Transcript Dirac Equation

The World
Particle content
Interactions
e, q
e, q
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.
The proper way to interpret KG equation is it is actually
a field equation just like Maxwell’s Equations.
Consider we try to solve this eq as a field equation with a
source.
     m2  j ( x)
We can solve it by Green Function.
    G( x, x' )  m2 ( x, x' )   x  x'
G is the solution for a point-like source at x’.
By superposition, we can get a solution for source j.
 ( x )   0 ( x )   d 4 x ' G ( x, x ' ) j ( x ' )
Green Function for KG Equation:
    G( x, x' )  m2G( x, x' )   4 x  x'
By translation invariance, G is only a function of coordinate
difference:
G ( x, x ' )  G ( x  x ' )
The Equation becomes algebraic after a 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
2 4
4

~
 p  m  G ( p)  1
2
2
~
G ( p) 
1
p 2  m2
This is the propagator!
x'
KG Propagation
x
Green function is the effect at x of a source at x’.
That is exactly what is represented in this diagram.
The tricky part is actually the boundary condition.
For those amplitude where
time 2 is ahead of time 1,
propagation is from 2 to 1.
For those amplitude where
time 1 is ahead of time 2,
propagation is from 1 to 2.
B
B
B
2
B
1
1
C
C
A
A
A
2
A
is actually the sum of the above two diagrams!
~
To accomplish this, G ( p) 
1
p 2  m2
~
G ( p) 
1
p 2  m 2  i
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.
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:

 k  1  1  2  3 
交叉項抵銷
 k  1 -  1 -  2 -  3 
Plane wave solutions for KG Eq.
 ( x)  a  e
ipx
 ae

ip0t ip x
2
p02  a  e ipx  p  a  e ipx  m 2  a  e ipx
2
p  p  m2
p 2  m2c 2  0
2
0
 E
p0
There are two solutions for each 3 momentum p (one for +E and one for –E )
 ( x)   a  e

p
 ipx

 a e
ipx
   a  e

p
 
 iEt  ip x

 a e
 
iEt ip x

Expansion of a solution by plane wave solutions for KG Eq.
 ( x)   a  e
 ipx

p

 b e
ipx
   a  e
 
 iEt  ip x

p

 
iEt ip x


 
iEt ip x

 b e
If Φ is a real function, the coefficients are related:
 ( x)   a  e

p
 ipx

 a e
ipx
   a  e

p
 
 iEt  ip x
 a e
Plane wave solutions for Dirac Eq.
( x)  a  e
ipx
u  a e
Multiply on the left with
p
2

 m2c 2  u  0

ip0t ip x
 u A1 


u  u 
u   A    A2 
 u B   u B1 
u 
 B2 
u
  p  mc
p 2  m2c 2  0
p0
 E
There are two sets of solutions for each 3 momentum p (one for +E and one
for –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

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
Two solution (spin up spin down)

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:
  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

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
How about internal lines?
Find the Green Function of Dirac Eq.
    G( x, x' )  m2G( x, x' )   4 x  x'
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

Fermion Propagator
2
p
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
erv
e
e
r  p   L
  iA  i r 
r 

2 r
2
2m
2m
v
2
e 

L
2m

e
2
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