# Document ```1
1 2 2
𝜇
ℒKG Free = 𝜕𝜇 𝜙 𝜕 𝜙 − 𝑚 𝜙
2
2
Free scalar particle
meaning anything not quadratic in fields.
 3 ( x),  4 ( x),  ( x) ( x) ( x)
to Klein-Gordon Lagrangian.
These terms will add the following non-linear terms to the KG equation:
3 2 , 4 3 ,  ( x) ( x)
Superposition Law is broken by these nonlinear terms.
Travelling waves will interact with each other.
Interaction Hamiltonian:

 (x)

Ô

 

f ( x, p)  f  x,i   f ( xˆ , pˆ )
x 



Oˆ    * ( x)  Oˆ  ( x)dx


Dirac Notation

Ket
 (x)

Bra
 * ( x)


Ô
ˆ ( x)
O
Ô 


Oˆ    * ( x)  Oˆ  ( x)dx
Ô 




Oˆ   Oˆ 

Ô  o
Ô   o 
Schrodinger Picture
States evolve with time, but not the operators:  (t )
It is the default choice in wave mechanics.

Ô
&ouml;
&para;Y &aelig; 2 &para;2
i
= &ccedil;+V(x)&divide; Y = ĤY
2
&para;t &egrave; 2m &para;x
&oslash;
𝑑
𝜓 𝑡
𝑑𝑡 S
𝜓S 𝑡
Y(x, t) = x Y(t)

xˆ  x pˆ    i 

x 
= −𝑖𝐻 𝜓S 𝑡
= 𝑒 −𝑖𝐻𝑡 𝜓S 0
Evolution Operator: the operator to move states from 𝑡 = 0 to 𝑡.
In quantum mechanics, only expectation values are observable.
For the same evolving expectation value, we can instead ask operators to evolve.
O   S (t ) OS  S (t )  e  iHt S (0) OS e  iHt S (0)
  S (0) e iHt OS e iHt  S (0)   S (0) Ot   S (0)
We move the time evolution to the operators:
OH t   e iHt OS e  iHt
Now the states do not evolve.
 H   S (0)
Heisenberg Picture
How does the operator evolve?
d
OH t   iH  e iHt OS e iHt  i  e iHt OS e iHt  H
dt
Heisenberg Equation
 iHOH t   OH t H   iH , OH t 
The rate of change of operators equals their commutators with H.
In Schrodinger picture, field operators do not change with time, looking not Lorentz invariant.
Fields operators in Heisenberg Picture are time dependent, more like relativistic classical fields.
For KG field without interaction:
Combine the two equations

2

 m2   0
The field operators in Heisenberg Picture satisfy KG Equation.
The field operators with interaction satisfy the non-linear Euler Equation, for example

2

 m 2   3 3  0
Interaction picture (Half way between Schrodinger and Heisenberg)
There is a natural separation between free and interaction Hamiltonians:
H  H 0  H int
O   S (t ) OS  S (t )   S (0) e iHt OS e  iHt  S (0)


  I (t ) e iH 0t OS e iH 0t  I (t )   I (t ) OI  I (t )
Move just the free H0 to operators.
OI (t )  e iH 0t OS e  iH 0t
𝜓I 𝑡
= 𝑒 𝑖𝐻0𝑡 𝜓S 𝑡
The rest of the evolution, that from the interaction H, stays with the state.
States and Operators both evolve with time in interaction picture:
Evolution of Operators
OI (t )  e
iH 0t / 
OS e
 iH 0t / 
𝑑𝑂I
= 𝑖 𝐻0 , 𝑂I
𝑑𝑡
Operators evolve just like operators in the Heisenberg picture
but with the full Hamiltonian replaced by the free Hamiltonian
For field operators in the interaction picture:
𝑑𝜙I
= 𝑖 𝐻0 , 𝜙I
𝑑𝑡
Field operators are free, as if there is no interaction!
The Fourier expansion of a free field is still valid.
Evolution of States
𝜓I 𝑡
𝑑
𝜓 𝑡
𝑑𝑡 I
= 𝑖𝐻0 𝜓I 𝑡
= 𝑒 𝑖𝐻0𝑡 𝜓S 𝑡
− 𝑖𝑒 𝑖𝐻0𝑡 𝐻0 + 𝐻int S 𝜓S 𝑡
= 𝑒 𝑖𝐻0𝑡 ∙ 𝐻int S ∙ 𝑒 −𝑖𝐻0𝑡 𝜓S 𝑡
𝐻int I = 𝑒 𝑖𝐻0𝑡 ∙ 𝐻int S ∙ 𝑒 −𝑖𝐻0 𝑡
Hint I is just the interaction Hamiltonian Hint in interaction picture!
𝑑
𝜓 𝑡
𝑑𝑡 I
= −𝑖𝐻int I 𝜓I 𝑡
States evolve like in the Schrodinger picture but with full H replaced by Hint I.
That means, the field operators in Hint I are free.
Interaction Picture
Operators evolve just like in the Heisenberg picture but with
the full Hamiltonian replaced by the free Hamiltonian
𝑑𝜙I
= 𝑖 𝐻0 , 𝜙I
𝑑𝑡
States evolve like in the Schrodinger picture but with the full
Hamiltonian replaced by the interaction Hamiltonian.
𝑑
𝜓 𝑡
𝑑𝑡 I
= −𝑖𝐻int I 𝜓I 𝑡
The evolution of states in the interaction picture
𝑑
𝜓 𝑡
𝑑𝑡 I
= −𝑖𝐻int I 𝜓I 𝑡
Define U the evolution operator of states.
𝜓I 𝑡
= 𝑈 𝑡, 𝑡0 𝜓I 𝑡0
All the problems can be answered if we are able to calculate this operator
𝑈 𝑡, 𝑡0 𝜓 is the state in the future t which evolved from a state 𝜓 in t0
𝑈 ∞, −∞ 𝜓 is the state in the long future which evolves from a
state 𝜓 in the long past.
The amplitude for 𝑈 ∞, −∞ 𝜓 to appear as a state 𝜙 is their inner product:
𝜙 ∙ 𝑈 ∞, −∞
𝜓 = 𝜙 𝑈 ∞, −∞ 𝜓
The amplitude is the corresponding matrix element of the operator 𝑈 ∞, −∞ .
The transition amplitude for the decay of A: 𝐴 → 𝐵 + 𝐶
the amplitude of 𝐵𝐶 in the long future which evolves from 𝐴 in the long past.
𝐵𝐶 𝑈 ∞, −∞ 𝐴
The transition amplitude for the scattering of A: 𝐴 + 𝐴 → 𝐵 + 𝐵
𝐵𝐵 𝑈 ∞, −∞ 𝐴𝐴
𝑆 ≡ 𝑈 ∞, −∞
S operator is the key object in particle physics!
U has its own equation of motion:
The evolution of states in the interaction picture
𝑑
𝜓 𝑡
𝑑𝑡 I
= −𝑖𝐻int I 𝜓I 𝑡
𝑈 𝑡, 𝑡0 is the evolution operator of states.
𝜓I 𝑡
= 𝑈 𝑡, 𝑡0 𝜓I 𝑡0
Take time derivative on both sides and plug in the first Eq.:
𝑑
𝜓 𝑡
𝑑𝑡 I
=
𝑑
𝑈 𝑡, 𝑡0 𝜓I 𝑡0
𝑑𝑡
𝑑
𝑈 𝑡, 𝑡0 = −𝑖𝐻I 𝑈 𝑡, 𝑡0
𝑑𝑡
= −𝑖𝐻I 𝑈 𝑡, 𝑡0 𝜓I 𝑡0
Perturbation expansion of U and S
𝑑
𝑈 𝑡, 𝑡0 = −𝑖 𝐻I ∙ 𝑈 𝑡, 𝑡0
𝑑𝑡
Solve it by a perturbation expansion in small parameters in HI.
𝑈 = 𝑈 (0) + 𝑈 (1) + ⋯
𝑈 (0) = 1
𝑑 (1)
𝑈
𝑡, 𝑡0 = −𝑖 𝐻I ∙ 𝑈
𝑑𝑡
0
𝑡, 𝑡0 = −𝑖𝐻I 𝑡
𝑡
𝑈 (1) 𝑡, 𝑡0 = −𝑖
𝑑𝑡 ′ 𝐻I 𝑡′
𝑡0
∞
𝑈 (1) ∞, −∞ = 𝑆 (1) = −𝑖
𝑑𝑡′ 𝐻I 𝑡′
−∞
To leading order, S matrix equals
∞
𝑆 (1) = −𝑖
∞
𝑑𝑡 𝐻I 𝑡 = −𝑖
−∞
∞
𝑑𝑡
−∞
𝑑 3 𝑥 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 the following interaction term in the Lagrangian:
ℒI 𝑥 = 𝑔 𝐴 𝑥 𝐵 𝑥 𝐶 𝑥
The transition amplitude for A decay: 𝐴 𝑝1 → 𝐵 𝑝2 + 𝐶 𝑝3 can be computed as :
𝐵𝐶 𝑈 ∞, −∞ 𝐴 = 𝐵𝐶 𝑆 𝐴
∞
∞
𝑑4 𝑥 ℒI 𝑥 = 𝑖
𝑆=𝑖
−∞
𝐵 𝑝2 𝐶 𝑝3 𝑆 𝐴 𝑝1
𝑑4𝑥 𝑔 𝐴 𝑥 𝐵 𝑥 𝐶 𝑥
−∞
= 𝐵 𝑝2 𝐶 𝑝3
𝑖
∞
4𝑥
𝑑
−∞
𝑔𝐴 𝑥 𝐵 𝑥 𝐶 𝑥
𝐴 𝑝1
&ograve;
d3 p
2w P ( 2p )
â &times; e-ip&times;x + â+p &times; eip&times;x )
3( p
&ograve;
d3 p'
2w P' ( 2p )
&ograve;
(
3
(
b̂p' &times; e-ip'&times;x + b̂p'+ &times; eip'&times;x
d3 p''
2w P'' ( 2p )
3
(ĉ
p''
&times; e-ip''&times;x + ĉ+p'' &times; eip''&times;x )
)
&raquo; ( â+ â+ ) &times; b̂+ b̂+ &times; ( ĉ+ ĉ+ )
𝐵 𝑝2 𝐶 𝑝3
𝑖
∞
𝑑4𝑥
−∞
âp A( p1 ) = 0 &times; d 3 ( p- p1 )
𝑔𝐴 𝑥 𝐵 𝑥 𝐶 𝑥
𝐴 𝑝1
 
BC  p2  c p2  B
)

B p3  bˆp3  0
&ograve;
d3 p
2w P ( 2p )
-ip&times;x
+
ip&times;x
â
&times;
e
+
â
&times;
e
(
)
p
p
3
&ograve;
d3 p'
2w P' ( 2p )
&ograve;
𝐵 𝑝2 𝐶 𝑝3
𝑖
∞
4
𝑑
𝑥
−∞
3
(
b̂p' &times; e-ip'&times;x + b̂p'+ &times; eip'&times;x
d3 p''
2w P'' ( 2p )
𝑔𝐴 𝑥 𝐵 𝑥 𝐶 𝑥
3
)
(ĉ
-ip''&times;x
+
ip''&times;x
&times;
e
+
ĉ
&times;
e
)
p''
p''
𝐴 𝑝1
+
b̂+p2 ĉ p3
âp1
The remaining numerical factor is:
B
C
ig
A
Momentum Conservation
For a toy ABC model
Three scalar particle with masses mA, mB ,mC
pi
1
External Lines
qi
i
q 2j  m 2j  i
Internal Lines
Lines for each kind of particle with appropriate masses.
C
k2
Vertex
A
k1
-ig
k3
2 4  4 k1  k2  k3 
B
The configuration of the vertex determine the interaction of the model.
A
  a  a
ℒI 𝑥 = 𝑔 𝐴 𝑥 𝐵 𝑥 𝐶 𝑥
 aa C

interaction Lagrangian
B
  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
ℒI 𝑥 = 𝑔 𝐴 𝑥 𝐵 𝑥 𝐶 𝑥
∞
𝑑4𝑥 𝑔 𝐴 𝑥 𝐵 𝑥 𝐶 𝑥
𝑖
−∞
Interaction Lagrangian
B
C
vertex
The Interaction Lagrangian is integrated over the whole spacetime.
Interaction could happen anywhere anytime.
The amplitudes at various spacetime need to be added up.
The integration yields a momentum conservation.
In momentum space, the factor for a vertex is simply a constant.
ℒI = 𝜆𝜙 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!
Propagator
Solve the evolution operator to the second order.
d
U(t, t0 ) =- i H I &times;U(t, t0 )
dt
t
dU (2) (t, t0 )
(1)
= -i H I (t)&times;U (t, t0 ) = -H I (t)&times; &ograve; dt ' [ H I (t ')]
dt
t0
HI is first order.
t ''
&eacute;
&ugrave;
U (t, t0 ) = - &ograve; dt ''&ecirc; H I (t '') &ograve; dt ' [ H I (t ')]&uacute;
&ecirc;&euml;
&uacute;&ucirc;
t0
t0
t
(2)
The integration of two identical interaction Hamiltonian HI.
The first HI is always later than the second HI
t ''
&eacute;
&ugrave;
U (t, t0 ) = - &ograve; dt ''&ecirc; H I (t '') &ograve; dt ' [ H I (t ')]&uacute;
&ecirc;&euml;
&uacute;&ucirc;
t0
t0
t
(2)
The integration of two identical interaction Hamiltonian HI.
The first HI is always later than the second HI
t’ and t’’ are just dummy notations and can be exchanged.
t ''
t'
&eacute;
&ugrave; t &eacute;
&ugrave;&uuml;&iuml;
1 &igrave;&iuml; t
U (t, t0 ) = - &iacute; &ograve; dt ''&ecirc; H I (t '') &ograve; dt ' [ H I (t ')]&uacute; + &ograve; dt '&ecirc;H I (t ') &ograve; dt '' [ H I (t '')]&uacute;&yacute;
2 &iuml;&icirc; t0
&ecirc;&euml;
&uacute;&ucirc; t0 &ecirc;&euml;
&uacute;&ucirc;&iuml;&thorn;
t0
t0
(2)
t’’
t ''
t
t'
&uuml;&iuml;
1 &igrave;&iuml; t
= - &iacute; &ograve; dt '' &ograve; dt ' [ H I (t '')&times; H I (t ')] + &ograve; dt ' &ograve; dt '' [ H I (t ')&times; H I (t '')]&yacute;
2 &iuml;&icirc; t0
&iuml;&thorn;
t0
t0
t0
We are integrating over the whole square but always
keep the first H later in time than the second H.
t’
1 t
U (t, t0 ) = - &times; &ograve; dt ''
2 t0
(2)
t
&ograve; dt ' T [ H (t '') H (t ')]
I
I
t0
T ( A(t1 ) B(t 2 ))   (t1  t 2 ) A(t1 ) B(t 2 )   (t 2  t1 )B(t 2 ) A(t1 )
t
1
U (2) (t, t0 ) = - &times; &ograve; dt ''
2 t0
t’’
t
&ograve; dt ' T [ H (t '') H (t ')]
I
I
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!
t’
1
S(2) = U (2) (&yen;,-&yen;) = - &times; &ograve; d4 x1 d 4 x2 T [ LI (x1 ) LI (x2 )]
2
This notation is so powerful, the whole series of
operator U can be explicitly written:
(n)
U (t, t0
-i )
(
)=
n!
n
t
&times; &ograve; dt1 dt2
dtn T [ H I (t1 ) H I (t2 )
H I (tn )]
t0
The whole series can be summed into an exponential:
&igrave;&iuml; &eacute; t
&ugrave;&uuml;&iuml;
U(t, t0 ) = T &iacute;exp &ecirc;-i &ograve; dt ' H I (t ')&uacute;&yacute;
&uacute;&ucirc;&iuml;&thorn;
&iuml;&icirc; &ecirc;&euml; t0
1
S = U (&yen;, -&yen;) = - &times; &ograve; d4 x1 d 4 x2 T [ LI (x1 ) LI (x2 )]
2
Amplitude for scattering A  A  B  B
(2)
(2)
B(p3 )B(p4 ) S A(p1 )A(p2 )
= B(p3 )B(p4 )
&ograve; d x d x T [ gA(x ) B(x ) C(x ) &times; g A(x ) B(x ) C(x )] A(p )A(p )
4
4
1
2
1
1
1
2
2
2
1
2
  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
Propagator between x1 and x2
p1-p3 pour into C at x2 p2-p4 pour into C at x1
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
B(p3 )B(p4 ) S A(p1 )A(p2 ) = -g2 &ograve; d4 x1 d4 x2 e-i ( p4 -p2 )x1 e
-i ( p3 -p1) x2
B(p4)
B(p3)
B(p4)
B(p3)
x2
B(p4)
x1
C
A(p1)
B(p3)
0 T [ C(x1 ) C(x2 )] 0
x1
A(p2)
x2
A(p1)
C(p1-p3)
C
A(p2)
A(p1)
A(p2)
This construction ensures causality of the process. It is actually the
sum of two possible but exclusive processes.
Again every Interaction is integrated over the whole spacetime.
Interaction could happen anywhere anytime and amplitudes
need superposition.
This propagator looks reasonable in coordinate space but
difficult to calculate and the formula is cumbersome.
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 by definition it should be!
So an even more useful form is obtained by extending the
integration to 4-momentum. And in the momentum space,
it becomes extremely simple:
The Fourier Transform of the propagator is simple.
4
4
-i ( p4 -p2 )x1
d
x
d
x
e
e
&ograve; 1 2
-i ( p3 -p1 ) x2
0 T [ C(x1 ) C(x2 )] 0 =
B(p4)
B(p3)
B(p4)
B(p3)
x2
4
d
(p4 - p2 + p3 - p1 )
2
( p1 - p3 ) - mC
2
B(p3)
B(p4)
x1
C
A(p1)
i
x1
A(p2)
x2
A(p1)
C(p1-p3)
C
A(p2)
A(p1)
A(p2)
0 T   ( x)  ( y )  0
x0  y0
y0 &gt; x0
iw k ( x0 - y0 )
-ik ( y- x)
0 T   ( x)  ( y )  0
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)
For a toy ABC model
qi
Internal Lines
i
q 2j  m 2j  i
Lines for each kind of particle with appropriate masses.
1
S(2) = U (2) (&yen;, -&yen;) = - &times; &ograve; d4 x1 d 4 x2 T [ LI (x1 ) LI (x2 )]
2
Amplitude for scattering A  A  B  B
B(p3 )B(p4 ) S A(p1 )A(p2 )
= B(p3 )B(p4 )
4
4
d
x
d
&ograve; 1 x2 T [ gA(x1 ) B(x1 ) C(x1 ) &times; g A(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
 a  a
 a  a
 a  a



Every field either couple with another field to form a propagator or
annihilate (create) external particles!
Otherwise the amplitude will vanish when a operators hit vacuum!
For a toy ABC model
Three scalar particle with masses mA, mB ,mC
pi
1
External Lines
qi
i
q 2j  m 2j  i
Internal Lines
Lines for each kind of particle with appropriate masses.
C
k2
Vertex
A
k1
-ig
k3
2 4  4 k1  k2  k3 
B
The configuration of the vertex determine the interaction of the model.
Scalar Antiparticle
Assuming that the field operator is a complex number field.
L0 = &para;m F+&para;m F - m2F+F

d3p
( x)  
(2 ) 3
1
2
a e
ipx
p
 b p e ipx

The creation operator b+ in a complex KG field can create a different particle!
H=
P=
Q=
&ograve;
d3 p
(2p )3
&ograve;
d3 p
+
+
p&times;
a
a
+
b
b
(
p p
p p)
(2p )3
&ograve;
d3 p
+
+
a
a
b
b
(
p p
p p)
(2p )3
p + m2 &times; ( apa+p + bpbp+ )
2
The particle b+ create has the same mass but opposite charge.
b+ create an antiparticle.

d3p
( x)  
(2 ) 3
1
2
a e
ipx
p
 b p e ipx

Complex KG field can either annihilate a particle or create an antiparticle!

d3p
 ( x)  
(2 ) 3

1
2
b e
p
ipx
 a p e ipx

Its conjugate either annihilate an antiparticle or create a particle!
The charge difference a field operator generates is always the same!
So we can add an arrow of the charge flow to every leg that corresponds
to a field operator in the vertex.
LI = gF3 + gF+3

d3p
( x)  
(2 ) 3
1
2
a e
p
ipx

p
b e
ipx
incoming particle or
outgoing antiparticle


d3p
 ( x)  
(2 ) 3

1
2
b e
p
ipx
 a p e ipx
incoming antiparticle or
outgoing particle
charge non-conserving interaction

LI = l &times; (F F)
+
2
incoming antiparticle or
outgoing particle
incoming particle or
outgoing antiparticle
charge conserving interaction
f &raquo; c+ c+
LI = gf Y*Y
Y &raquo; b+ a+
incoming antiparticle or
outgoing particle
Y &raquo; a+ b+
incoming particle or
outgoing antiparticle
Y &raquo; a+ b+ can either annihilate a particle or create an antiparticle!
Y &raquo; b+ a+ can either annihilate an antiparticle or create a particle!
U(1) Abelian Symmetry
L0 = &para;m F+&para;m F - m2F+F
The Lagrangian is invariant under the field phase transformation
( x)  e iQ ( x)
       eiQ  eiQ  (x)    
LI = l &times; (F F)
+
2
LI = gF3 + gF+3
invariant
is not invariant
U(1) symmetric interactions correspond to charge conserving vertices.
If A,B,C become complex, they all carry charges!
C
The interaction is invariant only if QA  QB  QC  0
The vertex is charge conserving.
A
B
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 = l &times; B( A+ A)
B(p4)
B(p3)
B(p4)
B(p3)
x2
x1
C
A(p1)
x1
A(p2)
x2
A(p1)
B(p3)
B(p4)
B( A+ A)
B( A+ A)
C(p1-p3)
C
A(p2)
A(p1)
A(p2)
For a toy charged AAB model
Three scalar charged particle with masses mA, mB
pi
1
External Lines
qi
i
q 2j  m 2j  i
Internal Lines
Lines for each kind of particle with appropriate masses.
B
k2
Vertex
-iλ
2 4  4 k1  k2  k3 
A
A
k1
k3
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
LI  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
```