One particle states:
Wave Packets States
Heisenberg Picture
Combine the two eq.
KG Equation
Dirac field and Lagrangian
The Dirac wavefunction is actually a field, though unobservable!
Dirac eq. can be derived from the following Lagrangian.
L   

  

L L

     i     m     i     m  
 
i




   m   0   i     m  0
i




   m   0   i     m  0
Negative energy!
Anti-commutator!
A creation operator!
~
~
b  b, b   b
b annihilate an antiparticle!
a , a  0  a a


p


p
a p a p  0

a p 0


p
Exclusion Principle


p


p


p
a a


p
Now add interactions:
For example, we can add
 3 ( x),  4 ( x),  ( x) ( x) ( x)
to our Klein-Gordon or Dirac Lagrangian.
Interaction Hamiltonian:
Schrodinger Picture
Heisenberg Picture
We can move the time evolution t the operators:
Heisenberg Equation
Interaction picture
H  H 0  H int
States and Operators both evolve with time in interaction picture:
S
OI (t )  eiH0t /  OS eiH0t / 
Evolution of Operators
OI (t )  eiH0t /  OS eiH0t / 
I (t )  eiH t /  S eiH t / 
0
d i
 H 0 ,  
dt 
0
dOI i
 H 0 , OI 
dt

Operators evolve just like operators in the Heisenberg picture
but with the full Hamiltonian replaced by the free Hamiltonian
Field operators are free, as if there is no interaction!
Evolution of States
S
States evolve like in the Schrodinger picture but with
Hamiltonian replaced by V(t).
V(t) is just the interaction Hamiltonian HI in interaction picture!
That means, the field operators in V(t) are free.
Interaction Picture
Operators evolve just like in the Heisenberg picture but with
the full Hamiltonian replaced by the free Hamiltonian
d i
 H 0 ,  
dt 
States evolve like in the Schrodinger picture but with the full
Hamiltonian replaced by the interaction Hamiltonian.
d
 i
 I (t )     H I  I (t )
dt
 
U operator
d
 i
 I (t )     H I  I (t )
dt
 
Define time evolution operator U
 I (t )  U (t, t0 )  I (t0 )
All the problems can be answered if we are able to calculate
this operator. It’s determined by the evolution of states.
d
d
 i
 I (t )  U (t , t 0 )  I (t 0 )      H I U (t , t 0 )  I (t 0 )
dt
dt
 
d
 i
U (t , t 0 )     H I  U (t , t 0 )
dt
 

Perturbation expansion
d
 i
U (t , t 0 )     H I  U (t , t 0 )
dt
 
Solve it by a perturbation expansion in small parameters in HI.
U (t, t0 )  U (0) (t, t0 )  U (1) (t, t0 )  
To leading order:
d (1)
 i
 i
U (t , t 0 )     H I U ( 0) (t , t 0 )     H I
dt
 
 
 i
U (1) (t , t 0 )      dt ' H I (t ' ' )
   t0
t
U (0) (t , t0 )  1
Define S matrix:






3

S  i  dt H (t )  i  dt dx HI ( x, t )  i  d 4 x LI ( x)
It is Lorentz invariant if the interaction Lagrangian is invariant.
Vertex
In ABC model, every particle corresponds to a field:
A  A ( x)  A( x)
Add an interaction term in the Lagrangian:
The transition amplitude for the decay of A:
can be computed:
To leading order:
BC U I , A  BC S A
 a  a
Numerical factors remain
B
C
ig
A
Momentum Conservation
A
 aa

interaction Lagrangian
  a  a
B
C
  a  a
vertex
Every field operator in the interaction corresponds to one leg in the vertex.
Every field is a linear combination of a and a+
 a  a
Every leg of a vertex can either annihilate or create a particle!
This diagram is actually the combination of 8 diagrams!
A
B
C
interaction Lagrangian
vertex
There is a spacetime integration.
Interaction could happen anytime anywhere and their
amplitudes are superposed.
The integration yields a momentum conservation.
This is in momentum space.
L I   4
 a  a
interaction Lagrangian
vertex
Every field operator in the interaction corresponds to one leg in the vertex.
Every leg of a vertex can either annihilate or create a particle!
  a  a
L I  g 
  a  b
  b  a
interaction Lagrangian
vertex
Every field operator in the interaction corresponds to one leg in the vertex.
Every leg of a vertex can either annihilate or create a particle?
  a  a
L I  g 
  a  b
  b  a
interaction Lagrangian
vertex
Every leg of a vertex can either annihilate or create a particle?
  a  b
can either annihilate a particle or create an antiparticle!
  b  a
can either annihilate an antiparticle or create a particle!
The charge flow is consistent!
So we can add an arrow for the charge flow.
L I  g 
  a  b
  b  a
External line
When Dirac operators annihilate states, they leave behind a u or v !
 ( x)   a  u  eipx  b  v  eipx 
e  ( p1 )

p

 
3
a p ' p  2 p  2    p  p'  0
Feynman Rules for an incoming particle
 ( x)   b  v  eipx  a   u  eipx 

p
u p1 0
e  ( p1 )
v p1 0
Feynman Rules for an incoming antiparticle
A ( x)  a     a   
LI  gA  

g  
  b  a
u p2
  a  b
u p1
Propagator
A A B B
i
d
U (t , t 0 )  H I U (t , t 0 )
dt
t
dU ( 2 ) (t , t 0 )
(1)
i
 H I (t )  U (t , t 0 )  H I (t )   dt ' H I (t ' )
dt
t0
t ''


(t , t 0 )   dt' '  H I (t ' ' )  dt' H I (t ' )


t0
t0
t
U
( 2)
The integration of two identical interaction Hamiltonian HI.
The first HI is always later than the second HI
t
U
( 2)
t
1
(t , t 0 )    dt ' '  dt ' T  H I (t ' ' ) H I (t ' )
2 t0
t0
T ( A(t1 ) B(t 2 ))   (t1  t 2 ) A(t1 ) B(t 2 )   (t 2  t1 ) B(t 2 ) A(t1 )
This definition is Lorentz invariant!
S ( 2 )  U ( 2 ) (, ) 
1
  d 4 x1 d 4 x2 T  L I ( x1 ) L I ( x2 )
2
Amplitude for scattering
A A B B
B( p3 )B( p4 ) S A( p1 ) A( p2 )
 B( p3 ) B( p4 )  d 4 x1 d 4 x2 T g A( x1 ) B( x1 ) C ( x1 )  gA( x2 ) B( x2 ) C ( x2 ) A( p1 ) A( p2 )
  d 4 x1 d 4 x2 e i ( p1  p3 ) x2 e i ( p2  p4 ) x1  0 T  C ( x1 ) C ( x2 ) 0
Fourier Transformation
p1-p3 pour into x2
p2-p4 pour into x1
Propagator between x1 and x2
t1  t2

0 T C( x1 )  C( x2 )  0  0 C( x1 )  C( x2 ) 0  0 a  a
  a  a 

x1
A particle is created at x2 and later annihilated at x1.
B(p4)
B(p3)
B(p3)
B(p4)
x1
x2
A(p1)
C(p1-p3)
C
A(p2)
A(p1)
A(p2)
x2
0
t1  t2

0 T C( x1 )  C( x2 )  0  0 C( x2 )  C( x1 ) 0  0 a  a
  a  a 

x2
A particle is created at x1 and later annihilated at x2.
B(p4)
B(p3)
B(p4)
B(p3)
x2
B(p4)
x1
C
A(p1)
B(p3)
x1
A(p2)
x2
A(p1)
C(p1-p3)
C
A(p2)
A(p1)
A(p2)
x1
0
ipx1 iqx2
4
4
d
x
d
x
e
 1 2 e 0 T  C ( x1 ) C ( x2 ) 0 
B(p4)
B(p3)
B(p4)
B(p3)
x2
B(p3)
B(p4)
x1
C
A(p1)
i
4

( p  q)
2
2
q  mC
x1
A(p2)
x2
A(p1)
C(p1-p3)
C
A(p2)
A(p1)
A(p2)
0 T   ( x)  ( y) 0
0 T   ( x)  ( y) 0
0 T   ( x)  ( y) 0
0 T   ( x)  ( y) 0
This doesn’t look explicitly Lorentz invariant.
But it is!
0 T   ( x)  ( y) 0
x0  y0
 a  a
 a  a



Every field either couple with another field to form a propagator or
annihilate (create) external particles! Otherwise it will vanish!
Scalar Antiparticle
Antiparticles can be introduced easily by assuming that the field
operator is a complex number field.
L0        m2

d3p
( x)  
(2 ) 3
1
2
a e
p
ipx
 b p e ipx

Complex KG field can either annihilate a particle or create an antiparticle!

d3p 1

ipx
 ipx
 ( x)  
b
e

a
p
pe
(2 ) 3 2


Its conjugate either annihilate an antiparticle or create a particle!
The charge flow is consistent!
So we can add an arrow for the charge flow.
L I  g 3  g 3
vertex
Charge non-conserving

LI    

2
vertex
Charge conserving
Propagator:


0 T  ( x1 ) ( x2 ) 0
t1  t2



0 T  ( x1 )  ( x2 ) 0  0 ( x1 )  ( x2 ) 0  0 b  a
  a  b 
An antiparticle is created at x2 and later annihilated at x1.
B(p4)
B(p3)
B(p3)
B(p4)
x1
x2
A(p1)
C(p1-p3)
C
A(p2)
A(p1)
A(p2)

x1
x2
0
t1  t2



0 T  ( x1 )  ( x2 ) 0  0 ( x2 )   ( x1 ) 0  0 a  b
  b  a 

x2
A particle is created at x1 and later annihilated at x2.
B(p4)
B(p3)
B(p4)
B(p3)
x2
B(p4)
x1
C
A(p1)
B(p3)
x1
A(p2)
x2
A(p1)
C(p1-p3)
C
A(p2)
A(p1)
A(p2)
x1
0


ipx1 iqx2
4
4

d
x
d
x
e
e
0
T

( x1 ) ( x2 ) 0 
 1 2
B(p4)
B(p3)
B(p4)
B(p3)
x2
B(p3)
B(p4)
x1
C
A(p1)
i
4

( p  q)
2
2
q  mC
x1
A(p2)
x2
A(p1)
C(p1-p3)
C
A(p2)
A(p1)
A(p2)

LI    B 
B(p4)
B(p3)
B(p4)
B(p3)
x2
x1
C
A(p1)
x1
A(p2)
x2
A(p1)

B(p3)
B(p4)
  B 
C(p1-p3)
  B 
C
A(p2)
A(p1)
A(p2)
U(1) Abelian Symmetry
L0        m2
The Lagrangian is invariant under the phase transformation
of the field operator:
( x)  e iQ ( x)
    eiQ  eiQ  (x)  

LI     


2
invariant
If A,B,C become complex, they carry charges!
A
C
The interaction is invariant only if QA  QB  QC  0
U(1) symmetry is related to charge conservation!
B


L  i     m 
The Dirac Fermion Lagrangian is also invariant under U(1)
 ( x)  eiQ ( x)


L   eiQ  eiQ  i     m   L
SU(N) Non-Abelian Symmetry
Assume there are N kinds of fields
 1 


2 
  3 


  
 
 n
If they are similar, we have a SU(N) symmetry!
( x)  U  ( x)  e
L0        m2
i iT i
 ( x)

LI    
are invariant under SU(N)!

2
量子力學下互換群卻變得更大！
量子力學容許量子態的疊加
u
u
a
u
d
c
u
+ b
d
d
古典
+ d
d
量子
a  b 1
2
u-d 互換對稱
2
c  d 1
2
2
ac*  bd *  0
u 
u 
a b 
   U  , U  

d 
d 
c d 
*
a
b

a


   *
UU   
c d  b
c* 
 1
*
d 
( x)  U  ( x )
  ( x)  U    ( x)
L0          m 2   
     U  U      m 2  U  U  
         m2  





L I            U  U        
2
They are invariant under SU(N)!
2

2
Gauge symmetry
( x)  eiQ ( x) ( x)
Global Symmetry
( x)  e iQ ( x)
    eiQ  eiQ  (x)  

      e iQ ( x)

 e iQ   ( x)
    
      eiQ  e iQ    (x)
     
Gauge (Local) symmetry
( x)  eiQ ( x) ( x)
    eiQ ( x)  eiQ ( x)  ( x)  

      e iQ ( x ) ( x)

 e iQ ( x )   ( x)  iQ (  ) e iQ ( x ) ( x)
Kinetic energy is not invariant
under gauge transformation!
Could we find a new “derivative” that
works as if the transformation is global?
D ( x)  eiQ ( x) D ( x)
    eiQ   (x)

      e iQ ( x ) ( x)

 e iQ ( x )   ( x)  iQ (  ) e iQ ( x ) ( x)
To get rid of the extra term, we introduce a new
vector field:
A ( x)  A ( x)    ( x)
D       iQ  A 
D       iQA   e iQ ( x) D ( x)  iQ (  ) e iQ ( x) ( x)  iQ (  ) e iQ ( x) ( x)
 e iQ ( x) D ( x)
Global Symmetry
Gauge (Local) symmetry
( x)  eiQ ( x) ( x)
( x)  e iQ ( x)
D ( x)  eiQ ( x) D ( x)
    eiQ   (x)
    
D   D  
      eiQ  e iQ    (x)
 D    eiQ  e iQ  D  ( x)
     
 D   D  


Replacing derivative with D  D  is invariant under gauge
covariant derivative,
transformation!
D  D          iQ  A      iQ  A     Q2  A A 
The scalar photon interaction vertices


L  i     m 
To force it to be gauge invariant,
 ( x)  eiQ ( x) ( x)
you only need to replace derivative with
coariant derivative.
   D


L  i  D  m 
is gauge invariant!
This gauge invariant Lagrangian gives a definite
interaction between fermions and photons

  i


 m   Q  

L   i  D  m    i     Q    A  m 


LI  gA  

  A
g  
A ( x)  a     a   
LI  gA  

g  
  b  a
u p2
  a  b
u p1
LI  gA  
This form is forced upon us by gauge symmetry!
It is really a Fearful Symmetry! Tony Zee
Tyger! Tyger! burning bright
In the forests of the night
What immortal hand or eye
Could frame thy fearful symmetry!
William Blake
Let there be light!
In the name of gauge symmetry!
Hermann Weyl, 1885-1955
Yang and Mills
SU(N) Non-Abelian Symmetry
Assume there are N kinds of fields
 1 


2 
  3 


  
 
 n
If they are similar, we have a SU(N) symmetry!
( x)  U  ( x)  e
L0        m2
i iT i
 ( x)

LI    
are invariant under SU(N)!

2
Non-Abelian Gauge Symmetry
( x)  U ( x)  ( x)  e i ( x) T  ( x)
i
i
We need one gauge field for each generator.
D       igT i Ai 
Gauge fields transform as:
1
A T  U  A T  U  i U   U 
g
i
i
i
i



i
D       igT i Ai     U     ig  UT i Ai U 1  U U 1  U
g


 U     U    ig UT i Ai   U U 1U  U  D 
D   U  D 
D  D  is invariant under gauge
transformation!


L  i  D  m 

  i


 m   Q  

L   i  D  m    i     Q   T i Ai   m 


LI  gAi   T i

T i  Ai 
g   T i
 e 
 e 
   W ?  
e
e
3
W  
i 1

g   e

 0 1
i
2 × 2 matrices
 
e    Wi   i   e 
i
 e 
0  i
1 0 
  3  

0
0

1


  2  
 1  
1
0


i
i
W 
W  iW   W3
 W
i Wi   i  W  3iW 1 W 2    2  W

2
3
 1
 
W1  iW2
2
,W  
2 W  

 W3 
W1  iW2
2
1 
   2 
 

L          m 2       

V  m      
2



2

2

V  m 2        

2
Vacua happen at:

 
m2

 v2
Choose:

 0

v

 
For infinitesimal transformation:
0  U  0  0   iT i  0
T  0  0, (T 3  Y )0  0, (T 3  Y )0  0
SU(2)χU(1)Y is broken into U(1)EM

  gT
D   D    D  0 D   0  gT W  gT 3W3  g ' YB  0
   T   W
g 2 T  0


0

3
3
0
 g' B
 gW

3
 g' B


Z become massive
W become massive
Z
Photon is massless.
A
 
W    gT 3W  3  g ' YB   0

   T  gW

 T 3 0
W




1
g 'W 3  gB
2
2
g  g'

1
gW 3  g ' B
2
2
g  g'
