ SΛ 1 µ m ψ x = 12 Note that: µ = x µ = x ν } x {{ µ } Λ ν µ x ν µ = Λ ν µ ν, allowing 12 to be written: isλγ µ SΛ 1 Λ ν µ ν m ψ x =. In order for this to match the Dirac equation, we need: SΛγ µ SΛ 1 Λ ν µ = γ ν, i.e. Λ ν µγ µ = SΛ 1 γ ν SΛ, 13 this is the condition on S which gives a covariant Dirac equation. We find SΛ first for an infinitesimal Lorentz transformation Λ L +. Write: Λ ν µ = g ν µ + ων µ 1 + ω ν µ. 12

13 9 gives dropping terms of O ω 2 and greater: Λ T g Λ = g 1 + ω T g 1 + ω = g g + ω T g + g ω +O ω 2 = g ω T g + g ω = ω µ νg µρ + g νµ ω µ ρ = ω ρν + ω νρ = ω ρν = ω νρ, i.e. ω µν is antisymmetric under µ ν. Since it is a 4 4 matrix, this means there are 6 independent parameters. These represent the 6 generators of a Lorentz transformation: 3 rotations and 3 Lorentz boosts. Now write: SΛ = 1 i 4 σ µν ω µν +O ω 2 14 SΛ 1 = 1 + i 4 σ µν ω µν +O ω 2, σ µν is an unknown coefficient matrix. We can assume σ µν = σ νµ since any symmetric part would not contribute when contracted with ω µν. Insert these into 13: 1 + ω ν µγ µ = 1 + i 4 σ µν γ ν 1 + i 4 σ µν ω µν γ ν + ω ν µγ µ = γ ν + i 4 ωρσ σ ρσ γ ν γ ν σ ρσ ω ν µγ µ = i 4 ωρσ[ σ ρσ,γ ν]. Post-multiply by γ ν to give: ω ν µγ µ γ ν = i 4 ωρσ σ ρσ γ ν γ ν γ ν σ ρσ γ ν 15 What is γ ν γ ν? γ ν γ ν = g νµ γ ν γ µ = g νµ 2g νµ γ µ γ ν = 2g νµ g νµ g νµ γ µ γ ν γ ν γ ν = 2g ν ν γ ν γ ν γ ν γ ν = g ν ν = Claim: σ ρσ = 2[ i ] γρ,γ σ. This means that γ ν σ ρσ γ ν in 15 will contain a term γ ν γ ρ γ σ γ ν. γ ν γ ρ γ σ γ ν = γ ν γ ρ 2g ρν γ ν γ σ by 1 = 2γ ν γ ρ g ρν γ ν γ ρ γ ν γ σ = 2γ σ γ ρ γ ν 2g ρν γ ν γ ρ γσ = 2γ σ γ ρ 2γ ν g ρν γ σ + γ ν γ ν γ ρ γ σ = 2γ σ γ ρ 2γ ρ γ σ + 4γ ρ γ σ = 2γ σ γ ρ + 2γ ρ γ σ γ ν γ ρ γ σ γ ν = 4g ρσ. 13

14 γ ν γ ρ γ σ γ ν is hence symmetric in ρ σ. The term γ ν σ ρσ γ ν must vanish because of our definition of σ ρσ. Combining this fact with 16 allows 15 to be written: ω νµ γ µ γ ν = i 4 ωρσ 4σ ρσ. Note that γ µ γ ν = 1 [ ] } 2 γµ,γ ν + 1 2{ γµ,γ ν and that the latter term does not contribute when contracted with the antisymmetric ω ρσ, so: Hence our claim is justified. ω νµ 1 [ ] 2 γµ,γ ν = 1 2 ωρσ[ ] γ ρ,γ σ 1 2 ωµν[ ] γ µ,γ ν = 1 2 ωρσ[ ] γ ρ,γ σ. 8.2 Lorentz boost For an infinitesimal Lorentz transformation we can write: ω ν µ = 6 ω n I n ν µ where I n are the generators of the Lorentz transformation. For example: 1 boost in x-direction: I ν µ = rotation around z-axis: I ν µ = n= For a boost in the x-direction, write ω ν µ = ωi ν µ and set ω = ω N. This gives: x ν = Λ ν µx µ = lim g + ω N N I ν α1 g + ω N I α 1 α2 g + ω N I α N+1 αn x α N = e ωi ν µ x µ }{{} Λ ν µ Since I 3 = I, note that I n I = I 2 if n is odd, if n is even. Using this in the expansion for e ωi gives: e ωi ω n = 1 + n! I n n=1 ω 2n n=1 2n! + I n=1 = 1 + I 2 ω 2n 1 2n 1! = 1 + I 2 coshω 1 + I sinhω. 14

15 Writing this as a matrix: ω is the rapidity. coshω = γ = coshω sinhω e ωi = Λ ν µ = sinhω coshω 1 1 1, sinhω = γβ and tanhω = β. So: 1 β 2 x x 1 x 2 x 3 γ γβ = γβ γ 1 1 x x 1 x 2 x 3 We want to find S for general ω, based on 14 which applies for an infinitesimal ω. S = lim 1 i ω N N 4 N σ µνi µν = exp i4 ωσ µνi µν, 17 where I µν = g νρ I µ ρ as usual. S = exp i 4 ωσ 1I 1 + σ 1 I 1 = exp i 4 ωσ 1 σ 1 = exp i 2 ωσ 1 Since iσ 1 2 = 1, we can write: iσ 1 n iσ 1 if n is odd, = 1 if n is even. So, finally: ψ x = exp = ψx ω 2 iσ 1 cosh ω 2 iσ 1 sinh ω 2 ψx σ 1 can be expanded: This gives: iσ 1 = i 1 2 γ γ 1 γ 1 γ = 1 2 2γ γ 1 = γ γ 1. ψ x = cosh ω 2 + γ γ 1 sinh ω ψx. 2 15

16 8.3 Rotation For a rotation around the z-axis, Iz 12 = 1, Iz 21 = 1, and all other entries are. Using a rotation angle of α we obtain an analogous equation to 17: S = exp i 4 ασ µνi µν z = exp 1 2 ασ 12 Since σ 12 = i 2 [γ 1,γ 2 ] = iγ 1 γ 2 = σ3 σ 3, = cos α 2 iσ 12 sin α 2 ϕ x = e i 2 ασ 3 ϕx. So a 4π rotation is needed to get back to ϕ. This is an important property of spinors. So for a rotation, SΛ = e i 4 σ µνω µν, where ω µν is real with ω µν = ω νµ. σ µν = i 2 [γ µ,γ ν ] = i 2 [γ ν,γ µ ]. Recall 11: γ µ = γ γ µ γ. This allows us to write: σ µν = γ σ µν γ. In other words, This means: since for any A: σ µν = σ µν if µ,ν = 1,2,3 σ µν if µ = or ν = S 1 Λ = γ SΛ γ, and SΛ = γ S 1 Λγ, e γ Aγ = n γ Aγ n, but n! γ Aγ n = γ A γ γ Aγ... = γ A n γ, }{{} =1 e γ Aγ = γ e A γ. 8.4 Probability current Recall the probability current: j µ x = ψ ψ,ψ α k ψ = ψ ψ,ψ γ γ k ψ = ψγ ψ,ψγ k ψ = ψγ µ ψ, 16

17 remember γ k = βα k, β = γ, ψ = ψ γ. So: j µ x = ψx γ µ ψ x = ψ x γ γ µ ψ x = ψ x SΛ }{{} γ γ µ SΛψx =γ S 1 Λ = ψ xγ S 1 Λγ µ SΛ ψx }{{}}{{} =ψx =Λ µ νγ ν So: j µ x = Λ µ ν j ν x, i.e. j µ is a Lorentz vector field. 8.5 General bilinear form What about a general bilinear form ψxaψx with an arbitrary 4 4 matrix A? Observation: ψ x = SΛψx ψ x = ψxs 1 Λ ψ x Aψ x ψxs 1 Λ A SΛψx. Any 4 4 matrix A can be expanded in a basis set of 16 matrices; one particular choice is provided by the γ matrices. Define: γ 5 = γ 5 = iγ γ 1 γ 2 γ 3. This has properties {γ 5,γ µ } =, γ 5 2 = 1 and γ 5 = γ 5. This allows us to define 16 matrices: These have properties: Γ a 2 = ±1, for a = 1,...,16. Γ S = 1 1 Γ µ V = γ µ 4 Γ µ ν T = σ µν = i 2 [γ µ,γ ν ] 6 Γ µ A = γ 5 γ µ 4 Γ P = γ 5. 1 For any Γ a Γ S there is a Γ b such that {Γ a,γ b } =. trγ a = for Γ a Γ S, since: tr[γ a Γ b Γ b ] = tr[γ b Γ a Γ b }{{} = Γ b Γ a ] = tr[γ a Γ b Γ b ]. 17

18 For any Γ a, Γ b, a b, there is a Γ c Γ S such that Γ a Γ b = ηγ c with η = ±1,±i. For example, if Γ a = γ 5 γ 1 = Γ A 1, and γb = σ 1 = Γ T 1, Γa Γ b = iγ 5 γ = iγ A. {Γ a } are linearly independent: 16 a=1 λ a Γ a = λ a = a. This means that any 4 4 matrix A can be written as a linear combination of Γ a s. How do ψxγ a ψx, a = 1,...,16 transform? Γ S : ψ x Γ S ψ x = ψxs 1 ΛSΛψx = ψxψx, Γ S represent a scalar field. For Γ V µ, a vector field, see j µ in Sec Γ P : ψ x γ 5 ψ x = ψxs 1 Λγ 5 SΛψx. For a proper Lorentz transformation, S = e i 4 σ µνω µν, γ 5 σ µν = σ µν γ 5 γ 5 SΛ = SΛγ 5 So, ψ x γ 5 ψ x = ψxs 1 ΛSΛγ 5 ψ x = ψxγ 5 ψ x, just like a scalar field! In general, ψ x γ 5 ψ x = detλψxγ 5 ψx. Γ P represents a pseudoscalar field. ψ x γ 5 γ µ ψ x = detλλ µ νψxγ 5 γ ν ψx. Γ A represents an axial vector field. ψ x σ µν ψ x = Λ µ ρλ ν σσ ρσ ψx. Γ T represents an antisymmetric tensor field. 8.6 Parity symmetry We want to find SP, where P µ ν = diag1, 1, 1, 1, the parity transformation. Since detp = 1, P is not a proper Lorentz transformation. We know that: S 1 Pγ µ SP = P µ νγ ν. Take µ = : S 1 Pγ SP = γ γ SP = SPγ. 18

19 µ = k: S 1 Pγ k SP = γ k γ k SP = SPγ k. These two conditions give: SP = e iφ γ, where φ is an arbitrary phase factor. For convenience we set φ =. We can now see how γ 5 transforms under P: ψ x γ 5 ψ x = ψxs 1 Pγ 5 SPψx = ψxγ γ 5 γ ψx = ψxγ γ γ 5 ψx = ψxγ γ γ 5 ψx = ψxγ 5 ψx Here a factor of 1 appears. In general the factor is detp. 9 Lagrangian density for free fields All field equations e.g. Maxwell, Klein Gordon, Dirac, Yang Mills, etc. can be obtained from a minimum action principle, where the action is: S = d 4 x L L is the Lagrangian density, commonly called simply the Lagrangian. We make the assumption that L depends only on ϕx and µ ϕx, and not on any higher derivatives. ϕx can be any field component, e.g. A x, ψ 1 x, etc. We vary the field: ϕx ϕx + ϵδϕx, and require that δs[ϕ] =. δs[ϕ] def = d dϵ S[ ϕ + ϵδϕ ] ϵ= = d d 4 xl ϕx + ϵδϕx, µ ϕx + ϵ µ δϕx dϵ [ ] L = d 4 x ϕx δϕx + L µ ϕx µδϕx [ ] L = d 4 x ϕx L µ δϕx µ ϕx δs[ϕ] = L ϕx L µ µ ϕx =, the field equation, or equation of motion for the field ϕ. 19

20 9.1 Real scalar field For a real scalar field ϕx, L x = 1 2 µ ϕx µ ϕx 1 2 m2 ϕx 2 L ϕx = L m2 ϕx, µ µ ϕx = µ µ ϕx So the equation of motion reads: m 2 µ µ ϕx =, or + m 2 ϕx =, i.e. the Klein Gordon equation. 9.2 Complex scalar field For a complex scalar field Φx, L x = µ Φ x µ Φx m 2 Φ xφx L Φx = m2 Φ L x, µ µ Φx = µ µ Φ x = Φ x So the equation of motion reads +m 2 Φ x =. Similarly, +m 2 Φx =. Since Φ and Φ can be treated as independent, one complex scalar field is equivalent to two real scalar fields. Φ can be defined in terms of real scalar fields φ 1 and φ 2 : This gives the properties: Φx = 1 2 φ1 x + iφ 2 x Φ x = 1 2 φ1 x iφ 2 x µ Φ µ Φ = 1 2 µφ 1 µ φ µφ 2 µ φ 2 Φ Φ = 1 2 φ φ Spin- 1 2 field So for each α,d L x = ψxi / mψx 4 = ψ α xi / αβ mδ αβψ β x α,β=1 L 4 ψ α x = i / β mδ αβψ β x β=1 = i / mψx. 2

21 L µ ψ α x = These give the equation of motion: i / mψx =. In addition, L ψ β x = mψ β x, µ L µ ψ β x = µ 4 ψ α xiγ µ αβ α=1 = µ ψxiγ µ β = ψxiγ µ β µ = ψxi / β These give the equation of motion: ψxi / + m =. 9.4 Abelian vector field where F µν x = µ A ν x ν A µ x. L = 1 4 F µνf µν + j µ A µ, L A ν x = j ν L x, ρ ρ A σ x = ρf ρσ So the equation of motion is ρ F ρσ = j σ. 9.5 Summary The Lagrangian density must behave like a scalar field under proper Lorentz transformations; d 4 x = detλ d 4 x = d 4 x; so the action S = d 4 x L x is invariant under Λ L +. The equations of motion obtained from δs = are covariant; they take the same form for all observers. For the standard model, we need the Lagrangians of spin- bosons scalar fields, real or complex, spin- 1 2 fermions Dirac eqn., spin-1 bosons abelian vector fields and non-abelian vector fields W ±, Z, gluons 21

22 1 More on the Dirac equation 1.1 Free-particle solutions This has free particle solutions given by: i / mψx =. i px ψx = upe where up is a 4-component fixed spinor. Since µ e i p = i p µ e i px, we have: = i / mupe i px = p / i px mupe p / mup =. To distinguish positive and negative energy solutions, go back to H = α p + βm, with p = i. where E is the energy eigenvalue. When p = this gives: H up = α p + βmup = E up m H u = βmu = m m u = Eu. m This has two solutions with E = m and two with E = m. When p, H u = m σ p σ p m u A u B = E u A u B Note that this is a 4 4 matrix multiplying a 4-component column vector; u A and u B are 2-component spinors. σ pu B = E mu A σ pu A = E + mu B 1.2 Constructing a basis We can construct a basis for solutions with E > : u s A = χs ; χ 1 1 =,χ 2 = 1 u s B = σ p E + m χs 22

23 So for E >, and s = 1,2: For solutions with E < : So for E <, and s = 1,2: u s = N χ s σ p E+m χs u s B = χs u s A = σ p E m χs = σ p E + m χs u s+2 = N σ p E +m χs χ s Solutions are orthogonal: u r u s δ r s 1.3 Helicity Within each pair of solutions, one distinguishes spin up vs. down by the helicity operator: where ˆp = p p. 1 2 Σ ˆp = 1 2 σ ˆp σ ˆp Since [ Σ p, H ] =, helicity is a good quantum number. 1 2 Σ ˆp 2 = 1 4 σ ˆpσ ˆp σ ˆpσ ˆp = So 1 2 Σ ˆp has eigenvalues ± 1 2, the possible helicities of the particle. 1.4 Interpretation of negative energy solutions So far, we have attempted to obtain a relativistic wave equation with 1-particle wavefunctions and a probability interpretation as in non-relatvistic quantum mechanics. With the Klein Gordon equation we defined the probability current: j µ = iϕ µ ϕ ϕ µ ϕ = ρ,j, and the free-particle solution ϕ = Ne i px gave j µ = 2p µ N 2 ρ E. For E < this gave ρ < and so the Klein Gordon equation was abandoned. With the Dirac equation, j µ = ψγ µ ψ = ρ,j, so ρ = ψγ ψ = ψ γ γ ψ = ψ ψ. There are no more negative probability densities, but there are still negative energy solutions! 23

24 Energy E m Dirac s solution was to use the Pauli exclusion principle and assume that all negative energy states are filled with electrons; the Dirac sea. When an electron with negative energy E is excited to a state with energy E >, the result is: an electron with charge e, E > m the absence of an electron with charge e and energy E <. E Interpret the latter as the presence of an antiparticle with charge +e and energy E >. Figure 1: Pair production The net result is hence the creation of a pair e E +e + E with E + E > 2m. Dirac predicted the existence of the positron in this way in 1932; it is an acceptable if outdated theory of spin- 1 2 particles. In 1934 Pauli and Weisskopf revived the Klein Gordon equation by reinterpreting the probability current as charge current density: j µ = ieϕ µ ϕ ϕ µ ϕ, so having ρ < is not a problem. 11 Feynman Stueckelberg interpretation 11.1 Motivation Dirac s interpretation of negative energies works only for fermions. What about bosons? The Feynman Stueckelberg interpretation treats a negative energy solution propagating backwards in time as identical to a positive energy solution going forwards in time Example Consider the scattering of a π + particle and π antiparticle by a potential V : Both time orderings seen in Fig. 2 must be taken into account. In each case the initial and final situation is identical. The second case, shown in Fig. 2b, can be interpreted as shown in Fig. 3, as the creation of a π + π pair by one potential, and the absorption of a π + π pair by the other potential. 24

25 t t t 2 V t 2 t 1 V t 1 space a b Figure 2: π + scattering t t 2 annihilation t 1 creation Figure 3: The Feynman Stueckelberg interpretation of Fig. 2b 11.3 Currents Consider the electromagnetic current for a positive-energy π + : j emπ µ + = +e prob. current for π + = 2 N 2 p µ = 2 N 2 E,p The current for a π with negative energy is: j emπ µ = e 2 N 2 p µ = 2e N 2 E,p = e 2 N 2 E, p = j emπ µ + p µ p µ Consider a system A with p µ A and charge q A and the emission of a π with E >. This can be represented in either of two ways shown in Fig. 4. p A and q A change as: p µ A pµ A pµ π = p µ A + pµ π. q A q A e = q A + e. The emission absorption of an antiparticle of 4-momentum p µ is physically equivalent to the absorption emission of a particle with 4-momentum p µ. 25

26 t t, E >, E < A A a b Figure 4: Two representations of the system A 12 Electrodynamics of spin- particles We consider π and K particles as elementary particles and interacting only electromagnetically although this is highly unrealistic. Recall time-dependent non-relativistic scattering theory, and the Schrödinger equation for a free particle: H ϕ n = E n ϕ n, where {ϕ,ϕ 1,...} form an orthonormal basis: ϕ n,ϕ m = d 3 x ϕ n xϕ mx = δ mn. Our aim is to solve the Schrödinger equation for a particle in the potential V x, t: Expand: and insert into 18: H +V x, tψ = i ψ t. 18 ψx, t = a n tϕ n xe ie n t n= a n tv x, tϕ n xe ie n t = i da n t ϕ n xe ie n t n n dt Multiply this by ϕ x and integrate. f d 3 x a n tϕ f xv x, tϕ nxe ie n t = i da f t n dt da f t = i dt n e ie f t a n t d 3 xϕ f xv x, tϕ nxe ie f E n t Assume at t = T 2 the system is in state i: a i T 2 = 1, an T 2 = n i 26

27 da f T dt 2 = i d 3 xϕ f xv ϕ i xe ie f E i T /2 19 Assume that the potential is small, so there is no perturbation of the initial state up to a th -order approximation. Integrate 19: a f t = i t T /2 dt d 3 xϕ f xv ϕ i xe ie f E i t At t = T /2, after the interaction, T f i = a T f 2 T /2 = i dt d 3 x [ϕ ] f xe ie f t V x, t [ϕ ] i xe ie i t, }{{}}{{} T /2 ϕ f x ϕ i x where x t,x. So a covariant form for T f i is: T f i = i d 4 xϕ f xv ϕ i x Interpretation of T f i Consider a time-independent V x. with: T f i = iv f i V f i = T /2 T /2 dt e ie f E i t, d 3 xϕ f xv xϕ i x. Look at the limit as T : T /2 T /2 dt e ie f E i t 2πδE f E i. So as T, T f i 2 δ 2 E f E i. This is not well-defined! But we can define: 1 W = lim T T T f i 2 = Vf i 2 lim 1 T T = Vf i 2 lim T T /2 T /2 T /2 T /2 dt e ie f E i t = Vf i 2 2πδE f E i lim 1 T T dt e ie f E i t lim 1 T T T /2 T /2 2 T /2 T /2 dt e ie f E i t dt e ie f E i t 27

28 Now because of the delta function, we can set E f = E i in the integral, since δe f E i = when E f E i. The integral simply reduces to T, giving: W = Vf i 2 2πδE f E i Fermi s golden rule We have an initial state specified and a set of final states. Let ρe f de f represent the number of final states with energies between E f and E f + de f. W f i = 2π de f ρe f V f i 2 δe i E f W f i = 2πρE i V f i 2. This is known as Fermi s golden rule Determining T f i Use the Klein Gordon equation µ µ + m 2 ϕ = and couple to an e.m. potential A µ = A,A through the substitution p µ p µ + e A µ the particle has charge e. In other words i µ i µ + e A µ. So: [ µ ie A µ µ ie A µ + m 2] ϕ = gives: where: µ µ + m 2 ϕ = V ϕ, V = ie µ A µ + A µ µ e 2 A µ A µ. Note that V is an operator acting on ϕ, and µ A µ ϕ = µ A µ ϕ + A µ µ ϕ. Since α = e2 4π = 1 137, organise interactions in powers of e. Neglect the e2 A µ A µ term. T f i = i = e ϕ f xv xϕ i xd 4 x ϕ f xaµ µ + µ A µ ϕ i xd 4 x This can be integrated by parts: ϕ f µa µ ϕ i d 4 x = µ ϕ f Aµ ϕ i d 4 x. 28

29 12.4 Currents Now we can define j such that: T f i = i j f i µ xa µ xd 4 x. This requires: j f i µ x = ie ϕ f µϕ i µ ϕ f ϕ i, the electromagnetic transition current for i f. The free particle solutions are: ϕ i x = N i e i p i x ϕ f x = N f e i p f x, which gives: j f i µ x = en i N f p i + p f e ip f p i x To determine A µ we solve the Maxwell equation: µ F µν x = j ν x, where j ν x is the electromagnetic current produced by a K : j ν x = ie ϕ 4 ν ϕ 2 ν ϕ 4 ϕ 2 = en 2 N 4 p 2 + p 4 ν e ip 4 p 2 x Using the Lorenz gauge condition µ A µ =,we can write: So the Maxwell equation reads simply: µ F µν = µ µ A ν ν A µ = A ν ν µ A µ = A ν. A ν x = j ν x. Using the fact that e i px = p 2 e i px, so this is easily solved: A ν 1 x = p 4 p 2 2 en 2N4 p 2 + p 4 ν e ip 4 p 2 x 1 = p 4 p 2 2 j ν x = 1 q 2 j ν x, where q p 4 p 2, the 4-momentum transfer. Now, T f i = i [ ] j f i µ x 1 j µ K ; x d 4 x q 2 29

30 Label i 1 and f 3, giving: T = i jµ 31 1 x j µ q 2 42 xd4 x = N 1 N3 N 2N4 d 4 x iep 1 + p 3 1 ep q p 4 µ e ip 3 p 1 x e ip 4 p 2 x Recall that d 4 x e i px = 2π 4 δ 4 p. So: T = 2π 4 δ 4 p 1 + p 2 p 3 p 4 N 1 N3 N 2N4 im, 2 where M is the invariant amplitude, defined such that: im = [ iep 1 + p 3 µ][ i g ] µν [iep2 q 2 + p 4 ν] The second term here is the photon propagator. The process can be described using a Feynman diagram as in Fig. 5. The arrows on lines follow the flow of electric charge. t p 3 p 4, q, p 1 K, p 2 Figure 5: Feynman diagram for π -K interaction 12.5 Observations 4-momentum is conserved at each vertex. At vertex 1, p 1 + p 3 + q = p 1 p 3 = q At vertex 2, p 2 + p 4 + q = p 4 p 2 = q Combining these shows that p 1 p 3 = p 4 p 2, i.e. p 1 + p 2 = p 3 + p 4, which is enforced by the δ in 2. p Each vertex ϕϕγ gets a factor: p photon propagator in the Lorenz gauge. µ = iep + p µ, µ q ν = g µν q 2, the 3

31 13 Scattering 13.1 From scattering amplitude to cross-section T = 2π 4 δ 4 p 1 + p 2 p 3 p 4 N 1 N 2 N 3N 4 im This is the transition rate per unit volume per unit time. The normalisation condition that we will use for the probability density ρ = 2E N 2 is: ρ dv = 2E, i.e. there are 2E particles per volume V. This gives: allowing us to write: V N = N = 1 V, W = T T V = 1 V 4 2π4 δ 4 p 1 + p 2 p 3 p 4 M The δ 4 2 which appears in expanding 21 is dealt with exactly as in section Cross-section = W 12 3,4 no. of final states, 22 initial flux so we need to find the number of final states and the initial flux. There are 2E particles per volume V ; in a finite V the number of states with momentum in p,...,p + d 3 p is: V 2π 3 2E d 3 p. So the total number of available final states is: V d 3 p 3 2π 3 2E 3 V d 3 p 4 2π 3 2E To find the initial flux we use a frame where particle 2 is at rest. The number of beam particles passing through unit area per unit time is: where V 1 is the velocity and 2E 1 V per unit volume is: So the initial flux is: V 1 2E 1 V, is the particle density. The number of target particles 2E 2 V. F V 2 = V 1 2E 1 2E 2 V V 31 24

32 So using 23 and 24 in 22 gives: dσ = 1 4E 1 E 2 V 1 2π4 δ 4 p 1 + p 2 p 3 p 4 M 2 d 3 p 3 2π 3 2E 3 d 3 p 4 2π 3 2E [dσ] = area; dσ is the effective area over which particles 1 and 2 interact to produce particles 3 and 4, whose momenta lie in p 3,...,p 3 +d 3 p 3 and p 4,...,p 4 +d 3 p 4 respectively. 25 can be written as: dσ = 1 F M 2 dlips 26 where d Lips is the Lorentz invariant phase space factor: dlips = 2π 4 δ 4 d 3 p 3 d 3 p 4 p 1 + p 2 p 3 p 4 2π 3 2E 3 2π E 4 F is defined in 24 as 4E 1 E 2 V 1. This is not manifestly Lorentz invariant, but by noting that p 1 = γ 1 m 1 V 1, and γ 1 = E 1 m 1, we can write: an explicitly Lorentz invariant form. F = 4 [ p 1 p 2 2 m 2 1 m2 2] 1/2 28 Are the terms d 3 p in dlips Lorentz invariant? Write: 2π 3 2E d 4 p 2π 4 2πδp2 m 2 θp f p d 3 p dp = 2π 3 2π δ p 2 E 2 θp f p where θ is the Heaviside step function. The δ term can be expanded as: 1 [ δp E + δp + E ] 2p = 1 2p [ δp E ], 29 since we define E = p 2 + m 2. Hence 29 can be written as d 3 p 2π 3 2E f E,p E= p 2 +m 2 and so the terms are Lorentz invariant Elastic scattering 1,2 3,4 We will focus on the centre-of-mass system CMS; see Fig. 6. Since we are in the CMS, p 1 + p 2 =, so let p 1 = p 2 = p, similarly p 3 = p 4 = p. So we can write the 4-momenta: p µ 1 = E 1,p, p µ 2 = E 2, p, p µ 3 = E 3,p, p µ 4 = E 4, p. 32

33 p 3 CM p 1 p 2 p 4 Figure 6: Elastic scattering in CMS Since we are dealing with elastic scattering, the initial and final states contain the same particles, so p = p p. Hence: E 2 1 = p2 + m 2 1, E 2 2 = p2 + m 2 2, E 2 3 = p2 + m 2 3 = p2 + m 2 1, E 2 4 = p2 + m 2 4 = p2 + m 2 2. The expression for dlips in 27 contains p 4 explicitly. This is eliminated by integrating so that only the -component of the δ 4 remains: d 3 p 4 E 4 δ 4 p 3 + p 4 p 1 p 2 = 1 E 4 δe 3 + E 4 E 1 E 2, 3 where E 4 on the right-hand side is determined by E 4 = p m2 4 and p 4 = p 3 +p 1 +p 2, i.e. it is no longer independent. We can decompose d 3 p 3 = p 3 2 d p 3 dω. Since E3 2 = p2 3 + m2 3, this gives: So using this and 3, we get: E 3 de 3 = p 3 d p 3. dlips = 1 4π 2 dω p 3 E 3 de 3 E 4 δe 3 + E 4 E 1 E 2 31 where dlips now refers to the original dlips integrated over p 4. Now, bearing in mind that in the CMS E 3 de 3 = p dp = E 4 de 4, define: w E 3 + E 4 dw = de 3 + de 4 = 1 E E 4 p dp = E 3 + E 4 E 3 E 4 E 3 de 3 = w E 4 de 3 33

34 This gives, defining w E 1 + E 2 : p de 3 δ E 3 + E 4 E 1 + E 2 = dw E 4 p E 4 w δw w E 4 = p w Mandelstam variables At this stage it is convenient to introduce Mandelstam variables: s p 1 + p 2 2 t p 1 p 3 2 = q 2 u p 1 p 4 2 Note that s = p 1 + p 2 2 = E 1 + E 2 2 = w 2 in the CMS, so w = s. So by using 32 in 31 we can write the convenient expression: dlips = 1 4π 2 dω p s, where again the integration has been absorbed into the definition of d Lips Flux factor In 28 we defined: Using the fact that: F = 4 [ p 1 p 2 2 m 2 1 m2 2] 1/2. p 1 p 2 = E 1 E 2 p 1 p 2 = E 1 E 2 + p 2, we can expand: F 2 = 16 [ E1 2 E p2 E 1 E 2 + p 4 m1 2 ] m2 2 = 16 [ p 2 + m 2 1 p2 + m p2 E 1 E 2 m 2 1 m2 2 + p4] = 16 [ 2p 4 + p 2 m m p2 E 1 E 2 ] = 16 [ p 2 p 2 + m1 2 + p 2 + m p 2 ] E 1 E 2 }{{}}{{} =E1 2 =E2 2 = 16p 2 E 1 + E 2 2 = 16p 2 s = F = 4p s. 34

35 13.5 Putting things together Now we can rewrite 26: dσ = 1 F M 2 dlips = M π 2 p s 1 4p s dω This gives: dσ dω = 1 CM 64π 2 s M 2 We can use t = q 2 as a variable, so that dσ dt dω can be expanded: t = q 2 = p 1 p 3 2 is independent of reference frame: = p p2 3 2E 1E 3 2p 1 p 3 = 2m 2 1 2p2 + m 2 + 2p 2 cosθ CM = 2p 2 1 cosθ CM dt = 2p 2 dcosθ CM dω = dϕdcosθ CM = dϕ dt 2p 2. Since these are spinless particles there is no ϕ dependence: dσ = 1 64π 2 s M 2 dϕ dt 2p 2 = 1 dt M 2 64πs p 2 Use F 2 = 16p 2 s: a Lorentz invariant form! dσ dt = 1 64π 1 p 1 p 2 2 m 2 1 m2 2 M 2, 14 External spin-1 particles 14.1 Massive particles For example, vector bosons, W ±, Z, etc. For a massive spin-1 particle, M so we can go to the rest frame. A spin-1 particle has 3 polarisation states, one particular basis is: 1 ϵ 1 =, ϵ 2 = 1, ϵ 3 = 1 35

36 Or, we can single out the z-axis to obtain a more common basis: ϵλ = 1 = 1 1 i, ϵλ = 1 = 1 1 i, ϵλ = = These represent left circular polarisation around z, right circular polarisation and longitudinal polarisation along z respectively. This basis is orthonormal: ϵλ ϵλ = δ λλ for λ = 1,,1. To construct the corresponding 4-vectors ϵ µ λ, set ϵ λ = in the rest frame, and define ϵ µ λ = ϵ λ,ϵλ. In the rest frame, 4-momentum q µ = M, so q µ ϵ µ λ =. For a boost in the z-direction, so the ϵ µ transform as: using γ = E M, β = p p E, γβ = M. The 4-momentum transforms as: x µ = Λ µ νx ν x = γx + βx 3 x 3 = γx 3 + βx x 1 = x 1, x 2 = x 2, ϵ µ λ = ±1 = ϵ µ λ = ±1 ϵ µ λ = = γβ,,,γ = 1 M p,,,e, 33 q µ = γm,,,γβm = E,,, p and 34 give the relation: q µ ϵ µ =. The completeness relation for {ϵ µ } is: λ=,±1 ϵ µ p,λϵ ν p,λ = g µν + pµ p ν M 2, and is needed in calculating M 2 with external spin-1 particles Massless particles For example, γ, etc. The Maxwell equation is: µ F µν = A ν ν µ A µ =. 36

37 15 Gauge invariance and the gauge principle 15.1 Invariance of Lagrangian The electron-photon Lagrangian is: L x = ψxi / + e A / x mψx 1 4 F µνxf µν x Just like in QM, this L is invariant under the combined transformations: ψx ψ x = e iωx ψx ψx ψ x = e iωx ψx A µ x A µ x = A µx + 1 e µωx This is easily verified. For the first term: ψ xi / + e A / ψ x = ψxe iωx γ µ [ i µ + e A µ x + µ ωx ] e iωx [ ψx = ψxe iωx γ µ µ ωxe iωx + e iωx i µ + e A µ xe iωx + + µ ωxe iωx] e iωx ψx = ψxγ µ i µ + e A µ xψx. For the second term: F µν x = µa ν x νa µ x = µ A ν x ν A µ x + 1 e µ ν ωx ν µ ωx = µ A ν x ν A µ x = F µν x. e iωx U1, the unit circle in the complex plane, so this is called a U1 gauge theory Inverting the logic Now we start with the free Lagrangian density: L x = ψxi / mψx which is invariant under ψ e iω ψ, ψ e iω ψ with constant ω a global U1 invariance. Now promote this global invariance to a local one by replacing ω ωx. The free L is not invariant: ψxi / mψx ψxi / / ωx mψx. 37

38 This is because, for ω = ωx, in general: µ ψ e iω µ ψ So introduce the covariant derivative: D µ = µ + i q A µ x, and demand that: What does this imply? if ψx e iωx ψx, D µ ψx e iωx D µ ψx. 35 µ + i q A µ xψx µ + i q A µ xψ x i µ ωx + i q A µ x! = i q A µ x So 35 determines how A µ transforms. = e iωx µ + i µ ωx + i q A µ x ψx,! = e iωx µ + i q A µ x ψx. A µ x = A µx 1 q µωx Conclusion We may convert the global U1 symmetry of the free L x to a local one by introducing a gauge field A µ x with interacts with the fermion. The same approach can be used for complex scalar fields An observation D µ D ν ψx = µ + i q A µ x ν + i q A ν xψx = [ µ ν + i q µ A ν + i q A ν µ + i q A µ ν q 2 ] A µ A ν ψx. Note the symmetric terms. Taking the commutator gives: [D µ D ν ]ψx = i q µ A ν µ ψx = i qf µν ψx. Since there is now no acting on ψ, this can be taken as a definition of F µν : F µν = [D µ,d ν ] Non-abelian symmetries Suppose ϕ has 2 components, e.g. a proton and a neutron: ψ = ψ n ψ p, 38

39 where ψ n and ψ p are Dirac spinors. The free Lagrangian is: L = i / m n ψ n ψ p i / m p SU2 is the non-abelian group of complex 2 2 matrices U with U U = 1, detu = 1. When m n = m p = m, L x has a global SU2 invariance: ψx Uψx ψx ψxu. L x U 11 U 12 i / m ψ n ψ p U 21 U 22 i / U 11 U 12 m U 21 U 22 Or, in a more compact notation, L x ψu i / muψ = ψu Ui / mψ = ψi / mψ = L x. ψ n ψ p ψ n ψ p Now we render U local: U Ux. This gives: µ ψx µ Uxψx = µ Uxψx +Ux µ ψx So, using the gauge covariant derivative: µ D µ = µ + i g A µ x, D µ ψx UxD µ ψx ] µ + i g A µ xψx [ µ Ux +Ux µ + i g A µ x ψx! = Ux µ + i g A µ xψx = A µ x = UxA µxu 1 x + 1 i g Ux µu 1 x. Note: µ U U 1 = µ 1 =, so we can use µ UU 1 +U µ U 1 =. Suppose A µ x =. Then: Any U SU2 can be written: A µ x = 1 i g Ux µu 1 x. 36 Ux = exp 1 + i [ i 3 a=1 3 a=1 ] α a x τa 2 α a x τa 2 as α 39

40 Similarly: U 1 x 1 i 3 a=1 α a x τa 2. Here α a x are real numbers, and τ a are the Pauli matrices a basis for SU2. NB: [ µ exp i a because the τ a do not commute. ] α a x τa i 2 a [ µ α a τa exp i 2 a ] α a x τa 2 But we can use: Now look at A µ in 36. µ e M = A µ x = 1 i g = 1 i g 1 ds e sm µ Me 1 sm 1 + iα a τa µ 1 iα b τb 2 i µ α b x τb 2 = 1 g µα a x τa 2, 2 up to Oα 2. So A µ is Hermitian and traceless. In fact, this can be obtained without using an explicit representation for U: A = 1 i g e M µ e M = 1 i g 1 A = 1 i g = 1 i g 1 ds e s 1M µ Me 1 sm 1 ds e 1 sm µ M e s 1M ds e s 1M µ Me 1 sm = A. tra = 1 i g 1 ds tr e s 1M µ Me 1 sm = 1 i g tr µm =. An SU2 gauge field A µ x is a traceless Hermitian matrix field. This means that for each x, A µ x su2, the Lie algebra of SU2. 4

41 A Lie algebra is a linear vector space combined with a Lie bracket a commutator. A basis for this Lie algebra is T a = τa 2, a = 1,2,3. The scalar product T a,t b = 2trT a T b = δ ab. The Lie bracket [T a,t b ] = i f abc T c, where f abc are real constants called the structure constants. We know that: [ ] τ a 2, τb abc τc = iϵ 2 2, so for su2, f abc = ϵ abc. 41