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Time finally…………..
粒子的狀態會隨時間而演化!
Ket States S  evolve with time, but not the operators 
測量期望值會隨時間變化:   = S   S 
Schrodinger Picture
Hamiltonian is the generator of time evolution!

 
 S
= −  ∙ S 
Schrodinger Equation
It is the default choice in wave mechanics.
狀態
波函數
Y(x, t) = x Y(t)
波函數隨時間而演化!
測量
算子
Oˆ
xˆ  x
 

pˆ    i 


x


這些算子與時間無關。
 2 2


i
 Hˆ    

V
(
x
)
2

t
 2 m x


Schrodinger Equation 就是Schrodinger Wave Equation。

 
 S
= −  ∙ S 
猜測:Schrodinger Eq. 的解可以寫成指數函數乘起始態(猶如  是一個常數)
S 
=  − ∙ S 0
≡   ∙ S 0
Evolution Operator   : the operator to move states from  = 0 to .
將算子的指數函數以展開的數列來定義(again 猶如是一個常數)
 =  −
1
= 1 −  + −
2
1
2+
−
3!
∞
3
+⋯=
=0
1
−
!

因為此式中只有一個算子  ,與自己當然commute,因此與數無異。
此算子指數函數的微分與一般純數指數函數無異,由此可得到:


  =  − = − − = − 


這是Evolution Operator   必須滿足的方程式!
這正是 U 應該必須遵守的.
若是:

  = − 

S 
=   ∙ S 0
Take time derivative on both sides and plug in the first Eq.:

 
 S
=

  S 0

= −   S 0
= − S 
We recover Schrodinger Equation. 嚴格證明了前一頁的猜測是對的。
能量的本徵態

 

= −  ∙  
 =
= − ∙  
這個方程式立刻可以解出來:
 
=  − ∙  0
能量的本徵態幾乎不隨時間演化,它的演化就只是乘上一個相位 phase!
如此其任何測量期望值都不隨時間變化:
  ∙ ∙   =  0 ∙   ∙ ∙  − ∙  0
=  0 ∙ ∙  0
因此能量的本徵態稱為定態 Stationary State
In quantum mechanics, only expectation values are observable.
For the same evolving expectation value, we can instead ask operators to evolve.
  = S  ∙ ∙ S  = S 0 ∙  ∙  ∙  − ∙ S 0
= S 0 ∙   ∙  ∙  − ∙ S 0 ≡ S 0 ∙  ∙ S 0
Heisenberg Picture
Now the states do not evolve.
 ≡  S 0
We move the time evolution to the operators:
  ≡   ∙  ∙  −
Heisenberg Picture is totally equivalent to Schrodinger Picture!
如同古典物理,物理量會隨時間變化!
  ≡   ∙  ∙  −
The time-dependent operators satisfy Equations of Motion.

 
 
=  ∙    ∙  − +   ∙  ∙  − ∙
Heisenberg Equation

  =  , 
 
The rate of change of operators equals their commutators with H.
量子物理可以清楚平順地過渡到古典物理。
以Poisson Bracket取代Commutator即可。
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.
H  = H ,  =   ∙ S  ∙  −
H ()
=  , H ()

For KG field without interaction:
Combine the two equations

2
m
2

 0
The field operators in Heisenberg Picture satisfy KG Equation.
The field operators in Heisenberg Picture satisfy KG Equation.
Schrodinger Picture中的場的傅立葉分析:
  =
3
1
 ∙
− ∙ †


+



2 3 2
每一個傅立葉分量 在Heisenberg Picture就像一個頻率為ω的簡諧震盪子。
 ()
=  ,  = − 

 ,  =
3
1
 ∙  −   +  − ∙     †



2 3 2
 ,  =
3
1
−∙  +  ∙  †



2 3 2
0 =
Interaction finally…………..
ℒKG Free
1
1 2 2

=     −  
2
2
Free scalar particle
Now add interactions to Lagrangian: Called Interaction Lagrangian
ℒInt  
meaning anything not quadratic in fields.
For example, we can add
ℒInt
to the free Lagrangian ℒKG Free
=  4  +  ∙      
1
1 2 2

ℒ =     −   +  4 +  ∙ 
2
2
These terms will add non-linear terms to the KG equation:
  + 2   − 4 3  −     = 0
Superposition Law is broken by these nonlinear terms.
Travelling waves will interact with each other.
Interactions Lagrangian 對Hamiltonian 有貢獻
Interaction Hamiltonian:
H  H 0  H int
ℋInt  
= −ℒInt  
There is a natural separation between free and interaction Hamiltonians:
  = S  ∙  ∙ S 
= S 0 ∙  ∙  ∙  − ∙ S 0
我們可以選擇只要移動一部分的時間演化到算子上,留一部分在狀態上!
只移動0 所產生的時間演化到算子上
= S   −0 ∙  0 ∙  ∙  −0 ∙  0 S 
I  =  0 ∙ S ∙  −0
I 
≡  I  ∙   ∙  I 
=  0 S  ~   S 0
其餘由 int 所產生的的時間演化留在狀態上!
States and Operators both evolve with time in interaction picture:
Interaction picture (Half way between Schrodinger and Heisenberg)
Evolution of Operators in Interaction Picture
I  =  0 ∙ S ∙  −0
I
=  0 , I

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:
I
=  0 , I

Field operators are free, as if there is no interaction!
 ,  =
3
1
−∙  +  ∙  †



2 3 2
0 =
The Fourier expansion of a free field is still valid.
Evolution of States in Interaction Picture
I 

 
 I
= 0 I 
=  0 S 
−  0 0 + int S S 
=  0 ∙ int S ∙  −0 S 
int I =  0 ∙ int S ∙  −0 
Hint I is just the interaction Hamiltonian Hint in interaction picture!
The field operators in Hint I are free.

 
 I
= −int I I 
States evolve like in the Schrodinger picture but with full H replaced by Hint I.
Interaction Picture ~1949
Operators evolve just like in the Heisenberg picture but with
the full Hamiltonian replaced by the free Hamiltonian
I
=  0 , I

States evolve like in the Schrodinger picture but with the full
Hamiltonian replaced by the interaction Hamiltonian.

 
 I
= −int I I 
Freeman Dyson, 1923
Now! Action!
關鍵主角是Interaction Picture中的時間演化算子 , 0
The evolution of states in the interaction picture

 
 I
= −int I I 
Again define  as the evolution operator of states like in Schrodinger Picture:
I 
=  , 0 I 0
Evolution Operator   : the operator to move states from  = 0 to .
Take time derivative on both sides and plug in the first Eq.:

 
 I
=

 , 0 I 0

= −I  , 0 I 0
 , 0 has its own equation of motion, just like in Schrodinger Picture:

 , 0 = −I  , 0

All the problems can be answered if we are able to calculate this operator

 , 00  is the state in the future t that evolved from a state  in t0
 ∞,

∞, −∞
−∞  is the state in long future that evolves from a state  in long past.
The amplitude for  ∞, −∞  to be measured as a state  is their inner product:
 ∙  ∞, −∞

 =   ∞, −∞ 
 ∞, −∞ 

Detector detects
=∞
 ∞, −∞
 = −∞
The amplitude for this whole sequence equals  
−∞  .
 ∞, −∞
The transition amplitude for the decay of A:  →  + 
the amplitude of  in the long future which evolves from  in the long past.
  ∞, −∞ 
The transition amplitude for the scattering of A:  +  →  + 
  ∞, −∞ 
The amplitude is the corresponding matrix element of the operator  ∞, −∞ .
 ≡  ∞, −∞
S operator is the key object in particle physics!
S operator can not be solved exactly.
But it can be calculated in a perturbative expansion.
Perturbation expansion of U and S

 , 0 = − I ∙  , 0
例如 ℒ Int =  4  +  ∙      

Solve it by a perturbation expansion in small parameters like ,  in HI.
 =  (0) +  (1) + ⋯
 () ∝  or 
 (0) =   = 0 = 1
無交互作用時,粒子狀態不變。
To leading order in ,  :
 (1)

, 0 = − I ∙ 

 (1) , 0 = −

0
0
 ′ I ′
, 0 = −I 
Born Approximation
∞
 (1) ∞, −∞ =  (1) = −
′ I ′
−∞
S operator 在領先項等於Interaction Hamiltonian I 對時間的積分!
To leading order, S matrix equals
∞
 (1) = −
∞
 I  = −
−∞
∞

 3  ℋI ,  = 
−∞
 4  ℒI 
−∞
the spacetime integration of interaction Lagrangian
It is Lorentz invariant if the interaction Lagrangian is invariant.
Vertex
In ABC model, every particle corresponds to a field:
 →   ≡   etc
Add the following interaction term in the Lagrangian:
ℒI  =       
The transition amplitude for A decay:  1 →  2 +  3 can be computed as :
  ∞, −∞  =   
To leading order:
∞
∞
4  ℒI  = 
=
−∞
 2  3   1
4       
−∞
=  2  3

∞
4

−∞
     
 1
Aˆ 

3 
d p
2 P 2

3
aˆ

p
e
 ip  x

ip  x
 aˆ p  e

Bˆ 

3 
d p'
2 P ' 2
Cˆ 
(

3
bˆ

p'
e
3 
d p''

2  P ''  2 

3
 ip '  x
cˆ

p ''

ip '  x
 bˆ p '  e
e
 ip ' '  x


ip ' '  x
 cˆ p ''  e

)
» ( aˆ + aˆ + ) × bˆ + bˆ+ × ( cˆ + cˆ+ )
 2  3

∞
4 
−∞
aˆ p1
     
bˆ+p2

aˆ p A  p 1   0  
3

p 

p1 
cˆ+p3


c p 2 C  p 2   0


BC  p 2  c p '  B  
~ 0 0 ~1
 1

bˆ p 3 b  p 3   0
3


 p ' p 2 
其他的項都是零!


B  p 3  bˆ p ''  0  
3


 p ' ' p 3 
ò
d3 p
2w P ( 2p )
-ip×x
+
ip×x
ˆ
ˆ
a
×
e
+
a
×
e
(
)
p
p
3
ò
d 3 p'
2w P' ( 2p )
ò
 2  3

∞
4


−∞
     
aˆ p1
3
(
bˆp' × e-ip'×x + bˆp'+ × eip'×x
d 3 p''
2w P'' ( 2p )
3
(cˆ
)
-ip''×x
+
ip''×x
ˆ
×
e
+
c
×
e
)
p''
p''
 1
+
bˆ+p2 cˆ p3
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
Lines for each kind of particle with appropriate masses.
C
k2
Vertex
A
k1
-ig
k3
 2  4  4  k 1  k 2  k 3 
B
The configuration of the vertex determine the interaction of the model.
A   a  a
ℒI  =       
 aa

Interaction Lagrangian
B  aa
C
vertex
Every field operator in ℒI 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  =       
∞
4       

−∞
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 =  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.

 , 0 = − I ∙  , 0

 (2)

, 0 = − I ∙ 

1

, 0 = −I  ∙
0
 ′ I ′
Remember HI is first order.

 (2)
, 0 = −
′′
′′ I ′′ ∙
0
0
 ′ I ′
The integration of two identical interaction Hamiltonian HI.
But the first HI is always later than the second HI

 (2) , 0 = −
′′
 ′′
0
 ′ I ′′ ∙ I ′
 ′′ > ′
0
在 ′ − ′′平面上積分範圍是上方三角形,在此範圍內,
積分式中時間在後的I 永遠在左方,時間在前的I 在右方。
But t’ and t’’ are just dummy notations and can be exchanged.

(2)
, 0
1
=−
2

′′
 ′′
0

 ′ I ′′ ∙ I ′ +
0
 ′′ > ′
′
 ′
0
 ′′ I ′ ∙ I ′′
0
 ′ > ′’
第二項可以看成積分範圍是下方三角形。兩項合起來就是整個正方形
但積分式中時間在後的I 依舊放在左方。時間在前的I 放在右方
t’’
1
=−
2
t’


 ′′
0
 ′ I ′′ ∙ I ′
0
時間在後的I
永遠放在左方
−
1
2


 ′′
0
 ′ I ′′ ∙ I ′
0
時間在後的I永遠放在左方
定義 time order product T
  1 ∙  2
=  1 − 2  1 ∙  2 +  2 − 1  2 ∙  1
t’’
 (2) , 0 = −
t’
1
2


 ′′
0
 ′  I ′′ ∙ I ′
0
 (2) , 0 = −
1
2


 ′′
0
 (2) =  (2) ∞, −∞ = −
 ′  I ′′ ∙ I ′
0
1
2
∞
∞
 4 1
−∞
 4 2  ℒI 1 ∙ ℒI 2
−∞
This is Lorentz invariant since time ordering is invariant!
This notation is so powerful, the whole series of
operator U can be explicitly written:
 () , 0 =
−
!


1 2 ⋯   I 1 ∙ I 1 ⋯ I 
0
The whole series can be summed into an exponential:
∞
− 
!
 , 0 =
=1
∞
=
=1
∞
=
=1
− 

!
−

!


1 2 ⋯   I 1 ∙ I 1 ⋯ I 
0
1 2 ⋯  I 1 ∙ I 1 ⋯ I 
0



∙
 I 
0

=  exp −
 I 
0
 (2) =  (2) ∞, −∞ = −
1
2
∞
∞
 4 1
−∞
 4 2  ℒI 1 ∙ ℒI 2
−∞
Amplitude for scattering A  A  B  B
B(p3 )B(p4 ) S A(p1 )A(p2 )
 4∙1
 B ( p3 ) B ( p4 )
 3∙2
 d x1 d x 2 T  g A ( x1 ) B ( x1 ) C ( x1 ) g A ( x 2 ) B ( x 2 ) C ( x 2 )  A ( p1 ) A ( p 2 )
4
4
 −1 ∙2

 d x1 d x 2 e
4
4
 i ( p1  p 3 ) x 2
e
 i ( p 2  p 4 ) x1
 −2 ∙1
 0 T  C ( x1 ) C ( x 2 )  0
Fourier Transformation
Propagator between x1 and x2
p1-p3 pour into C at x2 p2-p4 pour into C at x1
Why is it called propagator?
t1  t 2
0 T C ( x1 )  C ( x 2 )
0

 0 C ( x1 )  C ( x 2 ) 0  0 a  a

  a  a 

x1
A particle is created at x2 and later annihilated at x1.
It is indeed a propagator?
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  t 2
0 T C ( x1 )  C ( x 2 )

0
 0 C ( x 2 )  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 ò d 4 x1 d 4 x2 e-i( p4 -p2 )x1 e
B(p4)
B(p3)
B(p4)
B(p3)
x2
B(p3)
0 T [ C(x1 ) C(x2 )] 0
B(p4)
x1
C
A(p1)
-i( p3 -p1) x2
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.
現在來計算這個 propagator!
0    ∙ 
0
0    ∙ 
0
0    ∙ 
0
0    ∙ 
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  3  to 4-momentum 4  =  0 ∙  3 .
With a Lorentz invariant integration, it becomes extremely simple.
我們要證明:
0    ∙ 
0
為了證明兩者相等,我們將第二式的 0 積分完成。
 0 的積分計算最簡單的方法是在複數 0 的平面上作。
在複數 0 的平面上, 0 的積分是沿實數軸,實數軸上下,積分式各有一個 pole。
 2 − 2 +  = 0

02
−

2
+ 2 +  = 0
 02 − 2 +  = 0
 0 = ± ∓ 
在複數平面上沿封閉曲線作積分等於曲線所包圍的pole的residue
x0  y 0
將積分路徑沿下方半圓回到實數軸
半圓上的積分為零,因此實數軸積分即等於下方pole的residue
y0 > x0 將積分路徑沿上方半圓回到實數軸
半圓上的積分為零,因此實數軸積分即等於上方pole的residue
iw k ( x 0 - y 0 )
-ik ( y - x )
0    ∙ 
0
x2
x1
C
x1
x2
1
C
C
The Fourier Transform of the propagator is simple.
 4 1 ∙  4 2 ∙  1∙1 ∙  2∙2 0    ∙  
0 =  4 1 − 2 ∙

12 − 2 + 
The Fourier Transform of the propagator is simple.
4
4
-i( p4 -p2 )x1
d
x
d
x
e
e
ò 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)
(1 − 3)
C
A(p2)
A(p1)
A(p2)
For a toy ABC model
qi
Internal Lines
i
q j  m j  i
2
2
Lines for each kind of particle with appropriate masses.
 (2) =  (2) ∞, −∞ = −
1
2
∞
∞
 4 1
−∞
 4 2  ℒI 1 ∙ ℒ I 2
−∞
Amplitude for scattering A  A  B  B
B(p3 )B(p4 ) S A(p1 )A(p2 )
4
4
d
x
d
ò 1 x2 T [ g A(x1 ) B(x1 ) C(x1 ) × g A(x2 ) B(x2 ) C(x2 )] A(p1 )A(p2 )
= B(p3 )B(p4 )

 d x1 d x 2 e
4
4
 i ( p1  p 3 ) x 2
e
 i ( p 2  p 4 ) x1
 0 T  C ( x1 ) C ( x 2 )  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
Internal Lines
q j  m j  i
2
2
Lines for each kind of particle with appropriate masses.
C
k2
Vertex
A
k1
-ig
k3
 2  4  4  k 1  k 2  k 3 
B
The configuration of the vertex determine the interaction of the model.
希格斯粒子的產生與衰變
51
∙  ∙   
將滿足規範對稱的動能項展開來:
 Φ†  Φ =  Φ†  Φ +  Φ†  Φ −   Φ† ∙ Φ +  2   Φ† Φ
 Φ†  Φ −   Φ† ∙ Φ
 2   Φ† Φ
希格斯粒子的產生與衰變
Vector Boson Fusion VBF
希格斯粒子的衰變
即使沒有 Higgs 粒子,質子質子碰撞,很容易就可以產生兩個夸克!
在兩個夸克的衰變態觀察,背景 B 會淹沒了信號 S
但希格斯粒子的發現是靠兩個光子的衰變(0.23%) 。
兩個光子的衰變,信號比起背景還算有觀測可能,兩者比約是1: 4
60
由一個粒子衰變產生的兩個光子,信號其特徵就是一個隆起 resonance
CMS 信號
63
Scalar Antiparticle
Assuming that the field operator is a complex number field.
L0 = ¶m F+¶m F - m2F+F
 ( x) 
3 
d p
 ( 2 )
1
3
2
a
p
e
 ipx

 bpe
ipx

The creation operator b+ in a complex KG field can create a different particle!
H=
P=
Q=
ò
d3 p
(2p )3
ò
d3p
+
+
p
×
a
a
+
b
b
(
p p
p p)
(2p )3
ò
d3p
+
+
a
a
b
b
(
p p
p p)
(2p )3
p + m 2 × ( a p a+p + bp bp+ )
2
The particle b+ create has the same mass but opposite charge.
b+ create an antiparticle.
 ( x) 
3 
d p
 ( 2 )
1
2
3
a
e
p
 ipx

 bpe
ipx

Complex KG field can either annihilate a particle or create an antiparticle!

 ( x) 
3 
d p
 ( 2 )
1
3
2
b
e
p
 ipx

 a pe
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
 ( x) 
3 
d p
 ( 2 )
1
3
2
a
p
e
 ipx

p
b e
ipx

incoming particle or
outgoing antiparticle

 ( x) 
3 
d p
 ( 2 )
1
3
2
b
p
e
 ipx
incoming antiparticle or
outgoing particle
charge non-conserving interaction

 a pe
ipx

LI = l × (F F)
+
2
incoming antiparticle or
outgoing particle
incoming particle or
outgoing antiparticle
charge conserving interaction
f » c + c+
LI = g f Y*Y
Y » b + a+
incoming antiparticle or
outgoing particle
Y » a + b+
incoming particle or
outgoing antiparticle
Y » a + b+ can either annihilate a particle or create an antiparticle!
Y » b + a+ can either annihilate an antiparticle or create a particle!
U(1) Abelian Symmetry
L0 = ¶m F+¶m F - m2F+F
The Lagrangian is invariant under the field phase transformation
 ( x)  e


    e
iQ 
 iQ 
e
 ( x)
 iQ 
LI = l × (F F)
+

  ( x)   
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
The vertex is charge conserving.
Q A  Q B  QC  0
A
B
Propagator:



0 T  ( x1 )  ( x 2 ) 0
t1  t 2




0 T  ( x1 )   ( x 2 ) 0  0  ( x1 )   ( x 2 ) 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  t 2





0 T  ( x1 )   ( x 2 ) 0  0  ( x 2 )   ( 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
 d x1 d x 2 e
4
4
 ipx 1
e
 iqx 2
B(p4)
B(p3)

B(p4)
B(p3)
x2
i
q  mC
B(p3)
2
x1
A(p2)
x2
A(p1)
2
B(p4)
C(p1-p3)
C
A(p2)
A(p1)
 ( p  q)
4
x1
C
A(p1)


0 T  ( x1 )  ( x 2 ) 0 
A(p2)
LI = l × 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
Internal Lines
q j  m j  i
2
2
Lines for each kind of particle with appropriate masses.
B
k2
Vertex
-iλ
 2  4  4  k 1  k 2  k 3 
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

i 


L



   




      i     m     i     m  


   m   0    i    m  0
i 


  m   0
Negative energy!
  i     m   0

Anti-commutator!
A creation operator!
~
b  b,
~

b b
b annihilate an antiparticle!
a


p
,a


p
 0  a




p

p 0
a p a p  0
a
Exclusion Principle


p


p


p
a a a


p
L I  g 
 ba
  ab


External line
When Dirac operators annihilate states, they leave behind a u or v !
 ( x) 
 a  u  e
 ipx

 b v e

p

a p ' p 
 ( x) 
2 p  2
 b  v  e

p


3   p  p '  
 ipx

ipx

0
 a u e

e ( p1 )
u p1 0
Feynman Rules for an incoming particle
ipx


e ( p1 )
v p1 0
Feynman Rules for an incoming antiparticle

A (x)  a 


 a 

L I  gA   



g 
 ba
u p2


  ab
u p1

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