the heart of particle physics
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The heart of particle physics
How do we predict?
The major phenomena we observe in particle physics are Decays and Collisions.
Decay 衰變
That which is not explicitly forbidden is guaranteed to occur.
Every particle will decay if it get a chance.
Everything which is not forbidden is allowed (a principle of English Law)
Relativity allows particles to decay by transforming mass into energy.
Heavier particles usually don’t exist in nature.
They decay soon after they are produced.
Decay chain will continue until it can decay no more, forbidden usually by conservation law (symmetry).
The only five stable particles in nature
Neutron Proton Electron Photon Neutrino
β Decay epn
但中子是會衰變的!
基本粒子是會消失而形成其他粒子的!
W
W
d
u
在交互作用的交點,粒子是會消失的。
eπ 子的衰變
Bubble Chamber picture of pions
我們無法預測單一一顆中子何時衰變,只能預測衰變發生的機率。衰變是一個不確定的機率過程。
單位時間內的衰變機率:衰變率 Γ
如果是處理一大群中子,知道衰變的機率就足夠了:
NdtdN
teNN 0 隨時間以指數遞減
Γ 即是一個中子每秒衰變的機率 !
不衰變的機率即不衰變的粒子數
單一一顆中子的衰變機率即對應一大群中子的衰變分布。
β Decay epn
衰變率(機率密度)是產物粒子的能量及動量的函數。
衰變的產物可以有連續分布的可能的動量。物理只能預測衰變為某一個動量組合的機率密度。
epeep EEppp ,,,,
甚至衰變產物也有一個以上的可能。不同的可能稱為 Channel
每一個 Channel 就對應一個衰變率。
總衰變率就是所有衰變率的和。
τ 稱為 Life Time 生命期
t
t eNeNN 00
未衰變的粒子數在 t > τ 後就很少了。
π± 的衰變是透過弱作用,生命期約 10-8s ,
π0 的衰變是透過電磁作用,生命期約 10-15s ,若粒子透過強作用衰變,生命期約只有 10-23s
這樣的粒子即使產生,都無法在實驗室內看到他的痕跡。
20
這樣的粒子會以共振曲線的形式出現在其衰變產物的的散射分布上
pp
pp 散射率對質心能量的分布
在質心能量等於共振態質量時,會被加強。Breit-Wigner Resonance
4
12
2
mE
強作用衰變 p
這個過程會對以上散射率增加一個共振。
強衰變的粒子會以共振曲線出現在其衰變產物的的散射分布上
4
12
2
mE
共振曲線中心即粒子質量。共振曲線寬度即衰變率。衰變率 Γ 又稱為 Decay Width
Collision
LdN
粒子束越強,單位面積粒子數越多,反應發生的次數越多!Event Rate
通量 L Luminosity( 亮度 ) 為單位時間通過單位面積的入射粒子數:
tvAN
通量 L 為單位時間通過單位面積的粒子數:
時間內通過面積 A 的粒子數
vAt
NL
LdNd
dσ 是一個與粒子束強弱無關的量,只由該反應決定。
粒子束越強,單位面積粒子數越多,反應發生的次數越多!
dσ 是此反應的內在性質。dσ 一計算出來,就可以用在所有的實驗。
dσ 是一個面積。稱為散射截面 Scattering Cross Section 。
dσ 之於 dN ,就如同比熱之於熱容量。
Event Rate LdN 定義 dσ
Classical Scattering
Impact parameter b 與散射角度 θ 有一對一對應
在古典散射真的是截面積
通過左方此一截面的粒子,將散射進入對應的散射角 之間
散射角在 θ 與 θ+dθ 間的粒子,其 b 必定在對應的 b 與 b+db之間
dLdN
dddbbdbbd )(2)(2
故散射角為 的散射粒子數:
dLdN
LdNd 散射截面 Scattering Cross Section
LdNd
粒子物理的粒子已不再有特定軌跡了。古典的計算已不適用。但舊稱仍沿用。在此定義下, dσ 依舊是此反應的內在性質。
L
Bunch Crossing 4 107 Hz
7x1012 eV Beam Energy 1034 cm-2 s-1 Luminosity 2835 Bunches/Beam 1011 Protons/Bunch
7 TeV Proton Proton colliding beams
Proton Collisions 109 Hz
Parton Collisions
New Particle Production 10-5 Hz (Higgs, SUSY, ....)
p pH
µ+
µ-
µ+
µ-
Z
Zp p
e- e
q
q
q
q1
-
g~
~
20~
q~
10~
Selection of 1 event in 10,000,000,000,000
7.5 m (25 ns)
Event recorded with the CMS detector in 2012 at a proton-proton centre-of-mass energy of 8 TeV. The event shows characteristics expected from the decay of the SM Higgs boson to a pair of Z bosons, one of which subsequently decays to a pair of electrons (green lines and green towers) and the other Z decays to a pair of muons (red lines).
fANNvL BA
dσ 是一個垂直於射向的面積。在沿射向的羅倫茲變換下是不變的!Fixed Target 實驗的 dσ 與 Colliding Beam 實驗一樣。
生成的粒子可以觀察到軌跡
因此生成的粒子可以看成波包。
Δk
Δx
衰變率必須對動量波函數積分:
21
kx
假設我們對粒子位置並未太精密測量Δp 就不會太大我們可以以 p0 的衰變率來近似!
2)()( pbpdp
)( 0p
討論時會近似使粒子都具有一個特定動量,而忽略動量分布!
粒子的動量波函數有一個分布!
先將一定會出現的 Factors 從衰變率及截面中提出來。
Phase Space Factors
這一些 Factors 與作用的細節無關,只和入射粒子及產生粒子的數量與身分有關!
n
jjjzjy
n
jjx pdMdpdpdpMd
2
32
2
2
But this integral is not Lorentz invariant!
Particle decay: )()(3)(2)(1 321 npnppp
px
py
dpx
dpy
衰變率 dΓ 應與所產生粒子的動量所佔相空間的區間大小成正比
衰變後第 i 個粒子的動量在 iii pdpp , 之間的衰變率記為 dΓ
1
axdx
axax if0
Dirac function
In an integration, enforce the equation that x = a.
afxfaxdx
10
222
axdx
0
22
0
22
0
222 221 axdxxaxdxxaxdx
120
1
axcdxx
22 ax 只有在 x = ± a 不為零。 )(2122 axcaxcax
12 1 ca
120
2
axcdxx 12 2 ca
)(2122 axaxa
ax jj j
xxxf
xf )('
1)( 0)( jxf
xk
kx 1
3
3
3
3
222 221
22
1
pdE
pd
cmp
We make it more complicated by allowing an indefinite p0
integration and then fixing it by requiring the on-shell condition:
0222
4
4
22
pcmppd
But in this form, we can be sure it is Lorentz invariant!We can perform the p0 integration to recover the 3 space form.
22220222 cmppcmp
)(2122 axaxa
ax
2220
222
022220
2
1 cmppcmp
pcmpp
This is Lorentz invariant.
n
jjpdMd
2
32
n
jjj
n
j j mcpE 2 2222 2
12
1
n
j
j
j
pdE
Md2
3
32
221
npppp 321442
No matter what, the overall 4-momenta are conserved!
n
j
j
jn
pdE
ppppMd2
3
3
321442
2212
我們從衰變率再拉出這個 Factor使積分是羅倫茲不變
從衰變率再拉出一個執行動量守恆的 δ function
All the factors are Lorentz Invariant. But is M 2 Lorentz invariant?
n
j
j
jn
pdE
ppppMd2
3
3
321442
2212
But M 2 is not Lorentz invariant since Γ is not.For a particle, Γ transforms like 1/t1. t1 transforms like E1.
2
2
1cv
t
2
2
2
1cv
mcE
1
1E
n
j
j
jn
pdE
ppppME
d2
3
3
321442
1 22121
Now we can be sure M 2 is Lorentz invariant. It’s called Feynman Amplitude.
我們可以從 M2 再提出一個 1/ E1
The Lorentz Invariance makes M 2 simple.
n
j
j
jn
pdE
ppppME
d2
3
3
321442
1 22121
We separate the kinematics and dynamics in such an elegant way that the still dynamic part M 2 is Lorentz invariant.
M 2
Dynamic Factors 由交互作用的細節決定。Kinematic factors 只和入射粒子及產生粒子的數量與身分有關!
Now we apply this to pion two photon decay. 20
Choose the rest frame of pion:
mE 1 01 p
3
33
332
3
2321
442
1 221
22121
pd
Epd
EpppM
Ed
Total Decay Rate:
)()( afaxxfdx
式子與角度無關!
ddpdppd sin23
若生成粒子有質量:見課本推導。
注意 M 2 是沒有因次的。
Two body scattering: n 321
n
j
j
jn
pdE
ppppMd3
3
3
321442
2212
If we consider only Lorentz transformation along the 1-2 colliding axis, the cross section dσ is invariant!
But we do want to pull out a Lorentz invariant factor that reflects the inverse luminosity that must appear in cross section:
M 2 is Lorentz invariant since dσ almost is.
LdNd
221
2214 mmpp
In the rest frame of particle 2
1121121221
212
221
221 444 vvEEpmmEmmmmE
E
cvx
2
1
11
Lorentz invariant factor that becomes luminosity in rest frame:
22 mE 02 p
n
j
j
jn
pdE
ppppMd3
3
3
321442
2212
221
2214
1
mmpp
n
j
j
jn
pdE
ppppMmmpp
d3
3
3
321442
221
221
2212
4
1
我們從截面再拉出這個一定要出現的羅倫茲不變的 Luminosity 。
M 2 is Lorentz invariant. It’s called Feynman Amplitude.
The Lorentz Invariance makes M 2 simple.
Two body scattering in CM
散射截面的角度分布
注意 M 2 是沒有因次的。
Carry out the p4 integrationEnforce the 3 momentum conservation
注意 M 2 是沒有因次的。
Two body scattering in CM
如果生成粒子與入射粒子質量相等 fi pp
2CM
2
EM
dd
Feynman Rules
To evaluate the Lorentz Invariant Feynman Amplitude M
Components of Feynman Diagrams
External Lines
Internal Lines
Vertex
ip
iq
Determined by Particle Content (Masses and spins)
Interactions
Every line contains a 4-momentum.
Every component corresponds to a specific factor!
External Lines
Internal Lines
Vertex
ip
iqpropagator
-ig
1 (For spinless particles)
imqi
jj 22
i
ip442
Coupling Constant
Momentum Conservation
Its momentum is on-shell.𝑝❑2 =𝑚2
Its momentum is not on-shell.
Draw all diagrams with the appropriate external lines, matching the incoming and outgoing particles with momenta fixed by the experiment.
Integrate over all internal momenta iqd 4
421
Take out an overall momentum conservation.
npppp 321442
That’s it! The result is -iM. It’s so simple.
The Lorentz Invariance is explicit at every step.
Multiply all the factors!
A toy ABC model
There are 3 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.
etc.
qe,
qe,
Vertices for real Interactions
CBA
Particle A decay
gM
The Feynman Amplitude is just a constant!
321442 pppigiM
Take out overall momentum conservation.
Consider first the diagrams with the fewest number of vertices.
Each extra vertex carries an extra factor of –ig, which is small.
There are other possible diagrams, in fact infinite number of them.
Feynman Rule is a perturbation theory.
Lifetime of A
CBA
gM
20 For cm
g16
2
0 CB mm22mmp A
As an example, consider that a particle is pion and B,C particles are photons.
Assume that the interaction of has a coupling constant g.
It’s similar to electron-electron scattering!
Scattering
It takes at least two vertices to draw a diagram with appropriate external lines.
The leading order diagram:
One propagator
and two coupling constants
2ig
24 ppq
Carrying out the momentum integration will enforce momentum conservation at one of the vertex:
24
444
4
22
ppqdq
Take out overall momentum conservation.
24 pp
The remaining momentum conservations can be enforced by immediately carrying out the integration in the beginning!
The result can be written down right away!
243144
2444 22 ppppppq
The overall momentum conservation is always there.
24
444
4
22
ppqdq
24 pp
After momentum conservation is enforced, the momentum of internal particle c is 24 pp
It does not satisfy momentum mass relation of a particle! 22224 cmpp c
The internal particle is not a real particle, it’s virtual.
24 pp
It does not satisfy momentum mass relation of a particle! 22224 cmpp c
The internal particle is not a real particle, it’s virtual.
24 pp
In a sense, the mass relation is only for particles when we “see” them!
Internal lines are by definition “unseen” or “unobserved”.
It’s more like:
It’s more a propagation of fields than particles!
24 pp
Field fluctuation propagation can only proceed forward that is along “time“ from past to future.
This diagram is actually a Fourier Transformation into momentum space of the spacetime diagram.
The vertices can happen at any spacetime location xμ and the “location” it happens need to be integrated over. After all you do not measure where interactions happen and just like double slit interference you need to sum over all possibilities.
B
A
B
A
For those amplitude where time 1 is ahead of time 2, propagation is from 1 to 2.
C1
2B
A
B
A
For those amplitude where time 2 is ahead of time 1, propagation is from 2 to 1.
C
1
2
24 pp is actually the sum of the above two diagrams!Feynman diagram in momentum space is much simpler than in space-time!
24 pp
23 pp
This is very similar to electron-electron scattering.
Scattering
To the leading order, there could be more than one diagrams!
In the diagrams, the B lines in some vertices are incoming particle.
The striking lesson: Lines in a vertices can be either outgoing or incoming, depending on their p0.
p0 for an observed particle is always positive.
So this vertex diagram is actually 8 diagrams put together! That’s the simplicity of Feynman diagram.
If all the momenta in the diagrams can be determined through momentum conservation, the diagram has no loop and is hence called tree diagram!
24 pp
If there is a loop in the diagram, some internal momentum is not fixed and has to be integrated over! These are called loop diagram.
Oops! Loops are infinite!
31 pp 31 pp
qpp 31
q
將所有的動量守恆事先執行
若有 loop 就會有某些動量無法確定無法確定的動量必須積分。
Renormalization
24 pp
Cut off the momentum q at Λ
22Physical,
2 mmm cc
δm 中的 ln Λ 正好使 Tree 圖與 Loop圖的 ln Λ抵消 結果會是有限的!
ln
imp
i
mpimpmp
i
mpimp
i
c
c
cc
c
c22
22
2222
22
22
1
1
1
Mass Renormalization
ln2mi
22Physical,
2 mmm cc
22cmp
i
2Physical,
2cmpi
c 的原始質量是無限大,加上無限大的修正,量到的質量是有限的 2Physical,cm
pp p
q
pq p pp p
2222cc mp
imp
i
222222
ccc mpi
mpi
mpi
δm 中的 ln Λ 正好與 Σ 的 ln Λ抵消
Charge Renormalization
lng
Physicalg
原始耦合常數是無限大,加上無限大的修正,量到的有限的耦合常數 Physicalg
ggg Physical
All the infinities can be cancelled out by a finite numbers of parameter renormalization.
Schrodinger Wave EquationHe started with the energy-momentum relation for a particle
he made the quantum mechanical replacement:
How about a relativistic particle?
Expecting them to act on plane waves
ipxrpiiEt eee
The Quantum mechanical replacement can be made in a covariant form. Just remember the plane wave can be written in a covariant form:
As a wave equation, it does not work.It doesn’t have a conserved probability density.It has negative energy solutions.
ipxrpiiEt eee
0222 cmp
0p E
There are two solutions for each 3 momentum p (one for +E and one for –E )
ipxxpitip eaeax 0
)(Plane wave solutions for KG Eq.
ipxipxipx eameapeap 2220
2220 mpp
p
xpiiEtxpiiEt
p
ipxipx ebeaebeax
)(
Time dependence can be determined.
It has negative energy solutions.
Expansion of the KG Field by plane :
p
xpiiEtxpiiEt
p
ipxipx ebeaebeax
)(
p
ipxipx eaeax
)(
If Φ is a real function, the coefficients are related:
The proper way to interpret KG equation is it is not a Wavefunction Equation but actually a Field equation just like Maxwell’s Equations.
Plane wave solutions just corresponds to Plane Waves.
It’s natural for plane waves to contain negative frequency components.
Add a source to the equation:
)(2 xjm
We can solve it by Green Function. ')',()',( 2 xxxxmxxG
G is the solution for a point-like source at x’.
By superposition, we can get a solution for source j.
)'()',(')()( 40 xjxxGxdxx
Green Function for KG Equation: ')',()',( 42 xxxxmxxG
By translation invariance, G is only a function of coordinate difference:
)'()',( xxGxxG
The Equation becomes algebraic after a Fourier transformation.
)(~
2)'( )'(
4
4
pGepdxxG xxip
1)(~22 pGmp
)'(4
44
2)'( xxipepdxx
22
1)(~mp
pG
This is the propagator!
)'(
4
422)'(
4
4
2)(~
2xxipxxip epdpGmpepd
'x x
Green function is the effect at x of a source at x’.That is exactly what is represented in this diagram.
KG Propagation
The tricky part is actually the boundary condition.
B
A
B
A
For those amplitude where time 1 is ahead of time 2, propagation is from 1 to 2.
C1
2B
A
B
A
For those amplitude where time 2 is ahead of time 1, propagation is from 2 to 1.
C
1
2
is actually the sum of the above two diagrams!
To accomplish this, 22
1)(~mp
pG
imp
pG
22
1)(~
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