Klein-Gordon Equation ● the Existence of Plane Waves

Home , Conserved current

Klein-Gordon Equation ● The existence of plane waves

φ(r, t) exp(ik r iωt) ∝ · − satisfying de Broglie and Einstein relations

p = !k , E = !ω implies the quantum operator interpretation ∂ p i! , E i! . → − ∇ → ∂t

● Then the relativistic energy-momentum equation

E2 = p2c2 + m2c4

implies the Klein-Gordon equation 2 2 ∂ φ 2 2 2 2 4 ! = ! c φ + m c φ − ∂t2 − ∇

1 ● In covariant notation (see handout)

2 µ mc ∂µ∂ + φ = 0 ! ! " # $ where 1 ∂2 ∂ ∂µ = 2 µ c2 ∂t2 − ∇

● KG wave function φ is a Lorentz-invariant (scalar) function; Lorentz

transformation r, t r!, t! implies φ φ! where → →

φ!(r!, t!) = φ(r, t) .

Hence it must represent a spin-zero particle (no orientation). ● Since φ 2 is also invariant, this cannot represent a probability density. A | | density transforms as time-like (0-th) component of a 4-vector, due to Lorentz contraction of volume element.

2 ● Correct deﬁnition of density follows from the continuity equation: ∂ρ = J ∂t −∇ · (J = corresponding current vector), i.e.

µ ∂µJ = 0 where J µ = (cρ, J) is the 4-current.

● We can obtain an equation of this form from the KG equations for φ and φ∗, 2 2 ∂ φ ∂ φ∗ 2 2 2 i! φ∗ φ = i!c φ∗ φ φ φ∗ , ∂t2 − ∂t2 ∇ − ∇ % & ' ( ∂ ∂φ ∂φ∗ 2 i! φ∗ φ = i!c (φ∗ φ φ φ∗) . ∂t ∂t − ∂t ∇ · ∇ − ∇ % & Hence

∂φ ∂φ∗ ρ = i! φ∗ φ ∂t − ∂t % & 2 J = i!c (φ∗ φ φ φ∗) − ∇ − ∇ µ 2 µ µ i.e. J = i!c (φ∗∂ φ φ∂ φ∗). − 3 iEt/ ● Normalization is such that the energy eigenstate φ = Φ(r)e− ! has ρ = 2E Φ 2. Thus Φ = 1 corresponds to 2E particles per unit volume | | | | (relativistic normalization). ● Compare with Schro¨dinger current

S i! 1 KG J = (φ∗ φ φ φ∗) = J −2m ∇ − ∇ 2mc2 which thus has in fact E/mc2 particles per unit volume.

4 Problems with Klein-Gordon Equation 1. Density ρ is not necessarily positive (unlike φ 2) equation was rejected | | ⇒ initially. 2. Equation is second-order in t need to know both φ and ∂φ at t = 0 in order ⇒ ∂t to solve for φ at t > 0. Thus there is an extra degree of freedom, not present in the Schro¨dinger equation. 3. The equation on which it is based (E2 = p2c2 + m2c4) has both positive and negative solutions for E.

iEt/ ● Actually these problems are all related, since a solution φ = Φ(r)e∓ ! has ρ = 2E Φ 2, and so for the general solution ± | | iEt/! +iEt/! φ = Φ+(r)e− + Φ (r)e −

i! ∂φ both φ(t = 0) = Φ+ + Φ and E ∂t = Φ+ Φ are needed in order to − t=0 − − specify Φ+ and Φ . ) − ) )

5 Electromagnetic Waves

● In units where %0 = µ0 = c = 1 (‘Heaviside-Lorentz’) Maxwell’s equations are ∂B E = ρ , E = ∇ · em ∇ × − ∂t ∂E B = 0 , B = J + ∇ · ∇ × em ∂t µ where (ρem, J em) = Jem is the electromagnetic 4-current. ● In terms of the scalar and vector potentials V and A, ∂A E = V , B = A . − ∂t − ∇ ∇ × So we ﬁnd ∂2A ∂V ( A) ( A) 2A = J ∇ × ∇ × ≡ ∇ ∇ · − ∇ em − ∂t2 − ∇ ∂t

● In terms of the 4-potential Aµ = (V, A)

(∂ ∂ν )Aµ ∂µ(∂ Aν ) ∂ F νµ = J µ ν − ν ≡ ν em 6 where the electromagnetic ﬁeld-strength tensor is

F νµ = ∂ν Aµ ∂µAν = F µν . − −

● E and B, and hence Maxwell’s equations, are invariant under gauge transformations µ µ µ µ A A! = A + ∂ χ → where χ(r, t) is an arbitrary scalar function.

µ µ ● Therefore we can always choose A such that ∂µA = 0 (Lorenz gauge). If µ µ µ µ µ ∂ A = f = 0, we can change to A! = A + ∂ χ where ∂ ∂ χ = f. µ ( µ − µ ν µ ● Then in free space (J = 0) we have ∂ν ∂ A = 0. ❖ Massless KG equation for each component of Aµ ❖ Aµ is ‘wave function’ of photon ❖ Aµ is a 4-vector photon has spin 1. ⇒ µ µ µ ik x µ ● Plane wave solutions A = ε exp(ik r iωt) ε e− · where ε = · − ≡ polarization 4-vector, k x kµx , kµ = (ω, k) = wave 4-vector. · ≡ µ 7 ● From wave equation k k = 0 hence ω2 = k2, i.e. E2 = p2c2 (massless · photons). ● From Lorenz gauge condition ε k = 0 ε0 = ε k/ω. · ⇒ · µ µ µ µ ● Polarization 4-vector ε! = ε + ak is equivalent to ε for any constant a. Hence we can always choose ε0 = 0. Then Lorenz condition becomes transversity condition: ε k = 0. · ● E.g. for k along z-axis we can express εµ in terms of plane polarization states

µ µ εx = (0, 1, 0, 0) , εy = (0, 0, 1, 0) ,

or circular polarization states εµ = (0, 1, i, 0)/√2. R,L ± N.B. only 2 polarization states for real photons.

8 Electromagnetic Interactions ● As in classical (and non-relativistic quantum) physics, we introduce e.m. interactions via the minimal substitution in the equations of motion:

E E eV , p p eA → − → − i.e. pµ pµ eAµ , ∂µ ∂µ + ieAµ → − →

● The Klein-Gordon equation becomes

µ µ 2 (∂µ + ieAµ)(∂ + ieA )φ + m φ = 0 ,

(∂ ∂µ + m2)φ = ie[∂ (Aµφ) + A (∂µφ)] + e2A Aµφ µ − µ µ µ The conserved current is now (! = c = 1) µ µ µ µ J = i(φ∗∂ φ φ ∂ φ∗) 2eA φ∗φ − −

9 Klein Paradox ● Consider KG plane waves incident on electrostatic barrier, height V , width a V e ipx−iEt A eip’x−iEt T e ipx−iEt

R e−ipx−iEt B e−ip’x−iEt

x=0 x=a KG equation for x < 0, x > a gives E2 = p2 + m2

p = + E2 m2 ⇒ − (sign from B.C.). *

µ 2 2 2 ● In 0 < x < a, A = (V, 0) and so (E eV ) = p! + m − p! = + (E eV m)(E eV + m) ⇒ − − − (sign choice is arbitrary since*we include p!). ± 10 ● Matching φ and ∂φ/∂x at x = 0 and a gives (as for Schro¨dinger equation)

2 2 i p p! − T = cos p!a + sin p!a | | − 2 p p ) % ! & ) ) ) ) ) ) ) ● Now consider behaviour as V is increased:

❖ eV < E m: p! is real, T < 1 ( T = 1 when p!a = nπ). − | | | | ❖ E m < eV < E + m: p! is imaginary, T < 1, transmission by tunnelling. − | | ❖ eV > E + m: p! is real again! T = 1 when p!a = nπ!? | | ● Note that when eV > E + m density inside barrier is negative:

2 2 ρ! = 2(E eV ) φ < 2m φ − | | − | |

2 ● Meanwhile, the current inside remains positive, J ! = 2p T (current x | | conservation). Hence when eV > E + m there is a negative density ﬂowing from right to left, giving a positive current. We interpret this as a ﬂow of µ µ antiparticles: Jem = eJ always.

11 ● When eV > E + m and T = 1, antiparticles created at the back of the barrier | | (x = a) travel to x = 0 and annihilate the incident particles. At the same time, particles created at x = a travel to x > a, replacing the incident beam. E E=0 E

π+ π π+

● Antiparticles are trapped inside the barrier, but ﬁeld is zero there, so there can be perfect transmission for any thickness.

12 ● Antiparticles are like particles propagating backwards in time

π+ t t

π+

− + π+ π+ π π

x x eV < E−m eV > E+m

13 Charge Conjugation ● If φ is a negative-energy plane-wave solution of the KG equation, with

momentum p, φ = exp(ip r + iEt) (E > 0), then φ∗ = exp( ip r iEt) is a · − · − positive-energy wave with momentum p. Furthermore, in e.m. ﬁelds, φ∗ − behaves as a particle of charge e: − µ µ 2 (∂µ + ieAµ)(∂ + ieA )φ + m φ = 0 µ µ 2 (∂ ieA )(∂ ieA )φ∗ + m φ∗ = 0 ⇒ µ − µ − ● Thus if φ is a negative-energy solution, we take it to represent an antiparticle

with wave function φ∗ (and hence positive energy, opposite charge and momentum).

● Correspondingly, KG equation is invariant w.r.t. φ φ∗, e e. This is → → − called charge conjugation, C. N.B. Under C, J µ J µ as expected. → −

14 Electromagnetic Scattering ● We assume (for the moment) the same formula as in NRQM for the scattering amplitude in terms of the ﬁrst-order perturbation due to e.m. ﬁeld:

µ µ 4 = i φ∗ ie[∂ (A φ ) + A (∂ φ )] d x Afi − f { µ i µ i } + µ µ 4 by parts = e A [φ∗ (∂ φ ) (∂ φ∗ )φ ]d x µ f i − f i + = ie A J µ d4x − µ fi + µ µ µ µ where J = i[φ∗ (∂ φ ) (∂ φ∗ )φ ] is generalization of J to φ = φ fi f i − f i f ( i (transition current). Note that to get to order e1 we only need J µ to Afi fi order e0. Similarly, for Aµ we can use the free-ﬁeld form

µ µ ik x A = ε e− ·

ipf,i x ● For plane waves, φf,i = e− · ,

µ µ i(pf pi) x Jfi = (pi + pf ) e − · 15 Hence

µ i(p p k) x 4 = ieε (p + p ) e f − i− · d x Afi − µ i f + = ie(2π)4ε (p + p ) δ4(p p k) − · i f f − i − ● This corresponds to the Feynman rules for the diagram p i

p k f

❖ An overall factor of (2π)4 δ4(p p k) (momentum conservation) f − i − ❖ εµ for an external photon line ❖ ie(p + p )µ for a vertex involving a spin-0 boson of charge e. − i f N.B. 4-momentum cannot be conserved in this process for free particles! But we shall see that it can occur as part of a more complicated process, e.g. particle-particle scattering by photon exchange. 16 ● We shall consider process ab ab as scattering of a in e.m. ﬁeld of b (both → spin-0). µ 4 fi = iea AµJ d x A − a!a + p ’ a pa q=p’−p b b p p’ b b ● Then (in Lorenz gauge) Aµ satisﬁes

ν µ µ ∂ν ∂ A = eb J b!b N.B. We assume correct source current is

µ µ i(p! pb) x J = (pb + p! ) e b− · b!b b

17 ● Solution for 4-vector potential is then

µ 1 µ iq x A = e (p + p! ) e · −q2 b b b

2 where q = p! p and q = q q. b − b · ● Hence

ieaeb i(p! +p! p p ) x 4 = (p + p! ) (p + p! ) e a b− a− b · d x Afi q2 a a · b b + µ igµν ν = [ ie(p + p! ) ] − [ ie(p + p! ) ] − a a q2 − b b ! $ 4 4 (2π) δ (p! + p! p p ) × a b − a − b N.B. symmetry in a, b. ● Thus we have the additional Feynman rule: ❖ ig /q2 for an internal photon line. − µν

18 ● In processes involving antiparticles, remember we use particles with opposite energy and momentum; pµ = p¯µ. p− a

q − pa “p ”=p , “p ”= p¯ , “k”= q, i a f − a − = ie (2π)4ε (p p¯ )δ4(q p p¯ ) Afi − a · a − a − a − a p b

q −p b “p ”= p¯ , “p ”=p , “k”=q, i − b f b = ie (2π)4ε (p p¯ )δ4(p + p¯ q) Afi − b · b − b b b − 19 ● Annihilation process p p a b

q − p− pa b

e e = i a b (p p¯ ) (p p¯ )(2π)4δ4(p + p¯ q) Afi q2 a − a · b − b b b −

where q = pa + p¯a = pb + p¯b. ● Since we have already normalized to 2E particles per unit volume, we have = (2π)4δ4( p p ) Afi Mfi f − i where is the invariant matrix elemen,t (see han,dout). Mfi ● Thus e.g. for annihilation process e e = i a b (p p¯ ) (p p¯ ) Mfi q2 a − a · b − b

20 ● In terms of the Mandelstam variables

2 2 s = (pa + p¯a) = q t = (p p )2 = (p¯ p¯ )2 b − a a − b u = (p p¯ )2 = (p¯ p )2 a − b a − b we get e e = i a b (u t) Mfi s − and hence the invariant diﬀerential cross section is 2 2 2 dσ eaeb (u t) = 3 − 2 dt 64πs (pa∗)

2 where p∗ = s/4 m = c.m. momentum of a. a − a *

21 Dirac Equation ● Historically, Dirac (1928) was looking for a covariant wave equation that was ﬁrst-order in time, to avoid the above ‘problems’ of the Klein-Gordon equation:

∂ψ 2 i! = βmc ψ i!c α ψ HDiracψ ∂t − · ∇ ≡

● We want ψ also to satisfy KG equation β, α , α , α are matrices. Setting ⇒ x y z ! = c = 1: ∂2ψ ∂ψ ∂ψ = βmi + α − ∂t2 ∂t · ∇ ∂t = β2m2ψ im(βα + αβ) ψ (α )2ψ − · ∇ − · ∇ = m2ψ 2ψ (KG equation) − ∇ 2 2 2 2 Hence β = αx = αy = αz = 1 and βαj + αj β = αjαk + αkαj = 0 for all j = k = x, y, z. This means that β, α , α , α are (at least) 4 4 matrices. ( x y z ×

22 ● A suitable representation is

1 0 0 0  0 1 0 0  I 0 β =  0 0 1 0  ≡  0 I   −    −  0 0 0 1     −    0 σj αj =  σj 0    where σj are the Pauli matrices:

0 1 0 i 1 0 σx = , σy = − , σz =  1 0   i 0   0 1  −      

23 ● Then ψ is represented by a 4-component object called a spinor (not a 4-vector!)

ψ1

 ψ2  ψ =  ψ   3     ψ4      N.B. Each component ψ1,2,3,4 satisﬁes the KG equation.

● For a particle at rest, ψ = φ exp( imc2t/!), Dirac equation φ = βφ, and so − ⇒

φ1

 φ2  φ =  0       0      where φ1,2 tell us the spin orientation.

24 2 ● For antiparticle at rest, ψ = φe+imc t/! φ = βφ, so ⇒ − 0  0  φ =  φ   3     φ4      where φ3,4 now give spin orientation.

25 Spin of Dirac Particles ● How do we prove that Dirac equation corresponds to spin one-half? We must show that there exists an operator S such that J = L + S is a constant of 2 3 motion, and (! = 1) S = S(S + 1) = 4 I. ● Note ﬁrst that L = r p is not a constant of motion: × H = βm + α p · [L , H] = [x, H]p [y, H]p z y − x = iα p iα p . x y − y x In general, [L, H] = iα p = 0. × ( ● Thus we need [S, H] = iα p. 1 − × This is true if S = 2 Σ where

σj 0 Σj = = iαxαyαzα .  0 σj  −

2 1 2 2 2 3  1 Then S = 4 (Σx + Σy + Σz) = 4 I, proving that S = 2 .

26 Magnetic Moment ● In an electromagnetic ﬁeld we make the usual minimal substitutions:

H H eV , p p eA → − → − in the Dirac equation, to obtain

H = α (p eA) + βm + eV · −

● Note that we no longer get the KG equation when we “square”:

(H eV )2 = α α (p eA )(p eA ) + m2 − j k j − j k − k ,j,k = (p eA)2 + m2 e (α α p A + α α A p ) − − j k j k j k j k ,j,k Now for j = k, ( α α = iε Σ , p A = A p i A j k jkl l j k k j − ∇j k ε Σ A = Σ ( A) = Σ B jkl l∇j k · ∇ × · 27 Hence

(H eV )2 = (p eA)2 + m2 eΣ B − − − · 1 e H eV m + (p eA)2 Σ B − * 2m − − 2m · ● This corresponds to a magnetic moment e e µ = S = g S m e 2m " # where g = 2 (experiment 2.0023193. . . ). e ⇒

28 Dirac Density and Current ● Write Dirac equation as ∂ψ = imβψ α ( ψ) ∂t − − · ∇

● Transpose and complex conjugate:

∂ψ † = +imψ†β ( ψ†) α ∂t − ∇ · N.B. β, α are hermitian. Hence ∂ (ψ†ψ) = (ψ†αψ) ∂t −∇

● Thus we can take

2 2 2 2 ρ = ψ†ψ ψ + ψ + ψ + ψ ≡ | 1| | 2| | 3| | 4| J = ψ†αψ

29 N.B. Density ρ is positive deﬁnite! This is what Dirac wanted, but it is really a problem – what about antiparticles?! ● Answer will not come until we learn some quantum ﬁeld theory.

30 Covariant Notation ● Nobody uses α and β any more. Instead we deﬁne γ-matrices:

0 j γ = β , γ = βαj (j = 1, 2, 3)

γµγν + γν γµ γµ, γν = 2gµν . Also deﬁne ⇒ ≡ { }

ψ¯ ψ†β = (ψ∗, ψ∗, ψ∗, ψ∗) ≡ 1 2 − 3 − 4 in usual (‘Bjorken and Drell’) representation. Then

2 0 ρ = ψ†ψ = ψ†β ψ = ψ¯γ ψ 2 J = ψ†αψ = ψ†β αψ = ψ¯γψ

and J µ is a 4-vector: J µ = (ρ, J) = ψ¯γµψ

31 ● We can also show that

ψ¯ψ = ψ 2 + ψ 2 ψ 2 ψ 2 | 1| | 2| − | 3| − | 4| transforms like a scalar (invariant) under Lorentz transformations. ● Multiplying through by β, Dirac equation becomes ∂ψ iγ0 = mψ iγj ψ ∂t − ∇j Hence

µ (γ ∂µ + im)ψ = 0 (γµp m)ψ = 0 µ −

32 Free-Particle Spinors ● A positive-energy plane wave

ψ = u(E, p) exp(ip r iEt) · − satisﬁes (γµp m)u = 0. Writing µ −

φ1

φ  φ2  u = =  χ   χ   1       χ2      this means that E m σ p φ − − · = 0  +σ p E m   χ  · − −  σ p  Thus χ = · φ E + m

33 ● Remember that σ 0 1 1 S = 2 Σ = 2  0 σ  Hence  

1 φ = N for spin up (along z-axis)  0    0 = N for spin down  1    We have also

1 pz 0 px ipy σ p = , σ p = − ·  0   p + ip  ·  1   p  x y − z        

34 Thus 1 0  0   1  u↑ = N , u↓ = N . p px ipy  z   −   E+m   E+m   px+ipy   p     − z   E+m   E+m     

❖ Normalization is as usual ρ = ψ†ψ = u†u = 2E particles per unit volume. This gives 2 2 2 2 px + py + pz N 1 + 2 = 2E 3 (E + m) 4 Using p2 = E2 m2 gives N = √E + m. − ❖ Notice that the ‘small’ (3,4) components are (v/c) relative to ‘large’ ones O (1,2). ● For antiparticle of 4-momentum (E, p) we need solution with pµ ( E, p): → − − ψ = v(E, p) exp( ip r + iEt) − · 35 From the Dirac equation we now ﬁnd

E m σ p φ − − · = 0  σ p E m   χ  − · −     σ p Thus φ = · χ E + m

● Like 4-momentum, spin must be reversed, so

px ipy p − z E+m E+m pz px+ipy  E−+m   E+m  v↑ = N , v↓ = N .  0   1           1   0         

36 Charge Conjugation ● Like the KG equation, the Dirac equation has charge conjugation symmetry. If ψ is a negative-energy solution, there is a transformation

c ψ ψ = Cψ∗ → such that ψc is a positive-energy solution for charge e. To ﬁnd C: − µ γ (∂µ + ieAµ)ψ + imψ = 0 µ γ∗ (∂ ieA )ψ∗ imψ∗ = 0 ⇒ µ − µ − µ 1 c c Cγ∗ C− (∂ ieA )ψ + imψ = 0 . ⇒ − µ − µ µ 1 µ µ µ Hence we need Cγ∗ C− = γ , i.e. γ C = Cγ∗ . − −

37 ● Since all γµ are real except γ2 (which is pure imaginary) in our standard representation, we can take

0 0 0 1  0 0 1 0  C = iγ2 = −  0 1 0 0   −     1 0 0 0      c c Explicitly, for free particles, v↑ = u↑, v↓ = u↓ −

38 Parity Invariance ● Similarly if ψ(r, t) is a solution of the Dirac equation, there exists a transformation ψ(r, t) ψP (r, t) = P ψ( r, t) → − such that ψP is also a solution. Now ∂ γ0 γ + im ψ(r, t) = 0 ∂t − · ∇ % & ∂ γ0 + γ + im ψ( r, t) = 0 ⇒ ∂t · ∇ − % & 0 1 ∂ 1 P P γ P − + P γP − + im ψ (r, t) = 0 . ⇒ ∂t · ∇ % & 0 1 0 1 Hence we need P γ P − = γ , P γP − = γ , − i.e P γ0 = γ0P , P γj = γjP (j = 1, 2, 3) −

39 ● These relations are satisﬁed by

1 0 0 0  0 1 0 0  P = γ0 =  0 0 1 0   −     0 0 0 1   −   

● For a particle at rest,

imt P ψ = u(m, 0) e− , ψ = +ψ

but for an antiparticle at rest

ψ = v(m, 0) e+imt , ψP = ψ − Thus particle and antiparticle have opposite intrinsic parity. ● Notice that for KG equation the parity transformation is simply

φ(r, t) φP (r, t) = φ( r, t) → − 40 i.e. φ is a true scalar function, since ∂2 2 + m2 φ(r, t) = 0 ∂t2 − ∇ % & ∂2 2 + m2 φ( r, t) = 0 ⇒ ∂t2 − ∇ − % & ● For the Dirac equation, the scalar is not ψ but

0 Φ = ψ¯ψ = ψ†γ ψ

Check:

0 Φ(r, t) = ψ†(r, t)γ ψ(r, t) P 0 0 0 Φ (r, t) = ψ†( r, t)γ †γ γ ψ( r, t) − − 0 = ψ†( r, t)γ ψ( r, t) − − = Φ( r, t) −

41 ● Similarly, J µ is a true vector:

µ 0 µ J (r, t) = ψ†(r, t)γ γ ψ(r, t) P µ 0 0 µ 0 J (r, t) = ψ†( r, t)γ †γ γ γ ψ( r, t) − − 0 0 µ 0 µ 0 0 µ 0 µ But γ †γ γ γ = γ γ = γ γ for µ = 0, = γ γ for µ = 1, 2, 3. Hence, as − expected for a true vector,

J P 0(r, t) = J 0( r, t) , J P (r, t) = J( r, t) . − − −

● Weak interactions involve the axial current

µ ¯ µ 5 JA = ψγ γ ψ where 0 I γ5 = iγ0γ1γ2γ3 =  I 0  in our standard representation.  

42 µ ● Under parity transformations JA is an axial vector:

P µ µ 5 0 J (r, t) = ψ†( r, t)γ γ γ ψ( r, t) A − − Now γ5γ0 = γ0γ5 (actually γ5γµ = γµγ5 for all µ = 0, 1, 2, 3), so − − J P 0(r, t) = J 0( r, t) , J P (r, t) = J( r, t) A − − A − as expected for an axial vector.

5 ● Similarly ΦP = ψ¯γ ψ is a pseudoscalar

P 5 0 Φ (r, t) = ψ†( r, t)γ γ ψ( r, t) P − − = ψ¯( r, t)γ5ψ( r, t) − − − = Φ ( r, t) − P −

43 Massless Dirac Particles ● For m = 0 the positive-energy free particle solutions are

ψ = u(E, p) exp(ip r iEt) · − φ where E = p and so u = gives | |  χ    p σ p φ | | − · = 0  σ p p   χ  · −| |     Hence χ = Λφ where Λ = σ p/ p is the helicity operator: Λ = 1 for spin · | | ± aligned along/against direction of p (‘right/left-handed’) ● Note that if ψ represents a massless particle then

Λφ γ5ψ = = Λψ (Λ2 = 1)  φ    44 ● Hence γ5 is the helicity operator for massless particles (minus helicity for massless antiparticles). ● Weak interactions are ‘V–A’, i.e. they involve the current

(J µ J µ) = ψ¯ γµ(1 γ5)ψ − A fi f − i If i is a massless particle, then (1 γ5)ψ vanishes for helicity +1, i.e. only − i left-handed states interact. The same applies to particle f, since

µ 5 0 5 µ 5 0 µ ψ¯ γ (1 γ )ψ = ψ† γ (1 + γ )γ ψ = (1 γ )ψ † γ γ )ψ f − i f i − f i ❖ Thus if neutrinos are massless, only left-ha5nded neutrin6os (right-handed antineutrinos) interact. ❖ In the Standard Model, neutrinos are massless and right-handed neutrinos do not exist. ❖ This is consistent with relativity, because helicity is frame-independent for massless particles. ❖ In reality neutrinos do have mass, so both helicities must exist, but the right-handed states interact more weakly (as for electrons).

45 Electromagnetic Interactions ● We already saw that in an e.m. ﬁeld Dirac Hamiltonian is H = α (p eA) + βm + eV · − 0 µ = H0 + eγ γ Aµ

where H0 is the free-particle Hamiltonian. ● Hence ﬁrst-order perturbation theory gives a transition amplitude

0 µ 4 = i ψ† (eγ γ A )ψ d x Afi − f µ i + = ie J µ A d4x − fi µ + µ ¯ µ where Jfi = ψf γ ψi. ip x ● For plane waves, ψf,i = uf,ie− · , and so the only diﬀerence from the KG (spin zero) case is that we need a vertex factor of ieu¯ γµu − f i for spin one-half instead of ie(p + p )µ for spin 0. − f i 46 ● Invariant matrix element is ie e = a b (u¯ γµu )(u¯ γ u ) Mfi q2 a! a b! µ b

p ’ a pa q=p’−p b b p p’ b b Hence e2e2 2 = a b Lµν Lb |Mfi| t2 a µν where

µν µ ν La = (u¯a! γ ua)(u¯a! γ ua)∗ b Lµν = (u¯b! γµub)(u¯b! γν ub)∗

47 ● For given spin states of a, b, a! and b!, we can evaluate these tensors explicitly using the above expressions for free-particle spinors. However, we often consider unpolarized scattering, when we have to average over initial and sum over ﬁnal spin states. Then

µν 1 µ ν La = 2 (u¯a! γ ua)(u¯a! γ ua)∗ s,pins b and similarly for Lµν . This can be evaluated using the algebra of the γ-matrices.

48 Gamma Matrix Algebra ● The tensor µν 1 µ ν La = 2 (u¯a! γ ua)(u¯a! γ ua)∗ s,pins can be expressed in terms of traces of products of γ-matrices, using

µ uu¯ = u↑u¯↑ + u↓u¯↓ = γ pµ + m s,pins We also have

ν 0 ν ν 0 ν (u¯ γ u )∗ = (u† γ γ u )∗ = u† γ †γ u = u¯ γ u a! a a! a a a! a a! ν 0 0 ν since γ †γ = γ γ . ● Thus Lµν = 1 u¯ γµ(p + m )γν u a 2 a! ( a a a! a!,spins where we use Feynman’s notation p = γµp . Putting in Dirac matrix indices, ( µ u¯αΓαβuβ = Tr (uu¯Γ). Hence

µν 1 µ ν L = Tr (p! + m )γ (p + m )γ a 2 { ( a a ( a a } 49 µν 1 k k! L = Tr (p! + m )k(p + m )k! µ ν a 2 { ( a a ( ( a a ( } 1 1 2 = Tr p! kp k! + m Tr kk! 2 {( a( ( a( } 2 a {( ( } 2 = 2(p! kp k! + p kp! k! p p! k k! + m k k!) a · a · a · a · − a · a · a · (see examples sheet for last step).

● Removing the arbitrary vectors kµ and kν! ,

µν µ ν µ ν 2 µν L = 2 p p! + p! p (p p! m )g a a a a a − a · a − a and similarly 5 6

b 2 L = 2 p p! + p! p (p p! m )g µν bµ bν bµ bν − b · b − b µν so 5 6

µν b 2 2 2 2 L L = 8(p p p! p! + p p! p! p m p p! m p p! + 2m m ) a µν a · b a · b a · b a · b − a b · b − b a · a a b

50 ● Expressing this in terms of the Mandelstam invariants s, t and u, we ﬁnd an invariant diﬀerential cross section 2 2 dσ eaeb 2 2 2 2 2 2 2 = 2 2 s + u 4(ma + mb )(s + u) + 6(ma + mb ) dt 32πst (pa∗) − 5 6 ● For processes involving Dirac antiparticles, we should use the v-spinors in place of u’s: p a

q − pa µ “u ”=u , “u¯ ”=v¯! vertex factor iev¯! γ u . i a f a ⇒ − a a − −’ pa pa q

µ “u ”=v! , “u¯ ”=v¯ vertex factor iev¯ γ v! . i a f a ⇒ − a a 51 ● We also need

vv¯ = v↑v¯↑ + v↓v¯↓ = p m ( − s,pins N.B. Diﬀerent sign of m!

µν 2 ● Note, however, that the tensor La only involves ma. Replacing a by a¯ reverses sign of ma, which does not aﬀect the (unpolarized) scattering cross section. Hence ab, a¯b, a¯b and a¯¯b scattering (by single photon exchange) are all the same.

52 Compton Scattering ε ε’ ε ε’ k k ’ k k ’ + p+k p−k ’ p p’ p p’

● In the Compton scattering process γ + e γ + e, we need the propagator → factor for a virtual Dirac particle. This is i i(q + m) uu¯ = ( q2 m2 q2 m2 − s,pins −

● Compare with photon propagator

i i( gµν ) ε ε∗ “=” − q2 µ ν q2 s,pins

53 Thus the two graphs give = + where Mfi M1 M2

ν i(p + k + m) µ = ε! u¯!( ieγ ) ( ( ( ieγ )uε M1 ν − (p + k)2 m2 − µ − µ i(p k! + m) ν = ε u¯!( ieγ ) ( − ( ( ieγ )uε! M2 µ − (p k )2 m2 − ν − ! − ● The relative phase is +1 because the graphs diﬀer by exchange of identical bosons. ● For the unpolarized case, we want to average over initial spin states and sum over ﬁnal ones. We know how to do this for the electrons. For the photons,

consider the incoming polarization εµ. We can write schematically

2 µλ + = ε ε∗ M |M1 M2| µ λ ε=ε ,ε s,pins ,x y where the tensor M µλ is to be determined. However, we know that it must µλ µλ have the properties kµM = kλM = 0 to ensure gauge invariance, which allows us to replace ε ε + ak for any a. µ → µ µ

54 ● Choose z-axis along k: k = k (1, 0, 0, 1). Then above property implies µ | | − M 00 = M 03 = M 30 = M 33, while

µλ 11 22 εµελ∗ M = M + M ε=ε ,ε ,x y = M 11 + M 22 + M 33 M 00 − = M µ = g M µλ − µ − µλ ● Thus, due to gauge invariance, we can replace photon polarization sum by g . − µλ

● Applying the same trick to the outgoing photon polarization (εν! ) sum, we ﬁnd a contribution from the ﬁrst diagram of

4 1 2 e ν µ = Tr (p! + m)γ (p + k + m)γ (p + m)γ (p + k + m)γ 4 |M1| 4(s m2)2 { ( ( ( ( µ ( ( ν } s,pins −

55 ● In the extreme relativistic limit (s, t , u m2), this becomes (using results | | | | + on examples sheet) e4 e4 Tr p(p + k)p!(p + k) = 8 (p k)(p! k) s2 {( ( ( ( ( ( } s2 · · u = 2e4 − s ● Other diagram and interference terms are left as an exercise.