 5  i 0 1 2 3
L2 m  (  1) m
for s=1
S  2  1.41421
ms  1, 0,  1
S 2 msms  s( s  1) msms
u
for EM-waves
total energy density:
1
2

E
2 0

1 1 2
B
2 0
 0E 2

1
0
B2
 # of photons
1 E  B
0
s  21
0
E max Bmax
1 
 mc 
2
  
  0
2
2
c t
  
2
2
the Klein-Gordon equation:

s
wave number vector:


p  k
for FREE ELECTRONS
electrons i
(or any Dirac, i.e. spin ½ particle: muons, taus, quarks)
±1 or 1,2
 ( px )
s

 ( x, s )  e
v ( p)
 ( px )
s

 ( x, s )  e
u ( p)
a spinor
satisfying:

( p  mc)u  0
Note for each:
 E 
p   ; p
c 
i.e. we write
positrons i

( p  mc)uv  0
with
2 2
E m c  p c
v s ( p)  uss (E,
E,pp)
2 4
for
u 3, u 4
Notice: now we express all terms of the “physical” (positive) energy of positrons!
The most GENERAL solutions will be LINEAR COMBINATIONS

 (r , t )   
s 
s
3
3
dk
dk
(2 )3 2
e

k
k r
s
gu
g e
s it
h
hv ss e it

linear expansion coefficients
where gg(k,s), hh(k,s),
Insisting {(r,t), (r´,t)}= {†(r,t), †(r´,t)}=0
(in recognition of the Pauli exclusion principal), or, equivalently:
{(r,t), †(r´,t)} =d3(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
g (k , s ), h(k , s )  d (k  k )
the gg(k,s), hh(k,s) “coefficients” cannot simple be numbers!

 (r , t )   
s 
s
s 

 (r , t )   
s
3
3
dk
dk
(2 ) 2
3
3
3
dk
(2 ) 2
3
e
e

k
k r


kk r
gguse
s it
s  it
s
s
gu
hv e
g e h
†
s i t
†

s i t
 hh vs e

We were able to solve Dirac’s (free particle) Equation by looking for
solutions of the form:
This form automatically satisfied

the Klein-Gordon equation.
(x) = e-ix p/h

u
But the appearance of the Dirac spinors
means the factoring effort isolated what
very special class of particles?

u(p
1
0
cpz
E+mc2
c(px+ipy)
E+mc2
)
0
cpz
Emc2
c(pxipy)
Emc2
1
c(px+ipy)
Emc2
cpz
Emc2
c(pxipy)
E+mc2
1
1
cpz
Emc2
0
0
u(p) a “spinor” describing either spin up or down components
What about vector (spin 1) particles?
The fundamental mediators of
forces: the VECTOR BOSONS
Again try to look for solutions of the form
p /h

-ix
(x) =  (p)e 
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!)
1 
 mc 
2
  
  0
2
2
c t
  
2
2
the Klein-Gordon equation:
becomes:
1  2
2
   0
2
2
c t
or
 =0
2
Where the d’Alembertian operator:
2
1 2
     2 2  2
c t

 =0
2
is 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

B  0
 1 B
E 
0
c t
can both be guaranteed
by introducing the
scalar V and vector A
“potentials”


B   A


1 A
E  V 
c t
which form a 4-vector: (V;A)
along with the charge and current densities: (c;J)
Then the single relation:
 

 A  
4 
(  A )  J
c
completely summarizes:


  E  4
 1 E 4 
B 
 J
c t
c
Potentials can be changed by a constant for example, the arbitrary assignment
(of zero gravitational potential energy )
or even
A  A   leaving everything invariant.
In solving problems this gives us the flexibility to “adjust” potentials for our convenience

The Lorentz Gauge   A 
0
In the Lorentz Gauge:
4
 


   A   (  A ) 
c
0
and a FREE PHOTON satisfies:

The Coulomb Gauge   A  0

J

A 
4
c

J
a “vector particle” with 4 components (V;A)
A  0
The VECTOR POTENTIAL from E&M is
the wave function in quantum mechanics
for the free photon!
so continuing (with our assumed form of a solution)

A
ip x /  
( x )  ae
 (p)
like the Dirac u,
a polarization vector
characterizing spin
Substituting into our specialized Klein-Gordon equation:
(for massless vector particles)
A  0
p  p  0

E 2
c
2
 p 0
E2=p2c2
just as it should for a massless particle!

A
ip x /  
( x )  ae
 (p)
Like we saw with the Dirac u before,  has components!
How many? 4
The Lorentz gauge constrains
but not all of them are independent!
A0
p  =0
p0 0 .p = 0
while the Coulomb Gauge
A=0
which you should
  p = 0 recognize as the
familiar condition
only for
on em waves
free photons
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 , +1
antialigned
aligned
The m=0 imposes a harsher constraint
(adding yet another zero to all the
constraints on the previous page!)
v=c
The masslessness of our vector particle implies ???
In the photon’s own frame
longitudinal distances collapse.
How can you distinguish mspin = 1 ?
Furthermore: with no frame traveling
faster than c, can never change a ’s
spin by changing frames.
What 2 independent polarizations are then possible?
 s where s = 1, 2 or s = 1
The most general solution:
 
A(r , t )  
 
A(r , t )  
dk
3
(2 ) 2
3
dk
e
3
(2 ) 2
3

ik r
C  e
s  i t
1

ik r

e  C1e
s
s i t
 C2 e
 i t
 C2 e
moving forward
Notice here: no separate ANTI-PARTICLE
(just one kind of particle with 2 spin states)
Massless force carriers have no anti-particles.
i t


moving backward
Finding a Klein-Gordon Lagrangian
p  p  m2c 2  0

i    i
x
2

  2     m2c 2  0
2 2
m
c

or     
 0
2

Provided we can identify the appropriate
this should be derivable by
The Euler-Lagrange Equation
L
  L  L

 
0
  (  )  
 i 
i

The Klein-Gordon Equation
I claim the expression
2
2

1 1   
 mc  2 
  2           
2  c  t 
  

L
2

1    mc  2 
  
   
2  x x   

2


1
mc


           2 
2 
  

  
serves this purpose
2

1
 mc  2 
  0  0   x  x   y  y   z  z     
2 
  

L
L
L
L
 
 

 
0
  (  )  
 i 
i

L


1

  0   0    0   0
 ( 0 )
   2


 ct 
L
L


1

 ( x   x )   x   x
 ( x )
   2
 
 x 

1
m2c 2 
   2 2 
2
 2 
 
mc
L
 
       0
  

You can show (and will for homework!) show
the Dirac Equation can be derived from:
L

2


(
i



mc
)

DIRAC(r,t)
We might expect a realistic Lagrangian that involves systems of particles
L(r,t) = L
K-G
describes
photons
+
L
but each term
describes free
non-interacting
particles
DIRAC
describes
e+e objects

1
( r , t )  (      2 2 )   (i     m)
L
2
+
L
INT
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  ( V  J  A)
L
  J  A

J  (; J )


A  (V ; A)

current-field interactions
the fermion
(electron)
the boson
(photon) field
LINT  (e   ) A
from the Dirac
expression for J
particle
state
antiparticle
(hermitian conjugate)
state
Recall the “state functions”
Have coefficients that must
What does such a PRODUCT of states mean? satisfy anticommutation relations.
They must involve operators!