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All the particles are spin ½ fermions!
Schrodinger Wave Equation
He started with the energy-momentum relation for a particle
Expecting them to act on plane waves
e
 
 iEt ipr
e
he made the quantum mechanical replacement:
How about a relativistic particle？
 e ipx
The Quantum mechanical replacement can be made in
a covariant form. Just remember the plane wave can
 
 ipx
be written in a covariant form:  iEt ipr
e
e
e
As a wave equation, it does not work.
It doesn’t have a conserved probability density.
It has negative energy solutions.
Plane wave solutions for KG Eq.

ip0t ip x
 ( x)  a  e
 ae
ipx
2
p02  a  e ipx  p  a  e ipx  m 2  a  e ipx
2
p02  p  m 2 p 2  m2c 2  0
 E
p0
Time dependence can be determined.
There are two solutions for each 3 momentum p (one for +E and one for –E )
 ( x)   a  e

p
 ipx

 b e
ipx
It has negative energy solutions.
   a  e

p
 
 iEt  ip x

 b e
 
iEt ip x

The proper way to interpret KG equation is it is not a Wavefunction
Equation but actually a Field equation just like Maxwell’s Equations.
Plane wave solutions just corresponds to Plane Waves.
It’s natural for plane waves to contain negative frequency components.
Expansion of the KG Field by plane：
 ( x)   a  e

p
 ipx

 b e
ipx
   a  e
 
 iEt  ip x

p
If Φ is a real function, the coefficients are related:
 
 ( x)   a  e ipx  a   eipx 

p

 b e
 
iEt ip x

Dirac
This result is too beautiful to be false; it is more important to
have beauty in one's equations than to have them fit experiment.
Blaming the negative energy problem on the second time derivative of
KG Eq., Dirac set out to find a first order differential equation.
This Eq. still needs to give the proper energy momentum
relation. So Dirac propose to factor the relation!
For example, in the rest frame:
Made the replacement
i

 mc
t
First order diff. Eq.
Now put in 3-momenta:
Suppose the momentum relation can be factored into linear
combinations of p’s:
Expand the right hand side:

We get
and we need:
It’s easier to see by writing out explicitly:
Oops! What!
 0 1   1 0  0
or
No numbers can accomplish this!
Dirac propose it could be true for matrices.
 0 1   1 0
2 by 2 Pauli Matrices come very close
0  i
1 0 
  3  

0
 0  1
0 1
  2  
 1  
1 0
i
 ,  0
   1
i
i
j
 ,  
i
j
ij
i j
Dirac find it’s possible for 4 by 4 matrices
We need:
that is
He found a set of solutions:
Dirac Matrices
Dirac find it’s possible for 4 by 4 matrices
Pick the first order factor:
Make the replacement and put in the wave function:
If γ’s are 4 by 4 matrices, Ψ must be a 4 component column:
 


 
  mc  0
i  0
1
 2
 3
t
x
y
z 

It consists of 4 Equations.
The above could be done for 2 by 2 matrices if there is no mass.
Massless fermion contains only half the degrees of freedom.
A pure left-handed or a pure right-handed must be massless.
Now put in 3-momenta:
Suppose the momentum relation can be factored into linear
combinations of p’s:
Expand the right hand side:
p  p   k   pk p
Now β, γ do not need to be the same.
We can choose:
 k  1  1  2  3 
 k   p k p
交叉項抵銷
 k  1 -  1 -  2 -  3 
  p  0
  p  0
 k  1  1  2  3 
     0
 k  1 -  1 -  2 -  3 
     0
Massless fermions contain only half the degrees of freedom.
Expansion of a solution by plane wave solutions for KG Eq.
 ( x)   a  e

p
 ipx

 b e
ipx
   a  e
 
 iEt  ip x

p
If Φ is a real function, the coefficients are related:
 ( x)   a  e ipx  a   eipx 

p

 b e
 
iEt ip x

Plane wave solutions for Dirac Eq.
( x)  a  eipx  u  a  e
Multiply on the left with
p
2

 m2c 2  u  0

ip t ip x
0
 u A1 


u
u
  

u   A    A 2 
 u B   u B1 
u 
 B2 
u
  p  mc
p 2  m2c 2  0
p0
There are again two sets of solutions for each 3 momentum p
(one for +E and one for –E )
Negative Energy again! First order equation doesn’t escape it!
 E
How about u?
p0
p0
p0
p0
p0
p0
p0
We need:
p0
p0
You may think these are two conditions, but no.
Multiply the first by
p
p0
 p   2 
p 
0
0 2
p0
1

p
2
p
p0
0 2
So one of the above is not independent if
p 2  m2c 2  0
We need:
p
0
or
How many solutions for every p?
Go to the rest frame!


p  (m,0) or (m,0)
p0
Go to the rest frame!

p  ( m ,0 )
p0
p0
p0
p0
0  uA 
0
     0
 
 0  2m   u B 
uB  0
uA is arbitrary
1  0
u A   ,  
 0 1
u has two solutions corresponding to spin up and spin down in the rest frame.

p  (m,0)
p0
p0
p0
p0
  2m 0   u A 
     0
 
0   uB 
 0
uA  0
uB is arbitrary
1  0
u B   ,  
 0 1
Two solution (spin down and spin up antiparticle)
There are four solutions for each 3 momentum p (two for particle and
two for antiparticle)
p
or
0
p0
It’s not hard to find four independent solutions.
p0
p0
p0
p0

p0

p0
-
We got two positive and two negative energy solutions!
Negative energy is still here!
In fact, they are antiparticles.
Electron solutions:
  ae
 ipx
u  ae
 
 iEt  ipr
u
Positron solutions: u to v
  p  mc u  (E, p)  0
 p  mc v  (E, p)  0


  ae
 
iEt ipr
3, 4

1, 2

u
( 3, 4 )
 ae
 ipx (1, 2 )
v
 ae
 
iEt ipr
v
Expansion of a solution by plane wave solutions for KG Eq.
 ( x)   a  e

p
 ipx

 b e
ipx
   a  e
 
 iEt  ip x

p

 b e
 
iEt ip x

Expansion of a solution by plane wave solutions for Dirac Eq.
 ( x)   a  u  e ipx  b   v  eipx 

p
p0  E 
2
p  m2
4-rows
4-columns
How about internal lines?
Find the Green Function of Dirac Eq.
i    G( x, x' )  mcG( x, x' )  I   4 x  x'
i    G( x, x' )   mcG( x, x' )  I   4 x  x'

Now the Green Function G is a 4 ˣ 4 matrix
i    G( x, x' )  mcG( x, x' )  i  I   4 x  x'
Using the Fourier Transformation
d 4 p ip( x  x ') ~
G ( x  x' )  
e
G ( p)
4
2 
d 4 p ip( x  x ')
 ( x  x' )  
e
4
2 
4
~
 p  m  G( p)  i

~
G ( p) 

i
  p  m
i   p  m 
p m
2
2
Fermion Propagator
p
4 ˣ 4 matrices
Bilinear Covariants
Ψ transforms under Lorentz Transformation:
Interaction vertices must be Lorentz invariant.
The weak vertices of leptons coupling with W
W
W
-ig
-ig
e
νe
g  e W  e
 e 
 e 
   W ?   
e
e
μ
νμ
Bilinear Covariants
Ψ transforms under Lorentz Transformation:
Interaction vertices must be Lorentz invariant.
How do we build invariants from two Ψ’s ?
A first guess:
Maybe you need to change some of the signs:


It turns out to be right!
We can define a new adjoint spinor:
is invariant!
In fact all bilinears can be classified according to their
behavior under Lorentz Transformation:
    A
 ( x)   b  v  e ipx  a   u  eipx   ( x)   a  u  e ipx  b   v  eipx 


p
p
u 0
e( p1 )
u p3 a p3 u p1 0  u p3uupa1 pe( pe3()p1 )  u  0
1
Feynman Rules
for external lines
Photons:
It’s easier using potentials:
forms a four vector.
4-vector again
Charge conservation
Now the deep part:
E and B are observable, but A’s are not!
A can be changed by a gauge transformation
without changing E and B the observable:
So we can use this freedom to choose a gauge, a
condition for A:
4 
 A 
J
c


For free photons:
  A  0
Almost like 4 KG Eq.
A ( x)  a     e ipx
Energy-Momentum Relation
Polarization needs to satisfy Lorentz Condition:
Lorentz Condition does not kill all the freedom:
We can further choose
then
Coulomb Guage
The photon is transversely polarized.
For p in the z direction:
For every p that satisfy
there are two solutions!
Massless spin 1 particle has two degrees
of freedom.
    A
A ( x) 
 a
p ,i 1, 2
(i ) 
 ipx

(i ) 
ipx



e

a



e
p
p
p
p

 (q,  )
 (i )   0
Feynman Rules for external photon lines
Gauge Invariance
Classically, E and B are observable, but A’s are not!
A can be changed by a gauge transformation
without changing E and B the observable:

A'  A     (t , x )
Transformation parameter λ is a
function of spacetime.
  A ' A '    A        A         A   A
But in Qunatum Mechanics, it is A that appear in wave equation:
In a EM field, charged particle couple directly with A.
Classically it’s force that affects particles. EM force is written in E, B.
But in Hamiltonian formalism, H is written in terms of A.
2


1  e  



H
p

A
x
,
t

e

x
,t


2m 
c

Quantum Mechanics or wave equation is written by quantizing the
Hamiltonian formalism:


2 
   1 
i

  ieA    e  
t  2m

Is there still gauge invariance?
B does not exist outside.
Gauge invariance in Quantum Mechanics:


2 
   1 
i

  ieA    e  
t  2m

In QM, there is an additional Phase factor invariance:


i
( x , t )  e  ( x , t )
It is quite a surprise this phase invariance is linked to
EM gauge invariance when the phase is time dependent.

A  A    ( x, t )






ie ( x ,t )
( x , t )  e
 ( x , t )
This space-time dependent phase transformation is not an
invariance of QM unless it’s coupled with EM gauge transformation!

A  A    ( x, t )



ie ( x ,t )
( x , t )  e
 ( x , t )




   e

ie ( x ,t )


 e


ie ( x ,t )



  e

ie ( x ,t )
  e

ie ( x ,t )

 e

ie ( x ,t )
  
Derivatives of wave function doesn’t transform like wave function itself.

   1  2 
i

 
t  2m

Wave Equation is not invariant!
But if we put in A and link the two transformations:






 ieA   e

ie ( x ,t )




 ieA   e




 ieA   ie  e

ie ( x ,t )

ie ( x ,t )


     ie   e

 ieA  
This “derivative” transforms like wave function.

ie ( x ,t )


In space and time components:






ie ( x ,t )
  ieA   e
  ieA 




ie ( x ,t )  
  ie   e
  ie 
 t

 t

The wave equation:


2 
   1 
i

  ieA    e  
t  2m

can be written as


2 


 1 
i   ie    
  ieA  
 t

 2m



2 



ie   1
i e   ie    e 
  ieA  
 t

 2m

ie
It is invariant!

A  A    ( x, t )



ie ( x ,t )
( x , t )  e
 ( x , t )



This combination will be called “Gauge Transformation”
It’s a localized phase transformation.
Write your theory with this “Covariant Derivative”.





D     ieA   e

ie ( x ,t )


Your theory would be easily invariant.

 ieA  
There is a duality between E and B.
Without charge, Maxwell is invariant under:
Maybe there exist magnetic charges: monopole
Magnetic Monopole
The curl of B is non-zero. The vector potential does not exist.
If A exists,


  
  A  0


 
3   
 B  da   dx     A  0
there can be no monopole.
But quantum mechanics can not do without A.
Maybe magnetic monopole is incompatible with QM.
But Dirac did find a Monopole solution:
g 1  cos  
A 
r sin 
  1

A sin    A 
 A  r 
 
 r sin  
 1  1 Ar 
 1 
Ar 
 
 rA     rA  
r  sin   r
r  r
 

Dirac Monopole
g 1  cos  
A 
r sin 
It is singular at θ = π.
Dirac String
It can be thought of as an
infinitely thin solenoid that
confines magnetic field lines
into the monopole.
But a monopole is rotationally symmetric.
A 
g 1  cos   Dirac String doesn’t seem to observe the symmetry
r sin 
It has to!
In fact we can also choose the string to go upwards (or any direction):
g 1  cos  
They are related by a gauge transformation!
A'  
r sin 
Since the position of the string is arbitrary, it’s unphysical.
A 
g 1  cos  
r sin 
Using any charge particle, we can perform a Aharonov like
interference around the string. The effects of the string to
the phase is just like a thin solenoid:
  
 
 e   da    A  e   da  B  eg  4


Since the string is unphysical. e ieg 4
e
1
n
2g
Charge Quantization
Finally…. Feynman Rules for QED
e
e


e
e
4-rows
4-columns
4 ˣ 4 matrices
4 ˣ 4 matrices
Dirac
index flow,
from left
to right!
1 ˣ 1 in Dirac index
pk
p  k'
Numerator simplification
 p  m u( p, s)  0
using
 p  m u  0
The first term vanish!
In the Lab frame of e
Photon polarization has no time component.
The third term vanish!
Denominator simplification
Assuming low energy limit:
in low energy
 k  0
2nd term:
1
 
k
Finally Amplitude:
Amplitude squared

0

0
 0 0  0 0
k k  k2  0
旋轉帶電粒子所產生之磁偶極
磁偶極矩與角動量成正比
  iA  i r 2 
e 

L
2m

e
2 r
v
 r2 
erv
e
r  p   e L

2
2m
2m
帶電粒子自旋形成的磁偶極
e 
e 
s   S   g
S
m
2m

Anomalous magnetic moment
1
g theory  1  1159652187.9  8.810 12
2
1
g experi  1  1159652188.4  4.310 12
2