Now add interactions :
62
Now add interactions: For example, we can add to Klein-Gordon Lagrangian. These terms will add the following non-linear terms to the KG equation: meaning anything not quadratic in fields. ℒ KGFree = 1 2 ( ) ( ) − 1 2 2 2 Interaction Hamiltonian: Superposition Law is broken by these nonlinear terms. Travelling waves will interact with each other. Free scalar particle ) ( ) ( ) ( ), ( ), ( 4 3 x x x x x ) ( ) ( , 4 , 3 3 2 x x
description
Free scalar particle. Now add interactions :. meaning anything not quadratic in fields. For example, we can add. to Klein-Gordon Lagrangian . . These terms will add the following non-linear terms to the KG equation:. Superposition Law is broken by these nonlinear terms. . - PowerPoint PPT Presentation
Transcript of Now add interactions :
Now add interactions:
For example, we can add
)()()(),(),( 43 xxxxx to Klein-Gordon Lagrangian.
These terms will add the following non-linear terms to the KG equation:
meaning anything not quadratic in fields.
ℒKG Free=12 (𝜕𝜇𝜙 ) (𝜕𝜇𝜙 ) − 1
2𝑚2𝜙2
)()(,4,3 32 xx
Interaction Hamiltonian:
Superposition Law is broken by these nonlinear terms.
Travelling waves will interact with each other.
Free scalar particle
狀態 波函數物理量測量 運算子
)(x
O
dxxOxO )(ˆ)(ˆ *測量期望值
)ˆ,ˆ(,),( pxfx
ixfpxf
有古典對應的物理量就將位置算子及動量算子代入同樣的數學形式：
量子力學的原則
狀態波函數滿足疊加定律，因此可視為向量，所有的狀態波函數形成一個無限維向量空間，稱 Hilbert Space。
狀態 Ket
測量 算子
O
dxxOxO )(ˆ)(ˆ *測量期望值
)(x
Bra
)(* x
Dirac Notation
)(ˆ xO O
O 與 內積
OO ˆˆ
狀態可視為向量，因此可以以抽象的向量符號來代表。
oO oO測量值確定的本徵態
Schrodinger Picture
Evolution Operator: the operator to move states from to .
States evolve with time, but not the operators:It is the default choice in wave mechanics.
狀態 波函數測量 算子 O
x
ip ˆxx ˆ
)(t
|𝜓 S (𝑡 ) ⟩ =𝑒−𝑖 𝐻 𝑡|𝜓 S (0 ) ⟩
𝑑𝑑𝑡 |𝜓 S (𝑡 ) ⟩ =−𝑖𝐻|𝜓 S (𝑡 ) ⟩
Heisenberg Picture
We move the time evolution to the operators:
Heisenberg Equation
For the same evolving expectation value, we can instead ask operators to evolve.
Now the states do not evolve.
In quantum mechanics, only expectation values are observable.
The rate of change of operators equals their commutators with H.
How does the operator evolve?
)0()0()0()0(
)0()0()()(
SSSiHt
SiHt
S
SiHt
SSiHt
SSS
tOeOe
eOetOtO
)0(SH
iHtS
iHtH eOetO
tOHiHtOtHOi
HeOeieOeiHtOdtd
HHH
iHtS
iHtiHtS
iHtH
,
Fields operators in Heisenberg Picture are time dependent, more like relativistic classical fields.
In Schrodinger picture, field operators do not change with time, looking not Lorentz invariant.
For KG field without interaction:
Combine the two equations
The field operators in Heisenberg Picture satisfy KG Equation.
The field operators with interaction satisfy the non-linear Euler Equation, for example
03 322 m
022 m
)()()()(
)0()0()()(00 tOtteOet
eOetOtO
IIIItiH
StiH
I
SiHt
SiHt
SSSS
int0 HHH
Interaction picture (Half way between Schrodinger and Heisenberg)
tiHS
tiHI eOetO 00)(
States and Operators both evolve with time in interaction picture:
Move just the free H0 to operators.
There is a natural separation between free and interaction Hamiltonians:
The rest of the evolution, that from the interaction H, stays with the state.
|𝜓 I (𝑡 ) ⟩ =𝑒𝑖 𝐻0 𝑡|𝜓 S (𝑡 ) ⟩
// 00)( tiHS
tiHI eOetO
Evolution of Operators
Operators evolve just like operators in the Heisenberg picture but with the full Hamiltonian replaced by the free Hamiltonian
Field operators are free, as if there is no interaction!
For field operators in the interaction picture:
The Fourier expansion of a free field is still valid.
𝑑𝜙 I
𝑑𝑡 =𝑖 [𝐻 0 ,𝜙 I ]
𝑑𝑂I
𝑑𝑡 =𝑖 [𝐻0 ,𝑂I ]
Evolution of States
States evolve like in the Schrodinger picture but with full H replaced by Hint I.
Hint I is just the interaction Hamiltonian Hint in interaction picture!
That means, the field operators in Hint I are free.
|𝜓 I (𝑡 ) ⟩ =𝑒𝑖 𝐻0𝑡|𝜓 S (𝑡 ) ⟩𝑑𝑑𝑡 |𝜓 I (𝑡 ) ⟩ =𝑖𝐻0|𝜓 I (𝑡 ) ⟩ −𝑖𝑒𝑖𝐻 0𝑡 (𝐻0+𝐻 S )|𝜓S (𝑡 ) ⟩ =𝑒𝑖 𝐻0𝑡 ∙𝐻 S ∙𝑒− 𝑖𝐻 0𝑡|𝜓 S (𝑡 ) ⟩
𝐻 I =𝑒𝑖𝐻 0 𝑡 ∙𝐻S ∙𝑒−𝑖 𝐻0𝑡
𝑑𝑑𝑡 |𝜓 I (𝑡 ) ⟩ =−𝑖 𝐻I|𝜓 I (𝑡 ) ⟩
Operators evolve just like in the Heisenberg picture but with the full Hamiltonian replaced by the free Hamiltonian
States evolve like in the Schrodinger picture but with the full Hamiltonian replaced by the interaction Hamiltonian.
Interaction Picture
𝑑𝑑𝑡 |𝜓 I (𝑡 ) ⟩ =−𝑖 𝐻 I|𝜓 I (𝑡 ) ⟩
𝑑𝜙 I
𝑑𝑡 =𝑖 [𝐻 0 ,𝜙 I ]
All the problems can be answered if we are able to calculate this operator
Define U the evolution operator of states.
The evolution of states in the interaction picture
𝑑𝑑𝑡 |𝜓 I (𝑡 ) ⟩ =−𝑖 𝐻 I|𝜓 I (𝑡 ) ⟩
|𝜓 I (𝑡 ) ⟩ =𝑈 (𝑡 , 𝑡0 )|𝜓 I (𝑡0 ) ⟩
is the state in the future t which evolved from a state in t0
is the state in the long future which evolves from a state in the long past.
The amplitude for to appear as a state is their inner product:
=
The amplitude is the corresponding matrix element of the operator.
The transition amplitude for the scattering of A:
The transition amplitude for the decay of A:
𝑆≡𝑈 ( ∞,− ∞ ) S operator is the key object in particle physics!
⟨ 𝐵𝐶|𝑈 (∞ ,− ∞ )|𝐴 ⟩the amplitude of in the long future which evolves from the long past.
⟨ 𝐵𝐵|𝑈 (∞ ,−∞ )|𝐴𝐴 ⟩
is the evolution operator of states.
Take time derivative on both sides and plug in the first Eq.:
U has its own equation of motion:
The evolution of states in the interaction picture𝑑𝑑𝑡 |𝜓 I (𝑡 ) ⟩ =−𝑖 𝐻 I|𝜓 I (𝑡 ) ⟩
|𝜓 I (𝑡 ) ⟩ =𝑈 (𝑡 , 𝑡0 )|𝜓 I (𝑡0 ) ⟩
𝑑𝑑𝑡 |𝜓 I (𝑡 ) ⟩ = 𝑑
𝑑𝑡 (𝑈 (𝑡 , 𝑡0 )|𝜓 I (𝑡0 )⟩ )=− 𝑖 𝐻 I (𝑈 (𝑡 , 𝑡0 )|𝜓 I (𝑡0 ) ⟩ )
𝑑𝑑𝑡 𝑈 (𝑡 , 𝑡0 )=− 𝑖𝐻 I𝑈 (𝑡 , 𝑡0 )
Solve it by a perturbation expansion in small parameters in HI.
To leading order:
Perturbation expansion of U and S
𝑈 (1) ( ∞, − ∞ )=𝑆(1)=−𝑖− ∞
∞
𝑑𝑡 ′ 𝐻 I (𝑡 ′ )
𝑑𝑑𝑡 𝑈 (𝑡 , 𝑡0 )=− 𝑖𝐻 I ∙𝑈 (𝑡 ,𝑡 0 )
𝑑𝑑𝑡 𝑈
(1 )( 𝑡 ,𝑡 0 )=−𝑖 𝐻 I ∙𝑈 ( 0) (𝑡 , 𝑡 0 )=−𝑖 𝐻 I (𝑡 )
𝑈 (1) (𝑡 ,𝑡 0 )=−𝑖𝑡0
𝑡
𝑑𝑡′ 𝐻 I (𝑡 ′ )
𝑈=𝑈 (0)+𝑈 (1)+⋯𝑈 (0 )=1
To leading order, S matrix equals
It is Lorentz invariant if the interaction Lagrangian is invariant.
𝑆(1)=−𝑖− ∞
∞
𝑑𝑡 𝐻 I (𝑡 )=− 𝑖− ∞
∞
𝑑𝑡𝑑3 ��ℋ I ( �� , 𝑡 )=𝑖−∞
∞
𝑑4 𝑥ℒI (𝑥 )
Vertex
Add the following interaction term in the Lagrangian:
The transition amplitude for A decay: can be computed as :
To leading order:
In ABC model, every particle corresponds to a field:
)()( xAxA A
𝑆=𝑖− ∞
∞
𝑑4𝑥ℒ I (𝑥 )=𝑖− ∞
∞
𝑑4𝑥 𝑔 𝐴 (𝑥 )𝐵 (𝑥 )𝐶 (𝑥 )
ℒI (𝑥 )=𝑔 𝐴 (𝑥 )𝐵 (𝑥 )𝐶 (𝑥 )
⟨𝐵 ( ��2 )𝐶 (��3 )|𝑆|𝐴 ( ��1 ) ⟩=⟨𝐵 (��2 )𝐶 (��3 )|[𝑖−∞
∞
𝑑4 𝑥𝑔 𝐴 (𝑥 ) 𝐵 (𝑥 )𝐶 (𝑥 )]|𝐴 (��1 )⟩
⟨ 𝐵𝐶|𝑈 (∞ ,− ∞ )|𝐴 ⟩ =⟨ 𝐵𝐶|𝑆|𝐴 ⟩
⟨𝐵 (��2 )𝐶 (��3 )|[𝑖− ∞
∞
𝑑4 𝑥𝑔 𝐴 (𝑥 ) 𝐵 (𝑥 )𝐶 (𝑥 )]|𝐴 (��1 )⟩ 0ˆ
33 pbpB
BcpBC p 22
⟨𝐵 (��2 )𝐶 (��3 )|[𝑖− ∞
∞
𝑑4 𝑥𝑔 𝐴 (𝑥 ) 𝐵 (𝑥 )𝐶 (𝑥 )]|𝐴 (��1 )⟩
BC
ig
The remaining numerical factor is:
A
Momentum Conservation
For a toy ABC model
Three scalar particle with masses mA, mB ,mC
External Lines
Internal Lines
Vertex
ip
iq
-ig
1
imqi
jj 22
321442 kkk
A B
C
1k
2k
3k
Lines for each kind of particle with appropriate masses.
The configuration of the vertex determine the interaction of the model.
A
BC
Every field operator in the interaction corresponds to one leg in the vertex.
Every field is a linear combination of a and a+ aa
interaction Lagrangian vertex
Every leg of a vertex can either annihilate or create a particle!
This diagram is actually the combination of 8 diagrams!
aa
aa
aa
ℒI (𝑥 )=𝑔 𝐴 (𝑥 )𝐵 (𝑥 )𝐶 (𝑥 )
In momentum space, the factor for a vertex is simply a constant.
The integration yields a momentum conservation.
A
BCInteraction Lagrangian 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.
ℒI (𝑥 )=𝑔 𝐴 (𝑥 )𝐵 (𝑥 )𝐶 (𝑥 )
𝑖−∞
∞
𝑑4 𝑥𝑔 𝐴 (𝑥 ) 𝐵 (𝑥 )𝐶 (𝑥 )
Every field operator in the interaction corresponds to one leg in the vertex.
aa
Interaction Lagrangian Vertex
Every leg of a vertex can either annihilate or create a particle!
ℒI=𝜆𝜙4
Propagator
The integration of two identical interaction Hamiltonian HI. The first HI is always later than the second HI
Solve the evolution operator to the second order.
HI is first order.
The integration of two identical interaction Hamiltonian HI. The first HI is always later than the second HI
)()()()()()())()(( 1212212121 tAtBtttBtAtttBtAT
t’
t’’
We are integrating over the whole square but always keep the first H later in time than the second H.
t’ and t’’ are just dummy notations and can be exchanged.
)()()()()()())()(( 1212212121 tAtBtttBtAtttBtAT
This definition is Lorentz invariant! t’
t’’
This notation is so powerful, the whole series of operator U can be explicitly written:
The whole series can be summed into an exponential:
0)()(0 21)()(
24
14 142231 xCxCTeexdxd xppixppi
Amplitude for scattering BBAA
Propagator between x1 and x2 Fourier Transformationp1-p3 pour into C at x2 p2-p4 pour into C at x1
000)()(00)()(0212121 xx aaaaxCxCxCxCT
Cx2
x1
A(p1) A(p2)
B(p3)
C(p1-p3)
B(p4)
A(p1) A(p2)
B(p3) B(p4)
21 tt
A particle is created at x2 and later annihilated at x1.
000)()(00)()(0121221 xx aaaaxCxCxCxCT
C
x2
x1 Cx2
x1
A(p1) A(p2)
B(p3)
C(p1-p3)
B(p4)
A(p1) A(p1)A(p2) A(p2)
B(p3) B(p3) B(p4)B(p4)
21 tt
A particle is created at x1 and later annihilated at x2.
C
x2
x1 Cx2
x1
A(p1) A(p2)
B(p3)
C(p1-p3)
B(p4)
A(p1) A(p1)A(p2) A(p2)
B(p3) B(p3) B(p4)B(p4)
Again every Interaction is integrated over the whole spacetime.
Interaction could happen anywhere anytime and amplitudes need superposition.
This construction ensures causality of the process. It is actually the sum of two possible but exclusive processes.
0)()(0 yxT
This propagator looks reasonable in coordinate space but difficult to calculate and the formula is cumbersome.
0)()(0 yxT
0)()(0 yxT
0)()(0 yxT
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:
C
x2
x1 Cx2
x1
A(p1) A(p2)
B(p3)
C(p1-p3)
B(p4)
A(p1) A(p1)A(p2) A(p2)
B(p3) B(p3) B(p4)B(p4)
The Fourier Transform of the propagator is simple.
0)()(0 yxT
00 yx
0)()(0 yxT
C
x2
x1 Cx2
x1
A(p1) A(p2)
B(p3)
C(p1-p3)
B(p4)
A(p1) A(p1)A(p2) A(p2)
B(p3) B(p3) B(p4)B(p4)
For a toy ABC model
Internal Linesiq
imqi
jj 22
Lines for each kind of particle with appropriate masses.
0)()(0 21)()(
24
14 142231 xCxCTeexdxd xppixppi
Amplitude for scattering BBAA
aa
aa
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!
aa
For a toy ABC model
Three scalar particle with masses mA, mB ,mC
External Lines
Internal Lines
Vertex
ip
iq
-ig
1
imqi
jj 22
321442 kkk
A B
C
1k
2k
3k
Lines for each kind of particle with appropriate masses.
The configuration of the vertex determine the interaction of the model.
Assuming that the field operator is a complex number field.
ipxp
ipxp ebeapdx
21
)2()( 3
3
The creation operator b+ in a complex KG field can create a different particle!
Scalar Antiparticle
The particle b+ create has the same mass but opposite charge. b+ create an antiparticle.
ipxp
ipxp ebeapdx
21
)2()( 3
3
ipxp
ipxp eaebpdx
21
)2()( 3
3 Complex KG field can either annihilate a particle or create an antiparticle!
Its conjugate either annihilate an antiparticle or create a particle!
So we can add an arrow of the charge flow to every leg that corresponds to a field operator in the vertex.
The charge difference a field operator generates is always the same!
charge non-conserving interaction
ipxp
ipxp ebeapdx
21
)2()( 3
3 ipx
pipx
p eaebpdx 2
1)2(
)( 3
3
incoming particle or outgoing antiparticle
incoming antiparticle or outgoing particle
charge conserving interaction
incoming particle or outgoing antiparticle
incoming antiparticle or outgoing particle
can either annihilate a particle or create an antiparticle!
can either annihilate an antiparticle or create a particle!
incoming antiparticle or outgoing particle
incoming particle or outgoing antiparticle
U(1) Abelian Symmetry
)()( xex iQ
The Lagrangian is invariant under the field phase transformation
invariant
)(xee iQiQ
is not invariant
U(1) symmetric interactions correspond to charge conserving vertices.
A
BC
If A,B,C become complex, they all carry charges!
The interaction is invariant only if 0 CBA QQQ
The vertex is charge conserving.
000)()(00)()(0212121 xx baabxxxxT
Cx2
x1
A(p1) A(p2)
B(p3)
C(p1-p3)
B(p4)
A(p1) A(p2)
B(p3) B(p4)
21 tt
An antiparticle is created at x2 and later annihilated at x1.
0)()(0 21 xxT
Propagator:
000)()(00)()(0121221 xx abbaxxxxT
C
x2
x1 Cx2
x1
A(p1) A(p2)
B(p3)
C(p1-p3)
B(p4)
A(p1) A(p1)A(p2) A(p2)
B(p3) B(p3) B(p4)B(p4)
21 tt
A particle is created at x1 and later annihilated at x2.
)(0)()(0 422212
41
4 21 qpmq
ixxTeexdxdC
iqxipx
C
x2
x1 Cx2
x1
A(p1) A(p2)
B(p3)
C(p1-p3)
B(p4)
A(p1) A(p1)A(p2) A(p2)
B(p3) B(p3) B(p4)B(p4)
C
x2
x1 Cx2
x1
A(p1) A(p2)
B(p3)
C(p1-p3)
B(p4)
A(p1) A(p1)A(p2) A(p2)
B(p3) B(p3) B(p4)B(p4)
For a toy charged AAB model
Three scalar charged particle with masses mA, mB
External Lines
Internal Lines
Vertex
ip
iq
-iλ
1
imqi
jj 22
321442 kkk
A A
B
1k
2k
3k
Lines for each kind of particle with appropriate masses.
Dirac field and Lagrangian
The Dirac wavefunction is actually a field, though unobservable!
Dirac eq. can be derived from the following Lagrangian.
mimiLLL
00 mimi
Negative energy!
00 mimi
Anti-commutator!
A creation operator!
bbbb~
,~
b annihilate an antiparticle!
pppppp aaaaaa 0,
0ppaa
0 pap
Exclusion Principle
p
ipxipx evbeuax
)( )( 1pe
p
ipxipx euaevbx
)(
01pu
Feynman Rules for an incoming particle
gI L ba ab
External lineWhen Dirac operators annihilate states, they leave behind a u or v !
0'22 3' pppa pp
)( 1pe 01pv
Feynman Rules for an incoming antiparticle
1pu
gAI L
ba ab
2pu
g
aaxA )(
