Time finally…………..
粒子的狀態會隨時間而演化！
Ket States 𝜓S 𝑡 evolve with time, but not the operators 𝐴
測量期望值會隨時間變化： 𝐴 𝑡 = 𝜓S 𝑡 𝐴 𝜓S 𝑡
Schrodinger Picture
Hamiltonian is the generator of time evolution!
𝑑
𝜓 𝑡
𝑑𝑡 S
= −𝑖 𝐻 ∙ 𝜓S 𝑡
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。
𝑑
𝜓 𝑡
𝑑𝑡 S
= −𝑖 𝐻 ∙ 𝜓S 𝑡
猜測：Schrodinger Eq. 的解可以寫成指數函數乘起始態（猶如 𝐻 是一個常數）
𝜓S 𝑡
= 𝑒 −𝑖𝐻𝑡 ∙ 𝜓S 0
≡ 𝑈 𝑡 ∙ 𝜓S 0
Evolution Operator 𝑈 𝑡 : the operator to move states from 𝑡 = 0 to 𝑡.
將算子的指數函數以展開的數列來定義（again 猶如𝐻是一個常數）
𝑈 = 𝑒 −𝑖𝐻𝑡
1
= 1 − 𝑖𝐻𝑡 + −𝑖𝐻𝑡
2
1
2+
−𝑖𝐻𝑡
3!
∞
3
+⋯=
𝑛=0
1
−𝑖𝐻𝑡
𝑛!
𝑛
因為此式中只有一個算子 𝐻 ，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：
𝑑
𝑑
𝑈 𝑡 = 𝑒 −𝑖𝐻𝑡 = −𝑖𝐻𝑒 −𝑖𝐻𝑡 = −𝑖𝐻𝑈 𝑡
𝑑𝑡
𝑑𝑡
這是Evolution Operator 𝑈 𝑡 必須滿足的方程式！
這正是 U 應該必須遵守的.
若是：
𝑑
𝑈 𝑡 = −𝑖𝐻𝑈 𝑡
𝑑𝑡
𝜓S 𝑡
= 𝑈 𝑡 ∙ 𝜓S 0
Take time derivative on both sides and plug in the first Eq.:
𝑑
𝜓 𝑡
𝑑𝑡 S
=
𝑑
𝑈 𝑡 𝜓S 0
𝑑𝑡
= −𝑖𝐻 𝑈 𝑡 𝜓S 0
= −𝑖𝐻 𝜓S 𝑡
We recover Schrodinger Equation. 嚴格證明了前一頁的猜測是對的。
能量的本徵態
𝑑
𝐸 𝑡
𝑑𝑡
= −𝑖 𝐻 ∙ 𝐸 𝑡
𝐻𝐸 =𝐸𝐸
= −𝑖𝐸 ∙ 𝐸 𝑡
這個方程式立刻可以解出來：
𝐸 𝑡
= 𝑒 −𝑖𝐸𝑡 ∙ 𝐸 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：
𝐸 𝑡 ∙𝐴 ∙ 𝐸 𝑡 = 𝐸 0 ∙ 𝑒 𝑖𝐸𝑡 ∙𝐴 ∙ 𝑒 −𝑖𝐸𝑡 ∙ 𝐸 0
= 𝐸 0 ∙𝐴 ∙ 𝐸 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.
𝐴 𝑡 = 𝜓S 𝑡 ∙𝐴 ∙ 𝜓S 𝑡 = 𝜓S 0 ∙𝑒 𝑖𝐻𝑡 ∙ 𝐴 ∙ 𝑒 −𝑖𝐻𝑡 ∙ 𝜓S 0
= 𝜓S 0 ∙ 𝑒 𝑖𝐻𝑡 ∙ 𝐴 ∙ 𝑒 −𝑖𝐻𝑡 ∙ 𝜓S 0 ≡ 𝜓S 0 ∙𝐴 𝑡 ∙ 𝜓S 0
Heisenberg Picture
Now the states do not evolve.
𝜓𝐻 ≡ 𝜓 S 0
We move the time evolution to the operators:
𝐴𝐻 𝑡 ≡ 𝑒 𝑖𝐻𝑡 ∙ 𝐴 ∙ 𝑒 −𝑖𝐻𝑡
Heisenberg Picture is totally equivalent to Schrodinger Picture!
如同古典物理，物理量會隨時間變化！
𝐴𝐻 𝑡 ≡ 𝑒 𝑖𝐻𝑡 ∙ 𝐴 ∙ 𝑒 −𝑖𝐻𝑡
The time-dependent operators satisfy Equations of Motion.
𝑑
𝐴 𝑡
𝑑𝑡 𝐻
= 𝑖𝐻 ∙ 𝑒 𝑖𝐻𝑡 𝐴 ∙ 𝑒 −𝑖𝐻𝑡 + 𝑒 𝑖𝐻𝑡 ∙ 𝐴 ∙ 𝑒 −𝑖𝐻𝑡 ∙
Heisenberg Equation
𝑑
𝐴 𝑡 = 𝑖 𝐻, 𝐴𝐻
𝑑𝑡 𝐻
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.
𝜙H 𝑥 = 𝜙H 𝑥, 𝑡 = 𝑒 𝑖𝐻𝑡 ∙ 𝜙S 𝑥 ∙ 𝑒 −𝑖𝐻𝑡
𝜕𝜙H (𝑥)
= 𝑖 𝐻, 𝜙H (𝑥)
𝜕𝑡
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𝜙的傅立葉分析：
𝜙𝑆 𝑥 =
𝑑3𝑝
1
𝑖 𝑝∙𝑥
−𝑖 𝑝∙𝑥 †
𝑒
𝑎
+
𝑒
𝑎𝑝
𝑝
2𝜋 3 2𝜔𝑝
每一個傅立葉分量𝑎𝑝 在Heisenberg Picture就像一個頻率為ω的簡諧震盪子。
𝜕𝑎𝑝 (𝑡)
= 𝑖 𝐻, 𝑎𝑝 = −𝜔𝑝 𝑎𝑝
𝜕𝑡
𝜙𝐻 𝑥, 𝑡 =
𝑑3𝑝
1
𝑖 𝑝∙𝑥 𝑒 −𝑖𝜔𝑝 𝑡 𝑎 + 𝑒 −𝑖 𝑝∙𝑥 𝑒 𝑖𝜔𝑝 𝑡 𝑎 †
𝑒
𝑝
𝑝
2𝜋 3 2𝜔𝑝
𝜙𝐻 𝑥, 𝑡 =
𝑑3𝑝
1
−𝑖𝑝∙𝑥 𝑎 + 𝑒 𝑖𝑝∙𝑥 𝑎 †
𝑒
𝑝
𝑝
2𝜋 3 2𝜔𝑝
𝑝0 =𝜔𝑝
Interaction finally…………..
ℒKG Free
1
1 2 2
𝜇
= 𝜕𝜇 𝜙 𝜕 𝜙 − 𝑚 𝜙
2
2
Free scalar particle
Now add interactions to Lagrangian: Called Interaction Lagrangian
ℒInt 𝜙 𝑥
meaning anything not quadratic in fields.
For example, we can add
ℒInt
to the free Lagrangian ℒKG Free
= 𝜆𝜙 4 𝑥 + 𝑔 ∙ 𝜓 𝑥 𝜓 𝑥 𝜙 𝑥
1
1 2 2
𝜇
ℒ = 𝜕𝜇 𝜙 𝜕 𝜙 − 𝑚 𝜙 + 𝜆𝜙 4 + 𝑔 ∙ 𝜓𝜓𝜙
2
2
These terms will add non-linear terms to the KG equation:
𝜕𝜇 𝜕𝜇 + 𝑚2 𝜙 𝑥 − 4𝜙 3 𝑥 − 𝑔𝜓 𝑥 𝜓 𝑥 = 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 𝜙 𝑥
= −ℒInt 𝜙 𝑥
There is a natural separation between free and interaction Hamiltonians:
𝐴 𝑡 = 𝜓S 𝑡 ∙ 𝐴 ∙ 𝜓S 𝑡
= 𝜓S 0 ∙𝑒 𝑖𝐻𝑡 ∙ 𝐴 ∙ 𝑒 −𝑖𝐻𝑡 ∙ 𝜓S 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𝐻0 所產生的時間演化到算子上
= 𝜓S 𝑡 𝑒 −𝑖𝐻0𝑡 ∙ 𝑒 𝑖𝐻0𝑡 ∙ 𝐴 ∙ 𝑒 −𝑖𝐻0𝑡 ∙ 𝑒 𝑖𝐻0𝑡 𝜓S 𝑡
𝐴I 𝑡 = 𝑒 𝑖𝐻0𝑡 ∙ 𝐴S ∙ 𝑒 −𝑖𝐻0𝑡
𝜓I 𝑡
≡ 𝜓 I 𝑡 ∙ 𝐴𝐼 𝑡 ∙ 𝜓 I 𝑡
= 𝑒 𝑖𝐻0𝑡 𝜓S 𝑡 ~𝑒 𝑖𝐻𝑖𝑛𝑡 𝑡 𝜓S 0
其餘由 𝐻int 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
𝐴I 𝑡 = 𝑒 𝑖𝐻0𝑡 ∙ 𝐴S ∙ 𝑒 −𝑖𝐻0𝑡
𝑑𝐴I
= 𝑖 𝐻0 , 𝐴I
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!
𝜙 𝑥, 𝑡 =
𝑑3𝑝
1
−𝑖𝑝∙𝑥 𝑎 + 𝑒 𝑖𝑝∙𝑥 𝑎 †
𝑒
𝑝
𝑝
2𝜋 3 2𝜔𝑝
𝑝0 =𝜔𝑝
The Fourier expansion of a free field is still valid.
Evolution of States in Interaction Picture
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!
The field operators in Hint I are free.
𝑑
𝜓 𝑡
𝑑𝑡 I
= −𝑖𝐻int I 𝜓I 𝑡
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
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
Freeman Dyson, 1923
Now! Action!
關鍵主角是Interaction Picture中的時間演化算子𝑈 𝑡, 𝑡0
The evolution of states in the interaction picture
𝑑
𝜓 𝑡
𝑑𝑡 I
= −𝑖𝐻int I 𝜓I 𝑡
Again define 𝑈 as the evolution operator of states like in Schrodinger Picture:
𝜓I 𝑡
= 𝑈 𝑡, 𝑡0 𝜓I 𝑡0
Evolution Operator 𝑈 𝑡 : the operator to move states from 𝑡 = 0 to 𝑡.
Take time derivative on both sides and plug in the first Eq.:
𝑑
𝜓 𝑡
𝑑𝑡 I
=
𝑑
𝑈 𝑡, 𝑡0 𝜓I 𝑡0
𝑑𝑡
= −𝑖𝐻I 𝑈 𝑡, 𝑡0 𝜓I 𝑡0
𝑈 𝑡, 𝑡0 has its own equation of motion, just like in Schrodinger Picture:
𝑑
𝑈 𝑡, 𝑡0 = −𝑖𝐻I 𝑈 𝑡, 𝑡0
𝑑𝑡
All the problems can be answered if we are able to calculate this operator
𝑈
𝑈 𝑡, 𝑡00 𝜓𝜓 is the state in the future t that evolved from a state 𝜓 in t0
𝑈 ∞,
𝑈
∞, −∞
−∞ 𝜓𝜓 is the state in long future that evolves from a state 𝜓 in long past.
The amplitude for 𝑈 ∞, −∞ 𝜓 to be measured as a state 𝜙 is their inner product:
𝜙 ∙ 𝑈 ∞, −∞
𝜙
𝜓 = 𝜙 𝑈 ∞, −∞ 𝜓
𝑈 ∞, −∞ 𝜓
𝜓
Detector detects
𝑡=∞
𝑈 ∞, −∞
𝑡 = −∞
The amplitude for this whole sequence equals 𝜙𝜙 𝑈
−∞ 𝜓𝜓 .
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
The amplitude is the corresponding matrix element of the operator 𝑈 ∞, −∞ .
𝑆 ≡ 𝑈 ∞, −∞
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
𝑑
𝑈 𝑡, 𝑡0 = −𝑖 𝐻I ∙ 𝑈 𝑡, 𝑡0
例如 ℒ Int = 𝜆𝜙 4 𝑥 + 𝑔 ∙ 𝜓 𝑥 𝜓 𝑥 𝜙 𝑥
𝑑𝑡
Solve it by a perturbation expansion in small parameters like 𝜆, 𝑔 in HI.
𝑈 = 𝑈 (0) + 𝑈 (1) + ⋯
𝑈 (𝑛) ∝ 𝜆𝑛 or 𝑔𝑛
𝑈 (0) = 𝑈 𝜆 = 0 = 1
無交互作用時，粒子狀態不變。
To leading order in 𝜆, 𝑔 :
𝑑 (1)
𝑈
𝑡, 𝑡0 = −𝑖 𝐻I ∙ 𝑈
𝑑𝑡
𝑈 (1) 𝑡, 𝑡0 = −𝑖
𝑡
𝑡0
0
𝑑𝑡 ′ 𝐻I 𝑡′
𝑡, 𝑡0 = −𝑖𝐻I 𝑡
Born Approximation
∞
𝑈 (1) ∞, −∞ = 𝑆 (1) = −𝑖
𝑑𝑡′ 𝐻I 𝑡′
−∞
S operator 在領先項等於Interaction Hamiltonian 𝐻I 對時間的積分！
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
−∞
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:
𝐴 → 𝜙𝐴 𝑥 ≡ 𝐴 𝑥 etc
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
To leading order:
∞
∞
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
𝑖𝑔
∞
4𝑥
𝑑
−∞
𝑔𝐴 𝑥 𝐵 𝑥 𝐶 𝑥
𝐴 𝑝1
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+ )
𝐵 𝑝2 𝐶 𝑝3
𝑖𝑔
∞
𝑑4 𝑥
−∞
aˆ p1
𝑔𝐴 𝑥 𝐵 𝑥 𝐶 𝑥
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
𝐴 𝑝1

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
𝐵 𝑝2 𝐶 𝑝3
𝑖𝑔
∞
4
𝑑
𝑥
−∞
𝑔𝐴 𝑥 𝐵 𝑥 𝐶 𝑥
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''
𝐴 𝑝1
+
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 𝑥 = 𝑔 𝐴 𝑥 𝐵 𝑥 𝐶 𝑥
 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 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 𝐻I ∙ 𝑈 𝑡, 𝑡0
𝑑𝑡
𝑑 (2)
𝑈
𝑡, 𝑡0 = −𝑖 𝐻I ∙ 𝑈
𝑑𝑡
1
𝑡
𝑡, 𝑡0 = −𝐻I 𝑡 ∙
𝑡0
𝑑𝑡 ′ 𝐻I 𝑡′
Remember HI is first order.
𝑡
𝑈 (2)
𝑡, 𝑡0 = −
𝑡′′
𝑑𝑡′′ 𝐻I 𝑡′′ ∙
𝑡0
𝑡0
𝑑𝑡 ′ 𝐻I 𝑡′
The integration of two identical interaction Hamiltonian HI.
But the first HI is always later than the second HI
𝑡
𝑈 (2) 𝑡, 𝑡0 = −
𝑡′′
𝑑𝑡 ′′
𝑡0
𝑑𝑡 ′ 𝐻I 𝑡′′ ∙ 𝐻I 𝑡′
𝑡 ′′ > 𝑡′
𝑡0
在𝑡 ′ − 𝑡′′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𝐻I 永遠在左方，時間在前的𝐻I 在右方。
But t’ and t’’ are just dummy notations and can be exchanged.
𝑈
(2)
𝑡, 𝑡0
1
=−
2
𝑡
𝑡′′
𝑑𝑡 ′′
𝑡0
𝑡
𝑑𝑡 ′ 𝐻I 𝑡′′ ∙ 𝐻I 𝑡′ +
𝑡0
𝑡 ′′ > 𝑡′
𝑡′
𝑑𝑡 ′
𝑡0
𝑑𝑡 ′′ 𝐻I 𝑡′ ∙ 𝐻I 𝑡′′
𝑡0
𝑡 ′ > 𝑡′’
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𝐻I 依舊放在左方。時間在前的𝐻I 放在右方
t’’
1
=−
2
t’
𝑡
𝑡
𝑑𝑡 ′′
𝑡0
𝑑𝑡 ′ 𝐻I 𝑡′′ ∙ 𝐻I 𝑡′
𝑡0
時間在後的𝐻I
永遠放在左方
−
1
2
𝑡
𝑡
𝑑𝑡 ′′
𝑡0
𝑑𝑡 ′ 𝐻I 𝑡′′ ∙ 𝐻I 𝑡′
𝑡0
時間在後的𝐻I永遠放在左方
定義 time order product T
𝑇 𝐴 𝑡1 ∙ 𝐵 𝑡2
= 𝜃 𝑡1 − 𝑡2 𝐴 𝑡1 ∙ 𝐵 𝑡2 + 𝜃 𝑡2 − 𝑡1 𝐴 𝑡2 ∙ 𝐵 𝑡1
t’’
𝑈 (2) 𝑡, 𝑡0 = −
t’
1
2
𝑡
𝑡
𝑑𝑡 ′′
𝑡0
𝑑𝑡 ′ 𝑇 𝐻I 𝑡′′ ∙ 𝐻I 𝑡′
𝑡0
𝑈 (2) 𝑡, 𝑡0 = −
1
2
𝑡
𝑡
𝑑𝑡 ′′
𝑡0
𝑆 (2) = 𝑈 (2) ∞, −∞ = −
𝑑𝑡 ′ 𝑇 𝐻I 𝑡′′ ∙ 𝐻I 𝑡′
𝑡0
1
2
∞
∞
𝑑 4 𝑥1
−∞
𝑑 4 𝑥2 𝑇 ℒI 𝑥1 ∙ ℒI 𝑥2
−∞
This is Lorentz invariant since time ordering is invariant!
This notation is so powerful, the whole series of
operator U can be explicitly written:
𝑈 (𝑛) 𝑡, 𝑡0 =
−𝑖
𝑛!
𝑡
𝑛
𝑑𝑡1 𝑑𝑡2 ⋯ 𝑑𝑡𝑛 𝑇 𝐻I 𝑡1 ∙ 𝐻I 𝑡1 ⋯ 𝐻I 𝑡𝑛
𝑡0
The whole series can be summed into an exponential:
∞
−𝑖 𝑛
𝑛!
𝑈 𝑡, 𝑡0 =
𝑛=1
∞
=
𝑛=1
∞
=
𝑛=1
−𝑖 𝑛
𝑇
𝑛!
−𝑖
𝑇
𝑛!
𝑡
𝑡
𝑑𝑡1 𝑑𝑡2 ⋯ 𝑑𝑡𝑛 𝑇 𝐻I 𝑡1 ∙ 𝐻I 𝑡1 ⋯ 𝐻I 𝑡𝑛
𝑡0
𝑑𝑡1 𝑑𝑡2 ⋯ 𝑑𝑡𝑛 𝐻I 𝑡1 ∙ 𝐻I 𝑡1 ⋯ 𝐻I 𝑡𝑛
𝑡0
𝑛
𝑛
𝑡
∙
𝑑𝑡 𝐻I 𝑡
𝑡0
𝑡
= 𝑇 exp −𝑖
𝑑𝑡 𝐻I 𝑡
𝑡0
𝑆 (2) = 𝑈 (2) ∞, −∞ = −
1
2
∞
∞
𝑑 4 𝑥1
−∞
𝑑 4 𝑥2 𝑇 ℒI 𝑥1 ∙ ℒI 𝑥2
−∞
Amplitude for scattering A  A  B  B
B(p3 )B(p4 ) S A(p1 )A(p2 )
𝑒 𝑖𝑝4∙𝑥1
 B ( p3 ) B ( p4 )
𝑒 𝑖𝑝3∙𝑥2
 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
𝑒 −𝑖𝑝1 ∙𝑥2

 d x1 d x 2 e
4
4
 i ( p1  p 3 ) x 2
e
 i ( p 2  p 4 ) x1
𝑒 −𝑖𝑝2 ∙𝑥1
 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 𝑇 𝜙 𝑥 ∙𝜙 𝑦
0
0 𝑇 𝜙 𝑥 ∙𝜙 𝑦
0
0 𝑇 𝜙 𝑥 ∙𝜙 𝑦
0
0 𝑇 𝜙 𝑥 ∙𝜙 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 𝑑 3 𝑘 to 4-momentum𝑑 4 𝑘 = 𝑑𝑘 0 ∙ 𝑑 3 𝑘.
With a Lorentz invariant integration, it becomes extremely simple.
我們要證明：
0 𝑇 𝜙 𝑥 ∙𝜙 𝑦
0
為了證明兩者相等，我們將第二式的𝑑𝑘 0 積分完成。
𝑑𝑘 0 的積分計算最簡單的方法是在複數𝑘 0 的平面上作。
在複數𝑘 0 的平面上，𝑑𝑘 0 的積分是沿實數軸，實數軸上下，積分式各有一個 pole。
𝑘 2 − 𝜇2 + 𝑖𝜀 = 0
𝑘
02
−
𝑘
2
+ 𝜇2 + 𝑖𝜀 = 0
𝑘 02 − 𝜔𝑘2 + 𝑖𝜀 = 0
𝑘 0 = ±𝜔𝑘 ∓ 𝑖𝜀
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 𝑇 𝜙 𝑥 ∙𝜙 𝑦
0
x2
x1
C
x1
x2
𝑝1
C
C
The Fourier Transform of the propagator is simple.
𝑑 4 𝑥1 ∙ 𝑑 4 𝑥2 ∙ 𝑒 𝑖𝑝1∙𝑥1 ∙ 𝑒 𝑖𝑝2∙𝑥2 0 𝑇 𝜙 𝑥 ∙ 𝜙 𝑦
0 = 𝛿 4 𝑝1 − 𝑝2 ∙
𝑖
𝑝12 − 𝜇2 + 𝑖𝜀
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)
𝐶(𝑝1 − 𝑝3)
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.
𝑆 (2) = 𝑈 (2) ∞, −∞ = −
1
2
∞
∞
𝑑 4 𝑥1
−∞
𝑑 4 𝑥2 𝑇 ℒI 𝑥1 ∙ ℒ I 𝑥2
−∞
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
𝑦∙𝜙 𝑥 ∙𝑓 𝑥 𝑓 𝑥
將滿足規範對稱的動能項展開來：
𝐷𝜇 Φ† 𝐷𝜇 Φ = 𝜕𝜇 Φ† 𝜕𝜇 Φ + 𝑖𝑒𝐴𝜇 Φ† 𝜕𝜇 Φ − 𝑖𝑒𝐴𝜇 𝜕𝜇 Φ† ∙ Φ + 𝑒 2 𝐴𝜇 𝐴𝜇 Φ† Φ
𝑖𝑒𝐴𝜇 Φ† 𝜕𝜇 Φ − 𝑖𝑒𝐴𝜇 𝜕𝜇 Φ† ∙ Φ
𝑒 2 𝐴𝜇 𝐴𝜇 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
