32105 i

mmL )1(2

ss msmssmsmS )1(2

for s=141421.12 S

1,0,1 sm

for EM-wavestotal energy density:

21

2

21212

21

0

0

00

B

E

Eu B

maxmax

0

0

21

1

BE

BE

s

s

# of photons

wave number vector: kp

01

22

2

2

2

mc

tcthe Klein-Gordon equation:

)(),()(

puesx sxpi

for FREE ELECTRONS

electrons positrons

)(),()(

pvesx sxpi

±1 or 1,2

u va spinor

satisfying:

0)( umcp u 0)( umcp

v

Note for each: with

p

c

Ep

; 2242 cpcmE

i.e. we write ),()( pEupv ss s E,p for u3, u4

Notice: now we express all terms of the “physical” (positive) energy of positrons!

(or any Dirac, i.e. spin ½ particle: muons, taus, quarks)

tistisrk

s

ehveguedk

tr

2)2(),(

3

3

The most GENERAL solutions will be LINEAR COMBINATIONS

s

s sdk3 k g h

linear expansion coefficients

Insisting {(r,t), (r´,t)}= {†(r,t), †(r´,t)}=0(in recognition of the Pauli exclusion principal), or, equivalently:

{(r,t), †(r´,t)} =3(r – r)respecting the condition on Fourier conjugate fields:

forces the g, h to obey the same basic commutation relation (in the “conjugate” momentum space)

)(),(),,( 3 kkskhskg

where gg(k,s), hh(k,s),

the gg(k,s), hh(k,s) “coefficients” cannot simple be numbers!

tistisrk

s

evheugedk

tr

3

3

2)2(

),(

s

s sdk3 k g h

††

tistisrk

s

ehveguedk

tr

2)2(),(

3

3

s

s sdk3 k g h

(x) = ue-ixp/h

u(p)

1

0

cpz

E+mc2

c(px+ipy)E+mc2

0

1

c(pxipy)E+mc2

cpz

Emc2

1

0

cpz

Emc2

c(px+ipy)

Emc2

1

0

c(pxipy)Emc2

cpz

Emc2

This form automatically satisfied the Klein-Gordon equation.

But the appearance of the Dirac spinorsmeans the factoring effort isolated what

very special class of particles?

We were able to solve Dirac’s (free particle) Equation by looking forsolutions of the form:

u(p) a “spinor” describing either spin up or down components

What about vector (spin 1) particles?

Again try to look for solutions of the form

(x) = (p)e-i xp/h

Polarization vector (again characterizing SPIN somehow)

but by just returning to the Dirac-factored form of the Klein-Gordon equation, will we learn anything new?

What about MASSLESS vector particles? (the photon!)

01

22

2

2

2

mc

tc

01 2

2

2

2

tcbecomes: or

22

2

2

1

tc

2 = 0

Where the d’Alembertian operator: 2

The fundamental mediators offorces: the VECTOR BOSONS

the Klein-Gordon equation:

2 = 0is a differential equation you have already

solved in Mechanics and E & M

Classical Electrodynamics J.D.Jackson (Wiley)derives the relativistic (4-vector) expressions for Maxwells’ equations

0

0

1

tB

cE

B

can both be guaranteed

by introducing thescalar V and vector A

“potentials” tA

cVE

AB

1

which form a 4-vector: (V;A)along with the charge and current densities: (c;J)

Then the single relation:

JB

E

ctE

c

41

4

JAA

c

4)( completely summarizes:

Potentials can be changed by a constant for example, the arbitrary assignmentof zero gravitational potential energy( )

leaving everything invariant.or even AA

In solving problems this gives us the flexibility to “adjust” potentials for our convenience

The Lorentz Gauge The Coulomb Gauge A A

0 0

In the Lorentz Gauge:

JAAc

4)( JAc

40

and a FREE PHOTON satisfies:

0A

The VECTOR POTENTIAL from E&M isthe wave function in quantum mechanics

for the free photon!

a “vector particle” with 4 components (V;A)

so continuing (with our assumed form of a solution)/)( xipaexA

(p)

like the Dirac u,a polarization vectorcharacterizing spin

Substituting into our specialized Klein-Gordon equation:(for massless vector particles)

0A0

pp

022 p

cE

E2=p2c2

just as it should for a massless particle!

only for free photons

/)( xipaexA (p)

Like we saw with the Dirac u before, has components!

How many? 4

A = 0

The Lorentz gauge constrains A0 p = 0

p00.p = 0

while the Coulomb Gauge p = 0which you shouldrecognize as the

familiar conditionon em waves

but not all of them are independent!

Obviously only 2 of these 3-dim vectors can be linearly independent such that p = 0

Why can’t we have a basis of 3 distinct polarization directions?We’re trying to describe spin 1 particles! (mspin = 1, 0, 1)

spin 1 particles: mspin = 1 , 0 , +1alignedanti-

aligned

The m=0 imposes a harsher constraint (adding yet another zero to all the constraints on the previous page!)

The masslessness of our vector particle implies ???v = c

In the photon’s own framelongitudinal distances collapse.

How can you distinguish mspin = 1 ?

Furthermore: with no frame travelingfaster than c, can never change a ’s spin by changing frames.

What 2 independent polarizations are then possible?

tistisrki eCeCedk

trA

213

3

2)2(),(

The most general solution: s where s = 1, 2 or s = 1

titisrki eCeCedk

trA

213

3

2)2(),(

moving forward moving backward

Notice here: no separate ANTI-PARTICLE (just one kind of particle with 2 spin states)Massless force carriers have no anti-particles.

Finding a Klein-Gordon Lagrangian

022 cmpp

0222 cm

xii

2

The Klein-Gordon Equation02

22

cm

or

Provided we can identify the appropriatethis should be derivable by

The Euler-Lagrange Equation

0

)(

i

i

L

L L

2

22

21

2

1

mc

tcL

I claim the expression

2

2

2

1

mc

xx

2

2

2

1

mc

serves this purpose

2

2

002

1

mczzyyxx

0

)(

i

i

L LL

0000

0 2

1

)(

tc

xxxx

x

x

)(2

1

)(

L L

L L

2

222

2

1

cm L

02

mc

You can show (and will for homework!) show the Dirac Equation can be derived from:

)( 2mci LDIRAC(r,t)

We might expect a realistic Lagrangian that involves systems of particles

= LK-G + LDIRAC L(r,t)describes

e+e objectsdescribesphotons

)()(),( 2221 mitr

but each termdescribes free

non-interactingparticles

L+ LINT

But what does terms look like?How do we introduce the interactions the experience?

We’ll follow (Jackson) E&M’s lead:

A charge interacts with a field through:

)(INT AJV

L);(

);(

AVA

JJ

AJ

current-field interactions

the fermion(electron)

the boson(photon) field

Ae )(INT L from the Dirac

expression for J

antiparticle(hermitian conjugate)

state

particlestate

What does such a PRODUCT of states mean?

Recall the “state functions”Have coefficients that must

satisfy anticommutation relations.They must involve operators!