Klein-Gordon Equation● The existence of plane waves

!(r, t) ! exp(ik · r " i"t)

satisfying de Broglie and Einstein relations

p = !k , E = !"

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!

#t2= "!2c2$2! + m2c4!

1

● In covariant notation (see handout)!#µ#µ +

"mc

!

#2$

! = 0

where

#µ#µ =1c2

#2

#t2"$2

● KG wave function ! is a Lorentz-invariant (scalar) function; Lorentztransformation r, t# r!, t! implies !# !! where

!!(r!, t!) = !(r, t) .

Hence it must represent a spin-zero particle (no orientation).

● Since |!|2 is also invariant, this cannot represent a probability density. Adensity transforms as time-like (0-th) component of a 4-vector, due to Lorentzcontraction of volume element.

2

● Correct definition of density follows from the continuity equation:

#$

#t= "$ · J

(J = corresponding current vector), i.e.

#µJµ = 0

where Jµ = (c$, J) is the 4-current.

● We can obtain an equation of this form from the KG equations for ! and !",

i!%

!"#2!

#t2" !

#2!"

#t2

&= i!c2

'!"$2!" !$2!"

(,

i! #

#t

%!"

#!

#t" !

#!"

#t

&= i!c2$ · (!"$!" !$!") .

Hence

$ = i!%

!"#!

#t" !

#!"

#t

&

J = "i!c2 (!"$!" !$!")

i.e. Jµ = i!c2(!"#µ!" !#µ!").3

● Normalization is such that the energy eigenstate ! = !(r)e#iEt/! has$ = 2E|!|2. Thus |!| = 1 corresponds to 2E particles per unit volume(relativistic normalization).

● Compare with Schrodinger current

JS = " i!2m

(!"$!" !$!") =1

2mc2JKG

which thus has in fact E/mc2 particles per unit volume.

4

Problems with Klein-Gordon Equation

1. Density $ is not necessarily positive (unlike |!|2) % equation was rejectedinitially.

2. Equation is second-order in t % need to know both ! and !"!t at t = 0 in order

to solve for ! at t > 0. Thus there is an extra degree of freedom, not present inthe Schrodinger equation.

3. The equation on which it is based (E2 = p2c2 + m2c4) has both positive andnegative solutions for E.

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

! = !+(r)e#iEt/! + !#(r)e+iEt/!

both !(t = 0) = !+ + !# and i!E

!"!t

)))t=0

= !+ " !# are needed in order tospecify !+ and !#.

5

Electromagnetic Waves● In units where %0 = µ0 = c = 1 (‘Heaviside-Lorentz’) Maxwell’s equations are

$ ·E = $em , $&E = "#B

#t

$ ·B = 0 , $&B = Jem +#E

#t

where ($em, Jem) = Jµem is the electromagnetic 4-current.

● In terms of the scalar and vector potentials V and A,

E = "#A

#t"$V , B = $&A .

So we find

$& ($&A) ' $($ ·A)"$2A = Jem "#2A

#t2"$#V

#t

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

(####)Aµ " #µ(##A#) ' ##F #µ = Jµem

6

where the electromagnetic field-strength tensor is

F #µ = ##Aµ " #µA# = "Fµ# .

● E and B, and hence Maxwell’s equations, are invariant under gaugetransformations

Aµ # A!µ = Aµ + #µ&

where &(r, t) is an arbitrary scalar function.

● Therefore we can always choose Aµ such that #µAµ = 0 (Lorenz gauge). If#µAµ = f (= 0, we can change to A!µ = Aµ + #µ& where #µ#µ& = "f .

● Then in free space (Jµ = 0) we have ####Aµ = 0.

❖ Massless KG equation for each component of Aµ

❖ Aµ is ‘wave function’ of photon

❖ Aµ is a 4-vector % photon has spin 1.

● Plane wave solutions Aµ = 'µ exp(ik · r " i"t) ' 'µe#ik·x where 'µ =polarization 4-vector, k · x ' kµxµ, kµ = (", k) = wave 4-vector.

7

● From wave equation k · k = 0 hence "2 = k2, i.e. E2 = p2c2 (masslessphotons).

● From Lorenz gauge condition ' · k = 0 % '0 = ! · k/".

● Polarization 4-vector '!µ = 'µ + akµ is equivalent to 'µ for any constant a.Hence we can always choose '0 = 0. Then Lorenz condition becomestransversity condition: ! · k = 0.

● E.g. for k along z-axis we can express 'µ in terms of plane polarization states

'µx = (0, 1, 0, 0) , 'µ

y = (0, 0, 1, 0) ,

or circular polarization states 'µR,L = (0, 1,±i, 0)/

)2.

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 + ieAµ)(#µ + ieAµ)! + m2! = 0 ,

(#µ#µ + m2)! = "ie[#µ(Aµ!) + Aµ(#µ!)] + e2AµAµ!

The conserved current is now (! = c = 1)

Jµ = i(!"#µ!" ! #µ!")" 2eAµ!"!

9

Klein Paradox● Consider KG plane waves incident on electrostatic barrier, height V , width a

R e

A e T e

B e

e

−ipx−iEt −ip’x−iEt

ip’x−iEt ipx−iEtipx−iEt

x=ax=0

V

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

% p = +*

E2 "m2

(sign from B.C.).

● In 0 < x < a, Aµ = (V,0) and so (E " eV )2 = p!2 + m2

% p! = +*

(E " eV "m)(E " eV + m)

(sign choice is arbitrary since we include ±p!).

10

● Matching ! and #!/#x at x = 0 and a gives (as for Schrodinger equation)

|T |2 =))))cos p!a" i

2

%p

p!+

p!

p

&sin p!a

))))#2

● Now consider behaviour as V is increased:

❖ eV < E "m: p! is real, |T | < 1 (|T | = 1 when p!a = n().

❖ 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 = n(!?

● Note that when eV > E + m density inside barrier is negative:

$! = 2(E " eV )|!|2 < "2m|!|2

● Meanwhile, the current inside remains positive, J !x = 2p|T |2 (current

conservation). Hence when eV > E + m there is a negative density flowingfrom right to left, giving a positive current. We interpret this as a flow ofantiparticles: Jµ

em = eJµ 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.

EE =0E

ππ+ +π

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

12

● Antiparticles are like particles propagating backwards in time

x

t

π

π

π+

+

+

x

t

π+ π+π−

eV > E+meV < E−m

13

Charge Conjugation

● If ! is a negative-energy plane-wave solution of the KG equation, withmomentum p, ! = exp(ip · r + iEt) (E > 0), then !" = exp("ip · r " iEt) is apositive-energy wave with momentum "p. Furthermore, in e.m. fields, !"

behaves as a particle of charge "e:

(#µ + ieAµ)(#µ + ieAµ)! + m2! = 0

% (#µ " ieAµ)(#µ " ieAµ)!" + m2!" = 0

● Thus if ! is a negative-energy solution, we take it to represent an antiparticlewith wave function !" (and hence positive energy, opposite charge andmomentum).

● Correspondingly, KG equation is invariant w.r.t. !# !", e# "e. This iscalled charge conjugation, C.

N.B. Under C, Jµ # "Jµ 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:

Afi = "i

+!"f{ie[#µ(Aµ!i) + Aµ(#µ!i)]}d4x

by parts = e

+Aµ[!"f (#µ!i)" (#µ!"f )!i]d4x

= "ie

+AµJµ

fid4x

where Jµfi = i[!"f (#µ!i)" (#µ!"f )!i] is generalization of Jµ to !f (= !i

(transition current). Note that to get Afi to order e1 we only need Jµfi to

order e0. Similarly, for Aµ we can use the free-field form

Aµ = 'µe#ik·x

● For plane waves, !f,i = e#ipf,i·x,

Jµfi = (pi + pf )µei(pf#pi)·x

15

Hence

Afi = "ie'µ(pi + pf )µ

+ei(pf#pi#k)·xd4x

= "ie(2()4' · (pi + pf ) )4(pf " pi " k)

● This corresponds to the Feynman rules for the diagram

p

p

k f

i

❖ An overall factor of (2()4 )4(pf " pi " k) (momentum conservation)❖ 'µ for an external photon line❖ "ie(pi + pf )µ for a vertex involving a spin-0 boson of charge e.N.B. 4-momentum cannot be conserved in this process for free particles! Butwe 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 (bothspin-0).

Afi = "iea

+AµJµ

a!ad4x

ap p

p p

a

b

’

b’

q=pb’ b−p

● Then (in Lorenz gauge) Aµ satisfies

####Aµ = eb Jµb!b

N.B. We assume correct source current is

Jµb!b = (pb + p!b)

µei(p!b#pb)·x

17

● Solution for 4-vector potential is then

Aµ = " 1q2

eb(pb + p!b)µeiq·x

where q = p!b " pb and q2 = q · q.

● Hence

Afi =ieaeb

q2(pa + p!a) · (pb + p!b)

+ei(p!

a+p!b#pa#pb)·xd4x

= ["ie(pa + p!a)µ]!"igµ#

q2

$["ie(pb + p!b)

# ]

&(2()4)4(p!a + p!b " pa " pb)

N.B. symmetry in a, b.

● Thus we have the additional Feynman rule:

❖ "igµ#/q2 for an internal photon line.

18

● In processes involving antiparticles, remember we use particles with oppositeenergy and momentum; pµ = "pµ.

pa

pa

−q

“pi”=pa, “pf”="pa, “k”="q,

Afi = "iea(2()4' · (pa " pa))4(q " pa " pa)

p

pb

b

−q

“pi”="pb, “pf”=pb, “k”=q,

Afi = "ieb(2()4' · (pb " pb))4(pb + pb " q)19

● Annihilation process

bp

pbpa

pa

− −q

Afi = ieaeb

q2(pa " pa) · (pb " pb)(2()4)4(pb + pb " q)

where q = pa + pa = pb + pb.

● Since we have already normalized to 2E particles per unit volume, we have

Afi = Mfi (2()4)4(,

pf ",

pi)

where Mfi is the invariant matrix element (see handout).

● Thus e.g. for annihilation process

Mfi = ieaeb

q2(pa " pa) · (pb " pb)

20

● In terms of the Mandelstam variables

s = (pa + pa)2 = q2

t = (pb " pa)2 = (pa " pb)2

u = (pa " pb)2 = (pa " pb)2

we getMfi = i

eaeb

s(u" t)

and hence the invariant di"erential cross section is

d*

dt=

e2ae2

b(u" t)2

64(s3(p"a)2

where p"a =*

s/4"m2a = c.m. momentum of a.

21

Dirac Equation

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

i!#+

#t= ,mc2+ " i!c " ·$+ ' HDirac+

● We want + also to satisfy KG equation % ,, -x, -y, -z are matrices. Setting! = c = 1:

"#2+

#t2= ,mi

#+

#t+ " ·$#+

#t= ,2m2+ " im(," + ",) ·$+ " (" ·$)2+

= m2+ "$2+ (KG equation)

Hence ,2 = -2x = -2

y = -2z = 1 and ,-j + -j, = -j-k + -k-j = 0 for all

j (= k = x, y, z. This means that ,, -x, -y, -z are (at least) 4& 4 matrices.

22

● A suitable representation is

, =

-

...../

1 0 0 0

0 1 0 0

0 0 "1 0

0 0 0 "1

0

111112'

-

/ I 0

0 "I

0

2

-j =

-

/ 0 *j

*j 0

0

2

where *j are the Pauli matrices:

*x =

-

/ 0 1

1 0

0

2 , *y =

-

/ 0 "i

i 0

0

2 , *z =

-

/ 1 0

0 "1

0

2

23

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

+ =

-

...../

+1

+2

+3

+4

0

111112

N.B. Each component +1,2,3,4 satisfies the KG equation.

● For a particle at rest, + = ! exp("imc2t/!), Dirac equation % ! = ,!, and so

! =

-

...../

!1

!2

0

0

0

111112

where !1,2 tell us the spin orientation.

24

● For antiparticle at rest, + = !e+imc2t/! % ! = ",!, so

! =

-

...../

0

0

!3

!4

0

111112

where !3,4 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 ofmotion, and (! = 1) S2 = S(S + 1) = 3

4I.

● Note first that L = r & p is not a constant of motion:

H = ,m + " · p[Lz, H] = [x, H]py " [y, H]px

= i-xpy " i-ypx .

In general, [L, H] = i"& p (= 0.

● Thus we need [S, H] = "i"& p.This is true if S = 1

2! where

#j =

-

/ *j 0

0 *j

0

2 = "i-x-y-z" .

Then S2 = 14 (#2

x + #2y + #2

z) = 34I, proving that S = 1

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 = " · (p" eA) + ,m + eV

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

(H " eV )2 =,

j,k

-j-k(pj " eAj)(pk " eAk) + m2

= (p" eA)2 + m2 " e,

j,k

(-j-kpjAk + -j-kAjpk)

Now for j (= k,

-j-k = i'jkl#l , pjAk = Akpj " i$jAk

'jkl#l$jAk = ! · ($&A) = ! ·B

27

Hence

(H " eV )2 = (p" eA)2 + m2 " e! ·B

H " eV * m +1

2m(p" eA)2 " e

2m! ·B

● This corresponds to a magnetic moment

µ =e

mS = ge

" e

2m

#S

where ge = 2 (experiment % 2.0023193. . . ).

28

Dirac Density and Current

● Write Dirac equation as

#+

#t= "im,+ "" · ($+)

● Transpose and complex conjugate:

#+

#t

†= +im+†, " ($+†) ·"

N.B. ,, " are hermitian. Hence

#

#t(+†+) = "$(+†-+)

● Thus we can take

$ = +†+ ' |+1|2 + |+2|2 + |+3|2 + |+4|2

J = +†-+

29

N.B. Density $ is positive definite! This is what Dirac wanted, but it is really aproblem – what about antiparticles?!

● Answer will not come until we learn some quantum field theory.

30

Covariant Notation

● Nobody uses " and , any more. Instead we define .-matrices:

.0 = , , .j = ,-j (j = 1, 2, 3)

% .µ.# + .#.µ ' {.µ, .#} = 2gµ# . Also define

+ ' +†, = (+"1 , +"

2 ,"+"3 ,"+"

4)

in usual (‘Bjorken and Drell’) representation. Then

$ = +†+ = +†,2+ = +.0+

J = +†"+ = +†,2-+ = +#+

and Jµ is a 4-vector:Jµ = ($, J) = +.µ+

31

● We can also show that

++ = |+1|2 + |+2|2 " |+3|2 " |+4|2

transforms like a scalar (invariant) under Lorentz transformations.

● Multiplying through by ,, Dirac equation becomes

i.0 #+

#t= m+ " i.j$j+

Hence

(.µ#µ + im)+ = 0

(.µpµ "m)+ = 0

32

Free-Particle Spinors

● A positive-energy plane wave

+ = u(E, p) exp(ip · r " iEt)

satisfies (.µpµ "m)u = 0. Writing

u =

-

/ !

&

0

2 =

-

...../

!1

!2

&1

&2

0

111112

this means that -

/ E "m "$ · p

+$ · p "E "m

0

2

-

/ !

&

0

2 = 0

Thus & =$ · p

E + m!

33

● Remember that

S = 12! = 1

2

-

/ $ 0

0 $

0

2

Hence

! = N

-

/ 1

0

0

2 for spin up (along z-axis)

= N

-

/ 0

1

0

2 for spin down

We have also

$ · p

-

/ 1

0

0

2 =

-

/ pz

px + ipy

0

2 , $ · p

-

/ 0

1

0

2 =

-

/ px " ipy

"pz

0

2

34

Thus

u% = N

-

...../

1

0pz

E+mpx+ipy

E+m

0

111112, u& = N

-

...../

0

1px#ipy

E+m#pz

E+m

0

111112.

❖ Normalization is as usual $ = +†+ = u†u = 2E particles per unit volume.This gives

N2

31 +

p2x + p2

y + p2z

(E + m)2

4= 2E

Using p2 = E2 "m2 gives N =)

E + m.

❖ Notice that the ‘small’ (3,4) components are O(v/c) relative to ‘large’ ones(1,2).

● For antiparticle of 4-momentum (E, p) we need solution with pµ # ("E,"p):

+ = v(E, p) exp("ip · r + iEt)

35

From the Dirac equation we now find-

/ "E "m $ · p

"$ · p E "m

0

2

-

/ !

&

0

2 = 0

Thus ! =$ · p

E + m&

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

v% = N

-

...../

px#ipy

E+m#pz

E+m

0

1

0

111112, v& = N

-

...../

pz

E+mpx+ipy

E+m

1

0

0

111112.

36

Charge Conjugation

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

+ # +c = C+"

such that +c is a positive-energy solution for charge "e. To find C:

.µ(#µ + ieAµ)+ + im+ = 0

% ."µ(#µ " ieAµ)+" " im+" = 0

% "C."µC#1(#µ " ieAµ)+c + im+c = 0 .

Hence we need C."µC#1 = ".µ, i.e. .µC = "C."µ.

37

● Since all .µ are real except .2 (which is pure imaginary) in our standardrepresentation, we can take

C = i.2 =

-

...../

0 0 0 1

0 0 "1 0

0 "1 0 0

1 0 0 0

0

111112

Explicitly, for free particles, v%c = u%, v&c = "u&

38

Parity Invariance

● Similarly if +(r, t) is a solution of the Dirac equation, there exists atransformation

+(r, t)# +P (r, t) = P+("r, t)

such that +P is also a solution. Now%

.0 #

#t" # ·$+ im

&+(r, t) = 0

%%

.0 #

#t+ # ·$+ im

&+("r, t) = 0

%%

P.0P#1 #

#t+ P#P#1 ·$+ im

&+P (r, t) = 0 .

Hence we need P.0P#1 = .0 , P#P#1 = "# ,

i.e P.0 = .0P , P.j = ".jP (j = 1, 2, 3)

39

● These relations are satisfied by

P = .0 =

-

...../

1 0 0 0

0 1 0 0

0 0 "1 0

0 0 0 "1

0

111112

● For a particle at rest,

+ = u(m,0) e#imt , +P = ++

but for an antiparticle at rest

+ = v(m,0) e+imt , +P = "+

Thus particle and antiparticle have opposite intrinsic parity.

● Notice that for KG equation the parity transformation is simply

!(r, t)# !P (r, t) = !("r, t)

40

i.e. ! is a true scalar function, since%

#2

#t2"$2 + m2

&!(r, t) = 0

%%

#2

#t2"$2 + m2

&!("r, t) = 0

● For the Dirac equation, the scalar is not + but

! = ++ = +†.0+

Check:

!(r, t) = +†(r, t).0+(r, t)

!P (r, t) = +†("r, t).0†.0.0+("r, t)

= +†("r, t).0+("r, t)

= !("r, t)

41

● Similarly, Jµ is a true vector:

Jµ(r, t) = +†(r, t).0.µ+(r, t)

JPµ(r, t) = +†("r, t).0†.0.µ.0+("r, t)

But .0†.0.µ.0 = .µ.0 = .0.µ for µ = 0, = ".0.µ for µ = 1, 2, 3. Hence, asexpected for a true vector,

JP0(r, t) = J0("r, t) , JP (r, t) = "J("r, t) .

● Weak interactions involve the axial current

JµA = +.µ.5+

where

.5 = i.0.1.2.3 =

-

/ 0 I

I 0

0

2

in our standard representation.

42

● Under parity transformations JµA is an axial vector:

JPµA (r, t) = +†("r, t).µ.5.0+("r, t)

Now .5.0 = ".0.5 (actually .5.µ = ".µ.5 for all µ = 0, 1, 2, 3), so

JP0A (r, t) = "J0("r, t) , JP

A(r, t) = J("r, t)

as expected for an axial vector.

● Similarly !P = +.5+ is a pseudoscalar

!PP (r, t) = +†("r, t).5.0+("r, t)

= "+("r, t).5+("r, t)

= "!P ("r, t)

43

Massless Dirac Particles

● For m = 0 the positive-energy free particle solutions are

+ = u(E, p) exp(ip · r " iEt)

where E = |p| and so u =

-

/ !

&

0

2 gives

-

/ |p| "$ · p

$ · p "|p|

0

2

-

/ !

&

0

2 = 0

Hence & = $! where $ = $ · p/|p| is the helicity operator: $ = ±1 for spinaligned along/against direction of p (‘right/left-handed’)

● Note that if + represents a massless particle then

.5+ =

-

/ $!

!

0

2 = $+ ($2 = 1)

44

● Hence .5 is the helicity operator for massless particles (minus helicity formassless antiparticles).

● Weak interactions are ‘V–A’, i.e. they involve the current

(Jµ " JµA)fi = +f.µ(1" .5)+i

If i is a massless particle, then (1" .5)+i vanishes for helicity +1, i.e. onlyleft-handed states interact. The same applies to particle f , since

+f.µ(1" .5)+i = +†f.0(1 + .5).µ+i =

5(1" .5)+f

6†.0.µ)+i

❖ Thus if neutrinos are massless, only left-handed neutrinos (right-handedantineutrinos) interact.

❖ In the Standard Model, neutrinos are massless and right-handed neutrinosdo not exist.

❖ This is consistent with relativity, because helicity is frame-independent formassless particles.

❖ In reality neutrinos do have mass, so both helicities must exist, but theright-handed states interact more weakly (as for electrons).

45

Electromagnetic Interactions● We already saw that in an e.m. field Dirac Hamiltonian is

H = " · (p" eA) + ,m + eV

= H0 + e.0.µAµ

where H0 is the free-particle Hamiltonian.

● Hence first-order perturbation theory gives a transition amplitude

Afi = "i

++†

f (e.0.µAµ)+id4x

= "ie

+Jµ

fiAµd4x

where Jµfi = +f.µ+i.

● For plane waves, +f,i = uf,ie#ip·x, and so the only di"erence from the KG(spin zero) case is that we need a vertex factor of

"ieuf.µui

for spin one-half instead of "ie(pf + pi)µ for spin 0.46

● Invariant matrix element is

Mfi =ieaeb

q2(ua!.µua)(ub!.µub)

ap p

p p

a

b

’

b’

q=pb’ b−p

Hence

|Mfi|2 =e2ae2

b

t2Lµ#

a Lbµ#

where

Lµ#a = (ua!.µua)(ua!.#ua)"

Lbµ# = (ub!.µub)(ub!.#ub)"

47

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

Lµ#a = 1

2

,

spins

(ua!.µua)(ua!.#ua)"

and similarly for Lbµ# . This can be evaluated using the algebra of the

.-matrices.

48

Gamma Matrix Algebra● The tensor

Lµ#a = 1

2

,

spins

(ua!.µua)(ua!.#ua)"

can be expressed in terms of traces of products of .-matrices, using,

spins

uu = u%u% + u&u& = .µpµ + m

We also have

(ua!.#ua)" = (u†a!.0.#ua)" = u†

a.#†.0ua! = ua.#ua!

since .#†.0 = .0.# .

● ThusLµ#

a = 12

,

a! spins

ua!.µ((pa + ma).#ua!

where we use Feynman’s notation (p = .µpµ. Putting in Dirac matrix indices,u$%$%u% = Tr (uu%). Hence

Lµ#a = 1

2Tr {((p!a + ma).µ((pa + ma).#}49

kµk!#Lµ#a = 1

2Tr {((p!a + ma)(k((pa + ma)(k!}= 1

2Tr {(p!a(k(pa(k!}+ 12m2

aTr {(k(k!}

= 2(p!a · kpa · k! + pa · kp!a · k! " pa · p!ak · k! + m2ak · k!)

(see examples sheet for last step).

● Removing the arbitrary vectors kµ and k!# ,

Lµ#a = 2

5pµ

ap!#a + p!µa p#a " (pa · p!a "m2

a)gµ#6

and similarly

Lbµ# = 2

5pbµp!b# + p!bµpb# " (pb · p!b "m2

b)gµ#

6

so

Lµ#a Lb

µ# = 8(pa · pb p!a · p!b + pa · p!b p!a · pb "m2apb · p!b "m2

bpa · p!a + 2m2am2

b)

50

● Expressing this in terms of the Mandelstam invariants s, t and u, we find aninvariant di"erential cross section

d*

dt=

e2ae2

b

32(st2(p"a)25s2 + u2 " 4(m2

a + m2b)(s + u) + 6(m2

a + m2b)

26

● For processes involving Dirac antiparticles, we should use the v-spinors inplace of u’s:

pa

pa

−q

“ui”=ua, “uf”=v!a % vertex factor "iev!a.µua.

qpa− pa’−

“ui”=v!a, “uf”=va % vertex factor "ieva.µv!a.51

● We also need ,

spins

vv = v%v% + v&v& = (p"m

N.B. Di"erent sign of m!

● Note, however, that the tensor Lµ#a only involves m2

a. Replacing a by a

reverses sign of ma, which does not a"ect the (unpolarized) scattering crosssection. Hence ab, ab, ab and ab scattering (by single photon exchange) are allthe same.

52

Compton Scattering

p

ε’

p+k

’ε

+’

p’

εk k ’

p−k ’

p p’

kε

k

● In the Compton scattering process . + e# . + e, we need the propagatorfactor for a virtual Dirac particle. This is

i

q2 "m2

,

spins

uu =i((q + m)q2 "m2

● Compare with photon propagator

i

q2

,

spins

'µ'"# “=”i("gµ#)

q2

53

Thus the two graphs give Mfi = M1 +M2 where

M1 = '!# u!("ie.#)i((p + (k + m)(p + k)2 "m2

("ie.µ)u'µ

M2 = 'µu!("ie.µ)i((p" (k! + m)(p" k!)2 "m2

("ie.#)u'!#

● The relative phase is +1 because the graphs di"er by exchange of identicalbosons.

● For the unpolarized case, we want to average over initial spin states and sumover final ones. We know how to do this for the electrons. For the photons,consider the incoming polarization 'µ. We can write schematically

,

spins

|M1 +M2|2 =,

&=&x,&y

'µ'"'Mµ'

where the tensor Mµ' is to be determined. However, we know that it musthave the properties kµMµ' = k'Mµ' = 0 to ensure gauge invariance, whichallows us to replace 'µ # 'µ + akµ for any a.

54

● Choose z-axis along k: kµ = |k|(1, 0, 0,"1). Then above property impliesM00 = M03 = M30 = M33, while

,

&=&x,&y

'µ'"'Mµ' = M11 + M22

= M11 + M22 + M33 "M00

= "Mµµ = "gµ'Mµ'

● Thus, due to gauge invariance, we can replace photon polarization sum by"gµ'.

● Applying the same trick to the outgoing photon polarization ('!#) sum, we finda contribution from the first diagram of

14

,

spins

|M1|2 =e4

4(s"m2)2Tr {((p! + m).#((p + (k + m).µ((p + m).µ((p + (k + m).#}

55

● In the extreme relativistic limit (s, |t|, |u|+ m2), this becomes (using resultson examples sheet)

e4

s2Tr {(p((p + (k)(p!((p + (k)} = 8

e4

s2(p · k)(p! · k)

= "2e4 u

s

● Other diagram and interference terms are left as an exercise.

56