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F34TPP Theoretical Particle Physics notes by Paul Saffin Contents 1 Lecture one 1.1 Natural units. . . . . . . . . . 1.2 converting back to SI units . . 1.3 relativistic notation . . . . . . 1.3.1 raising/lowering indices 1.4 examples . . . . . . . . . . . . 1.4.1 energy-mass relation . 1.4.2 quantum operators . . 1.4.3 Klein-Gordon equation . . . . . . . . 5 5 6 7 7 7 7 8 8 2 Lecture two - Dirac equation 2.1 Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 solving the Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 9 11 3 Lecture three - spin, chirality 3.1 rotation generators . . . . . 3.2 rotating a fermion . . . . . . 3.3 chirality and helicity . . . . 3.4 Weyl fermions . . . . . . . . . . . . 12 13 14 15 16 . . . . 17 17 18 19 21 5 Lecture five - Dirac equation and magnetic fields 5.1 magnetic moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 magnetic moment of electron - Zeeman interaction . . . . . . . . . . . . . . . . . 21 21 22 6 Lecture six - Feynman rules 6.1 propogator theory . . . . . . . . . . . . 6.1.1 Schrödinger propogator . . . . . 6.1.2 Klein-Gordon propogator . . . . 6.2 Feynman rules - including interactions 6.3 diagramatic expansion . . . . . . . . . 23 23 24 24 25 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . and helicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Lecture four - gauge symmetry and electromagnetism 4.1 electromagnetism . . . . . . . . . . . . . . . . . . . . . . 4.2 gauge invariance of the Dirac equation . . . . . . . . . . 4.3 covariant derivatives . . . . . . . . . . . . . . . . . . . . 4.4 the standard model and gauge symmetries . . . . . . . . 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Lecture seven - action functionals and Feynman rules 7.1 action and equations of motion . . . . . . . . . . . . . . . . . . 7.1.1 Klein-Gordon action . . . . . . . . . . . . . . . . . . . . 7.1.2 Dirac action . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 the structure of an action . . . . . . . . . . . . . . . . . . . . . 7.2.1 example, real KG action I . . . . . . . . . . . . . . . . . 7.2.2 example, real KG action II . . . . . . . . . . . . . . . . . 7.2.3 example, complex KG action . . . . . . . . . . . . . . . . 7.3 example, Dirac particle coupled to photons . . . . . . . . . . . . 7.4 complex KG equation coupled to photons . . . . . . . . . . . . . 7.5 Feynman rules for the electroweak sector of the standard model 7.6 Fermi’s golden rule . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.1 ee scattering . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.2 µ decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 26 27 28 28 29 30 30 31 31 31 33 33 34 8 Lecture eight - group theory 8.1 definition of a group . . . . . . . . . . 8.2 direct product of groups . . . . . . . . 8.3 Abelian groups . . . . . . . . . . . . . 8.4 Lie groups (pronounced ”Lee groups”) 8.4.1 general linear group GL(n,R) . 8.5 special linear group SL(n,R) . . . . . . 8.6 orthogonal group O(n,R) . . . . . . . . 8.7 special orthogonal group SO(n,R) . . . 8.8 O(p,q,R) . . . . . . . . . . . . . . . . . 8.9 symplectic group Sp(2n,R) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 35 36 36 36 36 37 37 37 38 38 . . . . . . . . 38 38 39 39 39 40 40 40 41 10 Lecture ten - some representation theory 10.1 SU(2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 normalizing the states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 general SU(2) represenation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 42 45 46 . . . . . . . . . . 9 Lecture nine - non-Abelian gauge theory 9.1 unitary groups . . . . . . . . . . . . . . . 9.2 unitary group U(n) . . . . . . . . . . . . 9.3 special unitary groups . . . . . . . . . . 9.4 unitary groups and the standard model . 9.5 generators and Lie algebras . . . . . . . 9.5.1 o(n), the Lie algebra of O(n) . . . 9.5.2 su(n), the Lie algebra of SU(n) . 9.6 non-Abelian gauge theory . . . . . . . . 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Lecture eleven - isospin 11.1 neutrons, protons and isospin . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 pions and isospin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 using isospin in scattering calculations . . . . . . . . . . . . . . . . . . . . . . . 46 47 48 49 12 Lecture twelve - the quark model 12.1 more multiplets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 50 13 Lecture thirteen - charge conjugation 13.1 Charge conjugation . . . . . . . . . . 13.1.1 photons . . . . . . . . . . . . 13.1.2 fermion-antifermion pairs . . . 13.1.3 C violation . . . . . . . . . . 13.2 parity . . . . . . . . . . . . . . . . . 13.2.1 electromagnetism . . . . . . . 13.2.2 fermions . . . . . . . . . . . . 13.3 linearity of parity . . . . . . . . . . . 13.4 electron dipole moment (edm) . . . . . . . . . . . . . 54 54 55 55 56 56 56 57 58 58 14 Lecture fourteen - parity violation and time reversal 14.1 parity violation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 time reversal symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 59 60 15 Lecture fifteen - CP symmetry 15.1 CP symmetry . . . . . . . . . 15.2 neutral kaons . . . . . . . . . 15.3 kaons and C, P and CP . . . 15.4 Klong and Kshort decays . . . . 15.5 matter vs anti-matter . . . . . . . . . . 61 62 62 62 63 65 16 Lecture sixteen - quark mixing 16.1 three generations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 quark mixing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 65 66 17 Lecture seventeen - neutrino oscillations 17.1 Solar neutrinos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.2 neutrino oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.3 atmospheric neutrinos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 68 69 71 18 Lecture eighteen - Higgs mechanism and 18.1 symmetry restoration/breaking . . . . . 18.2 superconductivity . . . . . . . . . . . . . 18.3 massive fields . . . . . . . . . . . . . . . 18.4 breaking the symmetry . . . . . . . . . . 18.5 non-Abelian Higgs mechanism . . . . . . 18.6 Yukawa coupling and fermion masses . . 71 71 72 73 73 75 75 . . . . . . . . . . . . . . . . . . . . and parity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . breaking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Lecture nineteen - the real standard model 19.1 electroweak sector, part I . . . . . . . . . . . 19.2 chiral theories . . . . . . . . . . . . . . . . . 19.3 electroweak sector, part II . . . . . . . . . . 19.4 electroweak sector, part III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 76 76 77 77 List of Figures 1 2 3 4 The predicted flux of solar neutrinos, Bahcall et. al. ApJ, 621, L85 (2005). . . . The typical shower of particles produced by a cosmic ray. . . . . . . . . . . . . . This shows a superconductor hovering above a magnet. . . . . . . . . . . . . . . A typical wine bottle, where the dregs of the wine tend to form at the bottom, around the circular minimum. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 71 72 74 List of Tables 1 Table showing quark charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 54 F34TPP Theoretical Particle Physics 1 Lecture one Aim: To introduce the notion of natural units, dimensional analysis, and revise some relativistic notation. Learning outcomes: At the end of this lecture you should • be able to convert quantities between different systems of units • know how to use Einstein summation convention 1.1 Natural units. The first thing to note is that the numerical value of a dimensionful quantity does not have any intrinsic value. For example, if you were told that the speed of light was 7.9×1014 , that would be meaningless; the reason for this is that the numerical value depends on the system of measuring rods and clocks you use. If you use metres and seconds you would get c = 3 × 108 ms−1 , if you use miles and hours you get c = 6.7 × 108 miles hour−1 , and if you use cubits and fortnights you get c = 7.9 × 1014 cubits f ortnight−1 . Similarly, if we use kilograms, pounds or grains to measure measure mass we find that ~ = 10−34 m2 kg s−1 , 3 × 10−37 miles2 lb hour−1 , 9 × 10−24 cubits2 grain f ortnight−1 . So, by picking our units appropriately, we may essentially get any value we wish for dimensionful quantities. As we are interested here in relativistic quantum field theory, it makes sense to choose units where c = 1, ~ = 1, (1.1) (1.2) these are called natural units. For example we may use the year as our measure of time, and the light-year as our measure of distance, then light travels one light-year per year, i.e. c = 1 in these units. In effect, we should think of c as the quantity that allows us to relate distances to times, and ~ is the quantity that relates masses to times and distances. Using these natural units we find that the basic relations E 2 = m2 c2 + p2 c2 , E = ~ω 2π p = ~ λ (1.3) (1.4) E 2 = m2 + p2 , E = ω, 2π p = . λ (1.6) (1.7) (1.5) become 5 (1.8) Now, we still need to pick our basic unit, and in particle physics the convention is to use electron-Volts, eV, to measure energy. Then, from (1.6), we have that mass is measured in eV , as is momentum. Now (1.8) shows that distances are measured in (eV )−1 , and (1.7) gives that time is measured in (eV )−1 . 1.2 converting back to SI units At some point an experimenter will want to know how big their machine should be, and rulers don’t come with eV notches on them, so we need to able to convert from our eV numbers back to SI units. The best way to see how this is done is to use an example, so consider the following. The scattering cross section, σ, of a target particle, and the scattering rate, Γ, are related by Γ = nσv (1.9) where n is the number density of particles that are being sent toward the target, and v their speed, so Γ is the number of scattering events we expect per unit time, the dimensions are as follows [σ] = [L2 ], [Γ] = L−3 L2 LT −1 = T −1 (1.10) (1.11) So, suppose we are told that the cross section associated with Compton-scattering (ei γ → e− γ), in natural units, is σ = 8π 2 α2 . 3m2e (1.12) This clearly cannot be in SI units, because if it were, the left-hand-side would be m2 , while the right-hand-side is in kg −2 . Note that the fine structure constant α ∼ 1/137 is dimensionless. So, to make sense of this expression we need to sprinkle some factors of ~ and c around, so write 8π 2 α2 a b σ = ~ c. 3m2e (1.13) where a and b are unknown constants. Note that this gives the same expression as (1.12) in natural units where ~ = 1, c = 1. Now we do some dimensional analysis [L2 ] = [M L2 T −1 ]a [LT −1 ]b [M 2 ] (1.14) For this to be consistent, we see that a = 2, b = −2, so σ = 8π 2 α2 ~2 . 3m2e c2 6 (1.15) 1.3 relativistic notation First of all, recall that the object variously called the line element/proper distance/proper time/invariant interval... is given by ds2 = −dt2 + dx2 + dy 2 + dz 2 . (1.16) Rather than writing this out each time we introduce a matrix quantity called the metric, with components ηµν , where the first index µ labels the rows, and the second index ν the columns of the matrix. In (t, x, y, z) := (x0 , x1 , x2 , x3 ) co-ordinates the matrix is diagonal, with entries ηµν = diag(−1, 1, 1, 1). (1.17) Given this, the line element is ds2 = X ηµν dxµ dxν . (1.18) µν where we have introduced the spacetime four-vector with components dxµ = (dt, dx, dy, dz). The summation convention simply says that if an index appears twice in a single expression, then that index must be summed over. So, for example, the line element becomes (1.19) ds2 = ηµν dxµ dxν . P There is no need to explicitly write the µν symbol, because both µ and ν appear twice, so we know they must be summed over. 1.3.1 raising/lowering indices Along with the metric ηµν we have the inverse metric η µν , which is just the inverse matrix of the metric. So, in (t, x, y, z) co-ordinates we have η µν = diag(−1, 1, 1, 1). (1.20) This allows us to ”raise” and ”lower” indices of spacetime vectors, for example V µ = η µν Vν Vµ = ηµν V ν . (1.21) (1.22) This can be useful as it often further simplifies certain expressions, meaning that we don’t even need to write ηµν , on top of dropping the summation symbols. 1.4 1.4.1 examples energy-mass relation recall that the energy-momentum four-vector is P µ = (E, p) 7 (1.23) i.e. the zero-component of four-momentum is just the eneryg, P 0 = E. Using Pµ = ηµν P ν we also have Pµ = (−E, p) (1.24) Now we take the energy-momentum relation E 2 = p2 + m2 (1.25) and rewrite this as −E 2 + p2 + m2 = 0 −(P 0 )2 + (P 1 )2 + (P 2 )2 + (P 3 )2 + m2 η00 P 0 P 0 + η11 P 1 P 1 + η22 P 2 P 2 + η33 P 3 P 3 + m2 ηµν P µ P ν + m2 P µ Pµ + m 2 = = = = 0 0 0 0 (1.26) (1.27) (1.28) (1.29) (1.30) As this final expression contains only spacetime indices it is manifestly a relativistic equation. 1.4.2 quantum operators From quantum mechanics we know that Ê = i∂t P̂ x = −i∂x (1.31) (1.32) where we denote ∂x = ∂ . ∂x (1.33) Using the fact that P0 = −P 0 and P1 = P 1 we see P̂0 = −i∂0 , P̂1 = −i∂1 P̂2 = −i∂2 P̂3 = −i∂3 (1.34) or, using our relativistic notation 1.4.3 P̂µ = −i∂µ (1.35) P µ Pµ + m 2 = 0 (1.36) P̂µ = −i∂µ (1.37) Klein-Gordon equation We have already seen that and 8 so we may combine them to give ∂ µ ∂µ φ − m2 φ = 0 (1.38) This is the Klein-Gordon equation. If we unravel the index structure we find η µν ∂µ ∂ν φ − m2 φ = 0 −∂0 ∂0 φ + ∂1 ∂1 φ + ∂2 ∂2 φ + ∂3 ∂3 φ − m2 φ = 0 −φ̈ + ∇2 φ − m2 φ = 0 (1.39) (1.40) (1.41) where a dot denotes time derivative φ̇ = 2 ∂φ ∂t (1.42) Lecture two - Dirac equation Aim: To introduce the Dirac equation. Learning outcomes: At the end of this lecture you should • know how to derive the Dirac equation • know what the Dirac algebra is • be able to solve the Dirac equation 2.1 Dirac equation The Schrödinger equation ~2 2 ∂ φ − V (x)ψ = 0 i∂t ψ + 2m x (2.43) treats time and space on an unequal footing, and so cannot be relativistic. The Klein-Gordon equation on the other hand is manifestly relativistic, however, it is second-order in time derivatives, unlike the first-order time derivatives of the Schr”odinger equation. Dirac wanted to construct a relativistic wave equation that was first-order in time derivatives, as well as being relativistic, so he wrote down the following γ 0 ∂0 ψ + γ 1 ∂1 ψ + γ 2 ∂2 ψ + γ 3 ∂3 ψ + mψ = 0 (2.44) which is the most general linear, homogeneous, first-order equation, where the γ are as yet unknown. This equation is written in relativistic form as γ µ ∂µ ψ + mψ = 0 (2.45) In order to find out what the γ are we use ∂µ = iP̂µ and find (iγ µ P̂µ + m)ψ = 0 9 (2.46) Given that this vanishes, we can also act on it with (iγ ν P̂ν − m), which when acting on zero gives zero, so ⇒ (iγ ν P̂ν − m)(iγ µ P̂µ + m)ψ = 0 (2.47) ⇒ (−γ ν γ µ P̂ν P̂µ − m2 )ψ = 0 (2.48) 1 µ ν [γ γ + γ ν γ µ ]P̂µ P̂ν ψ + m2 ψ = 0 2 (2.49) Now, we denote the anti-commutator with curly braces {γ µ , γ ν } := γ µ γ ν + γ ν γ µ (2.50) 1 µ ν 2 {γ , γ }P̂µ P̂ν + m ψ = 0 2 (2.51) so we find which we should compare to the relativistic energy-momentum equation η µν Pµ Pν + m2 = 0 (2.52) {γ µ , γ ν } = 2η µν . (2.53) leading to the conclusion that So, how do we use this to find out what the γ µ are? Well, let’s start with {γ 0 , γ 0 } = 2η 00 ⇒ (γ 0 )2 = −1 ⇒ γ 0 = ±i (2.54) {γ 1 , γ 1 } = 2η 11 ⇒ (γ 1 )2 = 1 ⇒ γ 1 = ±1 (2.55) {γ 0 , γ 1 } = 2η 01 ⇒ γ 0 γ 1 = 0 ⇒ ±i = 0 (2.56) Now lets try And now we go for and we run into a contradiction. The reason this has gone wrong is that we assumed the γ µ are numbers, they are not, they are matrices, and so we should really write {γ µ , γ ν } = 2η µν I. (2.57) This is known as the Dirac algebra. In fact, the smallest set of matrices that can satisfy the Dirac algebra are 4 × 4 matrices, meaning that ψ is actually a four-component object - four complex components - as opposed to the Schrödinger wave function which is a single complex function. In fact, there are infinitely many choices of matrices that satisfy (2.57). This is not a problem, and in fact is a good thing; we can think of each of these different choices as being required for the infinite variety of observers, all travelling at different velocities. For example, 10 suppose that a set γ µ solves (2.57), then so will Γµ = −γ µ , and so will Γµ = ±γ µT , and so will Γµ = ±γ µ∗ . In fact, if we take a general invertible matrix R, we have that Γµ = Rγ µ R−1 will also satisfy the Dirac algebra, e.g. {Γµ , Γν } = = = = = = Γµ Γν + Γν Γµ Rγ µ R−1 Rγ ν R−1 + Rγ ν R−1 Rγ µ R−1 Rγ µ γ ν R−1 + Rγ ν γ µ R−1 R(γ µ γ ν + γ ν γ µ )R−1 R(2η µν )R−1 2η µν (2.58) (2.59) (2.60) (2.61) (2.62) (2.63) Although each of these different representations will lead to the same physical results, there are certain choices that can make calculations easier, two common choices are the: Dirac representation iI 0 0 iσ 0 I 0 5 γ = , γ= , γ = (2.64) 0 −iI −iσ 0 I 0 where the σ are the Pauli matrices. And the Weyl representation 0 iI 0 iσ −I 0 0 5 γ = , γ= , γ = iI 0 −iσ 0 0 I (2.65) Now, you will have noticed that I sneaked in a new matrix, γ 5 . It turns out that γ 5 plays a crucial role in much of the physics of the standard model, and it is defined as follows, γ 5 = iγ 0 γ 1 γ 2 γ 3 . (2.66) And the remarkable property that it satisfies, making it such a useful object, is γ 5 γ µ = −γ µ γ 5 . (2.67) We shall have more to say about this later. 2.2 solving the Dirac equation The Dirac equation is linear, so if we are to look for wave soluitons it is natural to look at the Fourier modes; that means we look for solutions of the form UA ipµ xµ ψ=e (2.68) UB where UA,B are two-component column vectors. Given this ansatz we note that i∂t ψ = −P0 ψ = Eψ 11 (2.69) and if we use the Dirac representation we find that EUA − σ.P UB + mUA = 0 −EUB + σ.P UA + mUB = 0 (2.70) (2.71) In particular, if we focus on the case of a wave in the z-direction, P = (0, 0, P z ), we have (E + m)UA = P z σ 3 UB (E − m)UB = P z σ 3 UA (2.72) (2.73) So that we find (E 2 − m2 )UB = (E + m)(E − m)UB = P z σ 3 (E + m)UA = P z σ 3 P z σ 3 UB = (P z )2 (σ 3 )2 UB = (P z )2 UB (E 2 − m2 − P 2 )UB = 0 (2.74) which is just the relativistic energy-momentum equation, as expected. As UB is a two-component object, we have two linearly independent solutions: solution one has z P /(E + m) Pz 0 1 1 ⇒ ψ↑ = eiP.x (2.75) UB = ⇒ UA = 1 0 E+m 0 0 solution two has UB 0 −P z /(E + m) Pz 0 0 = ⇒ UA = ⇒ ψ↓ = eiP.x 1 0 E + m −1 1 (2.76) In other words, for a particle travelling in the z− direction there are two possible solutions, these are just the spin-up and spin-down posibilities, as we shall see. 3 Lecture three - spin, chirality and helicity Aim: To see where angular momentum fits into the Dirac equation Learning outcomes: At the end of this lecture you should • know what the angular momentum operators for a Dirac fermion are • know what chirality and helicity are 12 3.1 rotation generators By now you will have come across the phrase ”a fermion is spin-half” a number of times over the course of your degree, here we shall see why. To do this we need to understand how to rotate objects, in particular fermions. The key is the angular momentum algebra, as it is the angular momentum operators that are resonsible for rotating objects, h i (3.77) L̂x , L̂y , = iL̂z This is typically derived using Lz = xPy − yPx → L̂z = −ix∂y + iy∂x (3.78) relevant for orbital angular momentum, but in fact the L̂ appearing in (3.77) can take a number of different forms. To see this, let’s consider how we rotate a three-vector V , V0 =R V (3.79) where R is a matrix satisfying RRT = I. For example, a is accomplished by cos θ − sin θ Rz = sin θ cos θ 0 0 rotation about the z-axis with angle θ 0 0 , 1 whereas a rotation about the x and y axes come from 1 0 0 cos θ 0 − sin θ , 1 0 Rx = 0 cos θ − sin θ , Ry = 0 0 sin θ cos θ sin θ 0 cos θ (3.80) (3.81) Now we know how to rotate vectors (we just use the R matrices), what about rotating fermions? We cannot use R to rotate a four-component fermion field ψ, as R is a 3x3 matrix. In order to progress we need something more general, and the route to generality comes by again looking at the rotation of vectors, but from a different perspective. We write, for example, Rz = e−iθT 3 (3.82) where T3 0 −i 0 = i 0 0 , 0 0 0 (3.83) and the exponential of a matrix is defined through the taylor expansion of the exponential function, 1 1 eM := I + M + M 2 + M 3 + ... 2 3! 13 (3.84) One makes similar definitions for Rx 0 1 0 T = 0 1 = e−iθT and Ry = e−iθT 0 0 0 0 −i , T 2 = 0 −i i 0 2 where 0 i 0 0 , 0 0 (3.85) This may seem like an obscure thing to do, but the point is that general group elements, g, α α can be written as g = e−iθ T , where the set of matrices T α are known as the generators of the group. The magic thing is that, by explicit computation, 1 2 (3.86) T ,T = iT 3 , which is just the angular momentum algebra, (3.77). The point is that it is the angular momentum algebra generators that are resposnsible for rotations, be they orbital angular momentum, in which case one should use the generators given by (3.78), or vector rotation, in which case one should use (3.83,3.85). Our task now is to find a set of 4x4 matrices that satisfy (3.77) so that we can rotate fermions. 3.2 rotating a fermion What we need to do is find some set of matrices, S that obey the angular momentum algebra 1 2 S ,S = iS 3 , (3.87) these should be 4x4 matrices, in order to be able to act on the four-component fermion field ψ. As we already have a set of 4x4 matrices - the Dirac matrices - this is the natural place to start looking. In fact, we shall simply state that these generators are given by defining a set of matrices Σjk , where Σjk := − S i := i j k γ ,γ , 4 1 ijk jk Σ 2 (3.88) (3.89) for example S3 = 1 312 12 1 321 21 i i Σ + Σ = Σ12 = − (γ 1 γ 2 − γ 2 γ 1 ) = − γ 1 γ 2 2 2 4 2 (3.90) along with i S 2 = − γ3γ1 2 i S 1 = − γ2γ3 2 (3.91) (3.92) One may check, by explicit computation using te Dirac algebra, that (3.87) is satisfied. So, we now know how to rotate a fermion. For example, in the representations given previously we find that the generator of a rotation about the z-axis is 1 σ3 0 3 S = , (3.93) 0 σ3 2 14 so, for example, our plane-wave solutions (2.75), (2.76) have 1 S 3 ψ↑ = + ψ↑ 2 1 S 3 ψ↓ = − ψ↓ 2 (3.94) (3.95) (3.96) which, as expected, shows that ψ↑ has spin + 12 , while ψ↓ has spin − 12 . By analogy with vectors, a rotation of a fermion about the z-axis through angle θ is accomplished by 3 ψ 0 = e−iθS ψ, (3.97) which means we may check explicitly the common phrase that ”a fermion picks up a minus sign when rotated through 2π”, by noting that −iθ/2 σ3 e 0 −iθS 3 e (3.98) = , 3 0 e−iθ/2 σ I cos(θ/2) − iσ 3 sin(θ/2) 0 = (3.99) 0 I cos(θ/2) − iσ 3 sin(θ/2) so, for θ = 2π, ψ 0 = −ψ 3.3 (3.100) chirality and helicity due to the presence of a γ 5 matrix we have a neat way of splitting up a Dirac fermion using the projectors PR = 1 1 (I + γ 5 ), PL = (I − γ 5 ) 2 2 (3.101) and we define the left and right chiral components by ψR = PR ψ, ψL = PL ψ (3.102) This can always be done, because 1 1 1 1 ψ + γ5ψ + ψ − γ5ψ 2 2 2 2 = ψR + ψL ψ = (3.103) (3.104) we then note that the chiral components are eigenfunctions of the γ 5 matrix γ 5 ψR = +ψR γ 5 ψL = −ψL 15 (3.105) (3.106) At first sight, there seems to be little reason to label these chiral components as left and right, why not white and black, chalk and cheese...? To understand this we need to introduce the notion of helicity. We now know that a fermion has two vector associated with it, its momentum, and its spin. Helicity tells us whether they are pointing in the same direction or not. If the momentum and spin are aligned (positive helicity) then the fermion is spinning in the direction given by the usual right-hand-rule, so we call it right-handed. We shall now see that this ties in exactly with the chirality of massless fields. The Dirac equation for a wave mode of a massless fermion may be manipulated as follows γ µ ∂µ ψ = 0 ⇒ γ µ Pµ ψ = 0 ⇒ γ 0 P0 ψ + γ.P ψ = 0 ×γ 0 ⇒ −P0 ψ − γ.P γ 0 ψ = 0 ×γ 5 ⇒ −P0 γ 5 ψ − γ.P γ 0 γ 5 ψ = 0 ⇒ −P0 γ 5 ψ + γ.P iγ 1 γ 2 γ 3 ψ = 0 ⇒ −P0 γ 5 ψ + iP x γ 2 γ 3 ψ + iP y γ 3 γ 1 ψ + iP z γ 1 γ 2 ψ ⇒ −P0 γ 5 ψ − 2P x S x ψ − 2P y S y ψ − 2P z S z ψ ⇒ Eγ 5 ψ − 2P .Sψ P .S ψ E = |P | ⇒ |P | = 0 = 0 = 0 1 5 = γ ψ 2 (3.107) so we define the helicity operator h := P .S |P | (3.108) which measure whether the momentum is aligned with the spin. So, we see that right and left massless chiral fields have 1 hψR = + ψR (3.109) 2 1 hψL = − ψL (3.110) 2 which is why they are called right and left, they follow the right hand rule, or left hand rule. In the above derivation we made critical use of the masslessness of the fermion, this is because helicity is not Lorentz invariant for massive fields. For example, suppose we see a fermion with helicity + 12 , we can simply convert it to − 12 by travelling faster than it in its direction of motion, then P will change sign, but S will remain the same. We cannot overtake massless fields, and so the helicity of massless particles is a Lorentz invariant. However, the notion of chirality, as defined by γ 5 , is Lorentz invariant. Chirality and helicity only coincide for massless particles, nevertheless, we still call chiral fermions left and right, irrespective of whether they are massless. 3.4 Weyl fermions We have just seen that mass plays an important role for chiral fermions, this can be seen in another context - Weyl fermions. Taking the Dirac equation and operating on it with PR , and 16 PL we find PR [γ µ ∂µ ψ + mψ] = 0 ⇒ γ µ ∂µ ψL + mψR = 0 (3.111) (3.112) PL [γ µ ∂µ ψ + mψ] = 0 ⇒ γ µ ∂µ ψR + mψL = 0 (3.113) (3.114) and If m = 0 then these are two independent equations, and we may consistently set either ψR = 0 and ψL 6= 0, or ψL = 0 and ψR 6= 0. However, if the fermion has non-zero mass we need both left and right components1 . 4 Lecture four - gauge symmetry and electromagnetism Aim: To see what electromagnetism has to do with symmetry Learning outcomes: At the end of this lecture you should • be able to turn a global symmetry into a local one • know what role gauge fields play in local symmetries. 4.1 electromagnetism In the standard model, the list of particles can be split up into matter particles and force carriers. Matter particles are the bits that stuff is made out of, such as electrons, protons and quarks, whereas force carriers are what bind the matter particles together, such as photons and gluons. Here we shall see that the origin of the photon is a symmetry principle. First of all, what do we mean by a symmetry? The type of symmetry that we will be looking at is the analogue of a symmetry in Newton’s second law, F = −∇V. (4.115) We can see quite clearly that this system of equations is unchanged if we take V → V + V0 , V0 ∈ R (4.116) i.e. V0 is a real constant. This is a symmetry; we can shift the potential by any number, and none of the physics changes. A similar thing happens in electrical circuits, the current between two points is determined by the difference in potential between those points, I= ∆V R 1 (4.117) There is another type of mass term that we will not be considering, a Majarona mass, for which these conclusions are a bit premature. 17 so we can again shift the potential by a constant amount, everywhere on the circuit, and the physics remains the same. In electromagnetism we have that B = ∇ × A, E = −∇φ − Ȧ. (4.118) This system is again governed by derivatives, so we are free to shift both φ and A by constants without changing the physics; in fact, we can do much more than that. Rather than just shifting by a constant, we can shift by an amount determined by an arbitrary function, α(t, x) φ → φ − α̇, A → A + ∇α (4.119) and the physics remains the same, i.e. under these shifts E → E, B → B. (4.120) These shifts are known as gauge transformations. The fact that φ and A are undefined up to this arbitrary function highlights that they are not physical quantities, only the electric and magnetic field can be measured. In line with our relativistic notation we shall now introduce the four-vector potential, by combining φ and A into Aµ = (φ, A), ⇒ Aµ = (−φ, A) (4.121) Then, in this unified language, we see that our gauge transformations become Aµ → Aµ + ∂µ α. (4.122) This will make it easier when we come to the relativistic Dirac equation - which is now. 4.2 gauge invariance of the Dirac equation The Dirac equation γ µ ∂µ ψ + mψ = 0 (4.123) has what is called a U(1) symmetry - more of which later. Simply put, this means the equation for ψ and ψ 0 are exactly the same if ψ 0 = e−iΛ ψ, Λ ∈ R (4.124) so, Λ is a real constant. This just means that we are allowed to perform a phase rotation by angle Λ, so long as we do it everywhere in spacetime. This is what is termed a global symmetry - we do it everywhere. Now we ask the question, can we make this symmetry local, i.e. let Λ → Λ(x)? The naive answer is no. For ψ 0 = e−iΛ(x) ψ 18 (4.125) we find that the Dirac equation for ψ implies γ µ ∂µ ψ 0 + mψ 0 = −iγ µ ∂µ Λ ψ 0 , (4.126) which is not the Dirac equation. Rather than just giving up, we notice that if, instead of starting with the Dirac equation, we start with γ µ (∂µ − iqAµ )ψ + mψ = 0 (4.127) and ψ 0 = e−iΛ(x) ψ, 1 A0µ = Aµ − ∂µ Λ q (4.128) (4.129) then we end up with γ µ (∂µ − iqA0µ )ψ 0 + mψ 0 = 0. (4.130) i.e. the (ψ, Aµ ) variables satisfy the same equation as the (ψ 0 , A0µ ) variables, moreover, the transformation of the four-vector Aµ is exactly what is required by electromagnetism if we identify α = −Λ/q. Let us recap. We now have something that looks a bit like the Dirac equation, but it generalizes the global U(1) symmetry of the basic Dirac equation to a local U(1), i.e. we can now perform a different U(1) transformation at each point in spacetime and the physics is unchanged. The penalty for achieving this is that we had to introduce a new quantity Aµ , but that’s just electromagnetism, so it’s more of a prize than a penalty! U(1) gauge symmetry implies the existence of a photon, Aµ . 4.3 covariant derivatives The best way to think about gauge invariance is through the notion of a covariant derivative. The reason that the original Dirac equation does not have a local U(1) symmetry is that when we take ψ → e−iΛ(x) ψ, ψ and ∂µ ψ transform differently. What we really want is some derivative, Dµ , that transforms as Dµ ψ → e−iΛ Dµ ψ, (4.131) because then the derivative piece of the enhanced Dirac equation (γ µ Dµ ψ) will transform in the same way as the mass term. This is just what we have constructed, Dµ = ∂µ − iqAµ (4.132) and we have ψ 0 = e−iΛ(x) ψ, A0µ = Aµ + ∂µ α (4.133) (4.134) (Dµ ψ)0 = e−iΛ(x) (Dµ ψ). (4.135) 19 With this covariant derivative we are now in a position to construct new objects, the most important of which is the field strength, Fµν . This is define by [Dµ , Dν ] ψ = −iqFµν ψ. (4.136) This is a nice quantity because it is gauge invariant, and so could be of physical relevance. To see that it is gauge invariant we note that as Dµ ψ transforms covariantly, do does Dν Dµ ψ, that is one of the strengths of using covariant derivatives. This implies that [Dµ , Dν ] ψ also transforms covariantly (i.e. [Dµ , Dν ] ψ → e−iΛ(x) [Dµ , Dν ] ψ). But, the right-hand-side of (4.136) also picks 0 up a e−iΛ(x) , meaning that Fµν must be unchanged, Fµν = Fµν . To be more explicit note that ψ 0 = U ψ, (Dµ ψ)0 = U (Dµ ψ), (Dµ Dν ψ)0 = U (Dµ Dν ψ) (4.137) where U = e−iΛ(x) . So, the primed field strength is just ([Dµ , Dν ] ψ)0 U [Dµ , Dν ] ψ U (−iqFµν ψ) U Fµν ψ = = = = 0 −iqFµν ψ0 0 −iqFµν Uψ 0 −iqFµν Uψ 0 Fµν U ψ (4.138) (4.139) (4.140) (4.141) Now, this is required to hold for all ψ so 0 U Fµν = Fµν U (4.142) and now note that U U † = 1 = U † so multiply on the right by U † . 0 U Fµν U † = Fµν (4.143) then note that in this case the gauge theory is Abelian, i.e. U(1), so the U , U † are just numbers and can move through the Fµν to cancel each other 0 Fµν = Fµν (4.144) This proves the gauge invariance of the field strength. Now we need to see what the field strength actually is, to do this, just calculate it −iqFµν ψ = Dµ Dν ψ − Dν Dµ ψ = (∂µ − iqAµ )(∂ν ψ − iqAν ψ) − (µ ↔ ν) = −iq(∂µ Aν − ∂ν Aµ )ψ (4.145) (4.146) (4.147) Fµν = ∂µ Aν − ∂ν Aµ (4.148) so we find and now we are nearly there, just a few examples should convince you that Fµν is nothing more than a re-packaging of the electric and magnetic fields. e.g. F01 = ∂0 A1 − ∂1 A0 = Ȧx + ∂x φ = −E x . e.g. F12 = ∂1 A2 − ∂2 A1 = ∂x Ay − ∂y Ax = B z . It is a useful excercise to check the other components. 20 4.4 the standard model and gauge symmetries In later lectures we shall come across more general symmetry groups than U(1); ones of particular importance are SU(2) and SU(3). The reason for their significance is that the standard model is a gauge theory of U(1)×SU(2)×SU(3), with the U(1)×SU(2) supplying the photon, the Z0 , and the W± gauge bosons, and the SU(3) giving us the gluons. 5 Lecture five - Dirac equation and magnetic fields Aim: To see how electromagnetism fits in with the Dirac equation Learning outcomes: At the end of this lecture you should • know what a magnetic moment of a particle is • know how to derive the magnetic moment of a non-relativistic Dirac particle 5.1 magnetic moments Recall from electromagnetism that the magnetic moment of a circular, planar loop of area A, with current I, is given by µ = IA. (5.149) Now, for a total charge of q running around the loop, in period T we have v (5.150) I = q/T = q 2πr so we find q q mvr = L µ = (5.151) 2m 2m where L is the angular momentum of the current, and m its mass. Now, suppose we place such a magnetic moment in a magnetic field, one finds that they prefer to align with the background field. This is because aligned magnets have a lower energy. In fact, classical electromagnetism says that the energy of a magnetic moment, µ, within a magnetic field B is given by the interaction Hamiltonian q Hint = −µ.B = − B.L (5.152) 2m This is the starting point for writing down the interaction energy of a particle of spin S within a magnetic field. As we expect there to be differences with the basic loop of wire we adopt a fudge-factor, g, to account for differences, giving the interaction Hamiltonian as q H = −g B.S (5.153) 2m where g for the electron is found, by experiment, to be gelectron ∼ 2.002 It is our task to derive the magnetic moment of the Dirac particle. 21 (5.154) 5.2 magnetic moment of electron - Zeeman interaction Now that we have learned all about gauge invariance, we shall see one of its consequences, namely a prediction for the magnetic moment of Dirac particles. We start with the gauged Dirac equation γ µ (∂µ − iqAµ )ψ + mψ = 0 and we write the Dirac field as ψ = ψ1 ψ2 (5.155) Taking the Dirac representation of gamma matrices (2.64), the definition of the four-vector potential (4.121), and the relation between momentum and derivatives (1.35) we find i∂t ψ1 = (−m + qφ)ψ1 + (p − qA).σψ2 i∂t ψ2 = (m + qφ)ψ2 + (p − qA).σψ1 . (5.156) (5.157) For a free wave, we have that the wave function varies as e−iEt , and now we take the nonrelativistic approximation that E ∼ m, m >> qφ (5.158) 1 (p − qA).σψ2 2m (5.159) and use this in (5.156) to find ψ1 ' showing that ψ1 is small compared to ψ2 , and also we find that (5.157) now becomes i∂t ψ2 = (m + qφ)ψ2 + 1 [(p − qA).σ][(p − qA).σ]ψ2 2m (5.160) and we then use that [(p − qA).σ]2 = (p − qA)2 − qB.σ (5.161) which gives the final equation (p − qA)2 q ψ2 + (m + qφ)ψ2 − B.Sψ2 = i∂t ψ2 2m m (5.162) which is just the Schrödinger equation for a particle in a magnetic field, whose magnetic interaction is q Hzeeman = − B.S (5.163) m giving a g-factor for the Dirac particle of gDirac = 2 22 (5.164) As this is so close to the observed value for the electron, we identify the electron as a Dirac particle. The corrections to the value of 2 is explained using quantum field theory - very precisely g − 2 = (1159.65218111 ± 0.00000074) × 10−6 (5.165) 2 experiment g − 2 = (1159.65213 ± 0.000003) × 10−6 (5.166) 2 theory (5.167) 6 Lecture six - Feynman rules Aim: To uncover the origin of a particuarly useful perturbative technique Learning outcomes: At the end of this lecture you should • know what a propogator is • be familiar with the route from Huygen’s principle to the propogator equation • be able to solve the propogator equation when there is a small perturbation present • be able to derive the propogtor for the Klein-Gordon equation 6.1 propogator theory Based on an idea of Dirac, Feynman decided to formulate his own version of quantum mechanics, and he based it on Huygen’s principle for calculating the propogation of wavefronts. This should not come as a great surprise, given that the Schrödinger equation is a wave equation. The question we want to ask is, what is the probability amplitude at some location xf , at time tf , given the amplitde at some earlier time t? To answer this, we introduce the notion of the propogator, G(tf , xf ; t, x), which tells us how an amplitude at (t, x) evolves into an amplitude at the late time (tf , xf ). X (6.168) ψ(tf , xf ) = iG(tf , xf ; t, x) ∆xψ(t, x), tf > t x What this equation does, is split space into blocks of size ∆x, then the block at location x contributes ∆xψ(t, x) to the amplitude at the initial time t. This disturbance then propogates to (tf , xf ) with the help of the propogator. We then have to add up all such contributions. The continuum version changes the summation to an integral, and we may incorporate the tf > t condition by including a theta-function Z Θ(tf − t)ψ(tf , xf ) = dx iG(tf , xf ; t, x)ψ(t, x), (6.169) To see how this works in practise, let’s take write the Schrödinger equation in the following form, introducing the operator Ô Ô(tf , xf )ψ(tf , xf ) = [i∂tf − H(tf , xf )]ψ(tf , xf ) = 0 23 (6.170) Now operate on both sides of (6.169) with Ô, using ∂tf Θ(tf − t) = δ(tf − t) (6.171) to find Z iδ(tf − t)ψ(tf , xf ) = i h dx Ô(tf , xf )iG(tf , xf ; t, x) ψ(t, x) (6.172) which implies that Ô(tf , xf )G(tf , xf ; t, x) = δ(tf − t)δ(xf − x). (6.173) So, this just tells us that the propogator, G(tf , xf ; t, x), is the Green-function associated to the Schrödinger equation. What this means is that the propogator is, in some sense, the inverse of the Schrödinger equation, i.e. ÔÔ−1 = I. 6.1.1 Schrödinger propogator The propogator for the free (no interaction terms in the Hamiltonian) Schrödinger equation comes from solving [i∂tf − H0 (tf , xf )]G0 (tf , xf ; t, x) = δ(tf − t)δ(xf − x) (6.174) As this is a linear equation we consider a solution using the Fourier transform by writing Z dE dp G0 (tf , xf ; t, x) = G̃0 (E, p)e−iE(tf −t)+ip(xf −x) (6.175) 2π 2π and then we use Z dxe−ix(y−a) = 2πδ(y − a) (6.176) to find G̃0 (E, p) = 1 E − p2 /2m (6.177) In fact, it is more common to call this, the Fourier transformed version, the propogator. 6.1.2 Klein-Gordon propogator The Klein-Gordon equation is (−∂µ ∂ µ + m2 )φ = 0 (6.178) and to find the KG propogator we simply replace φ with G0 (tf , xf ; t, x), and the zero on the right-hand-side by δ(tf − t)δ(xf − x). (−∂µ ∂ µ + m2 )G0 (tf , xf ; t, x) = δ(tf − t)δ(xf − x) 24 (6.179) Again, the equation is linear so we look for the Fourier solution Z d4 p G0 (tf , xf ; t, x) = G̃0 (p)eip.(xf −x) (2π)4 (6.180) and now we find the propogator for the KG equation to be G0 (p) = 1 . p2 + m2 (6.181) where p2 = −E 2 + p2 is the spacetime-square of the momentum four-vector. Now we start to spot a pattern, the (momentum-space) propogator is just 1/(equation of motion), both for the Schrödinger equation, and the KG equation - this is rather generic, especially given that we expect the propogator to be the inverse of the equation of motion, ÔÔ−1 = I. 6.2 Feynman rules - including interactions So far we have not discussed the propogators for interacting theories, only the free versions, now we shall add a small perturbation to the Schrödinger Hamiltonian to see how we may accomodate interactions. The starting point is Ô(tf , xf ) = i∂t − Ĥ = i∂t − Ĥ0 − VI = Ô0 (tf , xf ) − VI Ô(tf , xf )G(tf , xf ; t, x) = δ(tf − t)δ(xf − x) (6.182) (6.183) These may be rewritten as Ô0 (tf , xf )G(tf , xf ; t, x) = δ(tf − t)δ(xf − x) + VI (tf , xf )G(tf , xf ; t, x) (6.184) the solution of which is an integral equation Z G(tf , xf ; t, x) = G0 (tf , xf ; t, x) + dt1 dx1 G0 (tf , xf ; t1 , x1 )VI (t1 , x1 )G(t1 , x1 ; t, x) (6.185) This may not be apparent at first sight, and is easier to confirm by working backwards. If we operate on (6.185) with O0 (tf , xf ), the left-hand-side simply recovers the left-hand-side of (6.184), while the first term on the right-hand-side gives the first term on the right-hand-side of (6.184). To understand the rest, note that O0 (tf , xf ) only sees tf and xf , it’s not interested in t, t1 , x, x1 , so when O0 (tf , xf ) hits the integral on the right-hand-side of (6.185) it just turns the G0 (tf , xf ; t1 , x1 ) into δ(tf − t1 )δ(xf − x1 ), allowing the integral to be performed. We now use an unusual trick, that usually results in 0=0 when tried for most equations, we substitute the equation (6.185) into itself, i.e. the replace the G(t1 , x1 ; t, x) inside the integral on the right-hand-side of (6.185) by the solution for G(t1 , x1 ; t, x). Dropping the details we have Z G = G0 + G0 VI G Z Z G = G0 + G0 VI G0 + G0 VI G Z Z Z G = G0 + G0 VI G0 + G0 VI G0 VI G 25 Now we repeat what we have just done, by replacing the the G on the right-hand-side of this expression with the initial solution Z Z Z Z G0 VI G0 VI G0 + G0 VI G G = G0 + G0 VI G0 + Z Z Z Z Z Z = G0 + G0 VI G0 + G0 VI G0 VI G0 + G0 VI G0 VI G0 VI G at which point we start to see what’s going on. In principle we should do this an infinite amount of times, but as we assume VI is small, the later terms will be negligible, meaning that we have an expression for the full propogater G in terms of the much simpler free-propogater G0 . 6.3 diagramatic expansion As is common with such perturbative expansions, it is useful to use diagrams as mneumonics for the integrals, and objects appearing in the integrals. The diagramatic expansion for the above solution to the integral propogator equation is f ull propogator, G = + + + ... (6.186) where the line represents the free propogator G0 and the vertices (dots) represent the interaction term. We then think of a given diagram in physical terms as the particle moving for a bit, then interacting with the external potential, then moving a bit, and so on. Quantum mechanics is a single-particle theory, so the diagramatic technique is not really needed. But in the standard model we shall see that it allows for a much simpler visualization of what is going on. 7 Lecture seven - action functionals and Feynman rules Aim: To introduce action functionals for the basic theories, and see how to read off the Feynman rules. Learning outcomes: At the end of this lecture you should • know how to derive equations of motion from an action • know how to write down the Feynman rules, once an action has been given. 7.1 action and equations of motion A neat way of characterizing the dynamics of a system is through the action principle, which states that a system evolves in such a way as to extremize the action of a system. Such statements for static systems should be obvious by now, in that a static system has extremized its potential energy - a stationary ball on a slope must be either at the top of a hill, at the bottom of a valley, or at a saddle point. The action is the next step up, to include dynamical systems. This is not course on action principles, so we shall just proceed by example. 26 7.1.1 Klein-Gordon action The action for the real Klein-Gordon scalar field is given by Z 1 1 S[φ] = d4 x [− ∂µ φ∂ µ φ − m2 φ2 ] 2 2 (7.187) and it is this that we need to extremize. Now, when we find the extrema of a function f (x) we solve the following equation lim f (x + δx) − f (x) = 0 δx→0 (7.188) which is just the first-principles way of saying df /dx = 0. Here what we do is vary the field φ, rather than x, and so we write lim S[φ + δφ] − S[φ] = 0 δφ→0 (7.189) In the case of the real KG field we find Z Z 1 1 2 1 1 4 µ 2 lim d x [− ∂µ (φ + δφ)∂ (φ + δφ) − m (φ + δφ) ] − d4 x [− ∂µ φ∂ µ φ − m2 φ2 ] = 0 δφ→0 2 2 2 2 We now expand this out, dropping the δφ2 as they are subdominant in the limit. We find Z d4 x [−∂µ φ∂ µ δφ − m2 φδφ] = 0 (7.190) Now we integrate by parts, dropping the boundary term, i.e. use ∂µ φ∂ µ δφ = ∂ µ (δφ∂µ φ) − δφ∂µ ∂ µ φ (7.191) R and ignore the d4 x ∂ µ (δφ∂µ φ) term. We shall always assume this vanishes; it amounts to saying there is no current at infinity. Then Z d4 x [∂µ ∂ µ φ − m2 φ]δφ = 0 (7.192) Now, we want this integral to vanish for all possible deformations δφ, and the only way to achieve this is for the integrand itself to vanish, ∂µ ∂ µ φ − m2 φ = 0. (7.193) This is just the Klein-Gordon equation (1.38). The point is that the action contains the same dynamical information as the field equations, but it turns out that the action is easier to handle, especially when there are lots of field. For example, the standard model contains 37 fields, so if we wanted to write it down in terms of equations of motion, that would be 37 rather complicated equations. However, in terms of the action, there is just one. 27 7.1.2 Dirac action The Dirac action is given by Z S = − d4 x[ψ̄γ µ ∂µ ψ + mψ̄ψ] (7.194) where ψ̄ = −iψ † γ0 . This new object is because ψ is a fermion, and so it needs this peculiar version of Hermitian conjugation. Now, despite all the complications one would expect from the Dirac equation, the variation of the action is remarkably simple. We treat ψ and ψ̄ as independent objects, in which case the action must be extremized under variation of ψ and ψ̄ separately. If we vary ψ̄ then we find Z δSψ̄ = S[ψ, ψ̄ + δ ψ̄] − S[ψ, ψ̄] = − d4 x δ ψ̄[γ µ ∂µ ψ + mψ] = 0 (7.195) and this must hold for all variations δ ψ̄, in which case we must have γ µ ∂µ ψ + mψ = 0, (7.196) which is just the Dirac equation (2.45). One can check that varying ψ gives the same results, after a bit of work. 7.2 the structure of an action The action, S, is usually written in terms of a Lagrangian density, L, where they are related by Z d4 x L (7.197) S = and L takes the form L = T −V (7.198) where T is the ”kinetic” term, and V is the potential, which contins interaction term. There are a few useful things about the action that help to restrict the possible theories we may write down, they are • L is real • L is a Lorentz scalar - there are no free spacetime indices, they all come in pairs • L is a scalar with respect to internal symmetries From the examples given it is now clear that pieces in the action that are quadratic in fields will produce terms in the field equations that are linear. So, the interaction terms that will be the perturbation away from the linear (free) theory must be cubic (or higher) in the action, allowing us to write L = L0 − VI 28 (7.199) where VI is the interaction potential, and is cubic (or higher) in the fields. In terms of Feynman diagrams, it is the L0 that leads to the free propogator, and the VI that produces the vertices. The vertices are constructed rather simply by looking at each term in VI : if VI is cubic in the field, then the vertex has three free-propogators meeting at a point; (7.200) if VI is quartic in the field, then the vertex has four free-propogators meeting at a point; (7.201) if VI is quadratic in one field, but also quadratic in another, then the vertex has two freepropogators of each species. (7.202) In fact, each term in VI will come with a parameter in front of it, for example λ in VI = λφ4 , in which case we would associate λ with the vertex, just as we would associate the propogator with a basic line. 7.2.1 example, real KG action I 1 1 1 L = − ∂µ φ∂ µ φ − m2 φ2 − λφ4 2 2 4 (7.203) has VI = 1 4 λφ 4 (7.204) so the equation of motion for this theory is ∂µ ∂ µ φ − m2 φ = λφ3 and the Feynman rules are (7.205) 1 p2 + m2 (7.206) λ (7.207) Note that we shall not be worrying about factors of 2, 14 ,... for the vertices, we only want the parametric scale. 29 7.2.2 example, real KG action II 1 1 1 L = − ∂µ φ∂ µ φ − m2 φ2 − ζφ3 2 2 3 (7.208) has 1 3 ζφ 3 VI = (7.209) so the equation of motion for this theory is ∂µ ∂ µ φ − m2 φ = ζφ2 and the Feynman rules are 7.2.3 p2 1 + m2 ζ (7.210) (7.211) (7.212) example, complex KG action 1 L = −∂µ φ∂ µ φ? − m2 φφ? − λ(φφ? )2 2 has VI = and the Feynman rules are 1 λ(φφ? )2 2 (7.213) (7.214) 1 p2 + m2 (7.215) λ (7.216) here we see something new, arrows on the lines. This is because there is a conserved quantity associated to the field, due to the global U(1) symmetry. As we have seen before, such a U(1) symmetry is leads to an electric charge for the particle. When we have such a symmetry then, as charge cannot just dissapear, there must be the same amount of charge entering a vertex as leaving it, the arrows help us to keep track of where the charge is. 30 7.3 example, Dirac particle coupled to photons We have seen the Lagrangian for a basic Dirac particle, to couple in electromagnetism we simply replace ∂µ → Dµ (4.132), so L = −ψ̄γ µ Dµ ψ − mψ̄ψ = −ψ̄γ µ ∂µ ψ − mψ̄ψ + iq ψ̄γ µ ψAµ (7.217) VI = −iq ψ̄γ µ ψAµ (7.218) which is something different to those we have seen before, as there are now two fields involved. This just means that the vertex has two fermion lines, and one photon −iγ µ pµ + m p2 + m2 (7.219) q (7.220) Again, we put an arrow on the fermion line as there is a U(1) symmetry, and so charge is conserved. 7.4 complex KG equation coupled to photons To couple the complex KG equation to electromagnetism we simply replace ∂µ → Dµ , giving the Lagrangian L = −(Dµ φ)(Dµ φ)? − m2 φφ? = −∂µ φ∂ µ φ? − m2 φφ? + iqAµ (φ? ∂ µ φ − φ∂ µ φ? ) + q 2 Aµ Aµ φφ? (7.221) and then the Feynman rules are 7.5 p2 1 + m2 (7.222) q (7.223) q2 (7.224) Feynman rules for the electroweak sector of the standard model Now that we have a grasp of where the Feynman rules come from we shall write down the rules of one of the sectors of the standard model, the electrowek sector, i.e. the part that tells us 31 about the electromagnetic force and the weak nuclear force. The matter particles involved in this are the leptons νe νµ ντ , , , (7.225) e µ τ and the gauge bosons (force carriers) Z 0 , W ± , γ. (7.226) It is also believed to contain the Higgs scalar, but this has not been observed, yet. The basic set of diagrams are as follows e γ e (7.227) γ e (7.228) Z0 gw (7.229) W− gw (7.230) Z0 gw (7.231) e W− W+ νe νe νe e W+ W− 32 νe Z0 gw (7.232) Z0 gw (7.233) νe e e Although we have only explicitly given the rules for the νe e doublet, they also hold for the muon and the tau lepton doublets. 7.6 Fermi’s golden rule Now that we have our Feynman rules we need to do something with them. Each diagram tells us about the amplitude for a process to happen, for example the amplitude for an electron to anihilate an anti-electron producing a photon is given by diagram 7.227, and has amplitude e. However, we are interested in physical quantities, such as rates of events, and so we actually need the probability, rather than the amplitude, and this is given by the modulus-square of the amplitude. There is one extra factor to take into account, the density of final states ρ, which gives us the number of states that are available to be occupied. The final result is Fermi’s golden rule Γ(i → j) = 2π |Tij |2 ρ ~ (7.234) where Tij is the probability amplitude. 7.6.1 ee scattering e e ∼ e2 γ (7.235) e e This diagram shows the scattering of two electron by a photon. The diagram contains two vertices, each of strength e, and so the amplitude of the diagram is given parametrically by e2 . 2 1 so the rate of this process is Recall that the fine structure constant, α = 4πe0 ~c ' 137 Γ ∼ α2 33 (7.236) As the diagram contiains two vertices we say it is a second-order process. Another diagram that contributes to electron-electron scattering is e e γ ∼ e4 γ (7.237) e e but this is fourth-order (four vertices) and subleading so can be ignored at the first approximation. 7.6.2 µ decay The muon is the heavy version of the electron and, as it is heavy, is liable to decay. The leading order diagram responsible for the decay is νe µ W e ∼ gw gw 1 2 MW (7.238) νµ and here we also include the internal propogator of the W boson, which we have approximated as M12 , which is fine for low-momentum scattering. In fact, we can use this to give us an W expression for the Fermi coupling constant. Before the discovery of the W ± , and Z 0 it was known that a muon could decay into an electron and two neutrinos, so Fermi proposed an interaction term of the form VI (F ermi) ∼ GF (ēνe )(µ̄νµ ) which has the diagram (7.239) νe µ e ∼ GF (7.240) νµ It was this GF that was measured first (with a value of 10−5 (GeV )−2 ), and we can see from our 2 electroweak diagram that GF ∼ Mgw2 . W 8 Lecture eight - group theory Aim: To introduce the concept of a group, and some important examples of groups. Learning outcomes: At the end of this lecture you should 34 • know the definition of a group • know how to check whether a set, in combination with a composition law, forms a group • know what defines Abelian groups, Lie groups and orthogonal groups. 8.1 definition of a group A group is simply a set of objects, supplied with a rule that allows you to take two of these objects and make another one. This law of composition is denoted with a dot, g1 · g2 = g3 , (8.241) where gi are members of the set. For the set and composition law to be a group, G, it is required to satisfy the following • it must be closed: If g1 , g2 ∈ G, then g1 · g2 ∈ G • it must have an identity, denoted e: e · g = g · e = g, ∀g ∈ G. • each g must have an inverse, denoted g −1 : g −1 · g = g · g −1 = e. • the composition must be associative: g1 · (g2 · g3 ) = (g1 · g2 ) · g3 . Some examples should help to clarify this. • Consider the set {1, i}, where the composition law is multiplication. This does not form a group because i · i = −1, and -1 is not part of the set • Consider the set {1, i, −1, −i}, where the composition law is multiplication. This does form a group. • Consider the set {1, i, −1, −i}, where the composition law is addition. This does not form a group as, for example, 1 · 1 = 1 + 1 = 2 is not part of the set • Consider the set of integers, Z, where the composition law is addition. This does form a group • Consider the set of integers, Z, where the composition law is multiplication. This does not form a group as, for example, 2 does not have an inverse within the set, the inverse of 2 is 1/2, which is not an integer. The above examples should really be called representations of groups, which just means that they are explicit examples of the more abstract group itself. For example, we could say that the group we are interested in (called Z2 ) consists of elements {e, a}, such that a · a = e. In this definition we make no comment about what a, e actually are, nor do we say what the composition law is. An explicit representation of this group could be {e → 1, a → −1, · → multiplication} but there are many others. We shall see more consequences of different representions later. 35 8.2 direct product of groups In the standard model of particle physics there are a number of distinct groups, acting as independent internal symmetries, so we need a way of combining groups together, the direct product does just that. Suppose we have two groups G1 and G2 , we may form a group G3 = G1 ×G2 with elements g3 = (g1 , g2 ) and composition law g3 ·g30 = (g1 ·g10 , g2 ·g20 ), and it is simple to show that G3 is indeed a group. For example, consider the group D2 = Z2 ×Z2 , this is composed of elements {(e, e), (a, e), (e, a), (a, a)} where, for example (e, a) · (a, a) = (e · a, a · a) = (a, e). 8.3 Abelian groups Abelian groups are the simplest, they are the ones for which g1 · g2 = g2 · g1 . (8.242) For example, the group formed by multiplication within the rational numbers is Abelian; addition is also Abelian. However, there are important examples that are not Abelian, with the non-Abelian nature coming typically because matrix multiplication is not commutative, AB 6= BA. 8.4 Lie groups (pronounced ”Lee groups”) The previous examples were all discrete groups, meaning that there was a countable number of elements in the set; the next step is to think about non-countable sets. For example the set of rational numbers, excluding zero, forms a group under multiplication (zero is excluded because it has no inverse under multiplication). More common examples are those associated to some sort of rotation. Consider the unitary group U(1), or rather its fundamental representation, defined by the elements gα = eiα (8.243) with the composition law being multiplication (it is easily checked that this forms a group). This just corresponds to rotations in the complex plane by angle α. and as α is a continuous parameter this is a Lie group. 8.4.1 general linear group GL(n,R) The fundamental representation of GL(n,R) is defined to consist of elements A where • A is a real n × n matrix • det(A) 6= 0 The dimension of this group (the number of indepenedent numbers within an element) is n2 , as there are n rows and n columns. 36 8.5 special linear group SL(n,R) The fundamental representation of SL(n,R) is defined to consist of elements A where • A ∈GL(n,R) • det(A) = 1 The dimension of this group is n2 −1, as there are n rows and n columns, and one real constraint coming from det(A) = 1. 8.6 orthogonal group O(n,R) The fundamental representation of O(n,R) is defined to consist of elements A where • A ∈GL(n,R) • AAT = I = AT A The dimension of this group is 12 n(n − 1), which we now explain. Think of the matrix B, where B = AT A. By constraining B = I we are requiring the n terms along its diagonal to be 1, i.e. n constaraints. The same constraint also sets all n2 − n off-diagonal terms to zero, however, as B is symmetric this only imposes 12 (n2 − n) independent constraints. The total number of degrees of freedom is then n2 − [n + 21 (n2 − n)] = 12 n(n − 1). Note that AAT ⇒ det(AAT ) ⇒ det(A) det(AT ) ⇒ det(A) det(A) ⇒ det(A) 8.7 = = = = = I 1 1 1 ±1 (8.244) (8.245) (8.246) (8.247) (8.248) special orthogonal group SO(n,R) The fundamental representation of SO(n,R) is defined to consist of elements A where • A ∈O(n,R) • det(A) = 1 The dimension of this group is the same as that of O(N,R), 21 n(n−1), because imposing the unit determinant only selects a sector of O(N,R) rather than reducing its dimension, as we already have det(A) = ±1 for orthogonal matrices. 37 8.8 O(p,q,R) The notion of orthogonal matrices is easily extended to O(p,q,R) by thinking of the defining relation of O(n,R) as AIAT = I (8.249) Then we define elements of O(p,q,R) by requiring AηAT = η (8.250) where η − diag(−1, −1, −1, ..., 1, 1, 1...), with p minus ones, and q plus ones. This may seem like an obscure thing to do, but in fact this is just the Lorentz group of special relativity if we take p = 1, q = 3. 8.9 symplectic group Sp(2n,R) Define the symplectic matrix Ω := 0 I −I 0 (8.251) where I is the n × n identity matrix, then the symplectic group is defined to be those elements that satisfy • A ∈GL(n,R) • AT ΩA = Ω 9 Lecture nine - non-Abelian gauge theory Aim: To introduce the concept of a Lie algebras, and non-Abelian gauge theory Learning outcomes: At the end of this lecture you should • know the definition of unitary groups • know what group generators are, and how to calculate their basic properties • know how to gauge a non-Abelian symmetry 9.1 unitary groups Although the groups presented in the previous lecture all have their role to play in some area of physics, we are particularly interested in a different set of groups, the unitary groups, composed of complex matrices satisfying certain properties as follows 38 9.2 unitary group U(n) We have already seen a unitary group, U(1), here we generalize that to matrices by saying U ∈U(n) if • U †U = I = U U † where U † = (U T )∗ . (9.252) The dimension of U(n) is n2 , which we see as follows. First note that there are n2 complex entries in U , corresponding to 2n2 real degrees of freedom. Now the diagonal part of U † U = I gives us n real constraints, while the off-diagonal part gives 2 21 (n2 − 1) real constraints (the 12 is because U † U is symmetric and the 2 because the entries are complex). This gives a total of 2n2 − [n + (n2 − n)] = n2 real degrees of freedom for each U . 9.3 special unitary groups U ∈SU(n) if • U ∈U(n) • det(U ) = 1 The dimension of U(n) is n2 − 1, because the determinant imposes one real constraint, which we see by considering U †U ⇒ det(U † U ) ⇒ det(U † ) det(U ) ⇒ |det(U )|2 ⇒ det(U ) = = = = = I 1 1 1 eiα (9.253) (9.254) (9.255) (9.256) (9.257) so, if we impose det(U ) = 1, all we are doing is requiring the single real constraint α = 0. 9.4 unitary groups and the standard model In the standard model one finds that the paricles are described by their transformation properties under three different unitary groups, each with its own physical interpretation • U(1) describes electromegnetism. It has dimension one, and gives us the single photon • SU(2) describes the weak nuclear force. It has dimension three, and gives us the W ± and Z 0 bosons. • SU(3) describes the strong nuclear force. It has dimension eight, and gives us the eight gluons. 39 9.5 generators and Lie algebras We have already introduced the notion of generators when we were thinking about how to rotate a fermion field; the concept of generators is also useful for internal symmetries. Simply stated, the generators X a and the group elements g of a Lie group G are related by g = e−iθ aXa (9.258) . Once we have fixed our choice of generators, the parameters θa are what give us our different group elements. The reason that the generators are studied, rather than the group elements, is that for small θa this just corresponds to looking at group elements near the identity, g = e−iθ aXa ' I − iθa X a + ... (9.259) and that, it turns out, makes our life simpler. The generators X a form what is called a Lie algebra. 9.5.1 o(n), the Lie algebra of O(n) The defining property of O(n) is enough to tell us that the generators of O(n) are skewsymmetric, we see this as follows OT O ⇒ (I − iθa T a + ...)T (I − iθa T a + ...) ⇒ θa [T a + (T a )T ] ⇒ Ta = = = = I I 0 −(T a )T (9.260) (9.261) (9.262) (9.263) Now, recall from the question sheet that O(n) is just the set of matrices associated with rotations (actually also reflections), but we already have a set of generators for three-dimensional rotations given by (3.83), (3.85), which we note are indeed skew-symmetric. Although we chose these generators to do our calculations, we could have in fact chosen any set of linearly independent skew-symmetric matrices - it would just hve made the calculations harder and less intuitive. 9.5.2 su(n), the Lie algebra of SU(n) Again, we may use the defineing property of SU(n) to tell us something about its generators. U †U ⇒ (I − iθa X a + ...)† (I − iθa X a + ...) ⇒ θa [X a − (X a )† ] ⇒ Xa = = = = I I 0 (X a )† (9.264) (9.265) (9.266) (9.267) So, the generators are Hermitian. However, we also need to impose unit determinant det(I − iθa X a ) = 1 ⇒ 1 − iθa T r(X a ) = 1 ⇒ T r(X a ) = 0 40 (9.268) (9.269) (9.270) so, as well as being Hermitian, the generators of SU(n) are traceless. Now, again, we already know of a set of matrices that are Hermitian and traceless - the Pauli matrices. And they do indeed generate SU(2). However, any set of traceless and Hermitian matrices would do, it’s just that Pauli matrices make life easier. 9.6 non-Abelian gauge theory This is what the standard model is all about, and fortunately it’s not much different, conceptually, from electromagnetism - the maths is a little bit trickier though. To see how it works, consider the theory described by the Lagrangian L = −∂µ φ† ∂ µ φ − m2 φ† φ where φ is not just a complex scalar, but is a complex vector of scalars φ1 φ2 φ = .. . φn (9.271) (9.272) Now, given the definition of U(n), it is clear that this Lagrangian is invariant under φ → U φ, for U ∈SU(n), and where the matrix U does not depend on spacetime, i.e. it is a constant matrix. We can see this because if φ0 = U φ then φ0† φ0 = (U φ)† (U φ) = φ† U † U φ = φ† φ, ∂µ φ0† ∂ µ φ0 = ∂µ (U φ)† ∂ µ (U φ) = ∂µ φ† U † U ∂ µ φ = ∂µ φ† ∂ µ φ (9.273) (9.274) Now we make the same grand statements as we did for U(1) electromagnetism, and say that we want this SU(n) symmetry to be local, and not just global. To do that we again introduce a covariant derivative Dµ φ = ∂µ φ − iqAµ φ (9.275) (Dµ φ)0 = U (x)Dµ φ (9.276) and require when we promote the constant U matrix to a spacetime dependent matrix U (x). The only difference is that now the vector potential Aµ is actually a matrix, so we have to be a bit careful about how we manipulate expressions involving it, for example Aµ Aν 6= Aν Aµ and U Aµ 6= Aµ U . With the introduction of the gauge-covariant derivative Dµ we may easily construct a gauge invariant Lagrangian L = −(Dµ φ)† (Dµ φ) − m2 φ† φ 41 (9.277) The next step is to determine how Aµ changes under the gauge transformation, which we do with the following calculation (Dµ φ)0 ⇒ ∂µ φ0 − iqA0µ φ0 ⇒ ∂µ (U (x)φ) − iqA0µ U (x)φ ⇒ ∂µ U (x)φ + U (x)∂µ φ − iqA0µ U (x)φ = = = = U (x)Dµ φ U (x)(∂µ φ − iqAµ φ) U (x)∂µ φ − iqU (x)Aµ φ U (x)∂µ φ − iqU (x)Aµ φ i ⇒ A0µ U (x) = U (x)Aµ − ∂µ U (x) q i ⇒ A0µ = U (x)Aµ U † (x) − [∂µ U (x)]U † (x) q (9.278) (9.279) (9.280) (9.281) (9.282) (9.283) which joins φ0 = U φ to give us the full set of non-Abelian gauge trnsformations. We should compare this to the U(1) case (4.128, 4.129) to see that it does at least reduce to electromagnetism in the Abelian case. 10 Lecture ten - some representation theory Aim: To reintroduce the idea of SU(2) representations Learning outcomes: At the end of this lecture you should • be aware that there are many different guises for the same symmetry group • know how to use ladder operators to construct the states transforming in a given representation of SU(2) 10.1 SU(2) Because SU(2) is such an important group we shall spend a lecture going into some more of its properties. By now we have seen the algebra [j 1 , j 2 ] = ij 3 , +cyc. perm (10.284) many times - this is the definition of the algebra of the generators of SU(2). At this point however, we have made no statement about what the j a are, nor what there composition law is. We have seen examples where the j a are matrices, and the composition is matrix multiplication, but they need not be matrices. Another important representation of this algebra comes from the differential operators L̂z = −ix∂y + iy∂x , +cyc. perm (10.285) which also satisfy this algebra. This should not come as a surprise, because these are just the orbital angular momentum operators. Even if the j a are matrices, there are an infinite number of different matrix representations. the three-vector rotation generators (3.83,3.85) satisfy [T 1 , T 2 ] = iT 3 , +cyc. perm 42 (10.286) and the Pauli matrices satisfy [σ 1 /2, σ 2 /2] = iσ 3 /2, +cyc. perm (10.287) And, as these two matrix representations consist of 2x2 and 3x3 matrices, they are not trivially related, and form distinct representations. In fact, we can continue, and find a 4x4 representation √ 3/2 0 0 √0 3/2 0 1 √0 R1 = (10.288) 0 1 3/2 √0 3/2 0 0 0 √ 0 i 3/2 0 0 √ −i 3/2 0 i 2 √0 R = (10.289) 0 −i i 3/2 √0 0 0 −i 3/2 0 R3 = diag(−3/2, −1/2, 1/2, 3/2) (10.290) For later convenience we note that 1 2 2 2 3 2 (σ /2) + (σ /2) + (σ /2) (T 1 /2)2 + (T 2 /2)2 + (T 3 /2)2 (R1 /2)2 + (R2 /2)2 + (R3 /2)2 1 1 = +1 I 2 2 = 1 (1 + 1) I 3 3 +1 I = 2 2 (10.291) (10.292) (10.293) This is all very interesting, but it would not be very convenient if we had to go through all of this for each representation, we need a way of unifying the calcuation. The way to acheive this is to find a set of commuting operators, and use their eigenvalues to classify the representations. We have just seen that in the three cases we have looked at, (j 1 )2 + (j 2 )2 + (j 3 )2 is proportional to the identity, and so commutes with all the j a . In fact, one can show using the defining algebra that [(j 1 )2 + (j 2 )2 + (j 3 )2 , j a ] = 0. (10.294) This gives us our first commuting operator. The next one is a matter of choice, and it’s conventional to choose j 3 as the second commuting operator, giving us our commuting set of {j 2 = (j 1 )2 + (j 2 )2 + (j 3 )2 , j 3 } (10.295) We are not saying that these commute with everything, only that everything in the set {j 2 , j 3 } commutes. For example, we could not add j 1 to the set, as it does not commute with j 3 . As is well known, for example from quantum mechanics, we are free to assign simultaneous eigenvalues to operators that commute, and that is what we shall now do. We define j 3 to have eigenvalue m, and j 2 to have eigenvalue β, and the eigenstate will be denoted |β, mi. j 3 |β, mi = m|β, mi j 2 |β, mi = β|β, mi 43 (10.296) (10.297) Now, rather than working with j 1 and j 2 directly, we introduce the raising and lowering operators j ± defined by and note that the algebra gives us j ± := j 1 ± ij 2 (10.298) [j 3 , j ± ] = ±j ± (10.299) 2 The reason for calling these raising/lowering operators is that j 3 [j ± |β, mi] = = = = [j ± j 3 + (j 3 j ± − j ± j 3 )]|β, mi [j ± j 3 ± j ± ]|β, mi j ± [j 3 ± 1]|β, mi (m ± 1)[j ± |β, mi] (10.300) (10.301) (10.302) (10.303) Let’s explain what has just happened. We start with a state |β, mi that has a j 3 eigenvalue of m, then we showed that the state j ± |β, mi has a j 3 eigenvalue of m ± 1, i.e. j + raises the eigenvalue by one, and j − lowers the eigenvalue by one. What this means is that we now have a way of constructing all the eigenstates, once one has been given Now, suppose we are given a value of β, how should we go about constructing all the eigenstates? First of all we notice that if we choose the normalization hβ, m|β, mi = 1 then β = hβ, m|j 2 |β, mi = hβ, m|(j 1 )2 + (j 2 )2 + (j 3 )2 |β, mi > m2 (10.304) (10.305) (10.306) which tells us that there is a maximum, and minimum, value for m. So, at some value of m, usually denoted j, we will find j + |β, ji = 0, (10.307) because there are no more states with a higher eigenvalue than j by assumption. Now, a little algebra shows that j 2 = j − j + + j 3 (j 3 + 1) so we find j 2 |β, ji = [j − j + + j 3 (j 3 + 1)]|β, ji = j(j + 1)|β, ji (10.308) (10.309) but j 2 |β, ji = β|β, ji so β = j(j + 1). (10.310) Just as j is the largest possible eigenvalue for j 3 (by assumption), −j is the smallest eigenvalue. We see this by noting that j 2 = j + j − + j 3 (j 3 − 1), so j 2 |β, −ji = [j + j − + j 3 (j 3 − 1)]|β, ji = [j + j − + j(j + 1)]|β, ji 2 (10.311) (10.312) this technique has already been seen in the quantum mechanics of a harmonic oscillator, eg lec 20 of quantum World 44 but j 2 |β, −ji = j(j + 1)|β, ji so j − |β, −ji = 0. (10.313) To sum up, a given representation is specified by its j 2 eigenvalue, and a particular state within that representation is specified by its j 3 eigenvalue. As a matter of notation, it is more standard to denote states by |j, mi. Now, what are the possible values for m? We know that there are an integer number of states, with j 3 eigenvalues ranging from −j, −j + 1, ...j − 1, j, i.e. a total 2j + 1 states. As there is an integer number of states then 2j + 1 is an integer, then j must be either an integer, or half-integer. Here are some examples • j = 0 rep: The only possible value for m is zero. An example of this is the s-state of the hydrogen atom, corresponding to l = 0 angular momentum. • j = 12 rep: The possible states for this are m = ± 21 , an example being the two spin-states of an electron. We also note that this rep’ has j 2 = j(j + 1) = 12 ( 12 + 1), which matches the Pauli matrices (10.291) • j = 1 rep: The possible states for this are m = −1, 0, 1, with an example being the l = 1 p-state of the hydrogen atom. Again, note that (10.292) has j 2 = j(j + 1) = 1(1 + 1), and so is a j = 1 rep’. • j= 3 2 rep: the possible states are m = − 32 , − 12 , 21 , 23 , an example being the Ra rep (10.293) • j = 2 rep: this has m = −2, −1, 0, 1, 2, and an example is the l = 2 d-state of the hydrogen atom. • in fact, for orbital angular momentum we just make the identification j → l, and note that l must be integer 10.2 normalizing the states Although we have seen that j ± takes us from |j, mi to |j, m ± 1i, we have to be careful of normalization. From our eigenvalue calculation, all we can actually say is that j ± |j, mi ∝ |j, m ± 1i (10.314) It is convenient to choose a normalization such that hj, m|j, mi = 1 (10.315) for all m and j. So, we do this by |j, m + 1i = N j + |j, mi (j ± )† = f ∓ ⇒ hj, m + 1|j, m + 1i = |N |2 hj, m|j − j + |j, mi ⇒ 1 = |N |2 hj, m|j 2 − j 3 (j 3 + 1)|j, mi (10.316) (10.317) (10.318) ⇒ 1 = |N |2 [j(j + 1) − m(m + 1)]hj, m|j, mi (10.319) ⇒ 1 = |N |2 [j(j + 1) − m(m + 1)] (10.320) 45 we are at liberty to choose the phase of N , and we pick it to be zero, so we end up with p j + |j, mi = j(j + 1) − m(m + 1)|j, m + 1i (10.321) if we perform a similar calculation starting with |j, m − 1i = Ñ j − |j, mi we find p j − |j, mi = j(j + 1) − m(m − 1)|j, m − 1i 10.3 (10.322) general SU(2) represenation it may interest you, but most probably not, to have a general expression for SU(2) generators of all dimensions 0 dl 0 dl 0 dl−1 0 d 0 1 l−1 D1 = (10.323) , . . 2 . 0 d−l+1 d−l+1 0 0 −dl 0 dl 0 −dl−1 0 i 0 dl−1 D2 = (10.324) , . . 2 . 0 −d−l+1 d−l+1 0 l l−1 l − 2 D3 = (10.325) , .. . −l + 1 −l where dk = 11 p l(l + 1) − k(k − 1) Lecture eleven - isospin Aim: To introduce the concept of isospin Learning outcomes: At the end of this lecture you should • know how isospin relates to neutrons and protons • be able to use isospin considerations in basic scattering calculations 46 (10.326) 11.1 neutrons, protons and isospin At a very basic level, we note that the mass of a proton, p(938MeV), and neutron, n(940MeV), are very close. We also note that the excitation energy levels of mirror nuclei (where p and n 13 7 7 11 are swapped) are very similar, e.g. 13 6 C and 7 N are mirror, as are 3 Li and 4 Be, and also 5 B 11 and 6 C. Heisenberg suggested that this points to the idea that neutrons and protons are, from the perspective of nuclear dynamics, the same particle, or rather different states of the same particle. This is not such a wild idea, we do not think of the spin up and spin down states of an electron as separate particles. In fact, that analogy is rather apt, as the spin 1/2 property of an electron is going to become an isospin 1/2 property of the neutron/proton. So, we introduce the nucleon a |N i = (11.327) b with |a|2 + |b|2 = 1 to get the normalization, and identify 1 0 |pi = , |ni = 0 1 (11.328) So that |N i = a|pi + b|ni (11.329) |N⊥ i = −b? |pi + a? |ni (11.330) along with the orthogonal state which we can combine into |N i a b |pi |pi = =U |N⊥ i −b? a? |ni |ni (11.331) where we see that U is in fact an SU(2) matrix. And that is it. The point about doing all this is to show that, once again, SU(2) is at the heart of things. The proton and neutron form a doublet, i.e. a j = 12 rep, of SU(2); and we call that SU(2) isospin, with the proton being the m = + 21 state of isospin and the neutron being the m = − 12 isospin state. Because we have given this SU(2) a special name, it is also convential to use the notation j → I, m → I 3. (11.332) 1 1 |ni = | , − i. 2 2 (11.333) along with our standard notation for states 1 1 |pi = | , i, 2 2 47 11.2 pions and isospin So now we know what the I = 12 states of isospin are, what happens next? Well, from our analysis of SU(2) in general, we know that there are an infinitude of SU(2) representations, how do they make an appearnce here? For example, what do the I = 1 states correspond to? Well, the first thing we know is that there should be three particles in this multiplet, corresponding to I 3 = −1, 0, 1, and that they should all have roughly the same mass - as they are all related by an SU(2) transformation. Fortunately such a set of particles exists, the pions, π + (140MeV), π 0 (135MeV), π − (140MeV), and we identify these with the I 3 = 1, 0, −1 states of the I = 1 isospin representation. The isospin system does more for us than just classify particles, it tells us about the isospin composition of the elements within a multiplet. This concept will be of great value when we come to quarks, so we shall take a close look at how, from the isospin perspective, the pion triplet is constructed from the nucleon doublet. Given the two states |pi and |ni there are four independent combinations of their product that we may construct: |pi|pi, |pi|ni, |ni|pi, |ni|ni, and these may be arranged into the symmetric states |pi|pi, 1 √ (|pi|ni + |ni|pi), 2 |ni|ni, I=1 (11.334) to form the I = 1 triplet of pions, and 1 √ (|pi|ni − |ni|pi), 2 I=0 (11.335) to form the I = 0 singlet. From the point of view of group theory, what we have shown is that 2⊗2 = 1⊕3 (11.336) i.e. the direct product of two doublets gives us an SU(2) triplet and an SU(2) singet. It may not be obvious why we should look at the symmetric and anti-symmetric decompositions, but it is easily checked that they are the correct ones by using the generators on the product space 3 Iprod = I3 ⊗ I + I ⊗ I3 ± Iprod = I± ⊗ I + I ⊗ I± (11.337) (11.338) where the product of generators acts in the natural way, for example (I 3 ⊗ I + I ⊗ I 3 )|pi|pi = (I 3 ⊗ I)|pi|pi + (I ⊗ I 3 )|pi|pi = (I 3 |pi) ⊗ (I|pi) + (I|pi) ⊗ (I 3 |pi) 1 1 = ( |pi) ⊗ (|pi) + (|pi) ⊗ ( |pi) 2 2 1 = (|pi|pi + |pi|pi) 2 = 1 |pi|pi 48 (11.339) (11.340) (11.341) (11.342) (11.343) i.e. the state |pi|pi has eigenvalue I 3 = 1, which is what we claimed. All of this may be stated as follows for the pion triplet 1 1 1 1 |π + i = |1, 1i = | , i| , i = |pi|pi (11.344) 2 2 2 2 1 1 1 1 1 1 1 1 1 1 0 |π i = |1, 0i = √ | , i| , − i + | , − i| , i = √ (|pi|ni + |ni|pi)(11.345) 2 2 2 2 2 2 2 2 2 2 1 1 1 1 |π − i = |1, −1i = | , − i| , − i = |ni|ni (11.346) 2 2 2 2 (11.347) so, we have the nice results that the pions are constructed out of nucleons, but we get an extra state, the singlet. This is associated with the deuteron, |di, coming from the anti-symmetric product of nucleon states. 1 1 1 1 1 1 1 1 1 1 |di = |0, 0i = √ | , i| , − i − | , − i| , i = √ (|pi|ni − |ni|pi) (11.348) 2 2 2 2 2 2 2 2 2 2 (11.349) Or, equivalently |pi|pi = |π + i 1 |pi|ni = √ |π 0 i + |di 2 1 |ni|pi = √ |π 0 i − |di 2 − |ni|ni = |π i 11.3 (11.350) (11.351) (11.352) (11.353) using isospin in scattering calculations This is all very nice, but it is actually also quite useful, as it tells us something about the scattering of particles. If isopspin really is a useful quantum number, then it should be conserved, so we can use this to rule out, or constrain certain scattering rates. For example, let’s consider the two scattering processes pp → dπ + , pn → dπ 0 (11.354) The amplitude of a scattering event is given by amplitude(initial → f inal) = hf inal|Ĥint |initiali (11.355) so we need the bras and kets associated with pp, pn, dπ + , dπ 0 . 1 1 1 1 pp → |pi|pi = | , i| , i 2 2 2 2 1 1 1 1 pn → |pi|ni = | , i| , − i 2 2 2 2 + + dπ → |di|π i = |0, 0i|1, 1i dπ 0 → |di|π 0 i = |0, 0i|1, 0i 49 = |1, 1i, 1 = √ (|1, 0i + |0, 0i) , 2 = |1, 1i = |1, 0i (11.356) (11.357) (11.358) (11.359) Now, given that isospin is conserved, it commutes with the Hamiltonian, meaning that the initial and final states must have the same isospin eigenvalues, or else the amplitude will vanish. It is then immediately clear, for example, that we cannot have pp → dπ 0 , because the pp state and the dπ 0 state have different isospin eigenvalues. We can, however, have pn → dπ 0 , because the pn state contains a bit of |1, 0i. In fact, we can make a prediction for the ratio of cross sections as follows amp(pn → dπ 0 ) = amp(pp → dπ + ) √1 2 (h1, 0| + h0, 0|) Ĥint |1, 0i (11.360) h1, 1|Ĥint |1, 1i 1 = √ 2 h1, 0|Ĥint |1, 0i h1, 1|Ĥint |1, 1i + h0, 0|Ĥint |1, 0i ! h1, 1|Ĥint |1, 1i 1 h1, 0|Ĥint |1, 0i = √ 2 h1, 1|Ĥint |1, 1i 1 = √ 2 (11.361) (11.362) (11.363) the last step may not be obvious, but recall that (10.321) 1 |1, 1i = √ I + |1, 0i 2 2 − + I I = I − I 3 (I 3 + 1). (11.364) (11.365) Now that we have the ratio of the amplitudes, we find the ratio of cross sections by squaring, giving σ(pn → dπ 0 ) 1 = , + σ(pp → dπ ) 2 (11.366) which matches observations. 12 Lecture twelve - the quark model Aim: To introduce where quarks came from Learning outcomes: At the end of this lecture you should • know how to construct the baryon octet and decuplet • know the particles in the baryon octet and decuplet, and their quark content • be able to state the Gell-Mann Nishijima formula 12.1 more multiplets The success of isospin continued with the identification of the I = of particles ∆++ (1230M eV ), ∆+ (1231M eV ), ∆0 (1232M eV ), 50 3 2 quadruplet as the ∆-series ∆− (1232M eV ). (12.367) and it was noted that the known multiplets (nucleons, pions, deltas) safisfied 1 Q = I 3 + B, 2 (12.368) where Q is the electric charge, and B the baryon number. e.g. π − has I 3 = −1, B = 0, Q = −1. This was all very nice, until more particles came along. The troublesome particle was one called te lambda, Λ, which appeared without any charged partners, so was an isospin singlet with I = 0. That is fine, but it decayed via Λ → π−p (12.369) and the state on the right hand side has non-vanishing isospin (lec 11 questions). Actually it was not not a total disaster, because the decay time was ∼ 10−10 s, compared with the expected timescale for strong interactions of 10−23 s. So, although it was produced by the strong interaction, it did not decay by it. Nevertheless, as it could be produced by the strong interaction in π − p collisions it would have to be produced in pairs, so that the collision could conserve isospin. This meant that along with the Λ we also had another particle, the K 0 . In fact there were a host of pair-produced particles − + Σ K Σ0 K 0 π − p+ → (12.370) 0 ΛK Gell-Mann and Nishijima independently suggested that the new data could be explained by introducing a new quantum number, preserved by the strong interaction, strangeness, and assigned S(K + ) = 1, S(Σ− ) = −1. It was discovered that the lowest mass spin- 12 baryons fitted neatly into the following octet S=0 p(938M eV ) S = −1 Σ+ (1189M eV ) S = −2 ↑ I3 = 1 n(940M eV ) Σ− (1197M eV ) Σ0 (1192M eV ) Λ(1116M eV ) Ξ0 (1315M eV ) ↑ I 3 = 1/2 ↑ I3 = 0 Ξ− (940M eV ) ↑ I 3 = −1/2 ↑ I 3 = −3/2 This new quantum number extended (12.368) to the Gell-Mann Nishijima formula 1 Q = I 3 + (B + S). 2 (12.371) Along with the spin- 12 it was also discovered that the lowest-mass spin- 32 baryons fit nicely into 51 the following decuplet S=0 S = −1 ∆++ ∆+ Σ∗+ S = −3 Σ?0 Ξ?0 S = −2 ∆− ∆0 Σ?− Ξ?− Ω− ↑ ↑ ↑ ↑ ↑ ↑ ↑ I 3 = 3/2 I 3 = 1 I 3 = 1/2 I 3 = 0 I 3 = −1/2 I 3 = −1 I 3 = −3/2 In fact, the Ω− was actually a prediction, but they were right. Gell-Mann has subsequently commented that the next step was obvious, but he believed it to be more a mathematical trick, than a physical picture. The observation is that triangles are the base shape of both the hexagonal pattern for spin- 21 baryons, and the triangular pattern of the spin- 12 baryons; so he introduced three quarks3 . He gave them the labels u, d, c, standing for up, down and strange. Just as |ni and |pi were identified as the different states of the ”same particle”, with an SU(2) symmetry relating them, so |ui, |di, |si were considered the ”same”, but with an SU(3) symmetry relating them, called flavour; Baryons were then formed of three such quarks. We are then led to ask what we can form out of three quarks, and there are 27 possibilities: uuu,uud,udu,... However, rather than just list them in the basic form we should think bck to how our SU(2) isospin multiplets worked, for example Qu 5 of lecture 11. The multiplets form out of particular combinations of states that have symmetry under the interchange of pairs • totally anti-symmetric, ψf lavour (1): uds + dsu + sud − usd − sdu − dus (12.372) • anti-symmetric on the first pair, ψf lavour (8)12 : (ud − du)q, (us − su)q, (ds − sd)q (12.373) where q is one of u, d, s. • anti-symmetric on the second pair, ψf lavour (8)23 : q(ud − du), q(us − su), q(ds − sd) (12.374) where q is one of u, d, s. • totally symmetric, ψf lavour (10): uuu, ddd, sss, uud + udu + duu, uus + usu + suu, ddu + dud + udd, (12.375) dds + dsd + sdd, ssu + sus + uss, ssd + sds + dss 3 Zweig had a similar idea, calling them aces instead. 52 On the face of it, this gives us too many states, 29 instead of 27; however, we have over-counted. A linear combination of the states in the list that are anti-symmetric in the first pair is also toally anti-symmetric, but that one has already been counted, the same goes for the list of states that are anti-symmetric in the second pair. So, we find the final result 3 ⊗ 3 ⊗ 3 = 1 ⊕ 8 ⊕ 8 ⊕ 10 (12.376) and that is more like what we need, explaining the origin of the decuplet and the octet of baryons. However, we seem to have an embarrassment of riches, as there appears to be an extra octet and singlet that are not observed. To explain these we need to remember that quarks are spin 21 particles, and baryons are fermions so their overall wavefunction needs to be anti-symmetric. The total wavefunction consists of Ψtotal = ψspace ψspin ψcolour ψf lavour (12.377) and we know that physical states are colourless so the colour wavefunction is the totally antisymmetric colour singlet. For lowest enery states we have no angular momentum (l = 0), which is symmetric, so we are left with requiring ψspin ψf lavour to be totally symmetric. We calculated in question 5 of lecture 11 that the product of three spin- 12 states gave 2 ⊗ 2 ⊗ 2 = 4 ⊕ 2 ⊕ 2 totally anti − symmetric 1 anti − symmetric on f irst pair, ψspin ( )12 2 1 anti − symmetric on second pair, ψspin ( )23 2 3 totally symmetric, ψspin ( ) 2 : none : (↑↓ − ↓↑) ↑, (↑↓ − ↓↑) ↓ : ↑ (↑↓ − ↓↑), ↓ (↑↓ − ↓↑) : ↑↑↑, ↓↑↑ + ↑↓↑ + ↑↑↓, ↓↓↑ + ↓↑↓ + ↑↓↓, ↓↓↓ which means that we may form a totally symmetric ψspin ψf lavour out of 3 ψf lavour (10) × ψspin ( ) 2 (12.378) as each of the components are totally symmetric, or we can form the totally symmetric state 1 1 1 ψf lavour (8)12 ψspin ( )12 + ψf lavour (8)23 ψspin ( )23 + ψf lavour (8)31 ψspin ( )31 2 2 2 (12.379) and that is all. So, the ”extra” singlet and octet are in fact not allowed because the quarks are spin- 12 . Moreover, this explains why the baryon decuplet is spin- 32 (it is made out of the spin- 32 part of 2 ⊗ 2 ⊗ 2, i.e. the 4 in 2 ⊗ 2 ⊗ 2 = 4 ⊕ 2 ⊕ 2) and the baryon octet is spin- 12 (it is made out of the spin- 21 part of 2 ⊗ 2 ⊗ 2, i.e. the 2s in 2 ⊗ 2 ⊗ 2 = 4 ⊕ 2 ⊕ 2). Since the early days the quark model has been extended, as more particles were discovered due to higher energy colliders coming online, we are now at the stage where the quark model is given as in table 1 and the Gell-Mann Nishijima formula has extended to 1 Q = I 3 + (B + S + C + B 0 + T ). 2 53 (12.380) 13 Lecture thirteen - charge conjugation and parity Aim: To introduce discrete symmetries Learning outcomes: At the end of this lecture you should • know how to find the parity of photons and mesons • know the parity of fermions, anti-fermions and fermion-antifermion pairs • be able to argue that electric dipole moments of particles would violate parity 13.1 Charge conjugation We have seen that the standard model relies on continuous Lie symmetries to give us vector bosons, such as photons and gluons, but then we also need to think about discrete symmetries. For example, does the Universe care about a distinction between left and right - can we flip the Universe in a mirror, and leave it looking the same? Can we swap all particles in the Universe with their anti-particles without anyone noticing? The second of these is known as charge conjugation, C, and it acts on particle states as C|pi = |p̄i (13.381) where |p̄i is the anti-particle of |pi. We can find the eigenvalues of C by noting that C2 = 1 CC|pi = C|p̄i = |pi (13.382) so C has eigenvalues ±1. We also see that only particle that are there own anti-particles can be C eigenstates, i.e. the eigenvalue equation (13.383) C|pi = λ|pi implies that |p̄i = λ|pi. quark Q B I I3 S C d -1/3 1/3 1/2 -1/2 0 0 u 2/3 1/3 1/2 1/2 0 0 s -1/3 1/3 0 0 -1 0 c 2/3 1/3 0 0 0 1 b -1/3 1/3 0 0 0 0 t 2/3 1/3 0 0 0 0 B’ 0 0 0 0 -1 0 T 0 0 0 0 0 1 mass (MeV) 5 2 100 1,200 4,200 174,000 Table 1: Table showing quark charges 54 13.1.1 photons Recal that the photon is just the quantum version of electromagnetism, and that one, actually two, of Maxwell’s equations are ∂µ F µν = −µ0 j ν (13.384) Now, it is clear that under charge conjugation the sign of the current will change C : j µ → −j µ (13.385) so to maintain Maxwell’s equations we better have C : Aµ → −Aµ (13.386) Showing that the photon has negative intrinsic parity, and that electromagnetism preserves C. 13.1.2 fermion-antifermion pairs • Consider the decay of a lowest energy state scalar meson, i.e. l = s = 0, into two photons As electromagnetism preserves C, then the initial and final states must have the same C value, and we know that the photon has C = −1, so a final state of two photons has C = (−1)(−1) = 1, meaning that the l = s = 0 meson has C = 1. • Now suppose we consider the spin-flip of a meson, i.e. an s = 1 meson decays to an s = 0 meson plus a photon. As the photon has C = −1, and the s = 0 meson has C = 1, then the s = 1 meson must have C = (1)(−1) = −1 = (−1)s • Now suppose we have a meson in a zero spin, but non-zero angular momentum state, l, decaying to an l − 1 angular state via the emmision of a photon, the same argument as above shows that it has C = (−1)l . We combine all the above results into the single relation that C|f f¯, l, si = (−1)l+s |f f¯, l, si (13.387) This has important consequences for particle decays, and explains why certain decays do not occurr, for example the scalar meson π 0 in the l = 0 state has C = 1 so π 0 → γγ π 0 → γγγ is allowed is not allowed 55 (13.388) (13.389) 13.1.3 C violation Unfortunately, or fortunately depending on your point of view, the charge conjugation symmetry is not a symmetry of the full standard model. C is only a symmetry of the strong sector and the electromagnetic sector, with the weak nuclear force breaking charge conjugation symmetry. For example, the following decay is observed π + → µ+ (L) νµ(L) (13.390) with the muons always coming out left-handed. If we now perform charge conjugation on both sides we get π − → µ− (L) ν̄µ(L) (13.391) which looks fine, until you realize that experimentally all the muons from π − decay actually come out right-handed, so this weak sector decay (13.390) violates C. 13.2 parity The question of whether the Universe looks the same in a mirror is answered by considering parity symmetry, where we take the co-ordinates x → −x. When we do this we need to know how physical quantities change, for example the momentum p → −p, and so is an example of a polar vector - one that changes sign along with the co-ordinates. Angular momentum, however, changes according to L → L, because L = r × p; this is an example of a pseudo vector. This behaviour is not limited to vectors, scalars also split up into scalars and pseudo scalars. Electric charge is an example of a scalar as ρ → ρ, whereas the volume of a parallelipiped defined by vectors a, b and c, a · (b × c), is a pseudo scalar, changing sign under parity. 13.2.1 electromagnetism Electromagnetism is invariant under parity, and this can be seen by assigning the photon negative intrinsic parity as follows. From Maxwell’s equation ∇ · E = ρ/0 (13.392) we see that invariance of this equation under P : ρ → ρ, P : ∇ → −∇ requires P : E → −E. (13.393) However, E = −∇φ − ∂ A ∂t (13.394) showing that P : φ → φ, A → −A. 56 (13.395) Now recall that the four-vector potential Aµ = (φ, A) and we see that P : Aµ → −Aµ (13.396) so we associate a photon state with negative intrinsic parity P |γi = −|γi (13.397) Now that we know how the vector-potential behaves we can see that the magnetic field is a pseudo vector, as B = ∇ × A. 13.2.2 (13.398) fermions Fermions are a little trickier to figure out, but the result of quantum field theory is that fermions and anti-fermions satisfy P |f i = |f i, P |f¯i = −|f¯i (13.399) so fermion and anti-fermion have opposite parity. In particular, the s-state (vanishing angular momentum) of the scalar meson π 0 has negative intrinsic parity, so is actually a pseudo scalar meson. But now we seem to have a puzzle. If a meson has negative intrinsic parity, then why is the following allowed π 0 → γγ (13.400) as the final state, which consists of two photons, appears to have positive parity. The point is that the parity must take into account the wave function, as well as the intrinsic parity, so if the wavefunction of the two-photon state has negative parity, this leaves a negative parity state. For example, write a wavefunction in terms of spherical harmonics ψ(x) = φ(r)Ylm (θ, φ) (13.401) where parity corresponds to x → −x ≡ θ → π − θ, φ → φ + π. (13.402) then P : Ylm (θ, φ) → (−1)l Ylm (θ, φ). (13.403) For example, Y11 ∼ sin θeiφ → sin θ(−eiφ ). So, if the two photons are in an l = 1 state, then the overall parity is negative. Another way to think about it is to ask how the pions couple to the electromagnetic field. We know from our Feynman diagram work that if we want a single pion to decay to two photons then the Lagrangian needs to have a term containing a single pion field, and two photons, and that such a term must be a scalar. There are two choices L1int ∼ π 0 E · B L2int ∼ π 0 (E · E + BB) 57 (13.404) (13.405) In the first choice, as E is a vector and B is a pseudo vector, we must have the pion as a pseudo scalar for L1int to be a scalar. The second choice leads to the pion being a scalar, which is not what is observed. In fact, the E · B coupling actually tells us that the photons come out of the experiment with orthogonal polarizations, which is how it was figured out experimentally. All of this argument about including the spatial wavefunction into the parity assignment goes through for the fermion-antifermion pair, so in fact the parity for a meson state is P |f f¯i = (−1)l+1 |f f¯i 13.3 (13.406) linearity of parity We now do something that may seem strange, which is to calculate the parity of i, the square root of -1. Although it will give us what we think of as an obvious result, it is worth doing as we will see that things change when we look at time reversal. So, let’s start with the canonical commutator [x̂, p̂] = i~ (13.407) Now we perform a parity transformation, remembering that operators transform as P x̂P −1 if states transform as P |i. P x̂P −1 = −x̂, P p̂P −1 = −p̂ (13.408) to find P [x̂, p̂] P −1 = i~ [x̂, p̂] = P i~P −1 (13.409) (13.410) so for the commutation law to be preserved we must have P i = iP ⇒ P iP −1 = i (13.411) which is just the statement that the parity transformation acts linearly. As we usually think of i as a number this result is what we would normally write down without thinking, so that’s good, but we need to take care later on when we consider time reversal. 13.4 electron dipole moment (edm) We now think of ways that parity could be violated, or rather how we could observe its violation. Suppose that a fundamental particle, which has a non-zero spin, had an electric dipole moment, this would violate parity symmetry. There are now two possibilities, the electric dipole moment could either align or anti-align with the spin; we shall go through the argument for the aligned case, but the same holds for anti-aligned. So, we place such a particle in a state where the spin (and electric dipole moment) both point up, now perform a parity transformation. The parity transformed state would have the spin pointing up (spin is a pseudo vector so doesn’t change) and electric dipole moment pointing down. However, such a particle is not in our theory, we said 58 the particle had aligned spin and electric diole moment. So, the presence of an e.d.m. implies the violation of parity. For example, the edm of a neutrons and electrons are bounded by dneutron < 2.9 × 10−26 e cm dneutron < 1.6 × 10−27 e cm (13.412) (13.413) which, to put into perspective, compares to the size of a neutron of around 10−13 cm, i.e. dneutron < 2.9×10−13 e lneutron . So, imagine the neutron is the size of the Earth, then 10−13 e learth is 0.01e mm, corresponding to displacing an electron by 0.01mm in the centre of the Earth! 14 Lecture fourteen - parity violation and time reversal Aim: To introduce discrete symmetries, and their breaking Learning outcomes: At the end of this lecture you should • know how parity was found not to be a symmetry • know what time reversal symmetry is 14.1 parity violation In experiments that observed the decays of cosmic rays it was noted that two particles, which were otherwise identical, had two different decay channels, these were called the θ and τ θ+ → π + π 0 + 0 0 π π π + τ → π−π+π− (14.414) (14.415) If these were indeed distinct particles then this would be fine, but if not, there’s a problem with parity conservation. As pions have negative intrinsic parity then the parity of the θ+ final-state is positive, implying that θ+ has positive parity, but the parity of the τ + final-state is negative, implying the τ + parity is negative. For these to be the same particle (now caled K+ ), then the decays would have to violate parity - nobody wanted that. In 1956 Lee and Yang noted that there was an absence of data showing that parity ws conserved in the weak interaction, in contradistinction to data for the strong nuclear force and the electromagnetic force. In the same year Wu performed an experiment looking at the following decay of cobalt to nickel 60 Co → 60 N i e− ν̄e (14.416) Wu and collaborators aligned the spins of the cobalt nuclei by utilising their magnetic moment, which point in the same direction as their spin. The experiment revealed that most of the electrons were emitted in a direction opposite to the spin vector of the nucleus, a result which means that parity is not a symmetry in these decays. To understand that, think about the correlation between the direction of nuclear spin, and the momentum of the electrons, Corr = 59 hs · pi. If parity is a symmetry then Corr should have the same value before and after a parity transformation, however, P : Corr → −Corr as s is a pseudo vector, and momentum a polar vector. Another way to see this is to imagine aligning the nuclear spins pointing up, then the experiment says that the electrons are mostly emmitted downwards. If we perform a parity transformation the spin still points up, but now the electrons have reversed their momentum, and are also moving upwards. So, the actual experiment and its parity-reversed version are different, so explicitly violating parity4 . Another place that parity is seen to be broken by the weak sector is in the decay of pions to muons. π + → µ+ L νµ,L (14.417) where it is observed that the muon always comes out left-handed. If parity were a good symmetry of this decay then we would expect π + → µ+ R νµ,R , as this is the parity reversed version of the observed decay. Similarly, we observe π − → µ− R ν̄µ,R (14.418) but not its parity-reversed partner. The underlying reason for this is that all neutrinos are left-handed, and all anti-neutrinos are righthanded - that’s about as parity violating as you can get! 14.2 time reversal symmetry Although it may appear that at first-sight that nature does not have a time-reversal symmetry, shattered glass never recombines to become whole again, this is really just because of entropy, rather then an asymmetry of the fundamental interactions. If we think about Newton’s laws there is no problem in taking t → −t. Now we want to ask whether quantum mechanics has anything to say about it. Suppose that we have some map that takes un-primed states |ψi to primed states |ψ 0 i, for all states. Then Wigner showed that if the inner product is preserved |hφ|ψi|2 = |hφ0 |ψ 0 i|2 (14.419) we must have either |ψ 0 i = U |ψi, or |ψ 0 i = A|ψi (14.420) where U is a unitary and linear operator, and A is anti-linear and anti-unitary. The terms unitary and so on are defined by U [α|ψi + β|φi] A[α|ψi + β|φi] hU ψ|U φi hAψ|Aφi = = = = αU |ψi + βU |φi, α? A|ψi + β ? A|φi, hψ|φi, hφ|ψi = hψ|φi? , 4 linear anti − linear unitary anti − unitary (14.421) (14.422) (14.423) (14.424) note that a lot of texts get this the wrong way around, using arguments like the spin changes direction in a mirror - be careful. 60 So, the unusual thing about an anti-linear operator is that when you move it past a complex number, you have to complex-conjugate it, i.e. they don’t commute with numbers! To see why this makes an appearance in quantum mechanics consider again the canonical commutator [x̂, p̂] = i~ (14.425) and now we perform a time-reversal transformation with operator T̂ , T̂ x̂T̂ −1 = x̂, T̂ p̂T̂ −1 = −p̂ (14.426) so, whereas parity changes the sign of both x and p, time reversal changes only the momentum, meaning that T̂ [x̂, p̂] T̂ −1 = T̂ i~T̂ −1 [x̂, p̂] = −T̂ i~T̂ −1 (14.427) (14.428) So, if we are to maintain a time reversal symmetry in quantum mechanics, i.e. keep the commutation relations, then we must have T̂ iT̂ −1 = −i, T̂ i = −iT̂ (14.429) so the time reversal operator is anti-linear. This result also manifests itself in the behaviour of wavefunctions under time reversal, consider for example te plane wave T : ψ = e−i(Et−px) → ei(Et−px) = ψ ? (14.430) because t → −t and p → −p. So, we see that effectively i → −i. Under time reversal the spin changes sign, but the electric dipole moment does not so, just as a non-zero edm signals a breakdown of parity symmetry, it also implies a violation if time-reversal invariance. T may also be checked using a principle called detailed balance, for example, the scatterings np → Dγ, and Dγ → np should occur with the same rate if T is a symmetry. 15 Lecture fifteen - CP symmetry Aim: To introduce a product discrete symmetries, and its breaking Learning outcomes: At the end of this lecture you should • recognise how the electroweak sector (almost) preserves CP • know how C and P act on the neutral kaon system • be able to calculate the CP violation in te neutral kaon system. 61 (14.431) 15.1 CP symmetry We have seen that the weak sector violates both C and P individually, but there is still the possibility that together they are preserved. For example, the observed decay π + → µ+ L νµ,L becomes the following under C and P and the product CP P + + π + → µ+ L νµ,L −−−→ π → µR νµ,R yC yC (15.432) P − − π − → µ− L ν̄µ,L −−−→ π → µR ν̄µ,R And in fact the CP (or PC) transformed decay is observed, so maybe the weak sector preserves CP even though each discrete symmetry is broken. In fact, it was argued by Landau that CP, rather than P, is the correct version of the discrete symmetry associated to mirror reflections (see the questions associated with this lecture), and if we take that view, then maybe the Universe is mirror-symmetric afterall. 15.2 neutral kaons A good laboratory for studying CP symmetry is the neutral kaon system, it is interesting because the the K 0 is able transform to its anti-particle, K̄ 0 . This is achieved through the following processes, d s d s u W W , u W u u (15.433) W s s d d where we start with a K 0 (a ds̄ bound state), and evolve to a K̄ 0 . What this effectively means is that from the perspective of the weak interaction the K 0 and K̄ 0 are sort of the sme particle. However, as far as the strong force is concerned, they are definately different, for example K 0p → K +n (15.434) is allowed through the strong force as it conserves strangeness, however, the strong force does not allow K̄ 0 p → K + n, as this would require a change in strangeness. 15.3 kaons and C, P and CP as kaons are mesons (quark-antiquark pairs) then they have negative intrinsic parity, so we have P |K 0 i P |K̄ 0 i C|K 0 i C|K̄ 0 i = = = = 62 −|K 0 i −|K̄ 0 i −|K̄ 0 i −|K 0 i (15.435) (15.436) (15.437) (15.438) the sign change in the charge-conjugation operation is just convention. In particular, we note that kaons are not CP eigenstates CP |K 0 i = |K̄ 0 i CP |K̄ 0 i = |K 0 i (15.439) (15.440) We now consider the view that the weak interaction preserves CP and see if that is consistent with experiment. So, the picture is that kaons are produced by the strong force (so are eigenstates of strangeness), and then they decay via the weak force, which by assumption means they will be CP eigenstates. Although the kaons themselves are not CP eigenstates, it is easy to construct them, we call the CP eigenstates |K1 i and |K2 i, 1 |K1 i = √ (|K 0 i + |K̄ 0 i) 2 1 |K2 i = √ (|K 0 i − |K̄ 0 i) 2 (15.441) (15.442) so CP |K1 i = |K1 i CP |K2 i = −|K2 i (15.443) (15.444) These are the quantum states that are relevant for the weak decay of kaons, assuming the weak force preserves CP. Kaons decay mostly into pions, and CP conservation implies K1 → ππ K2 → πππ (15.445) (15.446) as pions have negative intrinsic parity. This is useful because the decay into three particles is slower (0.5 × 10−7 s) than into two pions (0.9 × 10−10 s). So, Cronin, Fitch and collaborators decided to set up a beam of kaons and watch what they decayed to. Given that the K1 decays so much faster than K2 , if we have a long enogh beam, then all of the k1 will have decayed by the end of the beam-pipe. So, we should expect only πππ decays at the end of the beam - if CP is conserved in the weak sector. However, experiment showed that about 1 in 500 of the kaon decays at the end of the pipe was into two pions, direct evidence of CP violation in the weak sector. 15.4 Klong and Kshort decays So, now that we know CP is violated, it does not make sense to analyse the system in terms of weak eigenstates, but rather we should use mass (or lifetime) eigenstates. Given the smallness of CP violation however, we know that the mass eigenstates are close to the CP eigenstates, so we write 1 |KS i = p (|K1 i − |K̄2 i) 1 + ||2 1 (|K2 i + |K̄1 i) |KL i = p 1 + ||2 63 (15.447) (15.448) Now we look for decays of KL into final states that are CP-conjugates of each other, for example the decays KL → π − e+ νe , KL → π + e− ν̄e , rate Γ1 rate Γ2 (15.449) (15.450) are allowed and, as we shall now see, allow us to distinguish between matter, and antimatter. • Γ1 : K 0 can decay to π − e+ νe , but not to π + e− ν̄e , via W s d 0 + − • Γ2 : K̄ can decay to π e ν̄e , but not to π − e+ νe , via W s e νe (15.451) u d e νe (15.452) u d d What all this means is that the rate Γ1 is controlled by the amount of K 0 in KL , and the rate Γ2 is controlled by the amount of K̄ 0 in KL , but we know that 1 1 |KL i = √ p (1 + )|K 0 i − (1 − )|K̄ 0 i 2 1 + ||2 (15.453) so if we define ∆ := Γ1 − Γ2 Γ1 + Γ2 (15.454) to distinguish between the rates we find ∆ = |1 + |2 − |1 − |2 ∼ 2 Re() |1 + |2 + |1 − |2 (15.455) experiments give ∼ 10−3 , so Γ1 > Γ2 64 (15.456) 15.5 matter vs anti-matter We now have a way of distinguishing matter from anti-matter. This is really quite remarkable, it would seem entirely natural to assume that our Universe could have been made from anti-matter rather than matter, without making any difference, but this is not the case - an anti-matter Universe is different to a matter Universe. The way we can define anti-matter uniquely is to say that it is the stuff associated with the more common lighter charged product of KL decay, i.e. e+ . In fact, that there is an anti-symmetry between matter and anti-matter is a good thing for us, otherwise we would expect our Universe to be half matter and half anti-matter, in fact we observe a baryon to photon ratio of bb − nb̄ ∼ 6 × 10−10 nγ (15.457) compared to around 10−40 that we would expect simply from statistical fluctuations. 16 Lecture sixteen - quark mixing Aim: To extend the idea of state mixing to quarks Learning outcomes: At the end of this lecture you should • know about the three generations of the standard model • know how the Cabbibo mechanism allows for generation-changing processes • be able to estimate the ratio of rates for certain basic processes. 16.1 three generations The lepton sector of the standard model neatly splits into three generations, ντ νµ νe . , , τ− e− µ− (16.458) and in terms of Feynman diagrams the basic interactions occur with strength gw νµ νe u gw W gw W gw . W (16.459) e µ d0 In particular, there is no weak decay that allows an electron to turn into a W and a νµ , weak interactions do not mix the lepton generations. The next observation is that the quark sector also has three generations u c t , , . (16.460) d s b 65 which makes us think there could be a correspondence, i.e. does the e− → W − νe , with strength gw , diagram imply a d → W − u diagram of strength gw ? Moreover, does the non-existence of weak-sector generation process imply the impossibility of d → W − c? It turns out that this intuition is close, but not quite right. For example, along with π − → W − → µ− ν̄µ (16.461) which just involves u, d quarks, we also find K − → W − → µ− ν̄µ (16.462) which shows that the W must couple to both u and s, as K − is a s̄u meson. 16.2 quark mixing We have already seen the importance of mixing quantum states in particle physics, this was how we explained the neutral kaon decays; here we just apply that notion to the quark sector. The point is that the weak and the strong forces are distinct, the interaction eigenstates of one need not correspond to the interaction eigenstates of the other. The correct quark weak-interaction eigenstates are actually a linear combination of the strong-interaction eigenstates, i.e. the weak nuclear force sees d0 , s0 , t0 rather than d, s and t. To keep things simpler, let’s just think about the first two generations, in which case one has 0 d d cos θc sin θc (16.463) = 0 s − sin θc cos θc s where θc is the Cabbibo angle, i.e. they are related by a unitary matrix. The correct Feynman diagrams are then c u gw gw W W (16.464) d0 s0 We may now, for example, test this idea through measurements of rates such as Γ(K − → W − → µ− ν̄µ ) Γ(π − → W − → µ− ν̄µ ) these rates involves the diagrams µ u µ u W (16.465) W (16.466) νµ νµ d0 sin θc + s0 cos θc d0 cos θc − s0 sin θc − The point being that the K is a particle in an eigen state of the strong interaction, K − = us̄ = u(d¯0 sin θc + s̄0 cos θc ), and the pion is also a strong-force eigenstate, π − = dū = (d0 cos θc − s0 sin θc )ū. Now, in the first diagram there is an us0 W vertex, but there is no such coupling, so 66 the first diagram contributes gw sin θc to the amplitude, as there is a ud0 W vertex in the standard model and it has strength gw . In the second diagram the ūW s0 vertex gives no contribution, meaning that the whole diagram contributes gw cos θc to the amplitude. So, Γ(K − → W − → µ− ν̄µ ) (gw2 sin θc )2 = Γ(π − → W − → µ− ν̄µ ) (gw cos θc )2 = tan2 θc (16.467) (16.468) and experiments show that θc ∼ 13◦ (16.469) Another way of stating this is to say the weak coupling strengths for quarks are gud = gcs = gw cos θc gus = −gcd = gw sin θc (16.470) (16.471) Now that we understand the coupling of quarks toνthe W boson, we use the weak interactions e e gz Z gz Z (16.472) gz Z (16.473) νe u e to predict the strong interactions d0 gz Z u d0 i.e. an interaction Lagrangian of the form Lint = ... + gz Z(uu + d0 d0 + cc + s0 s0 ) = ... + gz Z(uu + dd + cc + ss) (16.474) (16.475) and therefore Feynman diagrams of the form d gz Z d There are no neutral weak interactions that change, for example, strangeness. 17 Lecture seventeen - neutrino oscillations Aim: To extend the idea of state mixing to neutrinos Learning outcomes: At the end of this lecture you should • know about the solar neutrino problem • know about atmospheric neutrinos • be able to calculate neutrino oscillations due to their mass difference 67 (16.476) 17.1 Solar neutrinos Continuing the theme of state mixing we now examine neutrinos. Ever since their introduction by Pauli, in order to conserve energy, neutrinos hve been rather mysterious, and there are still some rather basic properties of neutrinos that are not understood. Our story here, however, goes back to 1968 with the Homestake experiment of Ray Davis. He and his team put in a herculean effort to measure the number of neutrinos coming from the nuclear fusion events in the Sun, in an effort to test the prediction of Bahcall and collaborators; a more recent version of the predicted flux is given in Fig. 1. There are a number of fusion processes in the Sun that Figure 1: The predicted flux of solar neutrinos, Bahcall et. al. ApJ, 621, L85 (2005). lead to neutrinos, the largest one being the pp fusion pp → de+ νe ppe− → dνe (17.477) (17.478) another key one, from an experimental point of view, is 8 B →8 Be∗ e+ νe (17.479) What we see from Fig. 1 is some rather extreme numbers, with around 60 billion neutrinos going through your small fingernail each second! And yet we do not notice them. The reason we do not feel some huge pressure due to this flux is that they don’t interact very strongly with 68 matter. To a very good approximation they al go straight through the Earth without noticing it is there - so a rather delicate experiment is need to measure them. Although most of the solar neutrinos do go through the planet, some of them hit stuff, and Davis’ team relied on the reaction νe 37 17 Cl → e− 37 18 Ar (17.480) to detect them. They used 615 tons of chlorine (essentially just cleaning fuid), waited for a few weeks, then measured how much of the chlorine had turned into argon. They ended up with 33 argon atoms. This is an astonishing acheivement, and when they reported that this was a factor of three smaller than predicted, nobody was particularly surprised - it would be quite easy to miss a few atoms in 615 tons of cleaning fluid. There were, however, some people who took the experiment seriously, one of whom was Pontecorvo who, in 1968, suggested that the neutrinos were all there, but the electron neutrinos had rotated to muon neutrinos. This is just the same basic physics as that behind the kaon decays, and the Cabbibo model - the physical neutrino states are really a superposition of electron, muon and tau neutrinos. And, if they have a different mass, then the state oscillates between them. Since the homestake experiment the tanks have got bigger, and the liquids have changed. There are two key experiments that really made people believe that Davis was right, the Solar Neutrino Observatory at Sudbury, using heavy water in its tank which is sensitive to all three neutrinos via νe d → ppe− νe,µ,τ d → npνe,µ,τ νe,µ,τ e− → νe,µ,τ e− (17.481) (17.482) (17.483) and the superK observatory in Japan, with 50,000 tons of water which is also sensitive to all three neutrino flavours through elastic scattering νe,µ,τ e− → νe,µ,τ e− 17.2 (17.484) neutrino oscillations For simplicity we shall shall only consider two neutrino flavours, electron and muon, and instead of writing the neutrinos as flavour eigenstates, we consider the Hamiltonian eigenstates ν1 and ν2 , as these are the objects whose evolution is simplest. ν1 cos θ − sin θ νe = (17.485) ν2 sin θ cos θ νµ then we know that ν1 (t) = ν1 (t = 0)e−iE1 t , ν2 (t) = ν2 (t = 0)e−iE2 t (17.486) We know that the solar neutrinos start life as electron neutrinos, as those are what the nuclear fusion events produce, so νe (t = 0) = 1, νµ (t = 0) = 0 ⇒ ν1 (t = 0) = cos θ, ν2 (t = 0) = sin θ 69 (17.487) (17.488) so we find νµ (t) = sin θ cos θ −e−iE1 t + e−iE2 t (E2 − E1 )t 2 2 2 ⇒ |νµ (t)| = sin (2θ) sin 2 (17.489) (17.490) so we see explicitly that the amount of muon neutrino oscillates with time. Now we may make some approximations, namely that because neutrinos are very light compared to their energy we have q q E2 − E1 = p2 + m22 − p2 + m21 (17.491) 1 1 ' |p|(1 + m22 /|p|2 ) − |p|(1 + m21 /|p|2 ) 2 2 2 2 m2 − m1 ' 2|p| (17.492) (17.493) As the neutrinos are highly relavistic we have |p| ∼ E, and t ∼ x, so P (νµ ) ' sin2 (2θ) sin2 L = π x 2L , 2πE ∆m2 (17.494) (17.495) where we should be careful to note that ∆m2 = m22 − m21 , not (m2 − m1 )2 . Thus we find that for oscillations to occur we need to properties • there must be mixing between the neutrino flavours, as happens in the quark sector, i.e. θ = 0. • the neutrinos must have different masses, ∆m2 6= 0 For three neutrinos the basic physics is the same, but the maths gets more complicated, in particular, the 2x2 matrix relating mass eigenstates to flavour eigenstates becomes a 3x3 unitary matrix. This matrix is called the MNS (Maki, Nakagawa, Sakata) matrix, and is parametrized by three angles5 cos θ13 0 sin θ13 cos θ12 sin θ12 0 1 0 0 0 1 0 − sin θ12 cos θ12 0 UM N S = 0 cos θ23 sin θ23 0 − sin θ23 cos θ23 − sin θ13 0 sin θ13 0 0 1 The results of all the experiments to date reveal that θ12 = 34.5◦ ± 1.4◦ , θ23 = 43.1◦ ± 4◦ , θ13 < 10◦ ∆m223 = (2.4 ± 0.5) × 10−3 eV 2 ∆m212 = (8 ± 0.5) × 10−5 eV 2 , 5 actually there’s also a phase, but I shall ignore it. 70 (17.496) (17.497) 17.3 atmospheric neutrinos As well as solar neutrinos there is another abundant source of freely-available neutrinos, sourced by cosmic rays. These arise from cosmic rays striking a nucleus of some atom in the atmosphere, producing a shower of (mostly) pions. These pions then decay to muons and muon neutrinos, and these muons then decay to electrons as well as muon and electron neutrino as shown in Fig. 2. The first, immediately obvious, thing to note from Fig. 2 is that there are twice as many cosmic ray pions stuff µ− νµ νe νµ e− Figure 2: The typical shower of particles produced by a cosmic ray. muon neutrinos as electron neutrinos, however, the observation of solar neutrinos by superK shows that roughly equal number of νe and νµ are actually observed, with roughly the correct amout of νe . Our knowledge of mixing then tells us that the muon neutrinos must be oscillating mostly into tau neutrinos. 18 Lecture eighteen - Higgs mechanism and symmetry breaking Aim: To discovery a way of giving particles a mass Learning outcomes: At the end of this lecture you should • be able to show how symmetry-breaking gives a mass to gauge bosons, and to fermions 18.1 symmetry restoration/breaking A nice intuitive way to think about symmetry breaking is to think back to ferromagnets, such as a bar magnet, which is a lump of material composed of many atoms, each of which is a tiny magnetic dipole. At high temperatures all these little dipoles point in random directions as the interaction energy between dipoles is overwhelmed by termal energy. This means no direction is picked out, and so we say that the system has a rotational symmetry - when averaged over a macroscopic volume. One way to phrase this is to ask what is the average angle of the dipoles, φ̄ = hφi, and then for this symmetric state we have hφi = 0. Now suppose we reduce the temperature, such that the thermal energy is below the energy scale of the dipole-dipole interaction. In this situation it is energetically favourable for the little 71 dipoles to point in the same direction and then, as a direction has been selected, we say the rotational symmetry is broken. In terms of the mean angle of the dipoles we thus have hφi = 6 0 in the symmetry-broken phase. This may not seem relevant to mass generation, but stick with me, I want to highlight the fact that symmetry breaking, which will eventually give us the standard model masses, is in fact a rather common effect. 18.2 superconductivity Another place where symmetry breaking plays an important role is in superconductors. Here, however, it is not the rotation angle of a magnetic moment that’s important, but a rather more abstract quantity - the boson made by pairing two electrons together. In a metal one has lots of electrons floating around and, as they are fermions, they occupy different quantum states. If we think of these electrons as being in a lattice, then there can actually be an attractive force between electrons, due to lattice vibrations6 . This allows pairs of electrons to become associated, forming a boson, which can then Bose-condense. We can loosely think of the order parameter in this case as φ̄ = heei, with a non-zero value in the superconducting phase, and a zero value in the normal phase. Here is where the hint comes that symmetry breaking, φ̄ 6= 0, has something to do with mass. It was known for some time that superconductors repel magnetic fields, they cannot penetrate a superconductor, this is why superconductors can hover above magnet, as in Fig. 3. The physical reason for this is that photons aquire an effective mass inside the superconductor, Figure 3: This shows a superconductor hovering above a magnet. and so it costs them energy to be inside. The idea behind the Higgs mechanism is to give gauge bosons, and other particles, a mass by exploiting a non-zero order parameter, just as photons get an effective mass inside a superconductor when φ̄ 6= 0. 6 Superconductivity is not part of this course, so I will not go into detail about the pairing mechanism. 72 18.3 massive fields Before we give a mass to particles we should figure out what we mean by mass. Going back to the Klein-Gordon Lagrangian we see that the mass of the KG field comes from the quadratic, non-derivative term of the Lagrangian, 1 1 LKG = − ∂µ φ∂ µ φ − m2 φ2 2 2 (18.498) For the complex KG field the mass also came form the quadratic, non-derivative piece LKG,cpx = −∂µ φ∂ µ φ? − m2 φφ? (18.499) And, finally, we have the Dirac field, whose mass similarly comes from the quadratic, nonderivative piece of the Dirac Lagrangian Lψ = −ψ̄γ µ ∂µ ψ − mψ̄ψ (18.500) So, the theme is that mass is just the quadratic, non-derivative piece of the Lagrangian. At which point we just write down 1 LA = ... − m2 Aµ Aµ 2 (18.501) in order to give the photon a mass. Easy. The problem with doing that is that this term breaks the gauge invariance that we worked so hard to acheive, i.e. if we take Aµ → Aµ − 1q ∂µ Λ, then this mass term is not invariant. However, we have sort of already solved this problem, in the form of covariant derivatives. Recall from lecture 4 that to construct a gauge invariant action, all we need to do is replace derivatives with their covariant counterparts, so the gauge invariant complex KG Lagrangian becomes LKG,cpx = −Dµ φDµ φ? − m2 φφ? − Vint = ∂µ φ∂ µ φ̄ − iqAµ (φ̄∂µ φ − φ∂µ φ̄) − (q 2 φ̄φ)Aµ Aµ − m2 φφ̄ − Vint (18.502) (18.503) and the important term is this one LKG,cpx = ... − (q 2 φφ? )Aµ Aµ ... (18.504) So, if we can arrange for φ̄φ 6= 0, this leads to an effective mass for the photon of 1 2 m = q 2 φφ? . 2 γ 18.4 (18.505) breaking the symmetry Before we break the symmetry in the complex KG system, we should make sure we understand what that symmetry is. Recall that the construction of electromagnetism involved a U(1) symmetry, φ → φ0 = e−iΛ φ 73 (18.506) so, for example, if we had φ = 0 then we would get φ0 = φ, as both vanish. This is what we mean by the symmetry not being broken. On the other hand, if we had φ = η, for some non-zero constant η, then clearly φ0 6= φ, which is what we mean by the symmetry being broken. So we see that, as with magnetism and superconductivity, the non-vanishing of the order parameter is just a statement about the symmetry being broken. The way we get the symmetry to break is by writing down a suitable interaction potential, the typical choice of the full potential, i.e. quadratic piece plus interaction piece, is λ V (φ) = (18.507) (|φ|2 − η 2 )2 4 which has a shape somewhat reminiscent of a wine bottle bottom as in Fig. 4. As we can see, Figure 4: A typical wine bottle, where the dregs of the wine tend to form at the bottom, around the circular minimum. this potential is positive, and is minimized at |φ| = η. (18.508) So, the potential is minimized at a particular value for the modulus of φ, but the angle is not given. This is just like the ferromagnet, the little dipoles in the broken phase could have pointed anywhere, but they must make a choice, and when they do the symmetry is broken. What we have found, then, is a way of creating φ? φ 6= 0, which is precisely what we needed in (18.505) to give the photon an effective mass - and we have done it in a way the preserves gauge invariance. 74 18.5 non-Abelian Higgs mechanism Once the Abelian - U(1) - version is understood it is easy to generalize it to the non-Abelian case, just use the covariant derivatives (9.275), where φ is now a column vector (9.272). Recall that the Lagrangian has the form L = −(Dµ φ)† Dµ φ + ... = −φ† A†µ Aµ φ + ... (18.509) (18.510) and again, for a non-zero value of φ this term just looks like a mass term for the gauge boson (i.e. quadratic in Aµ ). In the weak sector of the standard model one has to work a little bit harder because the gauge group is a product of groups, SU(2)×U(1), but the principle is the same. In fact it is the product structure that allows the W and Z to have a different mass, Mw ' 80GeV , Mz ' 91GeV , and the photon is not quite the U(1) that appears in SU(2)×U(1), but it lives partly in the U(1) and partly in the SU(2). The SU(3) gluons remain massless. 18.6 Yukawa coupling and fermion masses The Higgs mechanism is not just limited to giving mass to gauge bosons, it also provides a natural way of generating fermion mass. Consider, for simplicity, a model that has a real KG field and a Dirac field, and no gauge bosons. We could then couple the KG scalar to the fermion by an interaction term of the form Vint = gφψ̄ψ. (18.511) This type of interaction (linear in scalar and quadratic in fermion) is called a Yukawa interaction. Again, we see immediately that if we had φ = η, then mψ = gη. (18.512) so, once more, if φ 6== 0 we generate a mass term for some species - in this case a fermion. This highlights the importance of the Higgs field. In the standard model the Higgs field is responsible not just for the masses of the gauge bosons (the W and Z), but also for the fermions; it has a lot of work to do for something we don’t actually know exists! 19 Lecture nineteen - the real standard model Aim: To have a glimpse of what the standard model is actually like Learning outcomes: At the end of this lecture you should • be scared 75 19.1 electroweak sector, part I Over the course of this module we have introduced a number of different ideas, all of them with the purpose of understanding the standard model. As you may expect, although the standard model is simple compared to what it could have been, it is still rather complicated. In this lecture we shall bring together some of the concepts from earlier lectures and see how they fit into the standard model, or at least the electroweak sector. The first thing we note is that all the neutrinos that have been observed have been leftchiral, and anti-neutrinos are all right-chiral. We also know that there is a Feynman diagram connecting e νe W − , but this can’t be the full story if the neutrinos are only left-chiral, the ν e actual Feynman diagram connects e(L) νe(L) W − . This means that the doublet should e νe(L) and Re = e(R) . The fact that the right-hand component really be split up into Le = e(L) of the electron couples only to the photon and Z0 whereas the left-hand component couples to photon, W and Z, means that Le and Re have different covariant derivatives. Moreover, it also complicates the possible mass terms as we are not allowed to have L̄e Le or R̄e Re because they both vanish - mass terms must contain one of each chirality. However, we are also not allowed to have R̄e Le as Re contains two fermi fields, but Le contains only one so the coupling does not make sense. Given that the electron does have a mass there must be a resolution to this. 19.2 chiral theories The term chiral theory refers to theories where the left and right component of the same fermi field have different gauge symmetries. To see how this works let’s consider a fermion coupled to a complex KG scalar. The starting point is the ungauged Lagrangian g L = −ψ̄γ µ ∂µ ψ − ∂µ φ? ∂ µ φ − φψ̄ψ (19.513) 2 which we write as L = −ψ̄L γ µ ∂µ ψL − ψ̄R γ µ ∂µ ψR − ∂µ φ? ∂ µ φ − gφψ̄L ψR (19.514) Now that we have it in this form we start thinking about the symmetries. If it weren’t for the Yukawa term there would be three undependent U(1) symmetries, one each for ψL , φR and φ. However, for the Yukawa term to be invariant we find that one of these three is not independent, leading to the full set of symmetry transformations ψL → eiqL α ψL ψR → eiqR α ψR φ → eiqφ α φ (19.515) (19.516) (19.517) under which the Lagrangian remains invariant if qφ = qL − qR . So, we are free to give the left and right components a different charge when we gauge this symmetry, so long as the scalar has the correct charge. This means that if we promote α to α(x) then L = −ψ̄L γ µ (∂µ − iqL Aµ )ψL − ψ̄R γ µ (∂µ − iqR Aµ )ψR −(∂µ φ − iqφ Aµ φ)? (∂ µ φ − iqφ Aµ φ) − gφψ̄L ψR 76 (19.518) is invariant. This model then shows us how to make it work in the standard model, even though the left and right fermions transform differently, we soak up that difference in the mass term (the Yukawa piece) with a scalar field that transforms appropriately. 19.3 electroweak sector, part II We know follow the same proceedure as above, but for the standard model, and start by writing down the ungauged version L = −L̄e γ µ ∂µ Le − R̄e γ µ ∂µ Re − ∂µ Φ† ∂ µ Φ − g L̄e ΦRe (19.519) (19.520) where, at this point, we don’t really know what Φ needs to be. There are, again, two symmetries in this lagrangian, but now they are an SU(2) and a U(1) symmetry. The SU(2) comes from Le being a doublet, composed of both electron and neutrino fields; under this symmetry Re remains inert. The correct charge assignments for the standard model are SU (2) : U (1) : Φ → UΦ Le → U L e Re → Re Φ → e−iα/2 Φ Le → eiα/2 Le Re → eiα Re (19.521) where U ∈ SU (2), so that when we gauge the symmetry we need the following covariant derivatives i (19.522) Dµ Φ = ∂µ Φ + g 0 Bµ Φ + igAµ Φ 2 i Dµ Le = ∂µ Le − g 0 Bµ Le + igAµ Le (19.523) 2 Dµ Re = ∂µ Re − ig 0 Bµ Re (19.524) φ1 - this is the Higgs boson and we find that Φ is a two component complex scalar, Φ = φ2 that LHC is hoping to find. Note also that Aµ is the non-Abelian gauge boson associated to i SU(2) - and so is a matrix given by Aµ = σ2 Aiµ - while Bµ is the Abelian gauge boson. 19.4 electroweak sector, part III We are still not quite there as we haven’t come across the photon, the W± or the Z0 yet.; to do that we need to break the SU(2)×U(1) symmetry. This is achieved, as usual, by giving 2 2 an expectation value to the scalar field, conventionally taken as |Φ| = ν /2. So, if we take 0√ Φ= the symmetry will be broken, and we find ν/ 2 ν2 (Dµ Φ) Dµ Φ = ... + (gA3µ − g 0 Bµ )(gA3µ − g 0 B µ ) + g 2 (A1µ A1µ + A2µ A2µ ) 8 ? 77 and by introducing g 0 = q tan θw e = g sin θw = g 0 cos θw õ = cos θw Bµ + sin θw A3µ (19.525) (19.526) (19.527) Zµ = − sin θw Bµ + cos θw A3µ 1 Wµ± = √ (A1µ ± iA2µ ) 2 (19.528) (19.529) we see that this becomes 1 1 (Dµ Φ)? Dµ Φ = ... − m2z Z µ Zµ − m2w Wµ+ W −µ − mγ õ õ 2 2 (19.530) where 1 p 2 ν g + g 02 2 1 gν = 2 = 0 mA = (19.531) mW (19.532) mγ (19.533) i.e. we have just found a theory that gives us a massless photon, Ã, two W± vector bosons of the same mass, and a single heavier Z vector boson. Moreover, we may calculate the fermion covariant derivatives to find Dµ Re = ∂µ Re − ig 0 cos θw õ Re + ig 0 sin θw Zµ Re (19.534) 1 + − + − 2 i σ Wµ + Wµ iσ Wµ − Wµ √ √ Dµ Le = ∂µ Le − g 0 cos θw õ − sin θw Zµ Le + ig − Le 2 2 2 2 2 σ3 +ig (19.535) sin θw õ + cos θw Zµ Le 2 i g 1 sin2 θw I + cos2 θw σ 3 Zµ Le = ∂µ Le − ie (I − σ 3 )õ Le + 2 2 cos θw ig ig + √ (σ 1 − iσ 2 )Wµ+ Le + √ (σ 1 + iσ 2 )Wµ− Le (19.536) 2 2 2 2 and so we find that, as Re = eR , R̄e γ µ Dµ Re = ēR γ µ ∂µ eR − ieēR γ µ eR õ + ig 0 sin θw ēR γ µ eR Z µ (19.537) (19.538) showing that the right-handed fermion, Re couples to the photon with strength e, and to the Z-boson with strength g 0 sin θw , but not to the W, as required. We also have that, as νe(L) Le = , the left-chiral part gives e(L) g ν̄L γ µ νL Zµ (19.539) 2 cos θw g(2 sin2 θw − 1) g g +i ēL γ µ eL Zµ + i √ ēL γ µ νL Wµ+ + i √ ν̄L γ µ eL Wµ− 2 cos θw 2 2 L̄e γ µ Dµ Le = ν̄L γ µ ∂µ νL + ēL γ µ ∂µ eL − ieēL γ µ eL õ + i 78 and this one shows that the left-chiral part of the electron couples to the photon with the same strength as the right-chiral part, i.e. e. As well as this we see that the neutrino does not couple to the photon, as should be expected - it has no electric charge. The Z-boson couples to both the electron and the neutrino, and we have a coupling between electron-W-neutrino. What you should take away from this is that although it’s a bit scary at first sight, on reflection it just contains physics that we have seen before in the module: chirality; Abelian and non-Abelian gauge symmetry; Higgs mechanism. Given that the above Lagrangian is supposed to describe the Universe, I think we have got away with a remarkably simple theory, it could have been a lot worse. 79