UNIVERSITYOF ZURICH
NUCLEARAND PARTICLE PHYSICS Summary Contents
1 Recap1 1.1 Four vectors...... 1 1.1.1 Space-vector...... 1 1.1.2 Velocity-vector...... 2 1.1.3 Momentum-vector...... 2 1.1.4 Mandelstam variables...... 3 1.1.5 Rapidity...... 3 1.2 Quantum Mechanics...... 4 1.2.1 Clebsch-Gordon coefficients...... 5
2 Introduction8
3 Fermi’s golden rule8 3.1 Lifetime...... 8 3.2 Cross section...... 9 3.3 Luminosity...... 9 3.4 Mean free path...... 10 3.5 Decay rate...... 10 3.5.1 Lorentz invariance...... 13
4 Liquid Drop Model 15 4.1 Harmonic oscillator...... 16
5 Nuclear decay 17 5.1 α-decay...... 17 5.1.1 Gamov model...... 17 5.2 β-decay...... 18 5.3 γ-decay...... 19
6 Nuclear Fission and Fusion 22 6.1 Fission...... 22 6.2 Fusion...... 23
7 The Standard Model 24 7.1 Quarks...... 25 7.2 Leptons...... 25 7.3 Gauge bosons...... 25 7.4 Scalar bosons...... 25 8 Quantum Field Theory 26 8.1 The Klein-Gordon Equation...... 26 8.2 The Dirac Equation...... 27 8.3 Probability density...... 29 8.4 Covariant form...... 29 8.5 Solution of the Dirac Equation...... 30 8.6 Antiparticles...... 32 8.7 Antiparticle spinors...... 34 8.8 Charge conjugation...... 34
9 Spin 35 9.1 Helicity...... 35 9.2 Parity...... 37
10 Particle exchange 38 10.1 Time-ordered perturbation theory...... 38 10.2 Scattering in a potential...... 40
11 QED 41 11.1 Feynman rules for QED...... 43 11.2 t-channel Diagram...... 44 11.3 s-channel Diagram...... 45 11.4 u-channel Diagram...... 46 11.5 Example: Compton scattering...... 47 11.6 Example: Bhabha scattering...... 48 11.7 Electron-Positron annihilation...... 49 11.8 Spin sums...... 49
12 QCD 51 12.1 Global gauge symmetry...... 51 12.2 The local gauge principle...... 51 12.3 Strong interaction...... 52 12.4 Hadronization and jet...... 53 12.5 e+e− collisions...... 53 12.6 Charged current: Leptons...... 55 12.7 Charged current: Quarks...... 55 12.8 Neutral current: Quarks...... 56 12.9 Neutral current: Quarks...... 57 12.10Quark mixing...... 57 12.11Baryon number conservation...... 58 12.12Lepton number conservation...... 58 12.13Lepton flavor conservation...... 58 12.14Example: Beta decay...... 59 CONTENTS CONTENTS
13 Isospin 61
3 1 RECAP
1 Recap
1.1 Four vectors Four vectors are used in special relativity, as an example we use space and time coor- dinates in a four vector as well as energy and momentum coordinates. We write:
aµ = (a0, a1, a2, a3) for the contravariant 0 1 2 3 aµ = (a0, a1, a2, a3) = (a , −a , −a , −a ) for the covariant
1.1.1 Space-vector The space-vector in fourvector-form consists of the time-coordinate t as well as the space -coordiante ~r = (x, y, z). For the dimensions to match, time-coordinate gets multiplied by c: xµ = (ct, x, y, z) = (ct, ~r) xµ is a contravariant fourvector, because it is a coordinate-vector to an orthonormal base of the minkowski space and thus with a base-change it changes contravariant through a lorentz-transformation. In this metric of flat spacetime, t has a different sign to the space-coordiantes:
ds2 = c2dt2 − dx2 − dy2 − dz2
The contravariant fourvector a under the Lorentztransform Λ goes to
a0 = Λa
The covariant fourvector b goes to:
b0 = Λ−1T b where γ −γβ 0 0 −γβ γ 0 0 Λ = 0 0 1 0 0 0 0 1 Further we have Λ−1T = gΛg−1 with 1 0 0 0 0 −1 0 0 µν gµν = diag(1, −1, −1, −1) = = g 0 0 −1 0 0 0 0 −1
1 1.1 Four vectors 1 RECAP
Thus ν 0 1 2 3 aµ = (a0, a1, a2, a3) = gµνa = ga = (a , −a , −a , −a ) The product of two fourvectors in minkowski-space is given by: ν aµb = aµgµνbν = a0b0 − a1b1 − a2b2 − a3b3 If a is a fourvector we have: − a2 > 0 =⇒ aµ is called timelike − a2 < 0 =⇒ aµ is called spacelike − a2 = 0 =⇒ aµ is called lightlike If the distance between two events (∆xµ) is spacelike, they can not be in causal rela- tion.
1.1.2 Velocity-vector The velocity-vector uµ is given by differentiation xµ by proper time τ: dxµ uµ = dτ Where proper time τ is the smaller amount of time between two events and given by: 1 dτ = dt γ with 1 1 |~v | γ = = √ β = r 2 | ~v | 2 1 − β c 1 − c Thus d uµ = γ (ct, x, y, z) = γ(c, x,˙ y,˙ z˙) = γ( c,~v ) dt with its norm being q µ q µ ν q µ 2 2 2 |u | = u u gµν = uµu = γ (c − |~v | ) = c
1.1.3 Momentum-vector The momentum-vector in fourvector-form is defined as p µ = muµ = (γmc, γm~v ) where m is the rest mass of the body. Using E = γmc2 we get p µ = (E/c, ~p ) ~p = γm~v From where we can derive the energy-momentum relationship: E2 − | ~p |2c2 = m2c4
2 1.1 Four vectors 1 RECAP
1.1.4 Mandelstam variables In Particle physics there are three really useful Lorentz invariant quantities: s, t and u. We consider the scattering process:
1 + 2 → 3 + 4 we can then define
2 2 2 s = (p1 + p2) , t = (p1 − p3) , u = (p1 − p4) we note 2 2 2 2 s + t + u = m1 + m2 + m3 + m4 and we note that s is a scalar product of two four vectors:
2 2 2 s = (p1 + p2) = (E1 + E2) − (~p1 − ~p2)
And since these quantities are Lorentz invariant we can evaluate them in any frame, thus we choose the most convenient one (center of mass frame):
∗ ∗ ∗ ∗ ∗ ∗ p1 = (E1 , ~p ) p2 = (E2 , −~p )
∗ ∗ 2 =⇒ s = (E1 + E2 )
1.1.5 Rapidity ???
3 1.2 Quantum Mechanics 1 RECAP
1.2 Quantum Mechanics In quantum mechanics the state of a system is defined by a wave function |φi which is a complex number. This function is normalized as hφ|φi = 1. Measurements are represented by operators O, and the possible outcomes of a mea- surement are its eigenvalues λi. And all its eigenvectors |ψii are orthonormal. P 2 If a state φ = i αi|ψii is given, the quantity |αi| is the probability to measure the value λi when measuring O. Also αi = hψi|φi. Since we use bra-ket notation (dirac), we have
X ∗ hφ| = αi hψi| i
P ∗ and therefore: αiαi = 1. The expectation value of an operator O is defined as
hφ|O|φi
We could also use the function notation instead of the dirac notation, thus we have: Z hψ|φi = ψ∗(~x)φ(~x)d3x
The expectation value: Z ψ∗(~x)Oφ(~x)d3x
Some important operators: ∂ pˆ = −i ∇ Eˆ = ih xˆ = x ~ ∂t If two operators commute, then
[A, B] = AB − BA = 0
Then they can be measured in whatever order and they have a complete set of eigen- states in common. Hermitian operators O† = O have real eigenvalues, orthogonal eigenvectors and are associated with measurable quantities.
4 1.2 Quantum Mechanics 1 RECAP
1.2.1 Clebsch-Gordon coefficients In classical mechanics we know all the components of the angular momentum, but in quantum mechanics the various components of the angular momentum do not com- mute [Li,Lj] = i~ijkLk At best we can measure L2 and one of the components, moreover only certain values are allowed: 2 2 hψ|L |ψi = `(` + 1)~ ` = 1, 2, 3, ··· It might be that we are not interested in L and S separately but only in the total angular momentum J = L + S. Or you might want to add the spin of two particles to get the spin of a composite particle (in case the relative orbital angular momentum is zero). We cannot simply add component by component as they do not commute. But there is a handy little trick called Clebsch-Gordon coefficients, they are given in fancy tables, we use one small part of it:
L S state X 5/2 2 × 1/2 5/2 5/2 3/2
2 1/2 1 3/2 3/2
2 −1/2 1/5 4/5 5/2 3/2 X
1 1/2 4/5 −1/5 1/2 1/2 Xz
1 −1/2 2/5 3/5
0 1/2 3/5 −2/5
P
Lz Sz
Figure 1: Part of the Clebsch-Gordon coefficients table.
In the shaded rectangle above we have the possibilities (without the square root) for the states (given on top of the rectangle) and the associated z-component of the angular momentum and spin (given on the left of the rectangle). We will explain this using an example:
5 1.2 Quantum Mechanics 1 RECAP
Example 1: Angular momentum combination
We have a particle with spin S = 1/2 and angular momentum L = 2. In a mea- surement we discover that the particle is in the state of total angular momentum
3 1 X = , 2 2 We want to find out the probability to find the particle with spin up (+1/2) and to find it in the angular momentum state of Lz = 0.
How do we do this? Since we are dealing with a particle with spin 1/2 and angular momentum 2, we look for the Clebsch-Gordon table 2 × 1/2 (given above). And sin we are looking at the state X we search this column in the table (red arrow). Now we are able to write the combination for the state X:
|X,Xzi = P1|L1,L1,zi|S1,S1z i + P2|L2,L2,zi|S2,S2,zi
when we put in the numbers we get
s s 3 1 3 1 1 2 1 1 , = |2, 1i , − − |2, 0i , 2 2 5 2 2 5 2 2 | {z }
It is important to note that these coefficients Pi in the table (only the ones cir- cled by thin lines, in our case inside the shaded rectangle) are actually under a square-root, except for the minus sign, if there is a minus sign present it is always in front of the root. The possibility to find the particle with spin Sz = +1/2 we just square the co- efficient in front of the term (green brace and green arrow) in the combination, which gives us s 2 2 2 2 P = P = − = = 40% 2 5 5
And to find the possibility of finding the particle with angular momentum state Lz = 0 we again look in the combination and see that is again the second term (red brace), thus the possibility is also
s 2 2 2 2 P = P = − = = 40% 2 5 5
6 1.2 Quantum Mechanics 1 RECAP
state L
S1 S2 3/2 1 × 1/2 3/2 3/2 1/2 L
1 1/2 1 1/2 1/2 Lz 1 −1/2 1/3 2/3 3/2 1/2 0 1/2 2/3 −1/3 −1/2 −1/2 0 −1/2 2/3 1/3 −1 1/2 1/3 −2/3 S1,z S2,z
Figure 2: 1 × 1/2 Clebsch-Gordon coefficients table.
Example 2: Angular momentum combination
Two particles in states of spin
1 1 |1, 0i , + 2 2 are given. What is the possibility to find the system in total angular momentum L = 3/2?
Now this means S1 = 1 and S2 = 1/2 and in that case we can use the 1 × 1/2 given above. In this case we do not know what state the particle is in, we only know which table to use and we know S1,z = 0 and S2,z = 1/2, thus we search in these values in the table (green arrow). Now the exercise asked for the total angular momentum state of the system, this is L in L = |L, Lzi so we search the state L (red arrow). And voila, where these two arrows cross we get our coefficient (attention to the square root) which gives our possiblity:
s 2 2 2 2 P = P = = = 66.67% 3 3
7 3 FERMI’S GOLDEN RULE
2 Introduction
− Fermions are particles of half-integer spin and they obey the fermi-dirac statis- tics: their total wave function must be antisymmetric.
− Bosons are particles with integer spins and they obey the bose-einstein statistics: their total wave function must be symmetric
− If a particle is unstable, it is not an eigenstate of the hamiltonian (it has an imag- inary part in the energy)
3 Fermi’s golden rule
3.1 Lifetime The probability that a particle survives a time t + dt is given by
P (t + dt) = P (t)(1 − Γdt) where Γ is the decay probability per unit of time. We are basically saying that the prob- ability of a particle decaying at each time does not depend on the history of the particle.
P (t + dt) = P (t)(1 − Γdt) → P (t + dt) − P (t) = P (t) · (−Γdt) dP = −ΓP → P (t) = αe−Γt dt To find the coefficient α we just set P (0) = 1, meaning the particle exists at t = 0, we then get P (t) = e−Γt We can derive the probability density function (PdF) such that Z ∞ P(t)dt = 1 0 this implies that Z ∞ 1 A e−Γtdt = 1 ⇔ A = 1 0 Γ 1 =⇒ P(t) = e−Γt Γ The expectation value of t then is Z ∞ 1 hti = τ = P(t)dt = 0 Γ In general: − t − t P (t) = e γτ P(t) = τe γτ
8 3.2 Cross section 3 FERMI’S GOLDEN RULE
3.2 Cross section We define the cross section σ of a particle as the area of the target particles where the interaction occurs, in general this has nothing to do with the actual physical size of the particle. We define the differential cross section dσ dΩ as the quotient of the area element dσ in the plane of the impact and the differential angular range of the scattered particle dΩ = d(cos θ)dϕ. Thus we get Z dσ σ = dΩ dΩ As an example: In a time δt a particle of type a passes through a region containing nb(va + vb)Aδt particles of type b, then the interaction probability is obtained by the effective corss- sectional area occupied by particles of type b: n (v + v )Aδt P = b a b = n vδtσ v = v + v A b a b these implies the rate per particle of type a as: P = n vσ δt b
The cross-section is measured in barn (b).
1b = 100fm2 = 10−24cm2
3.3 Luminosity We define the luminosity as: dP W = = L· σ dt where P is the power and not the probability. We then see the integrated luminosity as Z Z L = L =⇒ P = W dt = Lσ
W = Nbϕaσ =⇒ L = Nbϕa where number of particles ϕ = At
9 3.4 Mean free path 3 FERMI’S GOLDEN RULE
3.4 Mean free path If we shoot particles at an infinitesimal rod with width dx, the number of interactions is given by nbvσdt = nbσdx = NAρbσdx and the mean free path is 1 l = nbσ If the scattering centers are nucleons then 1 l = NAρbσ with NA as the Avogadro number.
3.5 Decay rate In particle physics we are mainly concerned with particle interactions and decays, in other words: transitions between states. We use Fermi’s golden rule:
2 Γfi = 2π|Tfi| ρ(Ei) where
− Γf i is the number of transitions per unit time form initial state |ii to final state hf| (not Lorentz invariant)
− Tfi is the transition matrix element: ˆ ˆ ˆ X hf|H|jihj|H|ii Tf i = hf|H|ii + + ··· j6=i Ei − Ej
− Hˆ is the perturbing Hamiltonian
− ρ(Ei) is the density of states, defined as:
dn ρ(E ) = i dE Ei where dn is the number of accessible states in the energy range [E : E + dE] we can also define it using the Dirac-delta function:
dn Z dn = δ(E − E)dE dE dE i Ei
10 3.5 Decay rate 3 FERMI’S GOLDEN RULE
giving an alternative form of Fermi’s golden rule: Z 2 Γf i = 2π |Tf i| δ(Ei − E)dn
The matrix element contains the fundamental particle physics and the density of states is just kinematics. As an example we observe the decay rate or the process
a → 1 + 2 which can be calculated using Fermi’s golden rule. We calculate the transition matrix element to order the perturbation theory: ˆ Tfi = hψ1ψ2|H|ψai Z ∗ ∗ ˆ 3 = ψ1ψ2Hψad ~x V In the Born approximation, the perturbation is taken to be small and the initial and final state particles are represented by plane waves of the form
ψ(~x,t) = Aei(~p·~x−Et)
It is useful to adopt a scheme where each plane wave is normalised to one particle in a cubic volume of side a. Using the non-relativistic expression for probability density
ρ = ψ∗ψ gets us Z a Z a Z a ψ∗ψdxdydz = 1 =⇒ A2 = 1/a3 = 1/V 0 0 0 From quantum mechanics we know that the normalisation in a box of volume V = a3 implies that the wave-function satisfies the following periodic boundaries:
ψ(x + a, y, z) = ψ(x, y, z) =⇒ eipxxeipx(x+a) ···
This then implies for the momentum: 2π (p , p , p ) = (n , n , n ) n ∈ x y z x y z a i N Each state in momentum space occupies a cubic volume of
2π 3 (2π)3 d3p = dp dp dp = = x y z a V
11 3.5 Decay rate 3 FERMI’S GOLDEN RULE
The number of states dn with magnitude of momentum in the range [p, p+dp] is equal to the momentum space volume of the spherical shell at momentum p with thickness dp is divided by the average volume occupied by a single state: V dn = 4πp2dp · (2π)3 Thus the density of states gives us the density of states:
dn dn dp 4πp2 dp
ρ(E) = = = V dE dp dE (2π)3 dE The density of states corresponds to the number of momentum states accessible to a particular decay and increases with the momentum of the final-state particle. Hence, all other things being equal, decays to lighter particles, which will be pro- duced with larger momentum, are favoured over decays to heavier particles. Since the decay rate does not depend on the normalisation volume, it is convenient to normalise to one particle per unit volume by setting V = 1: d3p dn = i i (2π)3 We then get 4πp2 1 1 dp E ρ(E) = · = = (2π)3 β β dE p We know the delta function as infinitely narrow spike of unit area:
Z +∞ δ(x − a)dx = 1 −∞
Z +∞ f(x)δ(x − a)dx = f(a) −∞ Z y2 1 y < 0 < y δ(y)dy = 1 2 y1 0 else −1 df δ(f(x)) = δ(x − x ) dx 0 x0 This then gives us the density of states:
dn Z dn
ρ(Ef ) = = δ(E − Ef )dE dE dE Ef This gets us the golden rule as: Z 2 Γfi = 2π |Tfi| δ(Ef − E)dn
12 3.5 Decay rate 3 FERMI’S GOLDEN RULE
For a dn in a two-body decay, we only need to consider one particle as momentum conservation gives us the other particle. Thus Z d3~p Γ = 2π |T |2δ(E − E − E ) 1 fi fi f 1 2 (2π)3 We can include momentum conservation and integrate over the momenta of both par- ticles using another delta function:
Z 3 3 4 2 3 d ~p1 d ~p2 Γfi = (2π) |Tfi| δ(Ei − E1 − E2) δ (~pi − ~p1 − ~p2) 3 3 | {z } | {z } (2π) (2π) energy cons. momentum cons. | {z } density of states
3.5.1 Lorentz invariance When we consider a non-relativistic quantum mechanic wave function we normalise to one unit volume: Z ψ∗ψdV = 1 but when considering relativistic effects volume contracts by γ = E/m. Particle density thus increases by γ. So we usually normalise to Z ψ0∗ψ0dV = 2E
This implies √ ψ0 = 2Eψ We then define the Lorentz invariant matrix element: q 0 0 ˆ 0 0 Mfi = hψ1 · ψ2 · · ·|H|· · · ψn−1ψni = 2E1 · 2E2 · 2E3 ··· 2En · Tfi If we now consider the relation between the transition matrix matrix element and the lorentz invariant matrix element: 4 Z 3 3 (2π) 2 3 d ~p1 d ~p2 Γfi = |Mfi| δ(Ei − E1 − E2)δ (~pi − ~p1 − ~p2) 3 3 2Ea (2π) 2E1 (2π) 2E2
2 2 with |Mfi| = (2Ei2E12E2)|Tfi| . Because this integral is lorentz invariant,we can evaluate it in any frame we choose. It usually is best to choose the center of mass frame where: Ei = mi ~pi = 0 thus Z 3 3 1 2 3 d ~p1 d ~p2 Γfi = 2 |Mfi| δ(mi − E1 − E2)δ (~p1 + ~p2) 8π mi 2E1 2E2 integrating over ~p2 using the δ-function gives just 1, and now
2 2 2 E2 = m2 + |~p1|
13 3.5 Decay rate 3 FERMI’S GOLDEN RULE
because the δ-function imposes ~p2 = −~p1 we then write in spherical coordinates: 3 2 2 d ~p1 = p1dp1 sin θdθdϕ = p1dp1dΩ this leaves us with Z q q 2 1 2 2 2 2 2 p1dp1dΩ Γfi = 2 |Mfi| δ mi − m1 + p1 − m2 + p1 32π mi E1E2 This can be brought into the form Z 1 2 Γfi = 2 |Mfi| g(p1)δ ((p1)) dp1dΩ 32π mi with 2 p1 g(p1) = E1E2 q q 2 2 2 2 f(p1) = mi − E1 − E2 = mi − m1 + p1 − m2 + p1
The Dirac δ-function δ(f(p1)) imposes energy conservation and is only non-zero for ∗ ∗ ∗ p1 = p where p is the solution of f(p ) = 0. We then get ∗ Z p 2 Γfi = 2 2 |Mfi| dΩ 32π mi This is the general expression for any two-body-decay. The fundamental physics is contained in the matrix element and the additional factors arise form the phase space integral. We have p∗ given as q ∗ 1 2 2 2 2 p = [(mi − (m1 + m2) ][mi − (m1 − m2) ] 2mi For a scattering process 1 + 2 → 3 + 4 we now have the cross section: Γ σ = fi v1 + v2 if we insert Γfi from the Fermi golden rule we see that the velocities, Tfi and the integral over the momenta are parts that are not Lorentz invariant. Thus we need to use the lornetz invariant form: Z 3 3 1 2 3 d ~p3 d ~p4 σ = 2 |Mfi| δ(E1+E2−E3−E4)δ (~p1−~p2−~p3−~p4) 4E1E2(v1 + v2)(2π) 2E3 2E4 √ where Mfi = 4 E1E2E3E4Tfi and F = 4E1E2(v1 + v2) can be written in terms of a four-vector scalar and is thus lorentz invariant: q µ 2 F = 4 p1 p2µ) − m1m2 we then get ∗ Z 1 |~pf | 2 ∗ σ = 2 ∗ |Mfi| dΩ 64π mi |~pi |
14 4 LIQUID DROP MODEL
4 Liquid Drop Model
The liquid drop model treats the nucleus as a drop of incompressible nuclear fluid of very high density. Inside the nucleus there are protons and neutrons which are held together by the nuclear force (residual effect of the strong force). This is similar to the structure of a spherical liquid drop made of microscopic molecules. While this model will not explain all the properties of nuclei, it explains the spherical shape of most nuclei and their binding energy. A First we indicate a nuclide with A nucleons and Z protons as Z X. Two nuclides are called
− Isotopes if they have the same Z
− Isobars if they have the same A
− Isotone if they have the same N = A − Z
The mass of nuclides is given by
M(A, Z) = N · mn + Z · mp − B(A, Z)
B(A, Z) is the binding energy. The binding energy is composed of:
2 2 2/3 Z (A − 2Z) B(A, Z) = avA − aSA − aC 1/3 − aA − δ(A, Z) |{z} | {z } A A | {z } volume term surface term | {z } | {z } pairing term coulomb term asymmetry term − Volume term: Nuclei interact through the strong force with the neighbouring nuclei only, the binding energy per nucleon inside the nuclide is constant (about 15.85 MeV)
− Surface term: The nuclei on the surface interact also through the strong force but with less nuclei (about 18.34 MeV)
− Coulomb term: There is a repulsive term between all the protons, Each proton interacts with all the others (about 0.71 MeV)
− Asymmetry term: This effect is based on the Pauli exclusion principle. The heavier nuclei con- tain more neutrons than protons, these additional neutrons provide (through the attractive forces between neutrons and protons) some compensation for the re- pulsion between protons. (about 23.21 MeV)
15 4.1 Harmonic oscillator 4 LIQUID DROP MODEL
− Pairing term:
+δ0 for Z,N even δ(A, Z) = 0 for Z,N even and odd δ0 = 12MeV −δ0 for Z,N odd This captures the effect of spin-coupling. A paired spin is more stable and an odd number of protons and neutrons are mostly unstable.
An atom is particularly stable if we complete a shell.
− Atoms with one more electron tend to loose it, like alkali metals
− Atoms with one less electron tend to acquire it
− Atoms with the complete shell tend to not react, like noble gases
For nuclei there exist some magic numbers for which they are more stable, the numbers are: 2, 8, 20, 28, 50, 82, 126 Nuclei with double magic numbers are:
4 16 40 48 208 2He2, 8O8, 20Ca20, 20Ca28, 82Pb128
4.1 Harmonic oscillator We now define the Hamiltonian of the harmonic oscillator as pˆ2 1 H = + mω2xˆ2 2m 2 0 we also define the latter operators as
1 √ pˆ ! a+ =a ˆ = √ mωxˆ + √ 2~ω0 m 1 √ pˆ ! a− =a ˆ† = √ mωxˆ − √ 2~ω0 m Where we have a commutation relation: [ˆa, aˆ†] = 1. We can now write the hamiltonian as 1 H = ω aˆ†aˆ + ~ 0 2 ————- SHELL MODEL????
16 5 NUCLEAR DECAY
5 Nuclear decay
In this chapter we are looking at the nuclear decay of particles. There are only a hand- ful of stable nuclei, most of them are unstable and decay spontaneously in various ways. Isobars with a large surplus of neutrons gain energy by converting a neutron into a proton. If there is a surplus of protons, the inverse reaction may occur. These transformations are called β-decays, they are manifestations of the weak interactions. If a particle decays in daughter nuclei, and one of them is a 4He nucleus (A0 = 4, Z0 = 2) the decay is called α-decay and the Helium nucleus is called an α-particle.
The probability per unit time for a radioactive nucleus to decay is known as the de- cay constant λ. It is related to the lifetime τ and the half-life t1/2 by 1 ln(2) τ = and t = λ 1/2 λ
5.1 α-decay This is a type of radioactive decay where an atomic nucleus emits an α particle - e Helium nucleus - thus transforming into an atom with a mass number A reduced by four and an atomic number Z reduced by two: A0 = A − 4 Z0 = Z − 2
An α-particle is a 4He which consists of 2 protons, 2 neutrons and 2 electrons. Due to their relatively large mass, +2 electric charge and relatively low velocity, α-particles are very likely to interact with other atoms and thus can be stopped by just a few cen- timeters of air or a sheet of paper. Approximately 99% of Helium produced on earth is the result of alpha decay of ura- nium or thorium. The helium reaches the surface as a by-product of natural gas. This decay can happen if the binding energy of the α-particle Bα plus the binding energy of the daughter nucleus BY are larger than the binding energy of the initial nucleus BX : BX − BY < Bα ≈ 28.3MeV
5.1.1 Gamov model The Gamov-factor is a probability factor for two nuclear particles’ chance to overcome the coulomb barrier in order to undergo nuclear reactions. In the case of an α-decay we look at the nucleus as a Fermi-gas in a potential well, it is possible inside of that well that multiple nucleons gather in a bound state. This would release binding energy which increases the probability that the new particle (consisting
of the bound nucleons) can tunnel through the coulomb barrier.
17 5.2 β-decay 5 NUCLEAR DECAY
5.2 β-decay In the process of a β-decay a beta ray (electron or positron) and a neutrino are emitted from an atomic nucleus. A beta decay of a neutron transforms it into a proton by the emission of an electron, called β−-decay: 0 + − n → p + e + νe A beta decay of a proton transforms it into a neutron by the emission of a positron, called β+-decay: + 0 + p → n + e + νe The electron/positron and the neutrino do not exist prior to the emission but are created in the decay process. Nucleons are composed of up or down quarks and the weak force allows a quark to change type by the exchange of a W boson and the emission of an electron/antineutrino or a positron/neutrino pair. This example will be discussed later on.
18 5.3 γ-decay 5 NUCLEAR DECAY
5.3 γ-decay Gamma-decay usually occurs after other forms (α or β) of decay occur. An excited nucleus can decay by the emission of an α or β particle. The daughter nucleus is usually left in an excited state. It can then decay into a lower energy state by emitting a gamma ray photon. This emission typically requires 10−12 seconds. As an example we have 60Co decaying to excited 60Ni by emission of an electron of 0.31MeV. Then the excited 60Ni decays to the ground state by emitting gamma rays:
60 60 − 27Co → 28 Ni + e + νe +γ | {z } β-decay products These emitted γ-rays are quantized and we can classify them by there multipolarity. There are two kinds: electric and magnetic multipole radiation. The possible multipo- larities can be derived direcitly through the conservation laws of angular momentum and parity. The parity is given by:
` Πγ(E`) = (−1) electric multipole `−1 Πγ(M`) = (−1) magnetic multipole
Because of conservation of angular momentum
L = s + ` the equation |L − s|| ≤ ` ≤ L + s must be fulfilled. where L is the angular momentum quantum number, s = 1 is the spin of the photon and ` is the multipole order. This means ` is the delta, by which the angular momentum changes in transitions. The lifetime of a state depends strongly on the multipolarity of the γ-transitions through which it can decay. The lower the multipolarity is the higher is the possibility of a tran- sition. The possibility for M` is as high as the possibility for transition E(` + 1).
− electric dipole-, quadrupole- or octupoleradiation, in general 2`-pole-radiation, are called E1, E2, E3 in general: E`-radiation
− magnetic dipole-, quadrupole- or octupoleradiation, in general 2`-pole-radiation, are called M1, M2, M3 in general: M`-radiation
19 5.3 γ-decay 5 NUCLEAR DECAY
! − monopole-radiation (` = 0) does not exist, thus ` ≥ 1 and a transition between two spin-0-nuclei like 0+ → 0+ is not possible!
We also have a handy table which multipolarity connects to the angular momentum change ∆` and the parity change ∆Π:
Multipolarity ` ∆Π Multipolarity ` ∆Π M1 1 + E1 1 − M2 2 − E2 2 + M3 3 + E3 3 − M4 4 − E4 4 + M5 5 + E5 5 − Table 1: Handy table
Example 1: Gamma decay transition
Calculate the transition for the decay 4+ → 2+:
Step 1: We see that the parity is not changing, so ∆Π = + and the angular momentum change is ` = 2. Step 2: We look in the table and see that this corresponds to E2, thus
2 Πγ(E2) = (−1) = +1
hence E2 dominates.
Example 2: Gamma decay transition
Calculate the transition for the decay 2+ → 0+:
Step 1: We see that parity again does not change, thus ∆Π = + and the angular mo- mentum change is ` = 2. Step 2:
20 5.3 γ-decay 5 NUCLEAR DECAY
We look in the table and see that this corresponds again to E2:
2 Πγ(E2) = (−1) = +1
thus E2 dominates.
Example 3: Gamma decay transition
Calculate the transition for the decay 3+ → 2+:
Step 1: We see that parity does not change, thus ∆Π = + but angular momentum changes by one unit, thus ` = 1. Step 2: We look in the table that this cannot be E1, but is in fact M1, and:
2−1 Πγ(M1) = (−1) = −1
Hence M1 dominates.
21 6 NUCLEAR FISSION AND FUSION
6 Nuclear Fission and Fusion
Figure 3: B/A in MeV per nucleon against mass number A.
6.1 Fission We know that fission is the splitting of a heavy nucleus into two more stable daughter- nuclei, it is usually triggered by a slow neutron (hence the need of neutron-sources in nuclear power plants). Nuclides with Z > 40 can in principle split, but the barrier is so high that the tunneling effect is very improbable, so in practice only nuclei with very high A split. By looking at the deformation of a sphere and comparing the surface- and coulomb-energy we can see that the difference is only negative (hence energetically favorable) if:
Z > 114, A > 270
Only for atoms with these properties fission occurs. Spontaneous fission is very rare because it is suppressed by the tunneling of the coulomb
22 6.2 Fusion 6 NUCLEAR FISSION AND FUSION barrier. But we can perturb the nucleus such that it is unstable and breaks, this is called induced fission. In the case of induced fission we shoot a neutron at the nucleus, the nucleus gets split, beta and gamma radiation gets emitted and 2 or 3 neutrons get ejected. These ejected neutrons can then hit other nuclei, creating a chain reaction. Depending on the fission material, the shape and the mass, the reaction can be self- perturbing (hence the term critical mass). In a nuclear power plant the chain reaction gets controlled by so-called control rods which absorb more or less neutrons depending on the level to which they are put inside the reactor. We call nuclei that can fission for any incoming neutron fissile. The only naturally occurring fissile nucleus is 235U
Other fissile, but only synthetically created, nuclei are: 233U, 239Pu, 241Pu
Nuclei that can be induced to fission only by neutrons of energy higher than e certain threshold are called fissionable, examples are: 238U, 240Pu
The energy released by fission per nucleus is about ≈ 200MeV ≈ 3.2 · 10−11J, which is much, much greater than the energy released by combustion, but still only ≈ 0.09% of the mass energy of an uranium nucleus. The released energy from fission appears mostly as kinetic energy of the daughter nu- clei and only to a small part as kinetic energy of the neutrons and emitted radiation. The kinetic energy gets quickly reduced to heat which can then be used to power a generator.
6.2 Fusion In order for fusion to occur, two nuclei need to be brought close enough to each other for the strong force to come into action, they need to overcome the coulomb barrier. We need to give some energy into the system for that to happen. In stars this happens because the gravitational pressure is high enough to bring the nuclei close enough to- gether. Most interacting nuclei are isotopes of hydrogen (Z = 1) because as seen in the plot (figure ) the energy released there is maximized. A reaction that produces an α-particle which is particularly stable will also produce a large amount of energy ≈ 18 − 23 MeV.
23 7 THE STANDARD MODEL
7 The Standard Model
The Standard Model describes the fundamental particles that make up all our surround- ings.
Figure 4: The standard model.
It is now known that not all building blocks of an atom are fundamental particles:
− A proton is made up of two up and one down quark, thus p+ = (u,u,d).
− A neutron is made up of two down and one up quark, thus n0 = (u,d,d).
− An electron is a fundamental particle belonging to the Lepton category and thus part of the standard model
24 7.1 Quarks 7 THE STANDARD MODEL
7.1 Quarks There exist six types of quarks: up and down build the first generations of matter, charm and strange the second and bottom and top quarks build the third. They form Hadrons (like the proton or neutron). And they can only be observed in bound states, like inside hadrons, not as free particles. Hadrons can be
− Mesons: composed of a quark and an anti-quark
− Baryons: composed by three quarks
7.2 Leptons There exist three types of charged leptons: electrons e−, muons µ and taus τ. There exist three types of neutral leptons (neutrinos): νe, νµ, ντ .
7.3 Gauge bosons Leptons and Quarks are what we call matter, but there exist also the gauge bosons, which are responsible for the interaction of the strong, electromagnetic and the weak force. There are four gauge bosons: gluons g, photons γ, Z-bosons and W-bosons.
7.4 Scalar bosons And last but not least, all particles would be massless if there was not the Higgs field. Its quantum is the Higgs boson H.
25 8 QUANTUM FIELD THEORY
8 Quantum Field Theory
Quantum field theory says that every fundamental particle from the standard model is just an excitation or oscillation of its own field. It combines classical fields, special relativity and quantum mechanics to describe the behavior of fundamental particles.
8.1 The Klein-Gordon Equation The Klein-Gordon equation is obtained by writing the Einstein energy-momentum re- lationship E2 = m2c4 + p2c2 c==1 p2 + m2 in the form of operators acting on a wavefunction:
Eˆ2ψ(x, t) = pˆ2ψ(x, t) + m2ψ(x, t) with ∂ pˆ = −i∇ and Eˆ = i ∂t This leads us with the Klein-Gordon equation:
∂2 ψ = ∇2ψ − m2ψ ∂t2
This equation can be expressed in the Lorentz-invariant form:
µ 2 (∂ ∂µ + m )ψ = 0 where we have the Lorentz-invariant scalar product of two four vectors:
∂2 ∂2 ∂2 ∂2 ∂µ∂ = − − − µ ∂t2 ∂x2 ∂y2 ∂z2 The solution of the Klein-Gordon equation are the plane wave functions:
ψ(x, t) = Nei(p·x−Et) with probability density and current:
ρ = 2|N|2E and j = 2|N|2p
Which when inserted gives the Einstein energy-momentum relationship: q E = ± p2 + m2
26 8.2 The Dirac Equation 8 QUANTUM FIELD THEORY
In classical mechanics the negative energy solution can be dismissed as being unphys- ical, but in quantum mechanics all solutions are required to form a complete set of states. Even though it is not clear how the negative energy solutions should be interpreted, we cannot discard them. An even bigger problem is the probability density for such negative energy solutions, they are also negative (unphysical). This leads to the con- clusion that the Klein-Gordon equation does not provide a consistent description of single particle states for a relativistic system. However in quantum field theory the Klein-Gordon equation is used to describe the multi-particle excitations of a spin-0 quantum field, thus this problem does not exist.
8.2 The Dirac Equation The above mentioned problem with the Klein-Gordon equation is what motivated Dirac to search for an alternative formulation of relativistic quantum mechanics. The resulting wave function provided a solution for negative probability density-problems and also a natural description of the intrinsic spin and magnetic moments of spin-half fermions. Dirac looked for a wave function that was first order in both space and time derivatives:
Eψˆ = (α · pˆ + βm)ψ which can be written as: ∂ ∂ ∂ ∂ ! i ψ = −iα − iα − iα + βm ψ ∂t x ∂x y ∂y z ∂z For this equation to fulfill the Einstein energy-momentum relation it has to fulfill the Klein-Gordon equation, and if we "square" the above equation (meaning we only square the operators on both sides but not the function ψ on which we apply the oper- ator) we get a really big equation which I will not write out here (can be seen in Mark Thomsons book), but this equation only reduces to the Klein-Gordon equation ∂2 ∂2 ∂2 ∂2 ! ψ = + + − m2 ψ ∂t2 ∂x2 ∂y2 ∂z2 if the coefficients α and β satisfy:
2 2 2 2 αx = αy = αz = β = I
αjβ + βαj = 0
αjαk + αkαj = 0 j 6= k where I represents unity. From these equations we can clearly see that the αi and β are not normal numbers but
27 8.2 The Dirac Equation 8 QUANTUM FIELD THEORY instead need to be matrices. From the cyclic property of traces Tr(ABC) = Tr(BCA) and β2 = I we can see that the traces of matrices αi and β have to be zero:
Tr(αi) = Tr(αiββ) = Tr(βαiβ) = −Tr(αiββ) = −Tr(αi)
2 We can also get the eigenvalues for the matrix αi, using αi = I:
αiX = λX
2 2 αi X = λαiX =⇒ X = λ X therefore λ = ±1. Since the sum of the eigenvalues is equal to the trace, the only way this is possible is if the matrices are of even dimension. Since the eigenvalues are real, the matrices must be hermitian. † † αi = αi β = β Since there exist only three 2 × 2 matrices and we need 4, the lowest dimension object than can represent our αi and β are 4 × 4 matrices. Therefore the Dirac Hamiltonian ˆ HD = (α · pˆ + βm) is a 4 × 4 matrix of operators that must act on a four-component wavefunction known as a Dirac spinor ψ1 ψ 2 ψ = ψ3 ψ4 We use the Dirac-Pauli representation: ! ! I 0 0 σi β = αi = 0 −I σi 0 wherewe use the unity matrix I and the Pauli matrices σi: 1 0! 0 1! 0 −i! 1 0 ! I = σ = σ = σ = 0 1 x 1 0 y i 0 z 0 −1 Remember, we are dealing with 4 × 4 matrices, thus 1 0 0 0 0 0 0 1 0 0 0 −i 0 1 0 0 0 0 1 0 0 0 i 0 β = αx = αy = 0 0 −1 0 0 1 0 0 0 −i 0 0 0 0 0 −1 1 0 0 0 i 0 0 0 and analog for αz.
28 8.3 Probability density 8 QUANTUM FIELD THEORY
8.3 Probability density Through some calculations (again throughly noted in Mark Thomsons book) we get the probability density ρ = ψ†ψ and the probability current j = ψ†αψ † 2 2 2 2 where ψ ψ = |ψ1| + ψ2| + ψ3| + ψ4| > 0. Thus we have solved the problem from the Klein-Gordon equation: probability den- sities are always positive. Additionally the solutions to the Dirac equation are the four component Dirac Spinors, whose components give rise to the property of intrinsic spin. It can be shown that Dirac spinors represent spin-half particles with an intrinsic magnetic moment of q µ = Sˆ m where 1 σ 0! Sˆ = 2 0 σ
8.4 Covariant form So far we have expressed the Dirac equation in terms of α- and β-matrices. This brings out the connection with spin, but we can also express it in the form which emphasizes its covariance. This is: µ (iγ ∂µ − m)ψ = 0 with ∂ ∂ ∂ ∂ ! ∂ = , , , µ ∂t ∂x ∂y ∂z and γµ = (γ0, γ1, γ2, γ3) is defined through the Dirac γ-matrices:
0 1 2 3 γ ≡ β, γ ≡ βαx, γ ≡ βαy, γ ≡ βαz
The index µ is treated as the Lorentz index of a four-vector and summation over re- peated indices is implied. It is important to note that the Dirac γ-matrices are not four vectors, they are constant matrices which are invariant under Lorentz transformation. ! ! 0 I 0 k 0 σk γ = β = , γ = βαk = 0 −I −σk 0 Pauli-Dirac representation? Adjoint spinors?
29 8.5 Solution of the Dirac Equation 8 QUANTUM FIELD THEORY
8.5 Solution of the Dirac Equation We look at the plane wave solution of a free particle
ψ(x, t) = u(E, p)ei(p·x−Et) where u(E, p) is a four-component Dirac spinor and the overall wavefunction ψ(x, t) satisfies the Dirac equation µ (iγ ∂µ − m)ψ = 0 Using ∂ ∂ ∂ ∂ ! ∂ = , , , µ ∂t ∂x ∂y ∂z we consider the derivatives of the free particle solution: ∂ ∂ ∂ ψ = ψ = −iEψ, ∂ ψ = ψ = ip ψ, ··· 0 ∂t 1 ∂x x inserting these into the Dirac equations gives:
0 1 2 3 (γ E − γ px − γ py − γ pz − m)u = 0 which can be written as µ (γ pµ − m)u = 0 Which is the Dirac equation in "momentum", it contains no derivatives! For a particle at rest p = 0, and thus
ψ = u(E, 0)e−iEt
=⇒ Eγ0u − mu = 0 and thus 1 0 0 0 φ1 φ1 0 1 0 0 φ φ 2 2 E = m 0 0 −1 0 φ3 φ3 0 0 0 −1 φ4 φ4 This equation has four orthogonal solutions:
1 0 0 0 0 1 0 0 u1 = N , u2 = N , u3 = N , u4 = N 0 0 −1 0 0 0 0 −1 | {z } | {z } E=m E=−m in all cases N determines the normalisation of the wavefunction. u1 and u2 represent spin-up and spin-down positive energy solutions and u3 and u4
30 8.5 Solution of the Dirac Equation 8 QUANTUM FIELD THEORY represent spin-up and spin-down negative energy solutions. The four solutions for the Dirac equation for a particle at rest, including the time dependence are:
1 0 0 0 0 1 0 0 −imt −imt +imt +imt ψ1 = N e , ψ2 = N e , ψ3 = N e , ψ4 = N e 0 0 1 0 0 0 0 1
For p 6= 0 we get:
1 0 pz px−ipy E−m E−m px+ipy −pz 0 1 ψ = N p , ψ = N p −ip , ψ = N E−m , ψ = N E−m 1 1 z 2 2 x y 3 3 4 4 E+m E+m 1 0 px+ipy −pz E+m E+m 0 1
From the solutions for p = 0 we can say that u1, u2 are positive energy spinors with
q E = + p2 + m2
and u3, u4 are negative energy spinors with
q E = − p2 + m2
We cannot treat all four solutions as having E > 0 because then the exponent of
ψ(x, t) = u(E, p)ei(p·x−Et) would be the same for all four solutions. In that case the four solutions would no longer be independent of each other. Thus we cannot disregard the negative energy solutions as we would do in classical mechanics.
31 8.6 Antiparticles 8 QUANTUM FIELD THEORY
8.6 Antiparticles If negative energy solutions represented accessible negative energy particle states, we would expect that all positive energy electrons would fall into these lower energy states, but this does not occur. Dirac proposed that the vacuum corresponds to the state where all negative energy states are occupied. In this "Dirac sea" the Pauli exclu- sion principle presents positive energy electrons from falling into the fully occupied negative energy states. Furthermore a photon with energy E > 2me could excite an electron from a negative energy state, leaving a hole in the vacuum. A hole in the vacuum would then correspond to a state with more energy (less negative energy) and a positive charge relative to the fully occupied vacuum. In this way, holes in the Dirac sea correspond to positive energy antiparticles with the opposite charge to the particle states. This interpretation brings a lot of problems, for example antiparticle states for bosons are also observed but the Pauli exclusion principle does not apply. We now use the Feynman-Stückelberg interpretation. It is an experimentally proven fact that each fundamental spin-half particle possesses a corresponding antiparticle with opposite charge but otherwise the same behavior: it propagates forward in time after creation, it ionizes the gas in the detector, etc. The modern interpretation by Feynman and Stückelberg was developed with quantum field theory. The E < 0 solutions are interpreted as negative energy particles which propagate backwards in time. These particle solutions correspond to physical positive energy antiparticle states with opposite charge which propagate forwards in time. This can mathematically be validated by setting E → −E and t → −t.
e−(E > 0) e−(E > 0) γ γ
e−(E < 0) e+(E > 0) t t
Figure 5: The process of e+e− annihilation.
To the left we have a positive energy electron producing a photon and a negative en- ergy electron propagating backwards in time. To the right is the Feynman-Stückelberg interpretation with a positive energy positron (instead of negative Energy electron) propagating forwards in time. Time runs from the left to the right. In this case both the electron and the positron are traveling forwards in time. But in the Feynman diagram the arrow still points back in time (because it is an anti-particle). These first two Feynman diagrams were just a simplification, actually the electron- positron annihilation is a process where two photons are created (for the sake of mo-
32 8.6 Antiparticles 8 QUANTUM FIELD THEORY mentum conservation, these two photons go in opposite direction):
e−e+ → γγ
e− γ
e+ γ
Figure 6: The process of e−e+ → γγ .
Thus we have a fermion propagator in the middle and two photon branches on the right. Again we see that the antiparticle (positron) is pointing backwards in time, as noted before this is just convention to show that it is an antiparticle.
33 8.7 Antiparticle spinors 8 QUANTUM FIELD THEORY
8.7 Antiparticle spinors
We could in principle perform calculations with u3, u4, but we would always have to remember that the energy is negative of the physical energy. It would be much easier to work with antiparticle spinors written such that the physical momentum and the physical energy can be used inside the equation. This then leads to the two antiparticle solutions
−i(p·x−Et) ψi = vie with px−ipy pz E+m E+m −pz px+ipy √ √ v (p) = E + m E+m v (p) = E + m E+m 1 2 0 1 1 0 and the two particle solutions:
i(p·x−Et) ψi = uie with 1 0 √ 0 √ 1 u (p) = E + m p u (p) = E + m p −ip 1 z 2 x y E+m E+m px+ipy −pz E+m E+m The spinors have been normalized to 2E particles per unit volume with √ N = E + m
8.8 Charge conjugation Charge conjugation is an example of a discrete symmetry transformation, its effect is to replace particles with corresponding antiparticles and vice versa.
34 9 SPIN
9 Spin
For particles at rest, the spinors u1(E, 0) and u2(E, 0) are clearly eigenstates of 1 0 0 0 1 1 0 −1 0 0 ˆ Sz = Σz = 2 2 0 0 1 0 0 0 0 −1 and therefore represent "spin-up" (u1) and "spin-down" (u2) eigenstates of the z- component of the spin operator S. If we observe a motion only in ±z direction, we see that ui and vi are eigenstates and thus for the particles: 1 Sˆ u (E, 0, 0, ±p) = + u (E, 0, 0, ±p) z 1 2 1 1 Sˆ u (E, 0, 0, ±p) = − u (E, 0, 0, ±p) z 2 2 2 ˆ(v) ˆ For the antiparticle spinors the physical spin is given by the operator Sz = −Sz.Thus: 1 Sˆ(v)v (E, 0, 0, ±p) = −Sˆ v (E, 0, 0, ±p) = + v (E, 0, 0, ±p) z 1 z 1 2 1 1 Sˆ(v)v (E, 0, 0, ±p) = −Sˆ v (E, 0, 0, ±p) = − v (E, 0, 0, ±p) z 2 z 2 2 2
From the sign it is clear that u1, v1 represent the spin-up state and u2, v2 the spin-down state. ˆ But in general ui and vi are not eigenstates of Sz.
9.1 Helicity As already mentioned for particles not traveling in the z-direction, the spinors do not ˆ map on the spin states. Also because Sz does not commute with the Dirac hamiltonian ˆ ˆ [HD, Sz] 6= 0 ˆ ˆ it’s not possible to define a basis of simultaneous eigenstates of Sz and HD. This is why we introduce a new concept: helicity. S · p h ≡ p And for a four-component Dirac spinor, the helicity operator is
Σˆ · pˆ 1 σ · pˆ 0 ! hˆ = = 2p 2p 0 σ · p
35 9.1 Helicity 9 SPIN with that we see that ˆ [HD, Σˆ · pˆ] = 0 For a spin-half particle, the component of spin measured along any axis is quantized to be either ±1/2. From this we know that the eigenvalues of the helicity operator acting on a Dirac spinor are ±1/2. They are known as right-handed or left-handed helicity states. Helicity is not Lorentz-invariant!
S
p
S·p h = p
Figure 7: The helicity is the projection of the spin along its direction of motion.
The right handed helicity for a spin-half fermion is h = +1/2 and points in the direc- tion of motion, whereas the left handed helicity is h = −1/2 and points in the opposite direction. To find the eigenstates of helicity means to find the solutions of the dirac equation:
(Σ · pˆ)u↑ = +u↑ (Σ · pˆ)u↓ = −u↓
This gets us (after some calculation, which can be closer looked at in Thomsons book): For particles:
c −s iϕ iϕ √ se √ ce u↑ = E + m p u↓ = E + m p E+m c E+m s p iϕ p iϕ E+m se − E+m ce and for antiparticles:
p p E+m s E+m c p iϕ p iϕ √ − ce √ se v = E + m E+m v = E + m E+m ↑ −s ↓ c ceiϕ seiϕ where θ ! θ ! s = sin c = cos 2 2
36 9.2 Parity 9 SPIN
9.2 Parity Like charge conjugation, parity is also an example of a discrete symmetry transforma- tion. Parity is the spatial inversion through the origin:
x → −x y → −y z → −z t → −t
Interactions in QED and QCD always conserve parity! Lets take ψ as a solution of the dirac equation and ψ0 as the corresponding solution in the parity mirror, obtained from the parity operator Pˆ such that
ψ → ψ0 = Pˆ ψ
If we apply Pˆ twice to the wave-function, we should recover the original one, thus
ψ0 = Pˆ ψ =⇒ Pˆ ψ0 = ψ
Inserting ψ0 = Pˆ ψ into the dirac equation we obtain after some calculation
γ0Pˆ ∝ I and using Pˆ2 = I: Pˆ = ±γ0 Conventionally we choose Pˆ = +γ0 such that under parity transformation the form of the Dirac equation is unchanged provided the Dirac spinors transform as
ψ → Pˆ ψ = γ0ψ
The intrinsic parity of a fundamental particle is defined by the action of the parity operator Pˆ = γ0 on a spinor for a particle at rest. For example: The u1 spinor for a particle at rest is an eigenstate of the parity operator with 1 0 0 0 1 0 1 0 0 √ 0 ˆ 0 P u1 = γ u1 = 2m = +u1 0 0 −1 0 0 0 0 0 −1 0 analog: ˆ ˆ ˆ P u2 = +u2 P v1 = −v1 P v2 = −v2 Thus the intrinsic parity of a fundamental spin-half particle is opposite of that of a fundamental spin-half antiparticle. Also, the application of the parity operator on a dirac spinor corresponding to a particle with momentum p reverses its momentum but leaves the spin state unchanged: ˆ P u1(E, p) = +u1(E, −p)
37 10 PARTICLE EXCHANGE
10 Particle exchange
When particles interact, the exchange of momentum and energy happens through force carrying gauge bosons, which where introduced in the standard model. For that we use the transition rate Γfi from an initial state i to a final state f, which is 2 given by Fermi’s golden rule Γfi = 2π|Tfi| ρ(Ef ), where Tfi is the transition matrix element, given by the perturbation expansion:
X hf|V |jihj|V |ii Tfi = hf|V |ii + + ··· | {z } j6=i Ei − Ej I | {z } II Term I can be viewed as "scattering in a potential V " and term II as "scattering via an intermediate state j:
f f
Vfj j
Vfi Vji i i
Figure 8: Scattering: Term I (left) and term II (right).
This classical approach is unsatisfactory in quantum mechanics, because the scattered particle would transfer momentum to another without any other mediating body. Also the description of forces in terms of potentials would imply that if a distant particle were moved suddenly, the potential due to that particle would change instantaneously at all points in space, which would violate special relativity. In QFT the forces between particles result form the exchange of momentum.
10.1 Time-ordered perturbation theory We consider the particle interaction a + b → c + d which can occur via the exchange of particle X: In the image it is clear to see that for the left option we have the states: |ii : a + b |ji : c + b + X |fi : c + d This can be thought of as particle a emitting the exchange particle X, and then X being absorbed by b. And for the right: |ii : a + b |ji : a + d + X˜ |fi : c + d
38 10.1 Time-ordered perturbation theory 10 PARTICLE EXCHANGE
a c a c
Vfi Vfi
X X˜
Vfi Vfi b d b d i j f i j f t t
Figure 9: Two possible time orderings for a + b → c + d.
This can be thought of as b emitting X˜ which is then absorbed by a. It is assumed that X˜ has the same mass but opposite charge as X. This has to be the case if charge is conserved at each vertex. As a real-life example we take the process
− − e νµ → νeµ
These diagrams are not Feynman-diagrams, to get a Feynman diagram, we need to
e− νe e− νe
W− W+
νµ µ− νµ µ−
− − Figure 10: The two lowest time ordered diagrams for e νµ → νeµ . sum over all possible time-orderings. In a Feynman diagram, the left side is the initial state and the right the final state. The intermediate state represents the propagator. These propagators or exchange particles are referred to as virtual particles (mathemat- ical construct). Thus the Feynman diagram for the above process is: Now in the Feynman diagram energy and momentum are both conserved at each ver- tex. The four-momentum q which appears in the propagator can be determined from the conservation of four-momentum at the vertices. This means, for the s-channel annihi- lation: 2 2 q = p1 + p2 = p3 + p4 =⇒ q = (p1 + p2) = s hence the name s-channel (from the mandelstam variable s). For this case it follows that q2 > 0 and the exchanged virtual particle is "time-like".
39 10.2 Scattering in a potential 10 PARTICLE EXCHANGE
e− νe e− νe e− νe
W− + W+ = W
νµ µ− νµ µ− νµ µ−
− − Figure 11: Feynman diagram for e νµ → νeµ .
For the t-channel scattering diagram we have
2 2 q = p1 − p3 = p4 − p2 =⇒ q = (p1 − p3) = t
For this case q2 < 0 and the exchanged particle is "space-like".
10.2 Scattering in a potential The covariant formulation of a scalar interaction in terms of the exchange of (virtual) particles leads to a Lorentz-invariant matrix element:
gagb Mfi = 2 2 q − mX ——————
40 11 QED
11 QED
Quantum electrodynamics is the quantum field theory of the electromagnetic interac- tion (weak and electromagnetic forces). The Lorentz invariant matrix element for a scalar interaction is composed of three parts:
− The strength of interaction at each of the two vertices:
hψc|V |ψai hψd|V |ψbi
− The propagator for the exchanged virtual particle of mass mX : 1 2 2 q − mX it can be written as: 1 hψc|V |ψai 2 2 hψd|V |ψbi q − mX Previously the simplest Lorentz invariant choice for the interaction vertex was used, namely a scalar interaction of the form hψ|V |φi ∝ g. We consider the interaction o an electron e− and a tau lepton τ − by the exchange of a photon γ. The general ideas still hold but we need to account for the spin of the electron/tau lepton and the spin of the photon (polarization). To obtain the QED matrix element for a scattering process, the corresponding expres- sion for the QED interaction vertex is required.
e− e−
p1 µ p3 γ q
p2 ν p4 τ − τ −
Figure 12: Feynman diagram for the QED scattering e−τ − → e−τ −.
For the exchange of the photon γ, which is a spin-1 particle, it is necessary to sum over the quantum-mechanical amplitudes for the possible polarisation states. (λ) The free photon field Aµ can be written in terms of a plane wave and a four-vector ε for the polarisation state λ as:
(λ) i(p·x−Et) Aµ = εµ e
41 11 QED
For a real (not virtual) photon, the polarisation vector is always transverse to the direc- tion of motion, thus if the photon is propagating in the z-direction:
ε(1) = (0, 1, 0, 0) ε(2) = (0, 0, 1, 0)
Previously with a simple spin-less interaction we had 1 M = hψc|V |ψai 2 2 hψd|V |ψbi q − mX In QED we could again go through this procedure of summing the time-orderings using Dirac-spinors and ˆ 0 µ VD = qγ γ Aµ this would give us
∗ ελ ελ † 0 µ X µ ν † 0 ν M = ue(p3)qeγ γ ue(p1) 2 uτ (p4)qτ γ γ uτ (p2) | {z } λ q | {z } electron-photon interraction | {z } tau-photon interaction photon propagator polarization-sum
The propagator-sum in the middle gives
−gµν q2 thus −g M = u†(p )q γ0γµu (p ) µν u† (p )q γ0γνu (p ) e 3 e e 1 q2 τ 4 τ τ 2 and with ψ = ψ†γ0:
−g M = u (p )q γµu (p ) µν u (p )q γνu (p ) e 3 e e 1 q2 τ 4 τ τ 2
42 11.1 Feynman rules for QED 11 QED
11.1 Feynman rules for QED There are three basic elements to the matrix element corresponding to the Feynman diagram: − the Dirac spinors for the external fermions (initial and final state) − a propagator term for the virtual photon − a vertex factor at each interaction vertex For each of these elements of the Feynman diagram there is a Feynman rule for the corresponding term in the matrix element:
initial-state particle: u(p)
final-state particle: u(p)
initial-state antiparticle: v(p)
final-state antiparticle: v(p)
initial-state photon: εµ(p)
∗ final-state photon: εµ(p) ig photon propagator: − µν q2
2 i(gµν − qµqν/mW) W-boson propagator: − 2 2 q − mW 2 i(gµν − qµqν/mZ) Z-boson propagator: − 2 2 q − mZ iδabg gluon propagator: − µν q2 i(γµq + m) fermion propagator: − µ q2 − m2
QED vertex: −iQeγµ Where particles, antiparticles and the fermion propagator are spin 1/2 and photons and the photon propagator are spin 1.
43 11.2 t-channel Diagram 11 QED
11.2 t-channel Diagram Knowing that, we can calculate the matrix element for scattering process:
e−τ − → e−τ −
Since it is a t-channel process (scattering) we know that q = p1 − p3, for the two
e− e−
p1 µ p3
γ q = p1 − p3
p2 ν p4
τ − τ −
Figure 13: Feynman diagram for the QED scattering e−τ − → e−τ −. branches we are going against the arrows on the fermion branches (red arrow direction) and multiply the elements, first the branch on top, then the propagator, then the branch on the bottom, thus we get: −ig −iM = u(p )[ieγµ]u(p ) µν u(p )[ieγν]u(p ) 3 1 q2 4 2 the indices µ and ν label the two interaction vertices. Another example is the electron positron annihilation: Which gives us (going again in
e− γ
p1 µ p3
q = p1 − p3
p2 ν p4
e+ γ
Figure 14: Feynman diagram for the QED scattering e+e− → γγ. the direction of the red arrow): −i(γµq + m)! −iM = v(p )[ieγν]ε∗(p ) µ ε∗ (p )[ieγµ]u(p ) 2 ν 4 q2 − m2 µ 3 1
44 11.3 s-channel Diagram 11 QED
11.3 s-channel Diagram Now we try to find the matrix element for the Feynman diagram for the s-channel annihilation process e+e− → µ+µ−. Antiparticles are represented by lines in the negative time direction, reflecting the interpretation of the negative energy solutions as mentioned before. We get: −ig −iM = v(p )ieγµu(p ) µν u(p )ieγνv(p ) 2 1 q2 3 4 Where we again went in the direction of the red arrow, as seen below:
e− µ−
p1 p3 γ or Z µ ν q = p1 + p2 p2 p4 e+ µ+
Figure 15: Feynman diagram for the QED annihilation e+e− → µ+µ−.
In this diagram it is also possible to create a Z-boson, but only if the kinetic energy of the incoming particles is enough, otherwise it will just be a photon or a virtual Z-boson: − The Z-boson is real: there was enough kinetic energy available to create a Z- boson (high mass). − The Z-boson is virtual, this means that the kinetic energy of the incoming parti- cle is not enough to create a heavy Z-boson, however Heisenberg’s uncertainty relation ∆E∆t ≥ ~ allows for a high energy ∆E to be created (even though this would violate energy conservation) as long as it is just for a very short time ∆t. This means that it is possible that a Z-boson is created but just for a very short time, after that it decays into, for example, a lepton pair. The QED matrix element for the s-channel is very similar to that for the t-channel. Apart form the antiparticle spinors v the only difference is the order in which the par- ticles appear in the expressions for the currents:
The first particle encountered when following the line representing a fermion current from end to the start in the direction against the sense of the arrows, always appears as the adjoint spinor. In our example the incoming e− and the outgoing µ− are written as adjoint spinors.
45 11.4 u-channel Diagram 11 QED
11.4 u-channel Diagram Every u-channel diagram is actually based on a t- or a s-channel diagram, as an exam- ple we take the annihilation-process e−e+ → γγ.
e− γ e− γ
e+ γ e+ γ
Figure 16: t- and u-channel Feynman diagrams for e−e+ → γγ.
The u-channel decay diagram (on the right) is only to be drawn for processes where we have the two same outgoing indistinguishable particles.
− γ − γ e p3 e p1 p1 p3 µ ν
qt = p1 − p3 qu = p1 − p4 ν µ p4 p2 p4 p e+ γ e+ 2 γ
Figure 17: Labeled t- and u-channel Feynman diagrams.
Important note: as already said, every u-channel diagram is ’derived’ from either a t-channel or a s-channel diagram, in this case we switched the output branches of the t-channel. But we also have to switch the labels of either the vertices µ and ν or the momenta of the two photons p1 and p3, we must not switch both! In this case we just swapped the vertex labels. Thus the matrix element is: µ ! ν ∗ γ qtµ + m ∗ µ −iMt = v(p2)ieγ εν(p4) i 2 2 εµ(p3)ieγ u(p1) qt − m ν ! µ ∗ γ quν + m ∗ ν −iMu = v(p2)ieγ εµ(p3) i 2 2 εν(p4)ieγ u(p1) qu − m And then summing up: M = Mt + Mu This is a good example for the fermion propagator, which has a index in the formula, this index is the index of the vertex where the propagator starts.
46 11.5 Example: Compton scattering 11 QED
11.5 Example: Compton scattering The Compton scattering can be drawn as a s- or a t-channel diagram:
γ e− γ γ
e− γ e− e−
Figure 18: Feynman diagrams for the Compton scattering γe− → γe− .
e− II γ γ γ p1 p3 p1 µ p3 e e qt = p1 − p3 II a b I qs = p1 + p2 ν p2 p4 p2 p4 γ I e− e− e−
Figure 19: Labeled t- and s-channel diagrams for Compton-scattering.
In the case of these diagrams we have a ’fermion-path’, we just go in reverse (thus first we calculateI and thenII) when calculating the matrix element, this leaves us with:
µ ! ν γ qtµ + m ∗ µ −iMt = u(p4)ieγ εν(p2) i 2 2 εµ(p3)ieγ v(p1) qt − m
a ! b ∗ γ qsa + m a −iMs = u(p4)ieγ εb (p3) i 2 2 εa(p1)ieγ u(p2) qs − m and the total: M = Mt + Ms
47 11.6 Example: Bhabha scattering 11 QED
11.6 Example: Bhabha scattering In QED the process e+e− → e+e− is called Bhabha scattering. It is used as a luminosity monitor in electron-positron colliders. There are two leading-order Feynman diagrams contributing to the matrix element: t-channel and s-channel:
e− e− e− e− p1 p3 p1 µ p3 γ γ qt = p1 − p3 a b qs = p1 + p2 ν p2 p4 p2 p4 e+ e+ e+ e+
Figure 20: Feynman diagrams for e+e− → e+e−.
Now, going again with the red arrow and from top to bottom, we calculate the matrix elements, first for each channel separately: ! µ igµν ν −iMt = u(p3)[ieγ ]u(p1) − 2 v(p2)[ieγ ]v(p4) qt ! a igba b −iMs = v(p2)[ieγ ]u(p1) − 2 u(p3)[ieγ ]v(p4) qs
µ ν µ a b a For easier notation we use γ gµνγ = γµγ and γ gbaγ = γaγ and then the total matrix element then is the difference of the two (because of fermi-dirac statistics):
M = Mt − Ms ! 2 1 µ a 1 = −e u(p3)γµu(p1) 2 v(p2)γ v(p4) − v(p2)γ u(p1) 2 u(p3)γav(p4) qt qs
Concerning fermi-dirac statistics: Normally the total matrix element is the sum of the matrix elements of the different diagrams, but if we have asymmetry there is an exception: we include a minus sign between diagrams that differ only in the exchange of two identical fermions.We see that this is the case for the Bhabha scattering, if we flip v(p2) with u(p3) in the matrix elements we have two identical ones. This is not the case for Compton-scattering, there we take the sum.
48 11.7 Electron-Positron annihilation 11 QED
11.7 Electron-Positron annihilation The matrix element for the lowest-order diagram for the process
e+e− → µ+µ− was given above: −ig −iM = v(p )ieγµu(p ) µν u(p )ieγνv(p ) 2 1 q2 3 4
e2 M = − g [v(p )γµu(p )][u(p )γνv(p )] q2 µν 2 1 3 4 e2 = − g jµjν q2 µν e µ where µ µ ν ν je = v(p2)γ u(p1) jµ = u(p3)γ v(p4) The four-momentum of the virtual photon is determined by conservation of energy and momentum at the interaction vertex
q = p1 + p2 = p3 + p4 and thus 2 2 q = (p1 + p2) = s where s is the center-of-mass energy squared. Thus e2 M = − j · j s e µ Assuming that the electron and positron beam have equal energies, which is the case for most e+e−-colliders, the center of mass energy is simply twice the beam energy: √ s = 2Ebeam
11.8 Spin sums In order to calculate the cross section, the matrix element M needs to be evaluated taking into account the possible spin states of the particles involved. In the case of the
e+e− → µ+µ− process, each of the particles can be in one of two possible helicity states. There are four possible helicity configurations in the initial state: Since said process consists of 16 possible orthogonal helicity combinations, each of which constitutes a separate physical process
49 11.8 Spin sums 11 QED
e− e+ e− e+ e− e+ e− e+ RL RR LL LR
Figure 21: The four possible helicity combinations for the e+e− initial state.
− RR → RR
− RL → RR
− RL → RL
−··· where + − + − RR → RR : e↑ e↑ → µ↑ µ↑ We need to sum the four corresponding |M|2 terms, for example (colliding electron and positron are both in right-handed helicity states):
X 2 2 2 2 2 |MRR| = |MRR→RR| + |MRR→RL| + |MRR→LR| + |MRR→LL|
In most colliders the electron and positron beams are unpolarised, meaning there are equal numbers of positive and negative helicity electrons/positrons present in the initial state. In this case the helicity configuration or a particular collision is equally likely to occur in any of the four possible helicity states of the e+e− initial state. This is accounted for by defining the spin-average summed matrix element squared
1 h|M |2i = |M |2 + |M |2 + |M |2 + |M |2 fi 4 RR RL LR LL 2 2 2 = |MRR→RR| + |MRR→RL| + ··· + |MRL→RR| + ··· 1 = X |M|2 4 spins with e2 M = − j · j s e µ helicity amplitudes muon electron current cross section chirality
50 12 QCD
12 QCD
Before we discussed QED, thus only looking at the weak and the electromagnetic forces, QCD on the other hand focuses on the strong froce.
12.1 Global gauge symmetry We know from quantum mechanics that the wave function is determined up to a global phase, the physics is invariant for the transformations
φ → φe−iθ φ† → φ†eiθ
Both the Dirac and the Schroedinger equation are invariant under these transforma- tions, thus if φ is a solution then also φe−iθ is one. This is a result of θ not depending on the 4 coordinates.
12.2 The local gauge principle All interactions between fermions and spin-1 bosons in the standard model are speci- fied by the principle of local gauge invariance. Gauge invariance is a familiar idea from electromagnetism, where the physical electric and magnetic fields (defined through scalar φ and vector A potentials) do not change under the gauge transformation ∂ φ → φ0 = φ − A → A0 = A + ∇ ∂t This gauge transformation can be written more succinctly as
0 Aµ → Aµ = Aµ − ∂µ where Aµ = (φ, −A) and ∂µ = (∂0, ∇). So we search for transformations under which the dynamic of the particles stays in- variant. For that we use symmetries. There are global symmetries which are related to conser- vation laws and local gauge symmetries which are related to the interactions. The simplest symmetry U(1) (rotation in the complex plane) is related to QED. We will now apply the concept of symmetries to the quark flavors and distinguish between exact and approximate symmetries.
51 12.3 Strong interaction 12 QCD
12.3 Strong interaction In the table below we can see which particle feels which force, and which propagator corresponds to which force: EM-force Weak force Strong force Quarks × × × Charged Leptons × × Neutral Leptons × Z-Boson × W-Boson × × Gluon × Photon × We see that the neutral leptons (neutrinos) only interact through the weak force, this is the reason why they cannot be stopped easily and travel through pretty much every- thing with barely a change in momentum. In QED: − the electron has charge −e
− the positron has anti charge +e
− the force is mediated by a massles gauge boson - the photon γ
− photons do not carry the charge of the EM interaction while in QCD: − quarks carry color charge r, g, b
− anti quarks carry anti color charge r, g, b
− the force is mediated by massless gluons
− gluons do carry the color charge This means quarks interact by exchanging virtual massless gluons, which can carry color and anti-color: There are 8 possible physical gluons: 1 1 rg, rb, gr, gb, br, bg, √ (rr − gg), √ (rr + gg − 2bb) 2 6 It is believed (not yet proven) that all observed free particles are colorless, implying that there has never been observed a free quark. They are always found in bound states (hadrons).
52 12.4 Hadronization and jet 12 QCD
qb qr qb qr qb qr
rb + br = g
qr qb qr qb qr qb
Figure 22: Interaction of quarks through gluons.
12.4 Hadronization and jet If we were to consider a quark and anti-quark produced in electron positron annihila- tion, then initially quarks separate at high velocity, the color flux tube forms between the quarks. The energy stored in the flux tube is sufficient to produce quark and antiquark pairs. This process, called "hadronisation", continues until the quarks pair up into jets of col- orless hadrons. It is not yet calculable. This means that at collider experiments, quarks and gluons are observed as jets of par- ticles.
12.5 e+e− collisions e+e− colliders are an excellent place to study QCD. As mentioned above, when the quark anti-quark pair is created, they quickly transform to hadron jets.
e+ hadrons
q γ
q
e− hadrons
Figure 23: Feynman diagram for the quark antiquark pair creation. e+e− collisions produce all quark flavors, as they produce hadron jets, you usually can’t tell which came from the quark and which came from the anti-quark. This means
53 12.5 e+e− collisions 12 QCD that when this quark-antiquark pair is not forming a meson (which it rarely does) they form two hadron jets, in which the quarks could further "multiply" by hadronization and change flavor through charged currents (discussed later), thus creating the possi- bility for such a process: e+e− → p+p−
The angular distribution of the jets is proportional to (1 + cos2 θ). This means quarks are spin 1/2. Color is conserved and quarks are produced as
rr, gg, bb
For a single quark flavor and single color
4πα2 σ(e+e− → q q ) = G2 i i 3s q
54 12.6 Charged current: Leptons 12 QCD
12.6 Charged current: Leptons For Leptons, there exists a way to change family, but only with the corresponding neutrino, for example: − − e νe → µ νµ here are two examples:
− − νµ µ νe µ
W− W−
− − µ νµ e νe
Figure 24: Feynman diagrams for a Lepton family change.
Other examples are: − − µ → e νµνe − − τ → µ νµντ
12.7 Charged current: Quarks The same goes for quarks, there it is called flavor changing.
down-type (d,s,b) up-type (u,c,t) down-type (d,s,b) W
W
anti-up-type (u, c, t)
Figure 25: Feynman diagrams for a quark flavor change.
But it is important to note that inter-generation transformations are far more likely than trans-generation, for example: charm to strange is more likely than charm to bottom. Also: top to strange is more likely than top to down.
55 12.8 Neutral current: Quarks 12 QCD
12.8 Neutral current: Quarks Additionally, for an down-type quark to change to another down-type quark it has to first transform into a down-type quark, for example:
s → c → d it is clear to see that these are just two charged current processes happening after an- other. In the diagram we connect the two W-bosons which then form an arc (hence the name ’neutral current’):
s c d
W
Figure 26: Diagram for a down-type to down-type generation change.
This is only a diagram to show the single stages of this process, in a general Feynman diagram it would look more like that (we see a transition from a down-type to another down-type quark via an up-type quark):
γ or Z u, c, t
d, s, b d, s, b W
Figure 27: Feynman diagram, also called penguin diagram (see B-meson decay).
There is also a photon or Z-boson ejecting from the down-type quark, which was not shown in the first diagram (the reason why is more advanced), this gamma or Z-boson can either be a virtual or a real one, depending on the energy available in this process, as discussed above in the chapter about Feynman rules. It is important to note that this is a rather rare thing to happen (it is not a first order diagram), thus the possibility of processes involving this quark flavor/generation trans- formation to happen is not high.
Also: these changes can only happen between particles or between antiparticles, but we cannot transform an antiquark into a quark and vice versa.
56 12.9 Neutral current: Quarks 12 QCD
12.9 Neutral current: Quarks The same thing can happen with leptons, for example:
γ or Z ν
τ e− W
Figure 28: Feynman diagram, also called penguin diagram for leptons.
12.10 Quark mixing At the time when Cabibbo was doing research on quarks, there were only the first and second generation of quarks known:
up down charm strange
In the Cbibbo hypothesis, the weak interactions of quarks have the same strength as the leptons, but the weak eigenstates of quarks differ from the mass eigenstates. The weak eigenstates d0, s0 are related to the mass eigenstates d, s by
0! ! ! d cos θc sin θc d 0 = s − sin θc cos θc s with θc being the cabibbo angle. With the discovery of the top and bottom quark the 3 dimensional CKM-matrix had to be introduced.
57 12.11 Baryon number conservation 12 QCD
12.11 Baryon number conservation The baryon number is given by n − n B = q q 3 where
− nq is the number of constituent quarks
− nq is the number of constituent antiquarks In QED a quark can emit a photon or a quark-antiquark pair can annihilate. In QCD a quark can emit a gluon or a quark-antiquark pair can annihilate. In both cases B is constant.
12.12 Lepton number conservation The lepton number is given by
L = n` − n` where all leptons have a value +1 assigned, all antileptons −1 and non-leptonic parti- cles 0. In QED a lepton can emit a photon or a lepton-antilepton pair can annihilate. In QCD a lepton can emit a gluon or a lepton-antilepton pari can annihilate. In both cases L is constant. − + Charge weak interaction relates ` to ν` and ` to ν`, but L stays conserved.
12.13 Lepton flavor conservation In the lepton sector charged leptons are related to their relative neutrinos, family (called lepton flavor) has to be conserved. In addition also electric charge and color must be conserved. For example, this is allowed: − − e → W + νe while this is not − − e → W + νµ,τ E
58 12.14 Example: Beta decay 12 QCD
12.14 Example: Beta decay
e+ e−
νe νe
W+ W−
u d d u p+ d d n0 n0 d d p+ u u u u
Figure 29: Feynman diagrams for β+ and β−-decays.
Only the W-boson can change a down quark into an up quark and vice-versa. On the lower vertex we have (for the β+-decay) an up quark (+2/3) going in and a down quark (−1/3) going out, thus:
decay lower vertex upper vertex
β+ 2/3 − (−1/3) = 1 +1 + 0 = 1
β− −1/3 − 2/3 = −1 −1 + 0 = −1
Note: on the upper vertex, both the positron/electron and its corresponding neutrino are drawn at an angle pointing in the positive time direction (the positron arrow still points backwards in time - convention). This means, we have to look at both of them as outgoing particles. So first we check what gauge boson we need:
incoming − outgoing = gauge boson to check if charge is conserved we can use the following formula:
incoming = outgoing
The gauge boson now counts as outgoing on the lower vertex and as incoming on the upper vertex (it is also drawn at an angle to indicate the direction).
So our gauge-boson has to have the same charge as the two vertices it connects, since charge is not 0, there cannot be a Z-boson but has to be either a W+ or a W−-boson.
59 12.14 Example: Beta decay 12 QCD
We see that the baryon number is conserved: n − n B = q q 3 Since β-decay only changes neutrons to protons and vice versa, the number of indi- vidual quarks doesn’t change. It is only the baryon flavor that changes (labelled as the isospin). Up and down quarks have total isospin I = 1/2 and isospin projections: 1 1/2 up quark Iz = (nu − nd) = 2 −1/2 down quark all other quarks have I = 0. Also the Lepton number is conserved:
L = n` − n` where all leptons have a value +1 assigned, all antileptons −1 and non-leptonic parti- cles 0. Thus the lepton number in both decays is 0 before and after the decay.
60 13 ISOSPIN
13 Isospin
Isospin is a flavor-quantum number that describes an internal symmetry under the strong force. It is used to classify hadrons. Proton and neutron have very similar masses and the strong force is found to be ap- proximately charge-independent, thus
Vpp ≈ Vnp ≈ Vnn
To reflect this symmetry, Heisenberg proposed that if you could switch off the electric charge of the proton there would be no way to distinguish between a proton and a neutron and he proposed the proton should be considered as two states of a single entity; the nucleon: 1! 0! p+ = n0 = 0 1 Analogous to the spin-up/spin-down states of a spin-1/2 particle this would be called the isospin. The neutron and the proton form an isospin doublet with total isospin I = 1/2 and third component I3 = ±1/2.
We now extend this idea to quarks, knowing that all quark flavors are equally treated by the strong interaction. Because we know that mu ≈ md, we choose the basis
1! 0! u = = | 1 , + 1 i d = = | 1 , − 1 i 0 2 2 1 2 2
Where: |I,I3i from this we get the charge of the particle:
q = e(I3 + 1/2)
With the proton as |1/2, 1/2i and the neutron as |1/2, −1/2i we get the charges as
qp+ = e qn0 = 0
61