# Document

```Time finally…………..

Ket States S  evolve with time, but not the operators

Schrodinger Picture
Hamiltonian is the generator of time evolution!

S
= −  ∙ S
Schrodinger Equation
It is the default choice in wave mechanics.

Y(x, t) = x Y(t)

Oˆ
xˆ  x
 

pˆ    i 


x



 2 2


i
 Hˆ    

V
(
x
)
2

t
 2 m x


Schrodinger Equation 就是Schrodinger Wave Equation。

S
= −  ∙ S

S
=  − ∙ S 0
≡   ∙ S 0
Evolution Operator   : the operator to move states from  = 0 to .

=  −
1
= 1 −  + −
2
1
2+
−
3!
∞
3
+⋯=
=0
1
−
!

=  − = − − = −

= −

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。

= −  ∙
=
= − ∙

=  − ∙  0

∙ ∙   =  0 ∙   ∙ ∙  − ∙  0
=  0 ∙ ∙  0

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!

≡   ∙  ∙  −
The time-dependent operators satisfy Equations of Motion.

=  ∙    ∙  − +   ∙  ∙  − ∙
Heisenberg Equation

=  ,

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.
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

()
=  ,  = −

,  =
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.
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

= S   −0 ∙  0 ∙  ∙  −0 ∙  0 S
I  =  0 ∙ S ∙  −0
I
≡  I  ∙   ∙  I
=  0 S  ~   S 0

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!

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

Solve it by a perturbation expansion in small parameters like ,  in HI.
=  (0) +  (1) + ⋯
() ∝  or
(0) =   = 0 = 1

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 :
∞, −∞  =
∞
∞
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 



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

But t’ and t’’ are just dummy notations and can be exchanged.

(2)
, 0
1
=−
2

′′
′′
0

′ I ′′ ∙ I ′ +
0
′′ > ′
′
′
0
′′ I ′ ∙ I ′′
0
′ > ′’

t’’
1
=−
2
t’

′′
0
′ I ′′ ∙ I ′
0

−
1
2

′′
0
′ I ′′ ∙ I ′
0

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.

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.

0    ∙
0

0 的積分計算最簡單的方法是在複數 0 的平面上作。

2 − 2 +  = 0

02
−

2
+ 2 +  = 0
02 − 2 +  = 0
0 = ± ∓

x0  y 0

y0 > x0 將積分路徑沿上方半圓回到實數軸

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.

51
∙  ∙

Φ†  Φ =  Φ†  Φ +  Φ†  Φ −   Φ† ∙ Φ +  2   Φ† Φ
Φ†  Φ −   Φ† ∙ Φ
2   Φ† Φ

Vector Boson Fusion VBF

60

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

```

– Cards

– Cards

– Cards

– Cards

– Cards