Lecture 8: Relativistic

Contents:

1. Notation and units.

2. Relativistic kinematics.

3. The Schr¨odinger equation.

4. Relativistic equation (Klein-Fock-Gordon equation); .

5. Electromagnetic interaction of -0 particles.

5.1 ; Fermi’s Golden Rule. 5.2 Elastic scattering of spin-0 particles.

6. Calculation of the ; phase space; Mandelstam variables.

7. The .

7.1 Derivation. 7.2 . 7.3 Covariant form of the Dirac equation. 7.4 Properties of the γ matrices. 7.5 Adjoint equation. 7.6 Plane wave solutions. 7.7 Antiparticles. 7.8 Completeness relations. 7.9 Helicity. 7.10 Bilinear covariants.

8. Electrodynamics of spin-1/2 particles.

8.1 Transition . 8.2 Trace theorems. 8.3 Completion of the calculation of the . 8.4 Cross section of elastic eµ scattering in the CMS. 8.5 Electron- into pairs. 8.6 Electron-muon elastic scattering in the LAB frame.

Figures 1a and 1b.

1 1. Notation and units.

We denote the components of the space-time 4-vector x by x0 = ct, x1 = x, x2 = y, x3 = z or, in compact form, x = (x0, x1, x2, x3) contravariant components The dual vector has the following covariant components:

x = ct, x = x, x = y, x = z. 0 1 − 2 − 3 − The invariant square of x is:

x2 x x = xµx = (x0)2 (x1)2 (x2)2 (x3)2 = invariant ≡ · µ − − − where summation over the repeated index µ, one upper and one lower, is implied (Einstein summation convention). We define the gµν:

gµν = diag(1, 1, 1, 1) − − − hence µ µν x = g xν , µ = 0, 1, 2, 3 this is called the raising of the subscripts. The dual metric tensor gµν is given by

g = diag(1, 1, 1, 1) µν − − − hence ν xµ = gµνx , µ = 0, 1, 2, 3 (lowering of the superscripts).

Lorentz boost with boost velocity v in x direction:

x0 = γ(x vt), y0 = y, z0 = z, (1) − ct0 = γ(ct vx/c) (2) − where γ = 1/ 1 v2/c2. − To get the Invqerse transformation change the sign of v and exchange the primed and unprimed coordinates.

The - 4-vector p of a particle of m is

p = (E/c, p~) and its Lorentz invariant square is

p2 = pµp = (E/c)2 p~ 2 = (mc)2 = invariant µ −

We get the rest energy E0 by setting the momentum equal to nought, hence:

E E = mc2 0 ≡ |p~=0 2 The relativistic is defined by

T = E mc2 − and we can easily check that in the nonrelativistic limit p~ mc we get the well known formula T = p2/2m. | | 

Energy and momentum expressed in terms of the particle velocity v are pc~ E = γmc2, p~ = γm~v, hence ~v = c . E

Units: We use , defined by c = 1 and h¯ = 1. To carry out the conversion to GeV units use the conversion factors

hc¯ = 197.327 MeV fm, (¯hc)2 = 0.3894 GeV2 mbarn where 1 fm = 10−15 m (femto-meter or Fermi), and 1 mbarn = 10−31 m2 (milli-barn). For numerical estimates use the following approximate values:

hc¯ = 200 MeV fm = 200 eV nm, (3) (¯hc)2 = 0.4 GeV2 mbarn (4)

Typical particle (approximate values):

Particle symbol Mass in MeV/c2 Electron e− 0.5 Muon µ− 106 Charged π 140 Neutral pion π0 135 Charged kaon K 494 Neutral kaon K0 498 Proton p 938 n 940

2. Relativistic kinematics. For a collision of particles a and b, which gives rise to the creation of particles c, d, . . ., we write the reaction equation a + b c + d + . . . → Particles a and b are the incident particles, particles c, d, etc. are the outgoing particles.

The four-momenta of particles a and b will be denoted by p1 and p2, respectively, and those of 2 2 the outgoing particles by p3, p4 etc. Their masses are given by pi = mi . The kinematics of collision processes is described in various reference frames of which we discuss here the laboratory frame (LAB) and the centre-of mass frame (CMS) in some detail.

LAB frame: particle a is the incident particle, particle b is the target particle. Their four- momenta are

p1 = (Elab, 0, 0, plab), p2 = (m2, 0, 0, 0) (5)

3 The square of the total four-momentum of the system is

2 2 2 s = (p1 + p2) = m1 + m2 + 2m2Elab (6)

Solving for plab we get

p = [s (m m )2][s (m + m )2]/2m (7) lab − 1 − 2 − 1 2 2 q CMS frame: by definition of the CMS, the incident particles have 3-momenta of equal magnitude and op- posite direction: pcms = (E∗, 0, 0, p∗), pcms = (E∗, 0, 0, p∗). (8) 1 1 2 2 − hence ∗ ∗ 2 s = (E1 + E2 ) . (9) Thus √s is the total CMS energy of the system. Solving for p∗ and comparing with Eq. (7) we get m p∗ = p 2 (10) lab √s

3. The Schr¨odinger equation

Fundamental for quantum mechanics is the concept of particle-wave duality. Formally particle-wave duality is expressed by the Einstein-de Broglie relations:

E = hω¯ , p~ = h¯~k, (11) which relate the particle characteristics E (energy) and p~ (momentum) to the wave character- istics ω (frequency) and ~k (). Here h¯ = h/2π, where h is Planck’s constant. Implied in the statement of particle-wave duality is also that the particle must be describable in terms of a wave . Moreover, we demand that the contains the complete information on the state of of the particle at time t. This implies that the wave function must satisfy an evolution equation in time. This is the Schr¨odinger equation ∂ψ(x, t) ih¯ = Hˆ ψ(x, t) (12) ∂t where Hˆ is the Hamiltonian or Hamiltonian.

4. Relativistic (Klein-Fock-Gordon equation); antiparticles.

To construct a relativistic wave equation we shall use the relations that provide the transition from to quantum mechanics:

∂ E ih¯ , (13) → ∂t ∂ ∂ ∂ p ih¯ , p ih¯ , p ih¯ , x → − ∂x y → − ∂y z → − ∂z

4 together with the relativistic energy-momentum relation

E2 = (pc)2 + m2c4 which immediately gives the Klein-Fock-Gordon equation:

h¯2∂2φ(x) = (¯hc)2 ∂2 + ∂2 + ∂2 + m2c4 φ(x) (14) − t − x y z h   i where we have used the shorthand form of the differential operators: ∂ ∂ ∂ ∂ ∂ , ∂ , ∂ , ∂ . t ≡ ∂t x ≡ ∂x y ≡ ∂y z ≡ ∂z We can rewrite the KFG equation in a relativistically symmetric form which clearly exhibits its relativistic invariance. To do this we define the four-dimensional generalization of the :

pˆ = pˆ0, pˆ1, pˆ2, pˆ3 = ih¯∂0, ih∂¯ 1, ih∂¯ 2, ih∂¯ 3  h¯    = i ∂t, ih¯∂x, ih¯∂y, ih¯∂z c − − − ! and hence the four-dimensional generalization of the Laplacian operator: 1 ∂ ∂µ = ∂2 2 µ c2 t − ∇ With these definitions the KFG equation takes on the manifestly invariant form

2 µ mc ∂µ∂ + h¯ φ(x) = 0 (15)     provided that the wave function φ(x) is a Lorentz scalar or . This latter point appears here as a mathematical requirement, but it should be stressed that it has an important physical implication: we can apply the KFG equation only to particles which are described by scalar or pseudoscalar wave functions. Such particles do exist, for instance and kaons, which are pseudoscalar particles, but are not scalar particles and their wave functions are correspondingly not scalars. Note that mc/h¯ is the inverse of the Compton of the particle of mass m.

To derive a continuity equation we write down the KFG equation for the wave function φ∗(x), multiply it by φ(x), multiply Eq. (14) by φ∗(x) and subtract the resulting equations. As a result we get

∂ ρ + ~ = 0 (16) t ∇· where

ρ(x) = i (φ∗∂ φ φ∂ φ∗) , t − t ~ = i (φ∗ φ φ φ∗) − ∇ − ∇ Important is that ρ(x) can be negative as well as positive, so it is not a density.

In the particular case of a plane wave, φ(x) = N exp(i(p~ ~r Et)/h¯), we have · − ρ = 2E N 2, ~ = 2p~ N 2 | | | | 5 where E = (pc)2 + m2c4  Thus the negative values of ρ are related toqnegative . These solutions cannot be dropped as unphysical because the remaining positive energy wave functions would no longer form a complete . Therefore a physical explanation for these states is needed. This was given independently by Feynman and by Stuc¨ kelberg. As before, in the nonrelativistic case, we can get the equivalent integral form of the continuity equation: d ρ(~r, t)d3r = ~ dS~ (17) dt ZV − IS · and with the usual requirement that the fields vanish at large distances sufficiently fast for the surface integral to vanish when the integral is taken over all space we get d ρ(~r, t)d3r = 0 (18) dt Zall space and hence ρ(~r, t)d3r = constant. This means in particular that the sign of the integral Zall space of ρ over all space is conserved. Therefore, if ρ is multiplied by the charge of the particle, then it can be interpreted as the . Similarly we must also multiply ~ by the particle’s charge giving us the (Pauli-Weisskopf interpretation).

Next we note that ρ(~r, t) transforms like time since the wave function φ(x) is a scalar and the operator ∂/∂t transforms like time. It is therefore natural to combine ρ and ~ into a four-vector:

jµ = (ρ,~) = ie (φ∗∂µφ φ∂µφ∗) (19) − − where we have included the charge factor e. In terms of this four-vector current density the − continuity equation (16) takes on the form

µ ∂ jµ = 0 (20)

Now let us streamline our notation by using from now on units such that h¯ = 1 and c = 1. In these units masses, energies and momenta all have the same of energy, and length has the dimension of inverse energy. Thus the scalar product p x of four-momentum and the space-time four-vector is dimensionless. · In these units the plane wave is of the form

φ(x) = Ne−ip·x (21) and the current density is jµ(e−) = 2e N 2(E, p~) (22) − | | where the argument (e−) indicates that this is for a negatively charged particle. If we now change the sign of the charge we get

jµ(e+) = +2e N 2(E, p~) = 2e N 2( E, p~) (23) | | − | | − − i.e. the particle with positive charge travelling with energy E and momentum p~ is the same as a particle with negative charge and with energy ( E) and momentum ( p~). Both particles have the same mass m because the currents are constructed− from wave functions− satisfying the same wave equation. Such a pair of particles, having equal mass but opposite charges, are called the

6 antiparticles of each other: one particle is arbitrarily designated the particle, and the other one is then the . Another useful way of seeing the relationship between particle and antiparticle is based on a consideration of the time-dependent factor of the wave function: if we have a particle of negative energy E, then we can write this factor as

e−iEt = e−i(−E)(−t) (24) where the expression on the right-hand side corresponds to a particle of positive energy ( E) travelling backwards in time! Together with the previous discussion we can therefore conclude− that an antiparticle travelling forward in time is identical with a particle travelling backward in time. This is the Feynman-Stuc¨ kelberg interpretation of the negative energy states. Later on we shall see that the same interpretation applies also in the case of Dirac’s relativistic equation of the electron.

5. Electromagnetic interactions of spin-0 particles.

5.1 Time-dependent perturbation theory; Fermi’s Golden Rule. We are going to study collision processes in which the physical state of a system after the collision differs from that before the collision. Problems of this kind require time dependent perturbation theory. TDPT can be applied in cases in which the Hamiltonian is explicitly time dependent but also in cases with explicitly time independent Hamiltonians, if the interaction that causes the transition from one state into another persists for only a finite duration of time. Such is the case in collisions on short range potentials, if the particle is initially at a large distance from the region of nonzero potential, and after passing through that region is again travelling as a , but in a different state than before the interaction. Thus we consider the Schr¨odinger equation ∂ψ ih¯ = Hψ, with H = H + V (t) ∂t 0 where it is assumed that the φn and eigenvalues En of H0 are known, and V (t) is a small perturbation. Expanding ψ into the complete set of eigenfunctions φn,

−iEnt/h¯ ψ = an(t)φne n X we get the equivalent form of the Schr¨odinger equation in E-representation:

dam(t) −i(En−Em)t/h¯ ih¯ = an(t)Vmne dt n X where Vmn = (φm, V (t)φn) is the matrix element of V (t) between the mth and nth eigenstates of H0. Assuming that up to the time T/2 the system was in the state φ , we have the initial conditions − i a ( T/2) = 1, a ( T/2) = 0 for n = i i − n − 6 7 which for some final state φ gives the approximate solution for times t T/2: f ≥

T/2 0 0 −i(Ei−Ef )t /h¯ 0 af (t) = i Vfi(t )e dt − Z−T/2 (it is assumed that the perturbation V (t) is switched off at t = T/2). In the particular case of a potential independent of time we get, in the limit of T , → ∞ a = 2πiV δ(E E ) f − fi i − f where the Dirac delta-function ensures energy conservation (this is, of course, meaningful only if the states φn are degenerate, since otherwise the system would persist in the initial state; for a more detailed discussion read, e.g., D.I. Blokhintsev, Osnovy kvantovoi mekhaniki, 3rd edition, 1961, Chapter XIV). The transition probability from φi to φf is

w a 2 = 4π2 V 2 [δ(E E )]2 ∞ ≡ | f | | fi| i − f T/2 2 2 1 i(Ei−Ef )t/h¯ = 4π Vfi δ(Ei Ef ) lim e dt T →∞ | | − " 2πh¯ Z−T/2 # 2π 2 = Vfi δ(Ei Ef ) lim [T ] h¯ | | − T →∞ and the transition probability per second is

w∞ 2π 2 wfi = lim = Vfi δ(Ei Ef ) T →∞ T h¯ | | − The usual situation in particle collisions is that one measures the transition rate into a group of closely spaced final states. Denoting the density of final states by ρ(Ef ), then ρ(Ef )dEf is the number of final states in the interval (Ef , Ef + dEf ). Multiplying wfi by this number and integrating over Ef we get, on account of the δ function, Fermi’s Golden Rule:

W = 2π V 2ρ(E ) (25) fi h¯ | fi| i

Examples: (i) Interaction of spinless charged particles with electromagnetic field. Assume the oscillating e.m. field to have frequency ω, then V (t) exp( iωt), and the transition amplitude is ≈ − T/2 −i(Ei−Ef +ω)t af lim e dt δ(Ef Ei ω) T →∞ ≈ Z−T/2 ≈ − − i.e. Ef = Ei + ω, which means that the particle has absorbed the energy ω from the e.m. field.

(ii) Interaction of spinless charged antiparticle with electromagnetic field. According to the Feynman interpretation, the incident antiparticle of positive energy Ei is equivalent to a particle of negative energy E travelling backwards in time. Correspondingly, − i the outgoing antiparticle of energy Ef is equivalent to a particle of energy Ef travelling backwards in time. With V (t) exp( iωt), as before, we get therefore − ≈ − T/2 ∗ −i(−Ei)t −iωt −i(−Ef )t af lim e e e dt ≈ T →∞ −T/2 Z   δ(E E ω) ≈ f − i − 8 and hence again Ef = Ei + ω, in other words the antiparticle has also absorbed the energy ω from the field, as expected. This shows that the Feynman interpretation is consistent with our physical expectation in the case of interactions. We can similarly check that we get the intuitively expected results in the cases of particle- antiparticle pair creation and annihilation.

5.2 Elastic scattering of spin-0 particles. Consider the collision of two unlike spin-0 particles. We shall call them π+ and K+. We will treat them as point-like particles, which is not quite correct as the real have a size of the order of one Fermi. The correct description takes this into account by assigning form factors to the mesons. The pion is described by the Klein-Fock-Gordon equation,

µ 2 ∂ ∂µ + m Φ(x) = 0   The interaction with electromagnetic field is taken into account by the replacement of the derivative operator by the covariant derivative

i∂µ i∂µ eAµ → − where Aµ is the electromagnetic four-vector potential. Thus

µ µ 2 (∂ + ieA ) (∂µ + ieAµ) + m Φ(x) = h µ 2 µ i µ +∂ ∂µ + m + ie (∂ Aµ + A ∂µ) h e2AµA Φ(x) = 0 − µ i We neglect the term proportional to e2 as being of second order of smallness. Then the KFG equation takes the form ∂µ∂ + m2 Φ(x) = V (x)Φ(x) µ − with   µ µ V (x) = ie (∂ Aµ + A ∂µ) The sign of the potential V was chosen such as to give the correct form of the Schr¨odinger equation in the nonrelativistic limit. Having identified the interaction energy, we can now use the result of first-order time- dependent perturbation theory, Eq. (25), and write down the transition amplitude:

4 ∗ Tfi = i d x Φf V Φi − Z 4 ∗ µ µ = e d x Φf (∂ Aµ + A ∂µ) Φi (26) Z The first term in brackets is transformed by integration by parts:

4 ∗ µ 4 µ ∗ d x Φ ∂ (AµΦi) = d x (∂ Φf ) AµΦi Z − Z where we have dropped the surface term. Thus the transition amplitude takes on the form

T = e d4x Φ∗ (∂ Φ ) ∂ Φ∗ Φ Aµ fi f µ i − µ f i Z h   i

9 The expression in brackets has the form of a current (cf. Eq. (16)), but involving different wave functions Φi and Φf . It is therefore a transition current:

jµ(π+) = ie Φ∗ (∂µΦ ) ∂µΦ∗ Φ f i − f i h   i and we can write the transition amplitude in the form of

4 + µ Tfi = i d x jµ(π )A − Z The particle in the initial and final states can be represented by plane :

−ipi·x −ipf ·x Φi = Nie , Φf = Nf e where Ni and Nf are normalization factors, hence

µ + µ i(pf −pi)·x j (π ) = eNiNf (pi + pf ) e (27)

The formalism so far is appropriate for the interaction of the π+ particle with an arbitrary e.m. field Aµ. In order to apply it to the reaction π+K+ π+K+ we must consider the e.m. field to be emitted by the kaon. Thus we write down the→wave equation for Aµ with a source term: ν µ µ + ∂ ∂νA = j (K )

µ + where j (K ) is the kaon current. Changing our notation, we assign 4-momenta p1 and p2 to the incident pion and kaon, respectively, and similarly p3 and p4 to the outgoing particles. We have therefore µ + µ iq·x j (K ) = eN2N4 (p2 + p4) e where q = p p , and hence 4 − 2 1 Aµ = (∂ν ∂ )−1 jµ(K+) = jµ(K+) ν −q2 The transition amplitude takes now the form of

4 + 1 µ + Tfi = i d x jµ(π ) 2 j (K ) − Z −q ! and the integration can be done trivially, giving a four-dimensional delta function with the obvious meaning of overall 4-momentum conservation:

T = iN N N N (2π)4 δ(4)(p + p p p ) fi − 1 2 3 4 1 2 − 3 − 4 M where the transition matrix is defined by M

µ gµν ν i = ( ie)(p + p ) i 2 ( ie)(p + p ) − M − 1 3 − q − 2 4   Writing the transition matrix in this form, we have identified the following factors: µ (i) ( ie)(p1 + p3) is associated with the interaction vertex of the pion with the , − ν (ii) ( ie)(p2 + p4) is associated with the interaction vertex of the kaon with the photon, (iii) (− ig /q2) is associated with the exchanged photon. This is called the photon . − µν

10 The denominator q2 is the square of the photon’s 4-momentum. For a real photon this would be zero, but in the present case we can check that

q2 = (p p )2 < 0 1 − 3 except at zero scattering angle, where q2 = 0. This can be seen most conveniently in the CMS, where q2 = 2p2(1 cos θ) − − (p is the CMS momentum and θ is the CMS scattering angle). Therefore the photon is virtual or off mass-shell. In deriving the result for the scattering matrix, we have arbitrarily considered the kaon to be the source of the e.m. field, absorbed by the pion. The form of our result suggests, and an explicit calculation confirms it, that this distinction is unimportant: the pion and kaon play completely symmetric roles in the reaction.

6.) Calculation of the cross section; phase space; Mandelstam variables.

Elastic π+K+ scattering is a particular case of a 2 2 collision. More generally, any reaction of the type → a + b c + d → belongs to this class of processes. The differential cross section dσ is related to the by M 1 dσ = 2dQ F |M| where dQ is the Lorentz invariant phase space factor,

3 3 4 (4) d p3 d p4 dQ = (2π) δ (p1 + p2 p3 p4) 3 3 (28) − − (2π) 2E3 (2π) 2E4 and F is the flux of incident particles, defined by F = 2E 2E ~v ~v , where ~v and ~v are the 1 2 | 1 − 2| 1 2 velocities of the incident particles 1 and 2, respectively. The invariant form of the flux factor is

F = 4 (p p )2 (m m )2 1· 2 − 1 2 q Useful are the expressions of F in the LAB and CMS frames:

Flab = 4plabm2; Fcms = 4pi√s

Because of Lorentz invariance, the three expressions of F are equal. The phase space factor must be further simplified in order to remove the δ function. Three 3 of the six integrations are done trivially, for instance over d p4, removing the three-dimensional (3) δ function δ (p~1 + p~2 p~3 p~4). This leaves us with an expression for dσ, in which momentum conservation is explicitly− imp− osed. For the remaining integrations we work in polar coordinates:

3 2 d p3 = p3 dp3 dΩ where the element of solid angle dΩ is given by dΩ = sin θ dθ dϕ, with polar angle θ and azimuth ϕ. Then, from 2 3 2 E3 = p3 + m3

11 we have p3 dp3 = E3 dE3 hence 1 p3 dE3 dQ = 2 δ(E1 + E2 E3 E4) dΩ (4π) − − E4 To do the integration over energy, it is convenient to define the total initial and final energies,

Wi = E1 + E2, Wf = E3 + E4 hence p3 dp3 p4 dp4 dWf = dE3 + dE4 = + E3 E4 A convenient trick is to continue the calculation in the CMS where, on account of momentum conservation, we have p3 = p4 = pf , hence p4 dp4 = p3 dp3 = pf dpf , and therefore

Wf dWf = dE3 E4 which gives 1 pf dWf dQcms = 2 δ(Wi Wf ) dΩ (4π) − Wf

Carrying out the last integration over Wf we impose energy conservation, after which we have in the CMS Wf = Wi = √s, i.e. 1 p dQ = f dΩ cms (4π)2 √s

This completes the phase space calculation for the quasi-elastic collision process. Collecting all terms we get therefore the differential cross section in the CMS

1 pf 2 dσ = 2 dΩ 64π s pi |M| where pi is the CMS momentum of the incident particles. In the particular case of elastic scattering this formula simplifies on account of pf = pi.

Mandelstam variables. The remaining calculation is the evaluation of which we do in invariant form. To this end we define the Mandelstam variables s, t and u:M

s = (p + p )2 = m2 + m2 + 2p p 1 2 1 2 1· 2 = (p + p )2 = m2 + m2 + 2p p 3 4 3 4 3· 4 t = (p p )2 = m2 + m2 2p p 1 − 3 1 3 − 1· 3 = (p p )2 = m2 + m2 2p p 2 − 4 2 4 − 2· 4 u = (p p )2 = m2 + m2 2p p 1 − 4 1 4 − 1· 4 = (p p )2 = m2 + m2 2p p (29) 2 − 3 2 3 − 2· 3 The three Mandelstam variables are not independent: 4-momentum conservation gives

2 2 2 2 s + t + u = m1 + m2 + m3 + m4

12 but it is frequently convenient to use all three variables. In keeping with the subject of high energy physics, we shall frequently consider the ultrarelativistic case where s + t + u = 0. Carrying out the contraction over the Lorentz indices in the expression of the matrix element we get M e4 2 = (p p + p p + p p + p p )2 |M| q4 1· 2 1· 4 2· 3 3· 4 which in the case of elastic scattering is

e4 2 2 = s m2 m2 + m2 + m2 u |M| t2 − 1 − 2 1 2 − e4 h   i = (s u)2 t2 − and in the ultrarelativistic case s u 2 2 = e4 − |M|  s + u  Substituting this into the expression for the elastic differential cross section we get finally

2 2 dσ α s−u dΩ = 4s s+u   where we have also expressed the charge in terms of the fine structure constant: α = e2/4π. Experimental data on elastic scattering are usually expressed in terms of the scattering angle, rather than the Mandelstam variables. Denoting the CMS scattering angle by θ, we have in the CMS the 4-momenta p1 = (E, 0, 0, E), p2 = (E, 0, 0, E) and p4 = (E, p4x, p4y, p4z) − 2 with p4z = E cos θ, and hence from the above definition we have u = 2E (1 + cos θ) and with s = 4E−2 we get the differential cross section in the form of −

dσ α2 3 + cos θ 2 = (30) dΩ 4s 1 cos θ ! − We see that the differential cross section has a singularity at θ = 0. This is a characteristic feature of photon exchange or, more generally, of the exchange of a zero-mass particle. This result is well known from Rutherford scattering. Because of this singularity we cannot get a total elastic cross section. However, we can get a meaningful quantity if we integrate over θ from some small angle θ0 to θ = π. This corresponds to the usual experimental procedure, which does not allow detection of particles scattered through very small angles. Thus we get

2 πα 1 + cos θ0 θ0 σ(θ θ0) = 1 + cos θ0 + 8 + 16 ln sin ≥ 2s 1 cos θ0 2 ! − To convert our results to ordinary units we must multiply the cross section formulæ by (¯hc)2 0.4 GeV2 mbarn. ≈

13 7.) The Dirac equation.

7.1 Derivation. In constructing the relativistic quantum mechanical equation of the electron we demand, fol- lowing Dirac,1 that the equation be of the form of the time-evolution equation of quantum mechanics, ∂ψ ih¯ = Hˆ ψ, (31) ∂t and that it be Lorentz covariant. Then, since the derivative w.r.t. time is of first order, the Hamiltonian Hˆ must also linearly depend on the derivatives w.r.t. the coordinates. The only other quantities, on which Hˆ can depend, are the fundamental constants c, h¯ and m, the electron mass. Thus, on dimensional grounds, in the absence of forces, Hˆ must be of the form

2 Hˆ = α1pˆ1c + α2pˆ2c + α3pˆ3c + βmc (32) ∂ with pˆ = ih¯ . The α’s and β are dimensionless coefficients. They must be dimensionless i − ∂xi 2 because pic and mc have the dimension of Hˆ . They must also be independent of pˆi to ensure the linear dependence of Hˆ on the momenta, and they must not depend on the coordinates as this would introduce forces. This implies that they commute with the momentum operators. Further properties of these coefficients can be found by requiring that the iteration of the operator yields the Klein-Fock-Gordon equation2:

∂2ψ 2 h¯2 = α pˆ c + α pˆ c + α pˆ c + βmc2 ψ − ∂t2 1 1 2 2 3 3 h i 2 2 2 2 2 = (pˆ1c) + (pˆ2c) + (pˆ3c) + mc ψ     In evaluating [. . .]2 we must allow for the possibility that the α’s and β do not commute with each other. In order to get agreement with the Klein-Fock-Gordon equation, i.e. to impose the right-hand equality , we find that the α’s and β must satisfy the following relations:

2 2 αi = 1, β = 1 α α + α α = 0 for i = j (33) i j j i 6 αiβ + βαi = 0, i = 1, 2, 3

This means that the different α’s anticommute with each other and with β. It must be possible therefore to represent them by matrices; they are called Dirac matrices.

To ensure that all matrix products of Eq. (33) are defined they must be square matrices.

The traces of all Dirac matrices must vanish. This can be seen, for instance, by rewriting 2 the last of Eqs. (33) in the form of αi = βαiβ (using β = 1), then taking the trace, and hence, remembering that Tr (AB) = Tr (BA−), we get Tr α = Tr α , i.e. Tr α = 0. i − i i From the hermiticity of Hˆ it follows that the Dirac matrices are hermitian. Therefore their eigenvalues are real. From the first of Eqs. (33) it follows that their only eigenvalues are +1

1P.A.M. Dirac, Principles of Quantum Mechanics, 4th edition, Oxford University Press, 1958, chapter 11 2this is because the Klein-Fock-Gordon equation is just the relativistic energy-momentum relation E 2 = 2 p~ + (mc2)2 together with the quantum mechanical replacement E ih∂¯ /∂t, p~ ih¯ → → − ∇ 14 and 1. Then, since the trace of a square matrix is the sum of its eigenvalues, and since the Dirac−matrices are traceless, it follows that they must be of even dimension.

For the sake of economy we attempt to find the lowest order in which the Dirac matrices can be represented. The representation in terms of two-by-two matrices is ruled out because we know that there are only three linearly independent traceless two-by-two matrices, for instance the . But a representation in terms of four-by-four matrices is possible. Indeed, the complete set of linearly independent four-by-four matrices consists of sixteen matrices, one of which can be chosen to be the unit matrix and the other fifteen matrices to be traceless. For most purposes one does not need an explicit representation of the Dirac matrices. However, when an explicit representation is needed, it is mostly convenient to use what has become known as the standard representation of the Dirac matrices; partitioned into two-by-two matrices this is of the following form: 0 σ σ 0 α = i , i = 1, 2, 3, β = 0 , (34) i σ 0 0 σ i ! − 0 ! where σ0 is the two-by-two unit matrix, 0 is the two-by-two null matrix and σi, i = 1, 2, 3, are the Pauli matrices, 0 1 0 i 1 0 σ = , σ = , σ = (35) 1 1 0 2 i −0 3 0 1 ! ! − !

Exercise 1. Verify that the Dirac matrices (34) are hermitian and satisfy the properties (33). † [Hint: Recall that σiσj = δij + iεijkσk and σi = σi]

Having represented the Hamiltonian by a four-by-four matrix we are forced to represent the wave function by a column matrix with four components,

ψ1 ψ ψ(x) =  2  ψ3    ψ   4    or in partitioned form ϕ ψ(x) = χ ! where ϕ and χ are two-component . Such a function is called a four-component .

7.2 Continuity equation. To deduce a continuity equation, we must write down a second wave equation for the hermitian conjugate wave function ψ†: ∂ψ†(x) ih¯ = (Hψ(x))† (36) ∂t − Then, if we premultiply Eq. (31) by ψ†(x), postmultiply Eq. (36) by ψ(x) and add the resulting equations, we get ∂ ψ†ψ ih¯ = ψ†Hψ (Hψ)† ψ = ih¯ ψ†α~ ψ ∂t − − ∇·   15 or, with ρ = ψ†ψ and ~ = ψ†α~ ψ, ∂ρ(x) + ~ = 0 ∂t ∇· The density ρ is positive definite; it can therefore be interpreted as a probability density. Then ~ is a density. It can be shown that ρ and ~ transform under Lorentz transformations like the time and space components of a 4-vector. They can therefore be combined into the 4-vector jµ: jµ = (ρ,~ ) Then, if we also define the four-dimensional derivative operator ∂µ = (∂ , ), we can write t −∇ the continuity equation in the manifestly invariant form

µ ∂ jµ = 0 (37)

7.3 Covariant form of the Dirac equation. It is customary to introduce the matrices

γ0 = β, ~γ = βα~ and to rewrite the Dirac equation in the covariant form

µ mc iγ ∂µ ψ(x) = 0  − h¯  and we note that the mass term appears again, as in the Klein-Fock-Gordon equation, in the form of the inverse Compton wave length. 3 µ Setting from now on h¯ = c = 1, and using the “Feynman slash” notation, p γ pµ, we can rewrite the Dirac equation in the following form: 6 ≡

(i ∂ m)ψ(x) = 0 (38) 6 −

7.4 Properties of the γ matrices. Without proof we list here for reference the following basic properties of the γ matrices:

γµγν + γνγµ = 2gµν 㵆 = γ0γµγ0 (39) Tr γµ = 0

In the standard representation, the γ matrices are of the following form:

σ 0 0 σ γ0 = 0 , γ = i , i = 1, 2, 3 (40) 0 σ i σ 0 − 0 ! − i !

7.5 Adjoint equation. The adjoint equation is obtained by taking the hermitian conjugate of the Dirac equation:

← [(i ∂ m)ψ(x)]† = ψ†(i ∂ m)† 6 − 6 −← = ψ†( i㵆∂ m) = 0 − µ − 3 p is pronounced “p slash”. 6 16 and if we use the hermiticity equation for the γ matrices from above and set ψ¯(x) ψ†(x)γ0, ≡ then we get ← ψ¯(i ∂ + m) = 0 (41) 6 ← where it is understood that the derivative operator ∂ acts to the left. 6 From the Dirac equation and its adjoint equation we can immediately get the continuity equation. To do this we premultiply Eq. (38) by ψ¯(x), postmultiply the adjoint equation (41) by ψ(x) and add the two resulting equations. This directly leads to Eq. (37) with jµ = ψ¯(x)γµψ(x). That jµ is a 4-vector can be established either by appealing to the quotient theorem of tensor analysis or directly by carrying out a (see, e.g., P.A.M. Dirac, Principles of Quantum Mechanics, 4th edition, section 68).

7.6 Plane wave solutions. We can find the plane wave solutions of the Dirac equation if we substitute

ψ(x) = u(p)e−ip·x (42) which yields (p m)u(p) = 0, p = γµp (43) 6 − 6 µ Using the standard representation of the γ matrices, Eq. (40), we have

E σ p~ p = − · 6 σ p~ E · − ! where the energy E is understood to be multiplied into the unit 2 2 matrix. The function u(p) is a column matrix with four components. Let×us write it in the following partitioned form: u u(p) = A uB ! Substituting this into Eq. (43) we get the following coupled equations:

~σ p~ u (E + m)u = 0 · A − B (E m)u ~σ p~ u = 0 − A − · B from which, if we eliminate first uB and then uA, we get (E2 p~ 2 m2)u = 0 − − A,B and hence the eigenvalues E = p~ 2 + m2. The occurrence of negative energy eigenvalues had  to be expected from our discussionq of the solutions of the Klein-Fock-Gordon equation. The corresponding states are again interpreted as antiparticle states of positive energy travelling backwards in time. The eigenvectors can be found by standard methods of matrix algebra. For E > 0 we find

ϕs us(p) = N ~σ·p~ , s = 1, 2 E+m ϕs ! 1 0 where ϕ1 = , ϕ2 = , and N is a normalization factor. Similarly we get the negative 0 ! 1 ! energy solutions: ~σ·p~ E−m ϕs us+2(p) = N , s = 1, 2 ϕs !

17 It can be shown that the four solutions of the Dirac equation are mutually orthogonal, i.e. that u†(p)u (p) = 0 if r = s r s 6 The standard choice of normalization is to demand that there be 2E particles in the unit volume: ψ†(x)ψ(x) dV = 2E Zunit vol. which, with appropriate choice of phase, yields the normalization factor N = √E + m.

7.7 Antiparticles. As was mentioned in the previous section, the negative energy solutions are interpreted as antiparticle states. Thus, let E < 0 and make the substitution E E, p~ p~ , hence, if → − → − us+2(p) is the negative energy solution, we have

( p m)u ( p) = 0 − 6 − s+2 − or, if we set v (p) = u ( p), then s s+2 − (p + m)v (p) = 0 6 s

7.8 Completeness relations. Important are the following completeness relations, which we can verify for instance by using the standard representation:

u (p)u¯ (p) = p + m, s s 6 sX=1,2 v (p)v¯ (p) = p m (44) s s 6 − sX=1,2

7.9 Helicity. As we have seen in our discussion of the plane wave solutions of the Dirac equation, there are two linearly independent solutions for E > 0 and two solutions for E < 0, i.e. we have a two-fold degeneracy of solutions. This degeneracy implies that there is an additional degree of freedom. This can be interpreted as being the electron spin. Formally the degeneracy means that there is another operator which commutes with the Hamiltonian and with the momentum operator. Such an operator is

~σ pˆ 0 Σ~ pˆ = · , where pˆ = p~ /p (45) · 0 ~σ pˆ · ! The operator 1 ~σ pˆ 2 · is the component of spin in the direction of p~ . It is called the helicity operator, its eigenvalues λ = 1/2 are the helicities of the electron.  7.10 Bilinear covariants. We have seen above that the bilinear form ψ¯γµψ is a 4-vector. One can construct other bilinear forms ψ¯Γiψ, i = 1, 2, . . . 16, which are scalars (V), (T), axial vectors (A) and (P):

18 matrix Γi 1 γµ σµν γ5γµ γ5 ψ¯Γiψ S V T A P

1 Here σµν = (γµγν γνγµ) and γ5 = iγ0γ1γ2γ3. The matrix γ5 plays an important role 2 − especially in the theory of weak interactions. The main properties of γ5 are the following:

γ5γµ + γµγ5 = 0 γ5† = γ5, (γ5)2 = 1

These properties can be checked by direct calculations. In standard representation γ5 is of the form

0 σ γ5 = 0 σ0 0 !

8. Electrodynamics of spin-1/2 particles.

8.1 Transition matrix. To describe the interaction of electrons with electromagnetic field we must replace the operator pµ in the Dirac equation by pµ + eAµ, thus

(p + e A m)ψ(x) = 0 6 6 − or (p m)ψ(x) = γ0V ψ(x) 6 − with V = eγ0 A. In this definition of the interaction potential V , the sign and the factor γ0 are chosen− suc6h as to be consistent with the corresponding expression in the nonrelativistic limit. Thus, we can immediately write down the transition amplitude in the lowest order of perturbation theory:

† 4 ¯ 4 Tfi = i ψf V ψi d x = ie ψf Aψi d x − Z Z 6 µ 4 = i j Aµ d x − Z where jµ = eψ¯ γµψ is the transition current. Assuming plane waves in the initial and final − f i states, we get jµ = e (u¯ γµu ) e−iq·x − f i where q = p p and u u(p , s ). i − f i,f ≡ i,f i,f It is interesting to note the following decomposition of the transition current: 1 jµ = u¯ [(p + p )µ iσµν(p p ) ] u 2m f i f − i − f ν i (Gordon decomposition). In the first term in brackets we recognize the transition current of spinless particles (cf. Eq. (27)), which describes the interaction of the electromagnetic field with the charge of the electron. The second term corresponds to the interaction with the electron’s .

19 Reasoning as previously in the case of electromagnetic scattering of spinless particles, we can write down the transition amplitude in lowest order of perturbation theory for collisions of two unlike spin-1/2 particles, say e− and µ−, i.e. we consider the reaction

e−µ− e−µ− (46) →

We denote the 4-momenta of the incident electron and muon by p1 and p2, respectively, their spins by s1 and s2, and those of the outgoing particles by p3, p4, s3 and s4. To simplify the notation we shall write the spinors as u u(p , s ), i = 1, 2, 3, 4. The electron and muon i ≡ i i currents are then of the form

jµ(e) = e (u¯ γµu ) e−iq·x, − 3 1 jµ(µ) = e (u¯ γµu ) eiq·x (47) − 4 2 where q = p p = p p , and for the transition amplitude we get 1 − 3 4 − 2

4 µ 1 Tfi = i d x j (e) 2 jµ(µ) − Z −q ! The integral gives us a four-dimensional δ function, and we get

T = i(2π)4δ(p + p p p ) fi − 1 2 − 3 − 4 M where the transition matrix element is defined by4 ig i = [ijµ(e)] µν [ijν(µ)] (48) − M "− q2 # The three factors in square brackets refer, respectively, to the electron transition current, photon propagator and muon transition current. These are the elements of the that provides a graphical representation of the collision process (see Fig. 1a). The remaining calculation to find the differential cross section proceeds by the same steps as we have done in the case of scattering of spinless particles, except that we must in addition pay attention to the spins of the particles. We begin by taking the mod-squared of the matrix element : M 1 2 = (jµ(e)j (µ))∗ (jν(e)j (µ)) |M| q4 µ ν 1 = [jµ ∗(e)jν(e)] j∗(µ)j (µ) q4 µ ν h i Consider separately the two factors in square brackets. They carry two Lorentz labels, and therefore they are tensors. The first one of these refers to the electron. We call it the electron µν tensor and denote it by Ls1s3 , where we have made the spin dependence explicit by labeling the tensor with the spins of the incident and the scattered electrons. The second factor is the µν muon tensor Ms2s4 . Both factors have the same structure, so it is enough to calculate one of them, for instance the electron tensor. Written in detail, the electron tensor is of the following form:

µν 2 µ † ν Ls1s3 = e (u¯3γ u1) (u¯3γ u1) 2 † µ † † ν = e u1γ u¯3 (u¯3γ u1)   4from now on the symbols jµ will be understood to represent uγ¯ µu without the exponential factors.

20 † but u¯† = u†γ0 = γ0u, and if we also use the identity γµ = γ0γµ †γ0, then we get   µν 2 µ ν Ls1s3 = e u¯1 iγiju3 j (u¯3 kγklu1 l) Xijkl   where we have written the matrix labels i, j, k and l explicitly and indicated the summation from 1 to 4. Thus the expression under the sum is a product of ordinary numbers whose order is arbitrary. We can, for instance, take the last factor, u1 l, to the front of the product. Then we recognize that the resulting expression is the following trace:

µν 2 µ ν Ls1s3 = e Tr u1u¯1 γ u3u¯3 γ Now we have a choice. We can continue the calculation assuming definite spin states of the particles. This is meaningful if we want to describe experiments in which the spins of the particles are measured. Alternatively we can consider the more usual experiment, where the incident beam and target are unpolarized and all scattered particles are counted independently of their spin states. In this case we have to apply the procedure called spin summation. To describe the unpolarized beam and target, assuming random spin orientations of the particles, we must average over the spin states of the incident electron and muon; to account for the counting of all particles in the final state irrespective of their spin states we must sum over the spins of the scattered particles. Thus we get the spin averaged electron tensor

2 µν 1 µν e µ ν L = Ls1s3 = Tr u1u¯1 γ u3u¯3 γ 2 s1s3 2 s1 ! s3 ! X X X e2 = Tr (p + m) γµ (p + m) γν 2 6 1 6 3 where in the last step we have used the completeness relation, Eq. (44), of the Dirac spinors. Opening the brackets we get

2 µν e L = Tr p γµ p γν + m (γµ p γν+ p γµγν) + m2γµγν (49) 2 6 1 6 3 6 3 6 1 h i

8.2 Trace theorems. To make further progress in evaluating the traces of products of γ matrices we need now the following trace theorems:

Tr γµ = 0 (50) Tr γµγν = 4gµν (51) Tr γµγν . . . γω = 0 for odd numbers of factors (52) Tr γµγν γλγρ = 4 gµνgλρ gµλgνρ + gµρgνλ (53) −  

Proof of trace theorem (50): use the γ5 matrix defined in section 7.10, remembering that γ5 anticommutes with γµ, µ = 0, 1, 2, 3, and that (γ5)2 = 1, hence Tr γµ = Tr γµ(γ5)2 = Tr γ5γµγ5, where in the last step we have used the property of traces: Tr AB = Tr BA, which is generally true if both products AB and BA are defined and are square matrices. If we now commute γµ with one of the γ5 factors, we get Tr γµ(γ5)2 = Tr γµ, and hence Tr γµ = Tr γµ = 0. Proof of−trace theorem−(51): −

21 1 we write γµγν = (γµγν + γµγν), apply the theorem Tr AB = Tr BA to one of the terms in 2 brackets, then use the commutation relation of the γ matrices and note that the metric tensor gµν is multiplied by the 4 4 unit matrix whose trace is equal to 4. Proof of trace theorem (52):× we postmultiply the product under the trace by (γ5)2, take the last one of the γ5 factors to the front, then commute it past the product. In each commutation we pick up a factor of 1, in total an odd number of such factors, hence Tr γµγν . . . γω = − Tr γµγν . . . γω = 0. The− proof of (53) is more involved but proceeds on similar lines as the proof of (51). Similarly one can prove trace theorems for products of more than four γ factors.

8.3 Completion of the calculation of the electron tensor. Applying the trace theorems we see that the terms linear in the electron mass m drops out on account of being multiplied by traces of products of three γ factors. The remaining terms give µν L = 2e2 p p gαµgβν gαβgµν + gανgβµ + m2gµν 1α 3β − h   i = 2e2 pµpν + pν pµ + m2 p p gµν 1 3 1 3 − 1· 3 h   i The factor multiplying the metric tensor is q2/2; indeed, we have q2 = (p p )2 = 2m2 2p p 1 − 3 − 1· 3 and hence the statement. Thus we have the following final form of the electron tensor:

µν 2 µ ν ν µ 1 2 µν L = 2e p1 p3 + p1p3 + q g (54)  2  The muon tensor has the same structure as the electron tensor. We can immediately write down the expression for the spin-averaged muon tensor:

µν 2 µ ν ν µ 1 2 µν M = 2e p2 p4 + p2 p4 + q g (55)  2  The next step is to contract the two tensors. This is straight forward and sufficiently simple, but we can simplify this calculation further by making use of current conservation, µ ∂ jµ = 0. Indeed, substituting the plane wave expressions for the currents, Eq. (47), we see that ∂µj = iqµj . Thus current conservation is expressed by qµj = 0, and this in turn µ − µ µ implies µν µν qµL = qνL = 0 (56)

We apply this result to our calculation by replacing in the muon tensor p4 by p2 + q, hence

µν 2 µ ν ν µ µ ν 1 2 µν M = 2e 2p2 p2 + p2q + p2 q + q g  2  and then applying Eq. (56). Therefore the muon tensor can be replaced by the effective muon tensor µν 2 µ ν 1 2 µν Meff = 2e 2p2 p2 + q g (57)  2  8.4 Cross section of elastic eµ scattering in the CMS. After contraction of the tensors we have the following expression for the spin averaged mod- squared scattering amplitude: 8e4 2 = p p p p + p p p p m2p p M 2p p + m2M 2 |M| q4 1· 2 3· 4 1· 4 3· 2 − 2· 4 − 1· 3   22 and if we express the scalar products in terms of Mandelstam variables we get

2e4 2 = s2 + u2 2 m2 + M 2 s + u 3m2 3M 2 |M| t2 − − − h    i We shall be interested mainly in the ultrarelativistic limit, which formally corresponds to m = M = 0, hence 2e4 2 = s2 + u2 (58) |M| t2   and hence the differential cross section in the CMS: dσ α2 s2 + u2 = (59) dΩ 2s t2 Expressing t and u in terms of the CMS scattering angle θ, i.e. t = s sin2(θ/2), u = − s cos2(θ/2), we get − dσ α2 1 + cos4(θ/2) = (60) dΩ 2s sin4(θ/2) Qualitatively the θ dependence is similar to the one we found for elastic π+K+ scattering, Eq. (30). In particlular, we find again the singularity at θ = 0, characteristic of photon exchange. Experimentally the reaction eµ eµ is of no interest except in the following sense. A → method to measure the electromagnetic form factors of pions is to expose a target to a pion beam. The elastic scattering of pions by the atomic electrons of the target material has a characteristic kinematics that allows the reaction πe πe to be separated from collisions → of pions with . However pion beams are always contaminated with , and the relatively small mass difference between pion and muon causes a serious problem of background which must be understood for a correct interpretation of the data.

8.5 Electron-positron annihilation into muon pairs. Closely related to the process e−µ− e−µ− is the reaction e+e− µ+µ−, i.e. electron-positron pair annihilation into muon pairs, →see Fig. 1b. In the latter reaction,→ the incoming positron is equivalent to an electron travelling backwards in time. We give it therefore a 4-momentum ( p2) (we call it ( p2) rather than ( p3) because it is an incoming particle). Similarly the outgoing− positive m−uon gets a 4-momen−tum ( p ). In other words, the replacement p p , − 3 2 → − 3 p3 p2 takes us from elastic collision to pair annihilation. This procedure is called . We→can− check that under crossing the Mandelstam variables s and t are exchanged, the third variable u remaining unchanged. We can therefore immediately apply this procedure to the mod-squared matrix element (58), thus

2e4 (e+e− µ+µ−) 2 = t2 + u2 (61) |M → | s2   and hence the differential cross section dσ α2 = 1 + cos2 θ (62) dΩ 4s   where we have used the CMS kinematical relations s = 4E2, t = 2E2(1 cos θ), u = 2E2(1+ cos θ). − − − The characteristic feature of this result, which is worth remembering, is the of the angular distribution about θ = 90o. This was to be expected since, after spin-averaging, the reaction is completely symmetric in the CMS. This symmetry breaks down at sufficiently

23 high energies where the contribution of the becomes noticeable, because the weak interaction violates . If we integrate (62) over the angles θ and φ from 0 to π and from 0 to 2π, respectively, we get the total cross section. To convert it to ordinary units we also multiply the result by (¯hc)2 0.4 GeV2 mbarn, thus ≈ 4πα2 σ(e+e− µ+µ−) = (¯hc)2 (63) → 3s This cross section has been measured on electron-positron colliders over a large range of energies, in particular at the collider PETRA with CMS energies √s from 10 to nearly 40 GeV. The experimental results in this energy range confirm the theoretical result quantitatively. Such a good agreement between theory and experiment is not trivial since we have used in our calculation only the lowest order of perturbation theory. Finally, a last comment concerning the procedure, i.e. crossing, by which we have arrived at the result for the pair annihilation process. This is a powerful method. However, in the present simple and straight forward case we would have had no difficulty in deriving the result by repeating the entire calculation from the beginning. The reader is encouraged to do that as an exercise.

8.6 Electron-muon elastic scattering in the LAB frame. Recall that the invariant expression of the differential cross section is given by 1 dσ = 2 dQ F |M| where F is the flux factor, 2 is the spin-averaged mod-squared matrix element and dQ is |M| the Lorentz invariant phase space factor. We shall tackle separately the three parts of the calculation, the matrix element, the phase space and the flux factor, in the LAB frame.

(i) Matrix element of electron-muon elastics scattering in the LAB frame. We go back to the exact formulas for the lepton tensors, neglecting only the electron mass but keeping the muon mass, which will be essential in this calculation. Thus we have

1 µν 2 = L M |M| q4 µν

µν µν with L given by Eq. (54) and M by Eq. (55). To emphasise that we are now working in the LAB frame, let us denote the 4-momenta of the incoming and outgoing electron by k and k0, respectively, and similarly those of the muon by p and p0. Then carrying out the contraction and remembering that q2 = 2m2 2k k0 = 2M 2 2p p0, we get − · − ·

8e4 2 = [k p k0 p0 + k p0 k0 p |M| q4 · · · · m2p p0 M 2k k0 + 2m2M 2 − · − · i or, if we set the electron mass m equal to nought, and hence k2 = k0 2 = 0, and therefore q2 = (k k0)2 = 2k k0, we get − − · 8e4 1 2 = 2k p k0 p0 + q2 M 2 (k k0) p |M| q4 · · 2 − − ·  h i 24 In the LAB frame the muon is at rest, hence p = (M, 0, 0, 0), and if we denote the energies of the incident and the outgoing electrons by E and E0, respectively, we get

8e4 q2 q2 M(E E0) 2 = 2M 2EE0 1 + − |M| q4 " 4EE0 − 2M 2 2EE0 #

We can get a more useful form of this formula if we express the energy of the scattered electron in terms of the scattering angle. To do this we consider 4-momentum conservation:

k + p = k0 + p0 hence p0 2 = (k k0 + p)2 = q2 + M 2 + 2(k k0) p − − · hence, with p0 2 = M 2, we get q2 = 2M(E E0) (64) − − 0 0 0 0 0 Furthermore, setting k = (E, 0, 0, E) and k = (E , kx, ky, kz), we have θ q2 = 2EE0(1 cos θ) = 4EE0 sin2 (65) − − − 2

0 0 where we have put kz = E cos θ. Using these relations we can cast the result for the matrix element in the following final form:

8e4 θ q2 θ 2 = 2M 2EE0 cos2 sin2 |M| q4 " 2 − 2M 2 2#

(ii) Two-body phase space in the LAB frame. The Lorentz invariant two-body phase space factor is

3 3 1 (4) d p3 d p4 dQ = 2 δ (p3 + p4 p1 p2) (4π) − − E3 E4 and we have to carry out four of the six integrations to get rid of the four-dimensional δ function.

We begin by integrating over the 3-momentum p~ 4; this removes the three-dimensional δ function δ(3) (p~ + p~ p~ p~ ) and leaves us with 3 4 − 1 − 2 2 1 p3dp3dΩ dQ = 2 δ (E3 + E4 E1 E2) (4π) − − E3E4

3 where we have expressed the three-dimensional differential d p3 in polar coordinates with dΩ = sin θdθdφ, where θ and φ are the polar angle and azimuth, respectively. So far the expression is valid in any frame. At this point we specify the LAB frame by setting

0 E1 = E, E3 = E , E2 = M, 0 2 2 and E4 = p~ + M q hence 0 0 1 0 E dE dΩ dQLAB = 2 δ (E + E4 E M) (4π) − − E4

25 We note that our integration over p~ 4 has already enforced momentum conservation. There- fore we have 0 p~ 0 = k~ k~ , hence p~ 0 2 = E2 + E0 2 2EE0 cos θ − − Now consider the argument of the δ function. Let us denote it by f(E0 ), i.e.

f(E0 ) = E0 + p~ 0 2 + M 2 E M q − − Its zero corresponds to energy conservation (in addition to the momentum conservation, which is already enforced), in other words, it corresponds to 4-momentum conservation which in the LAB frame is expressed by Eqs. (64) and (65). To evaluate the integral over E 0 we rewrite the δ function in the form of 0 −1 0 df(E0) 0 0 δ(f(E )) = δ(E E0 ) dE0 −

0 0 0 0 where E0 is given by f(E0 ) = 0. Differentiating f (E ) w.r.t. E we get 0 0 0 df(E ) E E cos θ E4 + E E cos θ 0 = 1 + − = − dE E4 E4

0 or, applying energy conservation, E4 + E = E + M, df(E0) M + E(1 cos θ) Mk 0 = − = 0 dE E4 E E4 where in the last step we have once more used Eqs. (64) and (65). Thus finally we have

E0E δ(f(E0 )) = 4 δ(E0 E0 ) Mk − 0 Now we can do the integral over E0 and get

1 E0 2 1 2E0 2 dQ = 0 dΩ = dΩ LAB (4π)2 ME (4π)2 s M 2 − 0 where in the final expression we have dropped the subscript of E0, which is now redundant, and used s = M 2 + 2ME Note that in its final form the expression remains valid also for the case of quasielastic collisions, such as electroproduction of , ep eN ∗, if the mass M in the → denominator is understood to be the mass of N ∗.

(iii) The flux factor in the LAB frame is obtained from the invariant expression

2 2 4 (p1 p2) (m1m2) q · − which on account of m1 = 0, p1 = (E, 00, E) and p2 = (M, 0, 0, 0) simplifies to 4EM.

(iv) Differential cross section in the LAB frame. Putting our results for the matrix element, the phase space factor and the flux factor together, we get for the differential cross section of elastic electron-muon scattering the following expres- sion: dσ α2 E0 θ q2 θ = cos2 sin2 (66) 2 4 θ 2 dΩ!LAB 4E sin 2 E 2 − 2M 2!

26

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Figure 1: (a) Elastic e−µ− scattering; (b) e+e− µ+µ− annihilation →

27