Dirac Field
1
2
Sec. 3 Now interaction and evolution!
Schrodinger and Heisenberg Pictures
3
Interaction and Interaction Picture
The interactions are introduced into QFT when we add additional terms into the
Lagrangian beyond what we have in a free field theory:
For example, we can add  3 ( x),  4 ( x),  ( x) ( x) ( x) to our Klein-Gordon or Dirac
Lagrangian. Notice they must be Lorentz invariant. These interactions will generate
additional terms in the Hamiltonian. In most cases (when interaction contains no
derivatives), the additional Hamiltonian is just the negative interaction Lagrangian :
To analyze the effects of the interaction, it’s convenient to work in a picture called
interaction picture which lie between Heisenberg picture and Schrodinger picture.
The full Hamiltonian can now be separate into two parts:
H  H 0  H int
In the transforming from Schrodinger picture to Heisenberg picture, we can stop in
the middle:
by replacing e iHt /  with e iH t /  . Hence both the states and the operators will evolve
0
with time:
OI (t )  eiH0t /  OS e iH0t / 
Operators evolve just like operators in the Heisenberg picture but with the full
Hamiltonian replaced by the free Hamiltonian:
dOI i
 H 0 , OI 
dt

It applies also to the field operators:
d i
 H 0 ,  
dt 
Field operators are free, as if there is no interaction! Hence all the things we talked
4
about Klein-Gordon free Fields in Heisenberg picture in Sec 2.4 still works here for
the field operators in the interaction picture! The expansion of field is valid:
For the evolution of states:
So the states in the interaction picture evolve like in the Schrodinger picture but with
Hamiltonian replaced by V(t). This term V(t) is just the interaction Hamiltonian HI in
interaction picture! That means the field operators in V(t) are free, ie. evolving as if
there is n interaction!
So the states now evolve by the interaction Hamiltonian with free fields:
d
 i
 I (t )     H I  I (t )
dt
 
Sum up the time evolution by a unitary operator U
 I (t )  U (t , t0 )  I (t0 )
All the problems can be answered if we are able to calculate this operator. From the
evolution of states, we can find the evolution equation of U:
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
 

In QFT, it is usually calculated by a perturbation expansion in small parameters in HI.
U (t , t 0 )  U ( 0) (t , t 0 )  U (1) (t , t 0 )  
Plug this expansion into the above evolution equation, we can solve it order by order.
Note that the zeroth order of U is actually 1. U ( 0) (t , t 0 )  1 .
To leading order:
5
d (1)
 i
 i
U (t , t 0 )     H I  U ( 0) (t , t 0 )     H I
dt
 
 
 i
U (t , t 0 )      dt ' H I (t ' ' )
   t0
t
(1)
is the operator that describes the evolution of the system from
to t .
Setting the initial time 0 instead to negative infinity and the final time
, it will
describe the scattering and decay process of particle physics that evolve a state in the
long past into another state in the long future. We define the resultant operator as S:
. For example, the leading order of S is:








S  i  dt H (t )  i  dt dx 3 H ( x , t )  i  d 4 x H ( x)
6
Vertex
We now consider the ABC model in Griffiths’s book Ch. 6. Each particle will be
associated with a field with its Free Scalar Lagrangian as described above. Let’s call
the field of A particle as
etc. Till now the theory is free, ie particle do not
interact. To introduce interaction, we need to add additional terms in the Lagrangian
beyond the free scalar Larangian in Eq (2.6). Let’s add a term:
.
This is Lorentz invariant. You can calculate that
.
Now we use this term to calculate the transition amplitude for
. The initial state is
while the final state is
. To the
leading order, S is
.
Look, again it’s Lorentz invariant. The transition amplitude is then
.
Now each field operator is a linear combination of a annihilation operator and a
creation operator. States with particles can be annihilated (created) by annihilation
(creation) operator:
Expanding ABC will give you 8 terms. But only one term will give you nonzero result
in the amplitude:
.
The momenta of the annihilation and creation operator must match the momenta of
the particles it annihilates or creates. After creation and annihilation, a numerical
factor is left:
.
This is the total momentum conservation. So roughly speaking, a interaction
Lagrangian term corresponds to a vertex, with each field operator in the interaction
corresponds to a leg in the vertex. So the vertex can be read directly from the
interaction term in the Lagrangian: . And since each field operator can either create
or annihilate a particle, a leg in the vertex can mean both creation or annihilation,
depending on the process you are studying.
7
The same consideration can apply to other interactions. For example , L I   4 would
give a vertex:
It is also true for interactions involving fermions. For example, the Yukawa coupling
LI  g  would give:

Here the Dirac operator   a  b could either annihilate a particle or create an

antiparticle, while   b  a either annihilate an antiparticle or create a particle!
Note that annihilating a particle and creating an antiparticle would both annihilate one
unit of charge. In each case, the charge flow is consistent and hence we can add an
arrow on the lines to indicate the flow of charge.
We can also obtain the Feynman rule for the external lines. When Dirac operators
annihilate states, they leave behind a u or v !
 ( x)  p  a  u  e ipx  b   v  eipx  p  u  e ipx  0
 ( x)  p  b  v  e ipx  a   u  eipx  p  v  e ipx  0
8
Propogator
Let’s now consider the scattering process A  A  B  B in the ABC model.
It won’t happen in the leading order of g. We have to calculate U to the second order.
In interaction picture: the time evolution operator U satisfies:
d
i U (t , t 0 )  H I  U (t , t 0 )
dt
To the second order of the g, we only need to consider the U to the right hand side to
the first order of g, which we solved already. Hence
t
dU ( 2) (t , t 0 )
(1)
i
 H I (t )  U (t , t 0 )  H I (t )   dt ' H I (t ' )
dt
t0
This can be easily solved:
t ''


U (t , t 0 )   dt ' '  H I (t ' ' )  dt ' H I (t ' )


t0
t0
It’s the integration of two interaction Hamiltonian HI. Notice that the first HI is always
later than the second HI. We can write this instead as:
t
( 2)
t
U
( 2)
t
1
(t , t 0 )    dt ' '  dt ' T  H I (t ' ' ) H I (t ' )
2 t0
t0
The notation “time ordered” T means the earlier operator will be automatically put to
the right, ie:
T ( A(t1 ) B(t 2 ))   (t1  t 2 ) A(t1 ) B(t 2 )   (t 2  t1 ) B(t 2 ) A(t1 )
This notation seems strange and artificial, but you will see what its physical meaning
later.
So the S matrix becomes:
S ( 2 )  U ( 2 ) (,) 
1
  d 4 x1 d 4 x2 T  L I ( x1 ) L I ( x2 )
2
Here LI’s are the interaction Lagrangian.
For the scattering A  A  B  B , the amplitude can be calculated as:
B( p3 ) B( p4 ) S A( p1 ) A( p2 )
 B( p3 ) B( p 4 )  d 4 x1 d 4 x 2 T g A( x1 ) B( x1 ) C ( x1 )  gA( x 2 ) B( x 2 ) C ( x 2 ) A( p1 ) A( p 2 )
  d 4 x1 d 4 x 2 e  i ( p1  p3 ) x2 e i ( p2  p4 ) x1 0 T  C ( x1 ) C ( x 2 ) 0
The particle A(p1) is annihilated at x2 and A(p2) is annihilated at x1. The particle B(p3)
9
is created at x2 and B(p4) is created at x1.The object 0 T  C( x1 ) C( x2 ) 0 is called the
propagator. When t1  t 2 ，the operator C(x2) appearing to the right will create a C
particle at x2 and it will be annihilated at x1 by C(x1) appearing to the left. When t 2  t1
and the order of operators is reversed, the operator C(x1) appearing to the right will
create a C particle at x1 and it will be annihilated at x2 by C(x2) appearing to the left.
This construction ensures causality of the process. The propagator in the momentum
space is much easier to calculate:
ipx iqx
4
4
 d x1 d x2 e 1 e 2 0 T  C ( x1 ) C ( x2 ) 0 
i
 4 ( p  q)
2
2
q  mC
The delta function serves to make sure the momentum flowing into the particle is just
the momentum flowing out.
B(p3)
B
B
B
B(p4)
B
x1
x2
C(p1-p3)
x2
C
A
A
x1
C
A
A
A(p1)
A(p2)
Calculating the propagator
Sandwiched between two vacua, only the third term will survive:
10
0 T   ( x)  ( y )  0
0 T   ( x)  ( y )  0
0 T   ( x)  ( y )  0
The propagator is Lorentz invariant. But this form is not obviously Lorentz invariant.
So a even more useful form is obtained by extending the integration to 4-momentum.
11
The same kind of calculation can be applied to the fermion propagators.
0 T  ( x)  ( y) 0
12
Sec. 4 Antiparticle, charge and symmetry
Antiparticles can be introduced easily by assuming that the field operator is a complex number
field.
Now     . There are two degrees of freedom and therefore the Fourier expansion of the field
contains one more independent coefficient.

d3p 1
 ( x)  
a p e ipx  b p e ipx
3
(2 ) 2

d3p 1
  ( x)  
b p e ipx  a p e ipx
3
(2 ) 2




It could be shown that the new operator bp will annihilate an antiparticle with momentum p and bp+
will create an antiparticle. This is neat since it means a field Φ will always add charge of one unit
(annihilate a particle and create an antiparticle) while its complex conjugate Φ+ will always
annihilate charge of one unit (annihilate an antiparticle and create a particle).
When you write Lagrangian for a complex field, you need to make sure it is real. So the free
Lagrangian:
L0          m2   
The interaction Lagrangian could start with:
L I  g 3  g 3
(1)
(2)
So for ABC model with U(1) symmetry, the interaction Lagrangian becomes:
L I  g A  B  C  g A  B  C
(3)
You can easily figure out what the vertex could do. The field operators in a vertex could annihilate
or create particles (or antiparticles). They could also form propagators with other field operators in


another vertex. And the propagator now is 0 T   ( x1 ) ( x2 ) 0 . When t1  t 2 ，the operator Φ(x2)
appearing to the right will create an antiparticle at x2 and it will be annihilated at x1 by Φ+(x1)
appearing to the left. When t 2  t1 and the order of operators is reversed, the operator Φ+(x1)
appearing to the right will create a particle at x1 and it will be annihilated at x2 by Φ(x2) appearing to
the left. You can see that the charge (assuming that the particle is charge 1) always flow from x2
to x1. So we can give it an arrow to denote the definite direction of charge flow:
x2
x1
Now we can discuss the symmetry of this model. The free Lagrangian L0 of Eq. (1) is invariant
under an U(1) phase transformation:
( x)  e iQ ( x)
Here θ is a continuous variable (that’s why this is U(1)) and Q could be any constant. So we can
assign Q to be the charge of the particle the field operator Φ annihilates. Now we can ask the
interaction Lagrangian LI to be invariant under U(1) transformation. The easiest choice is for the
13
interaction terms to depend only on the absolute value of Φ, which is U(1) invariant. For example:
L I   (   ) 2
The interaction in Eq. (2) is not U(1) invariant unless the charge Q=0. But the ABC model is
invariant if the charge is conserved at a vertex:
Q A  QB  QC  0
Hence the principle of U(1) symmetry can exclude other interaction Lagrangian terms which is not
invariant under U(1) phase transformation. For example: Li  g A  B  B is not allowed
unless QB  QC .
The symmetry could also be non-Abelian. If there are n fields in the model with similar property, we
can list them as a n-column:
 1 


2 
   3 


  
 
 n
The free Lagrangian is of the same form as Eq. (1) if you understand the notation as multiplications
of columns. We can now define SU(n) transformation as any n×n special unitary matrix U, which
transforms the field Φ as
( x)  U  ( x)  e i T  ( x)
i i
Here we have decomposed the matrix U in terms of the exponents of generators U  e i T . You
can easily check the free Lagrangian Eq. (1) is invariant. If the interaction terms are also invariant
under this transformation, the model is SU(n) symmetric. For example, LI   (  ) 2 is
i
symmetric.
14
i
Sec. 5 Gauge symmetry
In the above discussion about U(1) transformation, the continuous variable θ does not depend on
spacetime x, ie. it’s a constant with respect to x. This symmetry is called global, meaning the fields
transform the same way wherever it is. Is it possible for us to extend it to a local form? That is, if we
allow the variable θ to depend on spacetime x, is the Lagrangian still invariant under this extended
form of transformation? This kind of transformation is called gauge transformation:
( x)  eiQ ( x ) ( x)
At first sight, this answer seems an obvious “no”. No to mention interaction, even the free
Lagrangian is not invariant. The mass term is OK but the kinetic energy term is not. The derivative
on Φ can not commute with the phase factor and will generate an additional term:


      e iQ ( x) ( x)  e iQ ( x )   ( x)  iQ (  ) e iQ ( x ) ( x)
In order for the kinetic energy term to be gauge symmetric, we need to eliminate the additional term.
The way to do it is to introduce a field which is a vector:Aμ(x). This field is usually called gauge
field or gauge boson. This vector transform under U(1) as:
A ( x)  A ( x)     ( x)
And replace all the    ’s with a new “covariant” derivative D       iQA   . The
transformation property of the new “covariant” derivative is very simple:
D       iQA   e iQ ( x) D ( x)  iQ (  ) e iQ ( x) ( x)  iQ (  ) e iQ ( x) ( x)
 e iQ ( x ) D ( x)
You can easily check that D    D  is invariant under gauge transformation. Surprisingly this
invariant kinetic energy not only give kinetic energy for the field Φ, it also supply the interaction
between Φ and the gauge field A. When you expand this term, you find:
D    D          iQA       iQA       Q 2 A A   
The interaction terms correspond to the following vertices:
The gauge field also needs a gauge invariant kinetic energy term. It’s not difficult since we know in
electromagnetism electric field and magnetic field is gauge invariant and the Hamiltonian equals the
square of these fields. You may check the field written as
15
F    A   A
is gauge invariant. The Lagrangian:
1
L  F  F
4
is both Lorentz invariant and gauge invariant.
Note that the gauge boson mass term A A is not gauge invariant. The gauge boson is forced by
gauge symmetry to be massless.
We can easily extend the gauge symmetry to Fermions. The Dirac Lagrangian


L   i     m  is invariant under a global U(1) symmetry:  ( x)  e iQ ( x) . To make it
invariant under a gauge local U(1): ( x)  e iQ ( x ) ( x) , you only need to replace the derivative with
covariant derivative:    D . It’s easy to see that:


L   i  D  m 
is gauge invariant. This gauge invariant form also gives you a definite interaction between gauge
boson and fermion:
L   i  D  m    i     Q    A  m 
  i     m   Q     A
The interaction Lagragian is
 Q      A
which gives rise the famous vertex:
.
Let me emphasize the existence of a gauge boson and this interaction form is forced by the gauge
symmetry.
16
For non-Abelian gauge symmetry, the formalism is almost the same. The gauge transformation is:
( x)  U ( x)  ( x)  e i
i
( x) T i
 ( x)
Now for every generator T i , we need to introduce one gauge field Ai . Then we multiply each
generator (a n×n matrix) with its corresponding gauge field and sum them up. The covariant
derivative is defined as: D       igT i Ai  . The constant g is the coupling constant. The
transformation of gauge fields is more complicated:
1
Ai T i  U  Ai T i  U   i U   U  .
g
You can check that the covariant derivative transforms in a simple way:


i
D       igT i Ai     U     ig  UT i Ai U 1  U U 1   U
g


i i
1
 U     U    ig UT A   U U U  U  D 
D   U  D 
and hence the kinetic energy term D    D  is again gauge invariant.
Now we can find the interaction of the gauge fields from the kinetic energy Lagrangian. Let’s
concentrate on those terms with three fields since they are the ones that will survive in the quantum
field theory of fermions. We’ll have: g (    ) T i Ai  or g (    ) T i  Ai . It’s a row vector
times a matrix times a column vector, producing a number which is then multiplied by gauge field A.
The interaction of a gauge field Ai is characterized by its corresponding generator T i . The most
remarkable thing is that for all the gauge fields the coupling constant is the same g.
For fermions, it is even simpler:
L   i  D  m    i     Q   T i Ai   m 
  i     m   Q   T i  Ai 
The interaction is: L I  gAi   T i . The vertex is:
So the interaction with a gauge boson sandwich its corresponding generator between the two
interacting fermions.
17
Let’s illustrate this formalism by the weak SU(2) gauge theory. The lepton fields can be listed
v
as     . The generators of SU(2) are 1/2 times Pauli matrices  i or their linear combinations.
e
1 0 1
1  0 0
1 1 0 

, T  

, T 3  
 . The corresponding gauge
We can choose: T  
2  0  1
2  0 0
2 1 0
fields are W+, W-,W3. The interaction of W+ is:
g    T   W 
g
2
   e W .
Take away the derivative, which will disappear in fermion theory, and this vertex is just what we
introduced in the mid term exam problem about weak interaction. The interaction of W- is:
g
g    T   W 
e   W .
2
These two interactions give rise to the following vertices:
The interaction of W3 is:
g  T 3  W3 
g
g
   W3  e   e W3 .
2
2
18
Sec. 6 Spontaneous Symmetry Breaking
We need to break some of the gauge symmetries to give masses to W and Z bosons while keep
photon massless. The gauge symmetry will be SU(2)×U(1). The generators are T  , T 3 , Y with
corresponding gauge bosons W±,W3 for SU(2) and gauge boson B for U(1).
The idea of spontaneous symmetry breaking is simple. Instead of breaking the symmetry of the
Lagrangian, we can break the symmetry by the vacuum while keeping the Lagrangian symmetric.
The simplest way is to introduce a scalar field Φ which is a doublet under SU(2) and assign its U(1)
1 
charge Y as 1/2. Then we can write Φ as a column:    2 
 


Assume that its Lagrangian is: L          m 2         . The potential of the theory is

2

V  m 2         . Note that the mass is negative here. This Lagrangian is apparently SU(2)
2
×U(1) symmetric. But the vacuum, which is ground state, is not where all the fields vanish as we
would expect. The vacuum exists at where:
 
m2

 v2
So there are a whole series of vacua and neither vacuum is invariant. So symmetries are broken.
0
Consider the vacuum:    
v
The change of the vacuum under infinitesimal gauge transformation is the generator times the
vacuum:  0  U   0   0   i T i   0
You can check T   0  0, (T 3  Y ) 0  0, (T 3  Y ) 0  0 . So symmetries corresponding to
three generators are broken while one is unbroken. This is exactly what we need.
To see the effects explicitly, we look at the kinetic energy term of the scalar field. When the energy
is small, the field Φ can be approximated by its vacuum  0 ：

  gT
D   D    D  0 D   0  gT W  gT 3W3  g 'YB  0

 
W    gT 3W  3  g 'YB   0

Remember (T 3  Y ) 0  0,  T 3  0  Y 0 . We get:
   T   W
g 2 T  0


0
   T  gW
W   T 3 0



3
0
3
 g' B
 gW

3
 g' B


1
1
 g 2 v 2 WW    ( g 2  g ' 2 )v 2 Z  Z 
2
4
1
gW 3  g ' B . The photon is the
The Z boson is linear combination of W3 and B: Z  2
2
g  g'
orthogonal combination which has no mass.
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