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AKT II – Atomic, Nuclear and Particle Physics II

18.3.2021

Standard Model of Particle Physics

Generations

I II III

weak isospin

electr chrg

color chrg

Gluons are carrying both color and anti-color. They participate in strong interactions. There are 8 types:

𝑟𝑏+𝑏𝑟

√2 , 𝑟𝑔+𝑔𝑟

√2 , 𝑏𝑔+𝑔𝑏

√2 , 𝑟𝑟−𝑏𝑏

√2 , −𝑖 𝑟𝑏−𝑏𝑟

√2 ,

−𝑖 𝑟𝑔−𝑔𝑟

√2 , −𝑖 𝑏𝑔+𝑔𝑏

√2 , 𝑟𝑟+𝑏𝑏−2𝑔𝑔

√2

Fe rm ions , s pi n ½ , i nt ri ns ic pa ri ty + 1 Quarks

up u 2.2 MeV

charm c 1.3 GeV

top t 173 GeV

- ½ + ½ we ak int e ract ion -1 - +⅔ EM int er act ion rg b r gb st ro n g in te ra ct io n G aug e B o son s, spi n 1, int ri ns ic pa ri ty - 1 EM

photon γ

0 GeV spi n 0 higgs H 125 GeV down

d 4.7 MeV

strange s 0.1 GeV

bottom b

4.2 GeV st rong

gluon g 0 GeV

parity +1

Lifetime: muon 2 μs, tauon 290 fs Neutrinos ν e , ν μ , and ν τ are mixtures of 3 fundamental neutrino states with defined masses ν 1 , ν 2 , and ν 3 .

Le pt ons

electron e - 0.5 MeV

muon μ - 0.1 GeV

tau τ - 1.8 GeV

w e ak int e ract ion

W boson W ± 80 GeV Q= ±1

Flavor: u, d, c, s, t, b electron

neutrino ν e

<1.1 eV muon neutrino

ν μ

<0.2 MeV tau neutrino

ν τ

<18 MeV

Z boson Z 91 GeV

Hadrons are bound Quark states Baryons: Hadrons w. odd number of quarks e.g. p(uud), n(ddu), half-spin Mesons: Hadrons with even number of quarks (e.g. qq̅), integer spin

Interaction Vertices

Electromagnetism Strong Interaction Weak Charged Current Interaction Weak Neutral Interaction All charged

particles, never chan- ges flavor.

𝛼 = 1/137

Only Quarks and the gluon itself, never changes flavor. 𝛼 𝑆 = 1

W ± couples charged Leptons with corresp. neutrinos and all Quark combinations so that charge is conserved. Always changes flavor! 𝛼 𝑊 = 1/30

All Fermions Never changes flavor.

𝛼 𝑍 = 1/30 Coupling constant 𝑔 Determines strength of interaction between gauge boson and fermion = probability of fermion to emit or absorb

boson. Scattering process with two vertices: ℳ ∝ 𝑔 2 ⟹ Interaction probability 𝑝 = |ℳ| 2 ∝ 𝑔 4 Fine struc. const. 𝛼 𝛼 ∝ 𝑔; 𝛼 𝐸𝑀 = 𝑒 2

4𝜋𝜀 0 ℏ𝑐 . Intrinsic strength of weak interaction > QED, but because of W-boson’s large mass it’s smaller.

Natural Units

Physical Quantity [𝒌𝒈, 𝒎, 𝒔] [ℏ, 𝒄, 𝑮𝒆𝑽] ℏ = 𝒄 = 𝟏 conversion Further Units

energy 𝐸 [𝐽] = [ 𝑘𝑔 𝑚 2

𝑠 2 ] [𝐺𝑒𝑉] [𝐺𝑒𝑉] 𝐸[𝐽] = 𝐸[𝑒𝑉] ∙ 𝑒 Barn [𝑏] 1𝑏 = 10 −28 𝑚 2

momentum 𝑝⃗ [ 𝑘𝑔 𝑚

𝑠 ] [ 𝐺𝑒𝑉

𝑐 ] [𝐺𝑒𝑉] 𝑝⃗ [ 𝑘𝑔 𝑚

𝑠 ] = 𝑝⃗[𝑒𝑉]∙𝑒

𝑐 ℏc = 197MeV fm ≈ 0.2GeV fm ℏ ≈ 10 −34 𝐽𝑠, 𝑒 ≈ 10 −19 𝐶, 𝑐 ≈ 10 8 𝑚

𝑠 Heavyside-Lorentz: ℏ=ε 0 =μ 0 =1

mass 𝑚 [𝑘𝑔] [ 𝐺𝑒𝑉

𝑐 2 ] [𝐺𝑒𝑉] 𝑚[𝑘𝑔] = 𝑚[𝑒𝑉]∙𝑒

𝑐 2

time 𝑡 [𝑠] [

𝐺𝑒𝑉 ] [ 1

𝐺𝑒𝑉 ] 𝑡[𝑠] = 𝑡 [ 1

𝑒𝑉 ]

𝑒 = 𝑡 [ 1

𝐺𝑒𝑉 ] 0.2[𝐺𝑒𝑉]10 −15 [𝑚]

𝑐[𝑚/𝑠]

distance 𝑑 [𝑚] [ ℏ𝑐

𝐺𝑒𝑉 ] [ 1

𝐺𝑒𝑉 ] 𝑑[𝑚] = 𝑑 [ 1

𝑒𝑉 ] ℏ𝑐

𝑒 = 𝑑 [ 1

𝐺𝑒𝑉 ] 0.2[𝐺𝑒𝑉]10 −15 [𝑚]

area 𝐴 [𝑚 2 ] [( ℏ𝑐

𝐺𝑒𝑉 ) 2 ] [ 1

𝐺𝑒𝑉 2 ] 𝐴[𝑚 2 ] = 𝐴 [ 1

𝑒𝑉 2 ] ( ℏ𝑐

𝑒 ) 2 = 𝐴 [ 1

𝐺𝑒𝑉 2 ] (0.2[𝐺𝑒𝑉]10 −15 [𝑚]) 2 Special Relativity and Four-Vectors

Beta and

Gamma 𝛾 = 1

√1−𝛽 2 ; 𝛽 = 𝑣

𝑐 Lorentz- transformation

Let S‘ be the „moving“ system, and let S be the „rest“ system; i.e. velocity and direction of S‘ with respect to S determine magnitude and sign of 𝛽. 𝜂 𝜇𝜈 =𝜂 𝜇𝜈 =diag(1, −1, −1, −1) Active LT

Boost in x 𝑆′ → 𝑆:

𝛬 𝜇 𝜈 = [ 𝛾 𝛽𝛾 𝛽𝛾 𝛾

0 0 0 0

0 0 0 0 1 0 0 1 ]

How does "moving"

system S‘ look like in

"rest" system S?

𝒂 𝝁 = 𝜦 𝝁 𝝂 𝒂′ 𝝂

Passive LT Boost in x, 𝑆 → 𝑆′:

𝛬̃ 𝜇 𝜈 = [

𝛾 −𝛽𝛾

−𝛽𝛾 𝛾

0 0 0 0

0 0 0 0 1 0 0 1 ]

How does "rest"

system S look like in "moving" S‘?

𝒂′ 𝝁 = 𝜦 ̃ 𝝁 𝝂 𝒂 𝝂 4-vector

position 𝑥 𝜇 = ( 𝑐𝑡 𝑥⃗ ) = 𝑐=1 ( 𝑡

𝑥⃗ ) Proper Time 𝜏 in S‘ 𝑑𝑠 2 =𝑑𝑠′ 2 ⇒ 𝑐 2 𝑑𝑡 2 - 𝑑𝑥 2 - 𝑑𝑦 2 - 𝑑𝑧 2 = 𝑐 2 𝑑𝜏 2 ⇒ 𝑑𝜏 2 = 𝑑𝑠 2

𝑐 2 = (1 − 𝑣⃗⃗ 2

𝑐 2 ) 𝑑𝑡 2 ⇒ 𝑑𝜏 = 1

𝛾 𝑑𝑡 4-vector

veloicity 𝑢 𝜇 = 𝑑𝑥 𝜇

𝑑𝜏 = 𝑑𝑥 𝜇

𝑑𝑡 𝑑𝑡 𝑑𝜏 = 𝛾 𝑑𝑥 𝜇

𝑑𝑡 = 𝛾 ( 𝑐 𝑣⃗ ) = ( 𝛾𝑐

𝛾𝑣⃗ ) = 𝑐=1 ( 𝛾

𝛾𝑣⃗ ) 𝑢 𝜇 𝑢 𝜇 = 𝛾 2 (𝑐 2 − 𝑣⃗ 2 ) = 𝑐 2 > 0 ⇒ time-like, invariant 4-vector

momentum 𝑝 𝜇 = 𝑚 0 𝑢 𝜇 = 𝑚 0 ( 𝛾𝑐 𝛾𝑣⃗ ) = (

𝐸 𝑐

𝑝⃗ ) = ( 𝑚 0 𝑐 + 𝐸 𝑘𝑖𝑛

𝑐

𝑝⃗ ) = 𝑐=1 ( 𝑚 0 + 𝐸 𝑘𝑖𝑛

𝑝⃗ ) = ( 𝐸

𝑝⃗ ) 𝑝 𝜇 𝑝 𝜇 = 𝑝 0 2 − 𝑝⃗ 2𝐸 2

𝑐 2 − 𝑝⃗ 2 = 𝑚 0 2 𝑐 2 …invariant 𝑝 𝜇 𝑝 𝜇 𝑐=1 = 𝑝 0 2 − 𝑝⃗ 2 ≡ 𝐸 2 − 𝑝⃗ 2 = 𝑚 0 2 …invariant Derivations 𝜕 𝜇 = ( 1

𝑐 𝜕 𝑡 , ∇ ⃗⃗⃗) = 𝑐=1 (𝜕 𝑡 , ∇ ⃗⃗⃗) 𝜕 𝜇 = (

1 𝑐 𝜕 𝑡

−∇ ⃗⃗⃗ ) = 𝑐=1 ( 𝜕 𝑡

−∇ ⃗⃗⃗ ) 𝜕 𝜇 𝜕 𝜇 = ⎕ = 𝜕 2

𝜕𝑡 21

𝑐 2 ∇ ⃗⃗⃗ 2 𝑐=1 = 𝜕 2

𝜕𝑡 2 − ∇ ⃗⃗⃗ 2 𝐸 = 𝛾𝑚 0

𝑝⃗ = 𝛾𝑚 0 𝑣⃗ = 𝛾𝑚 0 𝛽⃗𝑐 = 𝑐=1 𝛾𝑚 0 𝛽⃗

𝑣⃗ = 𝑝⃗

𝛾𝑚 0 = 𝑝⃗

𝐸

Energy: massive particle: 𝐸 = √𝑚 0 2 𝑐 4 + 𝑝⃗ 2 𝑐 2 𝑐=1 = √𝑚 0 2 + 𝑝⃗ 2 massless particle: 𝐸 = |𝑝⃗|𝑐 = 𝑐=1 |𝑝⃗|

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Particle Accelerators and Detectors

→ 𝑝 ← Collider 𝑒 (e.g. HERA):

Center-of-mass frame: 𝑝 𝑝 𝜇 𝑐=1 = ( 𝑚 𝑝 + 𝐸 𝑘𝑖𝑛 𝑝 𝑝 𝑝 ) ≈ ( 𝐸 𝑘𝑖𝑛 𝑝

𝑝 𝑝 ) ≈ ( 920𝐺𝑒𝑉

920𝐺𝑒𝑉 ) ; 𝑝 𝑒 𝜇 𝑐=1 = ( 𝐸 𝑘𝑖𝑛 𝑒

𝑝 𝑒 ) ≈ ( 27.5𝐺𝑒𝑉

−27.5𝐺𝑒𝑉 ) 𝑝 𝑡𝑜𝑡 𝜇 = 𝑝 𝑝 𝜇 + 𝑝 𝑒 𝜇 = ( 947.5𝐺𝑒𝑉

892.5𝐺𝑒𝑉 ) Available Energy: √𝑠 = √𝑝 𝑡𝑜𝑡 𝜇 𝑝 𝜇 𝑡𝑜𝑡 = √(947.5 2 − 892.5 2 ) = 318 𝐺𝑒𝑉 Fixed Target

proton: what electron energy is required for same s?

proton rest frame: 𝑝 𝑝 𝜇 𝑐=1 = ( 𝑚 𝑝

0 ) ; electron moves: 𝑝 𝑒 𝜇 𝑐=1 = ( 𝐸 𝑘𝑖𝑛 𝑒

𝑝 𝑒 ) ; 𝑝 𝑡𝑜𝑡 𝜇 = 𝑝 𝑝 𝜇 + 𝑝 𝑒 𝜇 = ( 𝑚 𝑝 + 𝐸 𝑘𝑖𝑛 𝑒 𝑝 𝑒 )

Available Energy: √𝑠=√𝑝 𝑡𝑜𝑡 𝜇 𝑝 𝜇 𝑡𝑜𝑡 = √(𝑚 𝑝 + 𝐸 𝑘𝑖𝑛 𝑒 ) 2 − 𝑝 𝑒 2 ≈ √(𝑚 𝑝 + 𝐸 𝑘𝑖𝑛 𝑒 ) 2 − (𝐸 𝑘𝑖𝑛 𝑒 ) 2 = √𝑚 𝑝 2 + 2𝑚 𝑝 𝐸 𝑘𝑖𝑛 𝑒 + (𝐸 𝑘𝑖𝑛 𝑒 ) 2 − (𝐸 𝑘𝑖𝑛 𝑒 ) 2

√𝑠 = √𝑚 𝑝 2 + 2𝑚 𝑝 𝐸 𝑘𝑖𝑛 𝑒 ≈ √2𝑚 𝑝 𝐸 𝑘𝑖𝑛 𝑒 ⟹ 𝐸 𝑘𝑖𝑛 𝑒 = 𝑠

2𝑚 𝑝 ⟹ would require electron energy 𝐸 𝑘𝑖𝑛 𝑒 = 318 2

2∙1 = 50 500 𝐺𝑒𝑉 for same 𝑠 LHC resolution 𝐸 = ℎ𝜈 ≈ ℎ 𝑐

𝜆 ⟹ 𝜆 = ℎ𝑐

𝐸 ; 𝐸 𝐿𝐻𝐶 = 14𝑇𝑒𝑉 ⟹ 𝜆 = 10 −19 𝑚 (quarks: 10 −17 𝑚 ) Interactions at ATLAS, CMS, ALICE, LHCb LINAC A voltage generator induces EM field inside the RF cavities with 400MHz. LHC: 8 × 2 𝑀𝑒𝑉

Cyclic accel. 2 types of magnets: Dipol magnets for beam “bending”, quadrupole magnets for focusing (only in one axis!) Synchrotron Bremsstrahlung energy loss 𝐸 = 4𝜋 3 𝑒 2 𝛽 𝑅 2 𝛾 4

Detecting momentum

Measuring momentum of charged particle by detecting deflection through Lorentz force (easy compared to energy detection) 𝐹 𝑍 = 𝐹 𝐿 ⟹ 𝑚𝜔 2 𝑟 = 𝑒𝐵𝑣|𝑣 = 𝑟𝜔 ⟹ 𝜔 = 𝑣

𝑟 ⟹ 𝑚 𝑣 2

𝑟 2 𝑟 = 𝑒𝐵𝑣 ⟹ 𝑚 𝑣

𝑟 = 𝑒𝐵 ⟹ 𝑝

𝑟 = 𝑒𝐵 ⟹ 𝑝 = 𝑒𝐵𝑟

Tracking detector

(1) Particle moves through gaseous substance, liberates electrons, which drift in an electric field towards sense wires.

(2) Particle moves through doped silicon waver, and generate electron-hole pairs.

The holes drift in direction of the electric field and are collected by pn-junctions. Sensors are shaped in strips. One particle = 10000 electron-hole-pairs.

Detectors are placed in cylindrical surfaces. A homogenous 𝐵 ⃗⃗ field is applied.

𝑝 [ 𝐺𝑒𝑉

𝑐 ] cos(𝜆) = 0.3𝐵[𝑇]𝑅[𝑚]

Photons:

Čerenkov radiation

Charged article traverses dielectric medium

𝑣 > 𝑐

𝑛 ⟹ cos(𝜗) = 𝑐𝑡/𝑛

𝑣𝑡 = 𝑐𝑡/𝑛

𝛽𝑐𝑡 = 1

𝑛𝛽

Photo effect:

Small energies

𝛾 gets absorbed, 𝑒 is emitted.

𝐸 𝑒 = 𝐸 𝛾 − 𝐸 𝑏𝑖𝑛𝑑𝑖𝑛𝑔 𝜎 ∝ 1

𝐸 3

Compton effect:

large energies

𝜎 ∝ 1

𝐸

EM shower

A high-energy electron interacts in a medium and radiates bremsstrahlung, which turns into a 𝑒 𝑒 + pair. Also a primary interaction of a high-energy photon will produce 𝑒 𝑒 + and create a shower. The pair production process continues to produce a cascade of photons, electrons and positrons. The number of particles double after each radiation length 𝑋 0

Energy of a particle after x radiation lengths: 〈𝐸〉 = 𝐸

2 𝑥 Shower stops, when 〈𝐸〉 ≤ 𝐸 𝑐 ⟹ 𝐸 𝑐 = 𝐸

2 𝑥𝑚𝑎𝑥 ⇒ 2 𝑥 𝑚𝑎𝑥 = 𝐸

𝐸 𝑐 ⇒ ln(2 𝑥 𝑚𝑎𝑥 ) = ln ( 𝐸 𝐸

𝑐 ) ⇒ ln(2) 𝑥 𝑚𝑎𝑥 = ln ( 𝐸

𝐸 𝑐 ) ⇒ 𝑥 𝑚𝑎𝑥 = ln(𝐸/𝐸 𝑐 )

ln(2) EM

calorimeters

Measures Energy of 𝑒 , 𝑒 + , 𝛾 with 𝐸 > 100𝑀𝑒𝑉. Alternate layers of high-Z material (e.g. lead) and scintillator material. EM- shower in lead layers. Scintillator detects the created electrons. Energy resolution 𝜎 𝐸

𝐸 = 3%−10%

√𝐸/𝐺𝑒𝑉 . Hadron

calorimeter

Measures Energy of hadronic showers. Large. Again, sandwich structure with thick layers of high-density absorbers (eg steel) and thin layers of active material (eg plastic scintillators). Energy resolution 𝜎 𝐸

𝐸 = 50%

√𝐸/𝐺𝑒𝑉

Scintillators

Cost effective way to detect passage of charged particles when precise spatial info is not required. When passing, they leave some of the scintillator molecules in an excited state. The subsequent decay results in emission of UV photons. By adding fluorescent dye, the molecules of the dye absorb the UV photons and emit blue light, which is detected by photomultipliers.

Bethe-Bloch

Ionisation energy loss per unit length of relativistic charged particle passing through a medium:

𝑑𝐸

𝑑𝑥 = 𝐾𝑍 𝑒 2 𝑍 𝐴 1 𝛽 2 [ 1

2 ln ( 2𝑚 𝑒 𝑐 2 𝛽 2 𝛾 2

𝐼 0 2 ) − 𝛽 2 − δ(𝛽𝛾)]

𝐾 … constants, 𝑍 𝑒 … charge number particle, 𝑍 … charge number material,

𝐴 … mass number material, 𝐼 0 ≈ 10𝑍 𝑒𝑉 … ionization potential δ(𝛽𝛾) … Energy Correction - independent of particle mass

1

𝛽 2 , the ion does not carry electrons anymore The ion carries electrons and

has thus reduced charge relativistic rise

∝ ln(𝛽 2 𝛾 2 )

(3)

Fermi's Golden Rule Schrödinger: 𝑖ℏ 𝜕

𝜕𝑡 Ψ(𝑥, 𝑡) = 𝐻 ̂ Ψ(𝑥, 𝑡)| 𝐻 ̂ = 𝐻̂ 0 + 𝐻 ̂ (𝑥, 𝑡) ⟹ 𝑖ℏ 𝜕

𝜕𝑡 Ψ(𝑥, 𝑡) = (𝐻 ̂ 0 + 𝐻 ̂ (𝑥, 𝑡)) Ψ(𝑥, 𝑡) … (1) Ψ(𝑥, 𝑡) = ∑ c 𝑘 𝑘 (𝑡) ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ

(1) ⇒ 𝑖ℏ 𝜕

𝜕𝑡 ∑ c 𝑘 𝑘 (𝑡) ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ = (𝐻 ̂ 0 + 𝐻 ̂ (𝑥, 𝑡)) ∑ c 𝑘 𝑘 (𝑡) ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ ⟹ 𝑖ℏ ∑ 𝜕

𝜕𝑡 (c 𝑘 (𝑡) ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ )

𝑘 = ∑ 𝐻 𝑘 ̂ 0 (c 𝑘 (𝑡) ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ ) + ∑ 𝐻 𝑘 ̂ (𝑥, 𝑡) (c 𝑘 (𝑡) ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ ) 𝑖ℏ ∑ ( 𝜕

𝜕𝑡 c 𝑘 (𝑡) ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ − 𝑖 𝐸 𝑘

ℏ c 𝑘 (𝑡) ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ )

𝑘 = ∑ c 𝑘 𝑘 (𝑡) 𝐻 ̂ 0 ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ + ∑ 𝐻 𝑘 ̂ (𝑥, 𝑡) (c 𝑘 (𝑡) ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ ) | 𝐻 ̂ 0 ϕ 𝑘 =𝐸 𝑘 ϕ 𝑘 𝑖ℏ ∑ 𝜕c 𝑘 (𝑡)

𝜕𝑡 ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ

𝑘 + ∑ c 𝑘 𝑘 (𝑡) 𝐸 𝑘 ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ = ∑ c 𝑘 𝑘 (𝑡) 𝐸 𝑘 ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ + ∑ 𝐻 𝑘 ̂ (𝑥, 𝑡) (c 𝑘 (𝑡) ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ ) 𝑖ℏ ∑ 𝜕c 𝑘 (𝑡)

𝜕𝑡 ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ

𝑘 = ∑ 𝐻 𝑘 ̂ (𝑥, 𝑡) (c 𝑘 (𝑡) ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ ) |let Ψ(𝑥, 0) = ! ϕ 𝑖 (𝑥) ⟹ c 𝑘 (0)=𝛿 𝑖𝑘 ; for small perturbations also c 𝑘 (𝑡)≈𝛿 𝑖𝑘

𝑖ℏ ∑ 𝜕c 𝑘 (𝑡)

𝜕𝑡 ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ

𝑘 = ∑ 𝐻 𝑘 ̂ (𝑥, 𝑡) (𝛿 𝑖𝑘 ϕ 𝑘 (𝑥) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ ) = 𝐻 ̂ (𝑥, 𝑡) ϕ 𝑖 (𝑥) 𝑒 −𝑖𝐸 𝑖 𝑡/ℏ | ϕ 𝑘 (𝑥) = |𝑘⟩, ϕ 𝑖 (𝑥) = |𝑖⟩

𝑖ℏ ∑ 𝜕c 𝑘 (𝑡)

𝜕𝑡 |𝑘⟩𝑒 −𝑖𝐸 𝑘 𝑡

𝑘 = 𝐻 ̂ (𝑥, 𝑡) |𝑖⟩𝑒 −𝑖𝐸 𝑖 𝑡/ℏ | ⟨𝑓| ∙⟹ 𝑖ℏ ∑ 𝜕c 𝑘 (𝑡)

𝜕𝑡 ⟨𝑓|𝑘⟩𝑒 −𝑖𝐸 𝑘 𝑡/ℏ

𝑘 = ⟨𝑓| 𝐻 ̂ (𝑥, 𝑡) |𝑖⟩𝑒 −𝑖𝐸 𝑖 𝑡/ℏ | ⟨𝑓|𝑘⟩ = 𝛿 𝑓𝑘 𝑖ℏ ∑ 𝜕c 𝑘 (𝑡)

𝜕𝑡 𝛿 𝑓𝑘 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ

𝑘 = ⟨𝑓| 𝐻 ̂ (𝑥, 𝑡) |𝑖⟩𝑒 −𝑖𝐸 𝑖 𝑡/ℏ ⟹ 𝑖ℏ 𝜕c 𝑓 (𝑡)

𝜕𝑡 𝑒 −𝑖𝐸 𝑓 𝑡/ℏ = ⟨𝑓| 𝐻 ̂ (𝑥, 𝑡) |𝑖⟩𝑒 −𝑖𝐸 𝑖 𝑡/ℏ

𝜕c 𝑓 (𝑡)

𝜕𝑡 = 1

𝑖ℏ ⟨𝑓| 𝐻 ̂ (𝑥, 𝑡) |𝑖⟩𝑒 𝑖𝐸 𝑓 𝑡/ℏ 𝑒 −𝑖𝐸 𝑖 𝑡/ℏ = 1

𝑖ℏ ⟨𝑓| 𝐻 ̂ (𝑥, 𝑡) |𝑖⟩𝑒 𝑖(𝐸 𝑓 −𝐸 𝑖 )𝑡/ℏ ⟹ 𝑑c 𝑓 (𝑡) = 𝑖ℏ 1 𝑇 𝑖𝑓 𝑒 𝑖(𝐸 𝑓 −𝐸 𝑖 )𝑡/ℏ 𝑑𝑡 with Transition Matrix Element 𝑇 𝑖𝑓 = ⟨𝑓|𝐻 ̂ |𝑖⟩ ⟹ c 𝑓 (𝑡) = 1

𝑖ℏ ∫ 𝑇 0 𝑡 𝑖𝑓 𝑒 𝑖(𝐸 𝑓 −𝐸 𝑖 )𝜏/ℏ 𝑑𝜏 |assumption: 𝑇 𝑖𝑓 = 𝑇 𝑖𝑓 (𝑥) ⟹ c 𝑓 (𝑡) = 1

𝑖ℏ 𝑇 𝑖𝑓 ∫ 𝑒 0 𝑡 𝑖(𝐸 𝑓 −𝐸 𝑖 )𝜏/ℏ 𝑑𝜏 … (2)

|Ψ(𝑡)⟩ = ∑ c 𝑘 𝑘 (𝑡) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ |𝑘⟩ |⟨𝑓| ∙⟹ ⟨𝑓| Ψ(𝑡)⟩ = ∑ c 𝑘 𝑘 (𝑡) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ ⟨𝑓|𝑘⟩ = ∑ c 𝑘 𝑘 (𝑡) 𝑒 −𝑖𝐸 𝑘 𝑡/ℏ 𝛿 𝑓𝑘 ⟹ ⟨𝑓| Ψ(𝑡)⟩ = c 𝑓 (𝑡) 𝑒 −𝑖𝐸 𝑓 𝑡/ℏ … (3) Probability for a transition to state |𝑓⟩ after duration T: 𝑝 𝑓𝑖 = |⟨𝑓| Ψ(𝑇)⟩| 2 (3) = |c 𝑓 (𝑇) 𝑒 −𝑖𝐸 𝑓 𝑡/ℏ | 2 = c 𝑓 ∗ (𝑇) c 𝑓 (𝑇) (2)

𝑝 𝑓𝑖 (𝑇) = ( 1

−𝑖ℏ 𝑇 𝑖𝑓 ∫ 𝑒 𝑖(𝐸𝑓−𝐸𝑖)𝑡

′ ℏ 𝑑𝑡

𝑇

0 ) ( 1

𝑖ℏ 𝑇 𝑖𝑓 ∫ 𝑒 0 𝑇 𝑖(𝐸𝑓−𝐸𝑖)𝑡 𝑑𝑡 )| 𝜔 𝑓𝑖 ≝ 𝐸 𝑓 −𝐸 𝑖

ℏ … (4) ⟹ 𝑝 𝑓𝑖 = ( 1

−𝑖ℏ 𝑇 𝑖𝑓 ∫ 𝑒 0 𝑇 −𝑖𝜔 𝑓𝑖 𝑡 𝑑𝑡 ) ( 1

𝑖ℏ 𝑇 𝑖𝑓 ∫ 𝑒 0 𝑇 𝑖𝜔 𝑓𝑖 𝑡 𝑑𝑡 ) 𝑝 𝑓𝑖 (𝑇) = 1

2 |𝑇 𝑖𝑓 | 2 ∫ 𝑒 0 𝑇 −𝑖𝜔 𝑓𝑖 𝑡 𝑑𝑡 ∫ 𝑒 0 𝑇 𝑖𝜔 𝑓𝑖 𝑡 𝑑𝑡 | 𝑡̃ ≝ 𝑡 − 𝑇

2 ⟹ 𝑑𝑡 = 𝑑𝑡̃; 𝑡 = 𝑡̃ + 𝑇

2 , same for 𝑡 → 𝑡̃ 𝑝 𝑓𝑖 (𝑇) = 1

2 |𝑇 𝑖𝑓 | 2+ 𝑒 −𝑖𝜔 𝑓𝑖 (𝑡̃ + 𝑇 2 ) 𝑑𝑡̃

𝑇 2

𝑇 2

+ 𝑒 𝑖𝜔 𝑓𝑖 (𝑡̃+ 𝑇 2 ) 𝑑𝑡̃

𝑇 2

𝑇 2

= 1

2 |𝑇 𝑖𝑓 | 2 𝑒 −𝑖𝜔 𝑓𝑖 𝑇 2 𝑒 𝑖𝜔 𝑓𝑖 𝑇 2+ 𝑒 −𝑖𝜔 𝑓𝑖 𝑡̃ 𝑑𝑡̃

𝑇 2

𝑇 2

+ 𝑒 𝑖𝜔 𝑓𝑖 𝑡̃ 𝑑𝑡̃

𝑇 2

𝑇 2

… (5) 𝑝 𝑓𝑖 (𝑇) = 1 2 |𝑇 𝑖𝑓 | 2 2 sin(𝜔 𝑓𝑖 𝑇/2)

𝜔 𝑓𝑖 2 sin(𝜔 𝑓𝑖 𝑇/2)

𝜔 𝑓𝑖 = 4

2 |𝑇 𝑖𝑓 | 2 sin 2 (𝜔 𝑓𝑖 𝑇/2)

𝜔 𝑓𝑖 2 = 1

2 |𝑇 𝑖𝑓 | 2 sin 2 (𝜔 𝑓𝑖 𝑇/2)

(𝜔 𝑓𝑖 /2) 2 … (6) Transition Rate (probability of transition per unit time) for 𝑇 → ∞: 𝑑Γ 𝑓𝑖 = lim 𝑇→∞ 1

𝑇 𝑝 𝑓𝑖 (𝑇) (6) ⇒ 𝑑Γ 𝑓𝑖 = 1

2 |𝑇 𝑖𝑓 | 2 lim 𝑇→∞ 1

𝑇 ∫ 𝑒 𝑖𝜔 𝑓𝑖 𝑡̃+ 𝑒 −𝑖𝜔 𝑓𝑖 𝑡̃ 𝑑𝑡̃

𝑇 2

𝑇 2

2𝜋 δ(𝜔 𝑓𝑖 ) + 𝑇 𝑑𝑡̃

2

𝑇 2 = 2𝜋

2 |𝑇 𝑖𝑓 | 2 lim 𝑇→∞ 1

𝑇 ∫ + 𝑒 𝑖𝜔 𝑓𝑖 𝑡̃ δ(𝜔 𝑓𝑖 ) 𝑑𝑡̃

𝑇 2

𝑇 2 … (7)

If there are 𝑑𝑛 accessible states in the energy range [𝐸 𝑓 , 𝐸 𝑓 + 𝑑𝐸 𝑓 ], then the total not Lorentz invariant transition rate is Γ 𝑓𝑖 = ∫ 𝑑Γ 𝑓𝑖 𝑑𝑛 = ∫ 𝑑Γ 𝑓𝑖

𝑑𝑛

𝑑𝐸 𝑓 𝑑𝐸 𝑓 (7) = 2𝜋

2 |𝑇 𝑖𝑓 | 2 ∫ lim 𝑇→∞

1

𝑇 ∫ + 𝑒 𝑖𝜔 𝑓𝑖 𝑡̃ δ(𝜔 𝑓𝑖 ) 𝑑𝑡̃

𝑇 2

𝑇 2

𝑑𝑛

𝑑𝐸 𝑓 𝑑𝐸 𝑓 | 𝜔 𝑓𝑖𝐸 𝑓 −𝐸 𝑖

ℏ Γ 𝑓𝑖 = 2𝜋

2 |𝑇 𝑖𝑓 | 2 ∫ lim 𝑇→∞

1

𝑇 ∫ 𝑒 𝑖(𝐸 𝑓 −𝐸 𝑖 )𝑡̃/ℏ 𝑑𝑛

𝑑𝐸 𝑓 δ ( 𝐸 𝑓 −𝐸 𝑖

ℏ ) 𝑑𝑡̃

+ 𝑇 2

𝑇 2 𝑑𝐸 𝑓 | δ ( 𝐸 𝑓 −𝐸 𝑖

ℏ ) = ℏ δ(𝐸 𝑓 − 𝐸 𝑖 ) ⟹ Γ 𝑓𝑖 = 2𝜋

ℏ |𝑇 𝑖𝑓 | 2 ∫ lim 𝑇→∞

1

𝑇 ∫ 𝑒 𝑖(𝐸 𝑓 −𝐸 𝑖 )𝑡̃/ℏ 𝑑𝑛

𝑑𝐸 𝑓 δ(𝐸 𝑓 − 𝐸 𝑖 ) 𝑑𝑡̃

+ 𝑇 2

𝑇 2 𝑑𝐸 𝑓 = 2𝜋

ℏ |𝑇 𝑖𝑓 | 2 lim 𝑇→∞ 1

𝑇 ∫ 𝑒 0 𝑑𝑛

𝑑𝐸 𝑓 𝑑𝑡̃

+ 𝑇 2

𝑇 2 = 2𝜋

ℏ |𝑇 𝑖𝑓 | 2 𝑑𝑛

𝑑𝐸 𝑓 |

𝐸 𝑖

lim 𝑇→∞ 1

𝑇 ∫ + 𝑑𝑡̃

𝑇 2

𝑇 2

1

Γ 𝑓𝑖 = 2𝜋

ℏ |𝑇 𝑓𝑖 | 2 𝑑𝑛

𝑑𝐸 𝑓 |

𝐸 𝑖

⟹ Γ 𝑓𝑖 = 2𝜋

ℏ |𝑇 𝑓𝑖 | 2 ρ(𝐸 𝑖 ) with ρ(𝐸 𝑖 ) = 𝑑𝐸 𝑑𝑛

𝑓 |

𝐸 𝑖

... Fermi’s Golden Rule (to the first order)

Non- relati- vistic Phase Space

Final-State particles are represented by plane waves Ψ(𝑥⃗, 𝑡) = 𝐴𝑒 𝑖(𝑘 ⃗⃗∙𝑥⃗−𝜔𝑡) = 𝐴𝑒 𝑖( 𝑝 ⃗⃗⃗ ∙𝑥⃗− 𝐸 𝑡) ℏ=1 = 𝐴𝑒 𝑖(𝑝⃗∙𝑥⃗−𝐸𝑡) … (9) Normalization to one particle per one cubic volume of side 𝑎 ⟹ ∫ ∫ ∫ Ψ 0 𝑎 0 𝑎 0 𝑎 Ψ 𝑑𝑥 𝑑𝑦 𝑑𝑧 = 1 (9)

∫ ∫ ∫ 0 𝑎 0 𝑎 0 𝑎 𝐴 𝑒 −𝑖(𝑝⃗∙𝑥⃗−𝐸𝑡) 𝐴𝑒 𝑖(𝑝⃗∙𝑥⃗−𝐸𝑡) 𝑑𝑥 𝑑𝑦 𝑑𝑧 = 1 ⟹ |𝐴| 2 𝑎 3 = |𝐴| 2 𝑉 = 1 ⟹ |𝐴| = 1

√𝑉 … (10) is not Lorentz-invariant!

Boundary condition: Ψ(𝑥, 𝑦, 𝑧, 𝑡) = Ψ(𝑥 + 𝑎, 𝑦, 𝑧, 𝑡) etc. ⟹ 𝑒 𝑖𝑝 𝑥 𝑥 = 𝑒 𝑖𝑝 𝑥 (𝑥+𝑎) ⟹ 𝑝 𝑥 𝑥 = 𝑝 𝑥 (𝑥 + 𝑎) ± 2𝜋𝑛 𝑥 ⟹ 𝑝 𝑥 𝑥 = 𝑝 𝑥 𝑥 + 𝑝 𝑥 𝑎 ± 2𝜋𝑛 𝑥 ⟹ 𝑝 𝑥 𝑎 = 2𝜋𝑛 𝑥 ⟹ 𝑝 𝑥 = 2𝜋𝑛 𝑥

𝑎 , 𝑝 𝑦 = 2𝜋𝑛 𝑦

𝑎 , 𝑝 𝑧 = 2𝜋𝑛 𝑧

𝑎 … (11) ⟹ Volume of a single state in p-space: 𝑑𝑝 𝑥 = 𝑝 𝑥

𝑛 𝑥 = 2𝜋

𝑎 , 𝑑𝑝 𝑦 = 𝑝 𝑦

𝑛 𝑦 = 2𝜋

𝑎 , 𝑑𝑝 𝑧 = 𝑝 𝑧

𝑛 𝑧 = 2𝜋

𝑎 ⟹ 𝑑 3 𝑝 = ( 2𝜋

𝑎 ) 3 = (2𝜋) 3

𝑉 … (12) Number of states 𝑛 in a sphere of radius 𝑝 in p-space: 𝑛 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑝ℎ𝑒𝑟𝑒 𝑖𝑛 𝑝−𝑠𝑝𝑎𝑐𝑒

𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 𝑠𝑡𝑎𝑡𝑒 𝑖𝑛 𝑝−𝑠𝑝𝑎𝑐𝑒 = 4𝜋𝑝 3

3 1 𝑑 3 𝑝

(12) ⇒ 𝑛 = 4𝜋𝑝 3

3 𝑉 (2𝜋) 3𝑑𝑛

𝑑𝑝 = (2𝜋) 4𝜋𝑝 2 3 𝑉 … (13) Dens. of states: ρ(𝐸 𝑖 ) = 𝑑𝑛

𝑑𝐸 |

𝐸 𝑖 = 𝑑𝑛

𝑑𝑝 | 𝑑𝑝

𝑑𝐸 |

𝐸 𝑖 (13) = (2𝜋) 4𝜋𝑝 2 3 𝑉 | 𝑑𝑝

𝑑𝐸 |

𝐸 𝑖 … (14) 𝐸 2 = 𝑝 2 + 𝑚 2 ⟹ 𝑝 2 = 𝐸 2 − 𝑚 2 ⟹ 𝑝 = (𝐸 2 − 𝑚 2 ) 1 2𝑑𝑝

𝑑𝐸 = 1

2 (𝐸 2 − 𝑚 2 ) 1 2 2𝐸 = 𝐸

√𝐸 2 −𝑚 2 = 𝐸

𝑝 = 𝛾𝑚

𝛾𝑚𝛽 ⟹ 𝑑𝑝

𝑑𝐸 = 1

𝛽 … (15) Infinitesimal number of states 𝑑𝑛 𝑖 for the i-th particle in an cuboid with dimensions 𝑑𝑝 𝑥 𝑖 × 𝑑𝑝 𝑦 𝑖 × 𝑑𝑝 𝑧 𝑖 in p-space:

𝑑𝑛 𝑖 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑢𝑏𝑜𝑖𝑑

𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 𝑠𝑡𝑎𝑡𝑒 𝑖𝑛 𝑝−𝑠𝑝𝑎𝑐𝑒 = 𝑑𝑝 𝑥 𝑖 𝑑𝑝 𝑦 𝑖 𝑑𝑝 𝑧 𝑖

𝑑 3 𝑝

(4) ⇒ 𝑑𝑛 = 𝑑𝑝 𝑥 𝑖 𝑑𝑝 𝑦 𝑖 𝑑𝑝 𝑧 𝑖

(2𝜋) 3 /𝑉 | 𝑉 = ! 1 ⟹ 𝑑𝑛 = 𝑑𝑝 𝑥 𝑖 𝑑𝑝 𝑦 𝑖 𝑑𝑝 𝑧 𝑖

(2𝜋) 3 = 𝑑 3 𝑝⃗ 𝑖

(2𝜋) 3 … (16)

General non-relativistic expression for N-body phase space (with N-1 independent momenta bc. of momentum conservation):

𝑑𝑛 = 𝑑𝑛 1 𝑑𝑛 2 … 𝑑𝑛 𝑁−1 (16) = (2𝜋) 𝑑 3 𝑝⃗ 1 3𝑑 (2𝜋) 3 𝑝⃗ 𝑁−1 3 = (2𝜋) 𝑑 3 𝑝⃗ 1 3𝑑 (2𝜋) 3 𝑝⃗ 𝑁−1 3 δ 3 (𝑝⃗ 𝑎 − 𝑝⃗ 1 − ⋯ − 𝑝⃗ 𝑁 )𝑑 3 𝑝⃗ 𝑁 ⟹ 𝑑𝑛 = (2𝜋) 3 𝑑 (2𝜋) 3 𝑝⃗ 1 3𝑑 (2𝜋) 3 𝑝⃗ 𝑁 3 δ 3 (𝑝⃗ 𝑎 − 𝑝⃗ 1 − ⋯ − 𝑝⃗ 𝑁 ) … (17)

Lorentz- invariant matrix element

Non-relativistic: ∫ Ψ 𝑉 Ψ 𝑑 3 𝑥 = 1, Lorentz-invariant: ∫ Ψ 𝑉 ′∗ Ψ 𝑑 3 𝑥 = 2𝐸 … (18) ⟹ Ψ = √2𝐸Ψ ⟹ ℳ 𝑓𝑖 = ⟨Ψ 1 ′ Ψ 2 ′ … |𝐻 ̂ |Ψ 𝑎 ′ Ψ 𝑏 ′ … ⟩ = 𝑇 𝑓𝑖 √2𝐸 1 2𝐸 2 … 2𝐸 𝑎 2𝐸 𝑏 … … (19𝑎) 𝑇 𝑓𝑖 = ℳ 𝑓𝑖

1

√2𝐸 1 2𝐸 2 …2𝐸 𝑎 2𝐸 𝑏 … … (19𝑏)

(4)

Lorentz-Invariant Transition Rate for Two Body Decay

Decay 𝑎 → 1+2

(8) ⟹ Γ 𝑓𝑖 = 2𝜋

ℏ |𝑇 𝑓𝑖 | 2 ∫ δ(𝐸 0 𝑓 − 𝐸 𝑖 ) 𝑑𝑛 ℏ=1 ⇒ Γ 𝑓𝑖 = 2𝜋|𝑇 𝑖𝑓 | 2 ∫ δ(𝐸 0 𝑓 − 𝐸 𝑖 ) 𝑑𝑛| 𝐸 𝑖 = 𝐸 𝑎 , 𝐸 𝑓 = 𝐸 1 + 𝐸 2 ⟹ Γ 𝑓𝑖 = 2𝜋|𝑇 𝑓𝑖 | 2 ∫ δ(𝐸 1 + 𝐸 2 − 𝐸 𝑎 ) 𝑑𝑛 = 2𝜋|𝑇 𝑖𝑓 | 2 ∫ δ(𝐸 𝑎 − 𝐸 1 − 𝐸 2 ) 𝑑𝑛 (17)

Γ 𝑓𝑖 = (2𝜋) 4 |𝑇 𝑓𝑖 | 2 ∫ δ(𝐸 𝑎 − 𝐸 1 − 𝐸 2 ) δ 3 (𝑝⃗ 𝑎 − 𝑝⃗ 1 − 𝑝⃗ 2 ) 𝑑 3 𝑝⃗ 1

(2𝜋) 3 𝑑 3 𝑝⃗ 2 (2𝜋) 3

(19𝑏)

⇒ Γ 𝑓𝑖 = (2𝜋) 4

2𝐸 𝑎 |ℳ 𝑓𝑖 | 2 ∫ δ(𝐸 𝑎 − 𝐸 1 − 𝐸 2 ) δ 3 (𝑝⃗ 𝑎 − 𝑝⃗ 1 − 𝑝⃗ 2 ) 𝑑 3 𝑝⃗ 1

(2𝜋) 3 2𝐸 1 𝑑 3 𝑝⃗ 2

(2𝜋) 3 2𝐸 2 … (20) Lorentz invariant transition rate

d𝐿𝐼𝑃𝑆:

Lorentz invar.

phase space and transi- tion rate with 4- vectors

d𝐿𝐼𝑃𝑆 = 𝑑 3 𝑝⃗ 1

(2𝜋) 3 2𝐸 1𝑑 3 𝑝⃗ 𝑁

(2𝜋) 3 2𝐸 𝑁 … (21) (20) ⇒ 𝑑Γ 𝑓𝑖 = (2𝜋) 4

2𝐸 𝑎 |ℳ 𝑓𝑖 | 2 ∫ δ(𝐸 𝑎 − 𝐸 1 − 𝐸 2 ) δ 3 (𝑝⃗ 𝑎 − 𝑝⃗ 1 − 𝑝⃗ 2 ) d𝐿𝐼𝑃𝑆 Γ 𝑓𝑖 = (2𝜋) 4

2𝐸 𝑎 |ℳ 𝑓𝑖 | 2 ∫ δ 4 (𝑝 𝑎

𝜇 − 𝑝 1 𝜇 − 𝑝 2 𝜇 ) d𝐿𝐼𝑃𝑆 … (22)

∫ δ(𝐸 𝑖 2 − 𝑝⃗ 𝑖 2 − 𝑚 𝑖 2 ) 𝑑𝐸 𝑖 = ∫ δ(f(𝐸 𝑖 )) 𝑑𝐸 𝑖 = ∫ | f (𝐸 1

𝑟𝑜𝑜𝑡 ) | δ(𝐸 𝑖 − 𝐸 𝑟𝑜𝑜𝑡 ) 𝑑𝐸 𝑖 with f(𝐸 𝑖 ) = 𝐸 𝑖 2 − 𝑝⃗ 𝑖 2 − 𝑚 2 and f(𝐸 𝑟𝑜𝑜𝑡 ) = 0 … (23) f(𝐸 𝑟𝑜𝑜𝑡 ) = 0 (17) ⇒ 𝐸 𝑟𝑜𝑜𝑡 2 − 𝑝⃗ 𝑖 2 − 𝑚 𝑖 2 = 0 ⟹ 𝐸 𝑟𝑜𝑜𝑡 2 = 𝑝⃗ 𝑖 2 + 𝑚 𝑖 2 ⟹ 𝐸 𝑟𝑜𝑜𝑡 = √𝑝⃗ 𝑖 2 + 𝑚 𝑖 2 … (24)

f (𝐸 𝑟𝑜𝑜𝑡 ) = (23) 𝑑

𝑑𝐸 𝑖 (𝐸 𝑖 2 − 𝑝⃗ 𝑖 2 − 𝑚 𝑖 2 )|

𝐸 𝑖 =𝐸 𝑟𝑜𝑜𝑡

= 2𝐸 𝑖 | 𝐸 𝑖 =𝐸 𝑟𝑜𝑜𝑡 = 2𝐸 𝑟𝑜𝑜𝑡

(24) ⇒ f (𝐸 𝑟𝑜𝑜𝑡 ) = 2√𝑝⃗ 𝑖 2 + 𝑚 𝑖 2 } (23)

∫ δ(𝐸 𝑖 2 − 𝑝⃗ 𝑖 2 − 𝑚 𝑖 2 ) 𝑑𝐸 𝑖 = ∫ 1

2√𝑝⃗ 𝑖 2 +𝑚 2

δ(𝐸 𝑖 − √𝑝⃗ 𝑖 2 + 𝑚 𝑖 2 ) 𝑑𝐸 𝑖 = 1

2√𝑝⃗ 𝑖 2 +𝑚 2

| √𝑝⃗ 𝑖 2 + 𝑚 𝑖 2 =𝐸 𝑖 ⟹ ∫ δ(𝐸 𝑖 2 − 𝑝⃗ 𝑖 2 − 𝑚 𝑖 2 ) 𝑑𝐸 𝑖 = 1

2𝐸 𝑖 … (25) (21) ⇒ d𝐿𝐼𝑃𝑆 = 1

(2𝜋) 3𝑁 δ(𝐸 1 2 − 𝑝⃗ 1 2 − 𝑚 1 2 ) . . . δ(𝐸 𝑁 2 − 𝑝⃗ 𝑁 2 − 𝑚 𝑁 2 ) 𝑑𝐸 1 𝑑 3 𝑝⃗ 1 … 𝑑𝐸 𝑁 𝑑 3 𝑝⃗ 𝑁 | 𝑝 𝑖 𝜇 𝑝 𝜇 𝑖 = 𝐸 𝑖 2 − 𝑝⃗ 𝑖 2 d𝐿𝐼𝑃𝑆 = (2𝜋) 1 3𝑁 δ(𝑝 1 𝜇 𝑝 𝜇 1 − 𝑚 1 2 ) . . . δ(𝑝 𝑁 𝜈 𝑝 𝜈 𝑁 − 𝑚 𝑁 2 ) 𝑑 4 𝑝 1 … 𝑑 4 𝑝 𝑁

(22) ⇒

Γ 𝑓𝑖 = (2𝜋) 4

2𝐸 𝑎 1

(2𝜋) 6 |ℳ 𝑓𝑖 | 2 ∫ δ 4 (𝑝 𝑎 𝜇 − 𝑝 1 𝜇 − 𝑝 2 𝜇 ) δ(𝑝 1 𝜇 𝑝 𝜇 1 − 𝑚 1 2 ) δ(𝑝 2 𝜈 𝑝 𝜈 2 − 𝑚 2 2 ) 𝑑 4 𝑝 1 𝑑 4 𝑝 2 … (26) LI transition rate with 4-vectors 𝑎 → 1+2

in center of mass frame

𝐸 𝑎 = 𝑚 𝑎 , 𝑝⃗ 𝑎 = 0 (20) ⇒ Γ 𝑓𝑖 = (2𝜋) 4

2𝑚 𝑎 |ℳ 𝑓𝑖 | 2 ∫ δ(𝑚 𝑎 − 𝐸 1 − 𝐸 2 ) δ 3 (−𝑝⃗ 1 − 𝑝⃗ 2 ) 𝑑 3 𝑝⃗ 1

(2𝜋) 3 2𝐸 1 𝑑 3 𝑝⃗ 2 (2𝜋) 3 2𝐸 2 ⟹ Γ 𝑓𝑖 = 𝑝

32𝜋 2 𝑚 𝑎 2 ∫|ℳ 𝑓𝑖 | 2 𝑑Ω with 𝑝 = 1

2𝑚 𝑎 √(𝑚 𝑎 2 − (𝑚 1 + 𝑚 2 ) 2 )(𝑚 𝑎 2 − (𝑚 1 − 𝑚 2 ) 2 ) … (27) Interaction Rate and Interaction Cross-Section

Cross-

section 𝜎 = # 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑠

#𝑡𝑎𝑟𝑔𝑒𝑡_𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠×𝑡𝑖𝑚𝑒 ∙ 1

𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡_𝑓𝑙𝑢𝑥 = # 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑠

#𝑡𝑎𝑟𝑔𝑒𝑡_𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠×𝑡𝑖𝑚𝑒 ∙ 𝑡𝑖𝑚𝑒∙𝑎𝑟𝑒𝑎

#𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡_𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 diff cr.

sect.:

𝑑𝜎

𝑑Ω ; 𝑑Ω = sin(𝜗) 𝑑𝜗 𝑑𝜑 doub diff

𝑑 2 𝜎 𝑑Ω𝑑E Interaction

probability and rate

Interaction probability: 𝑑𝑃 = 𝑑𝑁 𝑏

𝐴 𝜎 = 𝑛 𝑏 𝑑𝑉

𝐴 𝜎 = 𝑛 𝑏 (𝑣 𝑎 +𝑣 𝑏 ) 𝑑𝑡 𝐴

𝐴 𝜎 = 𝑛 𝑏 (𝑣 𝑎 + 𝑣 𝑏 )𝜎𝑑𝑡 … (1) Interaction rate per particle of type a: 𝑟 𝑎 = 𝑑𝑃

𝑑𝑡 (1) = 𝑛 𝑏 (𝑣 𝑎 + 𝑣 𝑏 )𝜎 = 𝑛 𝑏 𝑣𝜎 … (2) Total interact. rate: rate = 𝑟 𝑎 𝑁 𝑎 = 𝑟 𝑎 𝑛 𝑎 𝑉 = (2) 𝑛 𝑏 𝑣𝜎𝑛 𝑎 𝑉 = (𝑛 𝑎 𝑣)(𝑛 𝑏 𝑉)𝜎 = 𝜙 𝑎 𝑁 𝑏 𝜎 … (3) with flux of particles type a: 𝜙 𝑎 = 𝑛 𝑎 𝑣 = 𝑛 𝑎 (𝑣 𝑎 + 𝑣 𝑏 ) … (4)

Lorenz invariant flux

Γ 𝑓𝑖 = rate = (3) 𝜙 𝑎 𝑁 𝑏 𝜎 = 𝜙 𝑎 𝑛 𝑏 𝑉𝜎 = (4) 𝑛 𝑎 (𝑣 𝑎 + 𝑣 𝑏 )𝑛 𝑏 𝑉𝜎 … (5)

normalizing wavefunctions to 1 particle per Volume ⟹ 𝑛 𝑎 = 𝑛 𝑏 = 1 (5) ⇒ Γ 𝑓𝑖 = (𝑣 𝑎 + 𝑣 𝑏 )𝑉𝜎 … (6) normalizing volume to 1⟹ Γ 𝑓𝑖 = (𝑣 𝑎 + 𝑣 𝑏 )𝜎 ⟹ 𝜎 = Γ 𝑓𝑖

𝑣 𝑎 +𝑣 𝑏 … (7) Γ 𝑓𝑖 = (2𝜋) 4

2𝐸 𝑎 2𝐸 𝑏 |ℳ 𝑓𝑖 | 2 ∫ δ(𝐸 𝑎 + 𝐸 𝑏 − 𝐸 1 − 𝐸 2 ) δ 3 (𝑝⃗ 𝑎 + 𝑝⃗ 𝑏 − 𝑝⃗ 1 − 𝑝⃗ 2 ) 𝑑 3 𝑝⃗ 1

(2𝜋) 3 2𝐸 1 𝑑 3 𝑝⃗ 2 (2𝜋) 3 2𝐸 2 ⟹ Γ 𝑓𝑖 = 1

4𝐸 𝑎 𝐸 𝑏 1

(2𝜋) 2 |ℳ 𝑓𝑖 | 2 ∫ δ(𝐸 𝑎 + 𝐸 𝑏 − 𝐸 1 − 𝐸 2 ) δ 3 (𝑝⃗ 𝑎 + 𝑝⃗ 𝑏 − 𝑝⃗ 1 − 𝑝⃗ 2 ) 𝑑 3 𝑝⃗ 1

2𝐸 1 𝑑 3 𝑝⃗ 2

2𝐸 2 (7) ⇒

𝜎 = 1

4𝐸 𝑎 𝐸 𝑏 (𝑣 𝑎 +𝑣 𝑏 ) 1

(2𝜋) 2 |ℳ 𝑓𝑖 | 2 ∫ δ(𝐸 𝑎 + 𝐸 𝑏 − 𝐸 1 − 𝐸 2 ) δ 3 (𝑝⃗ 𝑎 + 𝑝⃗ 𝑏 − 𝑝⃗ 1 − 𝑝⃗ 2 ) 𝑑 3 𝑝⃗ 1

2𝐸 1 𝑑 3 𝑝⃗ 2

2𝐸 2 … (8)

Lorentz invariant flux factor: 𝐹 = 4𝐸 𝑎 𝐸 𝑏 (|𝑣⃗ 𝑎 | + |𝑣⃗ 𝑏 |) = 4𝐸 𝑎 𝐸 𝑏 (𝑣 𝑎 + 𝑣 𝑏 ) = 4√(𝑝 𝑎 ∙ 𝑝 𝑏 ) 2 − 𝑚 𝑎 2 𝑚 𝑏 2 … (9) (8) ⇒ 𝜎 = 1

𝐹 1

(2𝜋) 2 |ℳ 𝑓𝑖 | 2 ∫ δ(𝐸 𝑎 + 𝐸 𝑏 − 𝐸 1 − 𝐸 2 ) δ 3 (𝑝⃗ 𝑎 + 𝑝⃗ 𝑏 − 𝑝⃗ 1 − 𝑝⃗ 2 ) 𝑑 3 𝑝⃗ 1

2𝐸 1 𝑑 3 𝑝⃗ 2

2𝐸 2 … (10)

Scattering in center- of-mass frame

𝐹 = 4𝐸 𝑎 ∗ 𝐸 𝑏 (𝑣 𝑎 ∗ + 𝑣 𝑏 ) = 4𝐸 𝑎 ∗ 𝐸 𝑏 ( 𝑝 𝑎

𝐸 𝑎 + 𝑝 𝑏

𝐸 𝑏 )| 𝑝⃗ 𝑎 ∗ = −𝑝⃗ 𝑏 ⟹ |𝑝⃗ 𝑎 ∗ | = |𝑝⃗ 𝑏 | = 𝑝 𝑖 ⟹ 𝐹 = 4𝐸 𝑎 ∗ 𝐸 𝑏 ( 𝑝 𝑖

𝐸 𝑎 + 𝑝 𝑖

𝐸 𝑏 ) = 4𝐸 𝑏 𝑝 𝑖 + 4𝐸 𝑎 ∗ 𝑝 𝑖 ⟹ 𝐹 = 4𝑝 𝑖 (𝐸 𝑎 ∗ + 𝐸 𝑏 ) (10) ⇒ Γ 𝑓𝑖 = 1

4𝑝 𝑖 (𝐸 𝑎 +𝐸 𝑏 ) 1

(2𝜋) 2 |ℳ 𝑓𝑖 | 2 ∫ δ(𝐸 𝑎 ∗ + 𝐸 𝑏 − 𝐸 1 ∗ − 𝐸 2 ∗ ) δ 3 (𝑝⃗ 𝑎 ∗ + 𝑝⃗ 𝑏 − 𝑝⃗ 1 ∗ − 𝑝⃗ 2 ∗ ) 𝑑 3 𝑝⃗ 1

2𝐸 1 𝑑 3 𝑝⃗ 2

2𝐸 2 | 𝑝⃗ 𝑎 ∗ = −𝑝⃗ 𝑏

𝜎 = 1

4𝑝 𝑖 (𝐸 𝑎 +𝐸 𝑏 ) 1

(2𝜋) 2 |ℳ 𝑓𝑖 | 2 ∫ δ(𝐸 𝑎 ∗ + 𝐸 𝑏 ∗ − 𝐸 1 ∗ − 𝐸 2 ∗ ) δ 3 (−𝑝⃗ 1 ∗ − 𝑝⃗ 2 ∗ ) 𝑑 3 𝑝⃗ 1

2𝐸 1 𝑑 3 𝑝⃗ 2

2𝐸 2 | 𝐸 𝑎 ∗ + 𝐸 𝑏 ∗ ≝ √𝑠 𝜎 = 1

4𝑝 𝑖 √𝑠 1

(2𝜋) 2 |ℳ 𝑓𝑖 | 2 ∫ δ(√𝑠 − 𝐸 1 ∗ − 𝐸 2 ) δ 3 (𝑝⃗ 1 ∗ + 𝑝⃗ 2 ) 𝑑 3 𝑝⃗ 1

2𝐸 1 𝑑 3 𝑝⃗ 2

2𝐸 2 = 1

16𝜋 2 𝑝 𝑖 √𝑠 𝑝 𝑓

4√𝑠 ∫|ℳ 𝑓𝑖 | 2 𝑑Ω ⟹ 𝜎 = 1

64𝜋 2 𝑠 𝑝 𝑓

𝑝 𝑖 ∫|ℳ 𝑓𝑖 | 2 𝑑Ω … (11) Mandelstam Variables

s-channel annihilation 𝑠 = (𝑝 1 𝜇 + 𝑝 2 𝜇 ) 2 𝑠 = (𝑝 3 𝜇 + 𝑝 4 𝜇 ) 2 𝑞 = √𝑠 c.o.m energy Lorentz-invariant⟹

t-channel scattering 𝑡 = (𝑝 1 𝜇 − 𝑝 3 𝜇 ) 2 𝑡 = (𝑝 2 𝜇 − 𝑝 4 𝜇 ) 2 𝑞 = √𝑡 Lorentz-invariant

u-channel

scattering

𝑢 = (𝑝 1 𝜇 − 𝑝 4 𝜇 ) 2

𝑢 = (𝑝 2 𝜇 − 𝑝 3 𝜇 ) 2

𝑞 = √𝑢

Lorentz-invariant

(5)

Klein-Gordon Equation

Derivation of equation

𝐸 2 𝑐=1 = 𝑝 2 + 𝑚 2 ⟹ 𝐸 2 − 𝑝 2 = 𝑚 2 ⟹ 𝐸̂ 2 − 𝑝̂ 2 = 𝑚 2 | ∙ Ψ ⟹ (𝐸̂ 2 − 𝑝̂ 2 )Ψ = 𝑚 2 Ψ|𝐸̂ = 𝑖 𝜕

𝜕𝑡 ⟹ 𝐸̂ 2 = − 𝜕 2

𝜕𝑡 2 ⟹ (− 𝜕 2

𝜕𝑡 2 − 𝑝̂ 2 ) Ψ = 𝑚 2 Ψ| 𝑝̂ = −𝑖∇ ⃗⃗⃗⟹ 𝑝̂ 2 = −∇ ⃗⃗⃗ 2 ⟹ (− 𝜕 2

𝜕𝑡 2 + ∇ ⃗⃗⃗ 2 ) Ψ = 𝑚 2 Ψ| ∙ (−1) ⟹ 𝜕 2

𝜕𝑡 2 Ψ − ∇ ⃗⃗⃗ 2 Ψ = −𝑚 2 Ψ ⟹

𝜕 𝜇 𝜕 𝜇 Ψ = −𝑚 2 Ψ ⟹ 𝜕 𝜇 𝜕 𝜇 Ψ + 𝑚 2 Ψ = 0 ⟹ (𝜕 𝜇 𝜕 𝜇 + 𝑚 2 )Ψ = 0 Dirac Equation

Start like Klein Gord

𝐸 2 = 𝑝 2 + 𝑚 2 ⟹ 𝐻 ̂ 𝐷 2 = 𝑝̂ 2 + 𝑚 2 = (−𝑖∇ ⃗⃗⃗) 2 + 𝑚 2 = −∇ ⃗⃗⃗ 2 + 𝑚 2 … (1) Schrödinger: 𝑖𝜕 𝑡 Ψ = 𝐻 ̂ 𝐷 Ψ|𝑖𝜕 𝑡 ∙ ⟹ −𝜕 𝑡 2 Ψ = 𝑖𝜕 𝑡 (𝐻 ̂ 𝐷 Ψ) ⇒

−𝜕 𝑡 2 Ψ = 𝑖(𝜕 𝑡 𝐻 ̂ 𝐷 )Ψ + 𝑖𝐻 ̂ 𝐷 𝜕 𝑡 Ψ|𝜕 𝑡 𝐻 ̂ 𝐷 = 0 ⟹−𝜕 𝑡 2 Ψ = 𝐻 ̂ 𝐷 𝑖𝜕 𝑡 Ψ|𝑖𝜕 𝑡 Ψ = 𝐻 ̂ 𝐷 Ψ ⟹ −𝜕 𝑡 2 Ψ = 𝐻 ̂ 𝐷 2 Ψ (1) ⇒ −𝜕 𝑡 2 Ψ = (−∇ ⃗⃗⃗ 2 + 𝑚 2 )Ψ… (2)

Ansatz Ψ(𝑟⃗, 𝑡) = (

Ψ 1 (𝑟⃗, 𝑡) Ψ 2 (𝑟⃗, 𝑡) Ψ 3 (𝑟⃗, 𝑡) Ψ 4 (𝑟⃗, 𝑡))

(spinor ∈ ℂ 4 ) In order to allow Lorentz-covariance, 𝐻 ̂ 𝐷 , in x-space, must be linear in spatial derivatives:

𝐻 ̂ 𝐷 = 𝛼 ̳ 𝑖 𝑝̂ 𝑖 + 𝛽 ̳𝑚 … (3) with 𝛼 ̳ 𝑖 and 𝛽 ̳ being 4x4 matrices acting on the components of Ψ

Derivation of Dirac Equation

(1) ⟹ 𝐻 ̂ 𝐷 2 = −∇ ⃗⃗⃗ 2 + 𝑚 2 ⟹ 𝐻 ̂ 𝐷 2 = −𝜕 𝑖 𝜕 𝑖 + 𝑚 2 ⟹ 𝐻 ̂ 𝐷 2 = −𝜕 𝑖 𝜕 𝑗 𝛿 𝑖𝑗 + 𝑚 2 (3) ⇒ (𝛼 ̳ 𝑖 𝑝̂ 𝑖 + 𝛽 ̳𝑚) (𝛼 ̳ 𝑗 𝑝̂ 𝑗 + 𝛽 ̳𝑚) = −𝜕 𝑖 𝜕 𝑗 𝛿 𝑖𝑗 + 𝑚 2

1 𝑖 𝛼 ̳ 𝑖 𝜕 𝑖

1

𝑖 𝛼 ̳ 𝑗 𝜕 𝑗 + 𝛽 ̳𝑚𝛽 ̳𝑚 + 𝛽 ̳𝑚 1 𝑖 𝛼 ̳ ⏟ 𝑗 𝜕 𝑗 𝑗→𝑖

+ 1

𝑖 𝛼 ̳ 𝑖 𝜕 𝑖 𝛽 ̳𝑚 = (−𝜕 𝑖 𝜕 𝑗 𝛿 𝑖𝑗 + 𝑚 2 )𝟙

−𝛼 ̳ 𝑖 𝛼 ̳ 𝑗 𝜕 𝑖 𝜕 𝑗 + 𝑚 2 𝛽 ̳ 2 + 1

𝑖 𝑚𝛽 ̳𝛼 ̳ 𝑖 𝜕 𝑖 + 1

𝑖 𝑚𝛼 ̳ 𝑖 𝛽 ̳𝜕 𝑖 = (−𝜕 𝑖 𝜕 𝑗 𝛿 𝑖𝑗 + 𝑚 2 )𝟙

−𝛼 ̳ 𝑖 𝛼 ̳ 𝑗 𝜕 𝑖 𝜕 𝑗 + 𝑚 2 𝛽 ̳ 2 − 𝑖𝑚 (𝛽 ̳𝛼 ̳ 𝑖 + 𝛼 ̳ 𝑖 𝛽 ̳) 𝜕 𝑖 = (−𝜕 𝑖 𝜕 𝑗 𝛿 𝑖𝑗 + 𝑚 2 )𝟙 Coefficients of -𝜕 𝑖 𝜕 𝑗 : 𝛼 ̳ 𝑖 𝛼 ̳ 𝑗 = 𝛿 𝑖𝑗 𝟙{

𝑖𝑓 𝑖=𝑗

⇒ 𝛼 ̳ 𝑖 𝛼 ̳ 𝑖 = 1

2 [𝛼 ̳ 𝑖 , 𝛼 ̳ 𝑖 ]

+ = 𝟙

𝑖𝑓 𝑖≠𝑗

⇒ 𝛼 ̳ 𝑖 𝛼 ̳ 𝑗 = 𝟘 = 𝛼 ̳ 𝑗 𝛼 ̳ 𝑖1

2 [𝛼 ̳ 𝑖 , 𝛼 ̳ 𝑗 ]

+ = 𝟘

} ⟹ [𝛼 ̳ 𝑖 , 𝛼 ̳ 𝑗 ] + = 2𝛿 𝑖𝑗 𝟙 Clifford algebra

Coefficients of 𝑚 2 : 𝛽 ̳ 2 = 𝟙 solved by 𝛽 ̳ = 𝛾 0 = ( 𝟙 2 𝟘 2

𝟘 2 −𝟙 2 ) Coefficients of 𝟘: 𝛼 ̳ 𝑖 𝛽 ̳ + 𝛽 ̳𝛼 ̳ 𝑖 = [𝛼 ̳ 𝑖 , 𝛽 ̳]

+ = 𝟘 solved by 𝛼 𝑖 = ( 𝟘 2 𝜎 𝑖

𝜎 𝑖 𝟘 2 ) with 𝜎 1 = ( 0 1

1 0 ) , 𝜎 2 = ( 0 −𝑖

𝑖 0 ) , 𝜎 3 = ( 1 0 0 −1 ) Free particle

Dirac Equa- tion with γ Matrices

𝑖𝜕 𝑡 Ψ = 𝐻 ̂ 𝐷 Ψ (3) ⇒ 𝑖𝜕 𝑡 Ψ = (𝛼 ̳ 𝑖 𝑝̂ 𝑖 + 𝛽 ̳𝑚) Ψ | 𝑝̂ 𝑖 = 1

𝑖 𝜕 𝑖 ⟹ i𝜕 𝑡 Ψ = ( 1

𝑖 𝛼 ̳ 𝑖 𝜕 𝑖 + 𝛽 ̳𝑚) Ψ ⟹ i𝜕 𝑡 Ψ − 1

𝑖 𝛼 ̳ 𝑖 𝜕 𝑖 Ψ − 𝛽 ̳𝑚Ψ = 0 ⟹ i𝜕 𝑡 Ψ + 𝑖𝛼 ̳ 𝑖 𝜕 𝑖 Ψ − 𝛽 ̳𝑚Ψ = 0 | 𝛽 ̳ ∙ ⟹ i𝛽 ̳𝜕 𝑡 Ψ + 𝑖𝛽 ̳𝛼 ̳ 𝑖 𝜕 𝑖 Ψ − 𝛽 ̳ 2 𝑚Ψ = 0| 𝛽 ̳ ≝ 𝛾 0 , 𝛽 ̳𝛼 ̳ 𝑖 = 𝛾 𝑖 , 𝛽 ̳ 2 = 𝟙

i𝛾 0 𝜕 0 Ψ + 𝑖𝛾 𝑖 𝜕 𝑖 Ψ − 𝟙𝑚Ψ = 0|𝛾 𝜇 = (𝛾 0 , 𝛾 𝑖 ) 𝑇 ⟹ (i𝛾 𝜇 𝜕 𝜇 − 𝑚)Ψ = 0 ⟹ (𝑖∂ / − 𝑚)Ψ = 0 with ∂ / ≝ 𝛾 𝜇 𝜕 𝜇 , 𝑚 … rest mass Properties of Gamma Matrices 𝜸 𝝁

Gamma Matrices (Dirac represent- tation)

𝛾 0 ≝ 𝛽 ̳ = (

1 0 0 0 0 1 0 0 0 0 −1 0 0 0 0 −1

) 𝛾 1 ≝ 𝛽 ̳𝛼 ̳ 1 = (

0 0 0 1 0 0 1 0 0 −1 0 0

−1 0 0 0

) 𝛾 2 ≝ 𝛽 ̳𝛼 ̳ 2 = (

0 0 0 −𝑖 0 0 𝑖 0 0 𝑖 0 0

−𝑖 0 0 0

) Clifford algebra

𝛾 3 ≝ 𝛽 ̳𝛼 ̳ 3 = (

0 0 1 0 0 0 0 −1

−1 0 0 0 0 1 0 0

) 𝛾 5 ≝ 𝑖𝛾 0 𝛾 1 𝛾 2 𝛾 3 = (

0 0 1 0

0 0 0 1

1 0 0 0

0 1 0 0

)

(𝛾 0 ) 2 = 𝛽 ̳ 2 = 𝟙 (𝛾 𝑖 ) 2 =−𝟙 𝛾 0† =(𝛾 0 ) −1 =𝛾 0 (𝛾 𝑖 ) =−𝛾 𝑖

[𝛾 𝜇 , 𝛾 𝜈 ] + =2𝑔 𝜇𝜈 𝟙 (𝛾 5 ) 2 =𝟙 𝛾 5† =(𝛾 5 ) −1 =𝛾 5 𝛾 5 𝛾 𝜇 =−𝛾 𝜇 𝛾 5

(𝛾 𝜇 ) =𝛾 0 𝛾 𝜇 𝛾 0 𝛾 0 𝛾 1 =−𝛾 1 𝛾 0 𝛾 0 𝛾 2 =−𝛾 2 𝛾 0 𝛾 0 𝛾 3 =−𝛾 3 𝛾 0 Proof

(𝛾 𝑖 ) 2 = −𝟙

(𝛾 𝑖 ) 2 = 𝛽 ̳𝛼 ̳ 𝑖 𝛽 ̳𝛼 ̳ 𝑖 = (𝛽 ̳𝛼 ̳ 𝑖 + 𝛼 ̳ 𝑖 𝛽 ̳ − 𝛼 ̳ 𝑖 𝛽 ̳) 𝛽 ̳𝛼 ̳ 𝑖 = ([𝛽 ̳, 𝛼 ̳ 𝑖 ]

+ − 𝛼 ̳ 𝑖 𝛽 ̳) 𝛽 ̳𝛼 ̳ 𝑘 | [𝛽 ̳, 𝛼 ̳ 𝑖 ]

+ = 𝟘 ⟹ (𝛾 𝑖 ) 2 = −𝛼 ̳ 𝑖 𝛽 ̳𝛽 ̳𝛼 ̳ 𝑖 | 𝛽 ̳𝛽 ̳ = 𝛽 ̳ 2 = 𝟙 ⟹ (𝛾 𝑖 ) 2 = −𝛼 ̳ 𝑖 𝛼 ̳ 𝑖 |𝛼 ̳ 𝑖 𝛼 ̳ 𝑖 = 𝟙 ⟹ (𝛾 𝑖 ) 2 = −𝟙

Proof 𝛾 0 hermitian

(𝛾 0 ) = 𝛽 ̳ = 𝛽 ̳ ⟹ (𝛾 0 ) = 𝛾 0 Proof 𝛾 𝑖 anti

hermitian (𝛾 𝑖 ) = (𝛽 ̳𝛼 ̳ 𝑖 ) = 𝛼 ̳ 𝑖 𝛽 ̳ = 𝛼 ̳ 𝑖 𝛽 ̳ = 𝛼 ̳ 𝑖 𝛽 ̳ + 𝛽 ̳𝛼 ̳ 𝑖 − 𝛽 ̳𝛼 ̳ 𝑖 = [𝛼 ̳ 𝑖 , 𝛽 ̳]

+ − 𝛽 ̳𝛼 ̳ 𝑖 | [𝛼 ̳ 𝑖 , 𝛽 ̳]

+ = 𝟘 ⟹ 𝛼 ̳ 𝑖 𝛽 ̳ = −𝛽 ̳𝛼 ̳ 𝑖 (𝛾 𝑖 ) = −𝛽 ̳𝛼 ̳ 𝑖 = −𝛾 𝑖 𝛼 ̳ 𝑖 2 , 𝛽 ̳ 2 𝛼 ̳ 𝑖 2 = 𝛽 ̳ 2 = 𝟙 ⟹ Eigenvalues = ±1 (𝛾 𝑖 ) 2 = −𝟙 ⟹ Eigenvalues = ±𝑖 tr(𝛾 𝜇 𝛾 𝜈 ) = 4𝑔 𝜇𝜈

tracelessness 𝛼 ̳ 𝑖 = 𝟙𝛼 ̳ 𝑖 = 𝛽 ̳𝛽 ̳𝛼 ̳ 𝑖 = −𝛽 ̳𝛼 ̳ 𝑖 𝛽 ̳ ⟹ tr(𝛼 ̳ 𝑖 ) = tr (−𝛽 ̳𝛼 ̳ 𝑖 𝛽 ̳) = tr (−𝛼 ̳ 𝑖 𝛽 ̳𝛽 ̳) = tr(−𝛼 ̳ 𝑖 ) ⟹ tr(𝛼 ̳ 𝑖 ) = 0 tr (𝛽 ̳) = 0 tr(𝛾 𝜇 ) = 0 Repr. with

Pauli matr.: 𝛼 ̳ 𝑖 = ( 𝟘 2 𝜎 𝑖

𝜎 𝑖 𝟘 2 ) , 𝛽 ̳ = ( 𝟙 2 𝟘 2

𝟘 2 −𝟙 2 ) , 𝛾 𝑖 = 𝛽 ̳𝛼 ̳ 𝑖 = ( 𝟘 2 𝜎 𝑖

−𝜎 𝑖 𝟘 2 ) Probabiliy Density of Dirac Equation

Adjoint Dirac equation

(i𝛾 𝜇 𝜕 𝜇 − 𝑚)Ψ = 0| ⟹ Ψ (−i(𝛾 𝜇 ) 𝜕 𝜇 − 𝑚) = 0|(𝛾 𝜇 ) = 𝛾 0 𝛾 𝜇 𝛾 0 ⟹ Ψ (−i(𝛾 0 𝛾 𝜇 𝛾 0 ) 𝜕 ← 𝜇 − 𝑚𝛾 0 𝛾 0 ) = 0 ⟹ Ψ 𝛾 0 (−i𝛾 𝜇 ← 𝜕 𝜇 − 𝑚)𝛾 0 = 0|Ψ 𝛾 0 ≝ Ψ ⟹ Ψ(−i𝛾 𝜇 ← 𝜕 𝜇 − 𝑚) = 0| ∙ (−1) ⟹ Ψ(i𝛾 𝜇 ← 𝜕 𝜇 + 𝑚) = 0|𝛾 𝜇 ← 𝜕 𝜇 ≝ ←∂ /

Ψ (i←∂ / + 𝑚) = 0 with ←∂/ ≝ 𝛾 𝜇 ← 𝜕 𝜇

and Ψ ≝ Ψ 𝛾 0

continuity equation

Dirac: (𝑖∂ / − 𝑚)Ψ = 0|Ψ ∙ ⟹ Ψ(𝑖∂ / − 𝑚)Ψ = 0 … (1), adjoint Dirac: Ψ(i←∂ / + 𝑚) = 0|∙ Ψ ⟹ Ψ (i←∂ / + 𝑚) Ψ = 0 … (2) (2) + (1) ⟹ iΨ←∂ /Ψ + Ψ𝑚Ψ + iΨ∂ /Ψ − Ψ𝑚Ψ = 0|: 𝑖 ⟹ Ψ(←∂ / + ∂ /)Ψ = 0 ⟹ Ψ(𝛾 𝜇 ← 𝜕 𝜇 + 𝛾 𝜇 𝜕 𝜇 )Ψ = 0 ⟹

(Ψ𝛾 𝜇 ) 𝜕 ← 𝜇 Ψ + Ψ𝛾 𝜇 𝜕 𝜇 Ψ = 0|(Ψ𝛾 𝜇 ) 𝜕 ← 𝜇 = 𝛾 𝜇 Ψ 𝜕 ← 𝜇 = 𝜕 𝜇 Ψ𝛾 𝜇 ⟹ 𝜕 𝜇 Ψ𝛾 𝜇 Ψ + Ψ𝛾 𝜇 𝜕 𝜇 Ψ = 0 ⟹ 𝜕 𝜇 (Ψ𝛾 𝜇 Ψ) = 𝜕 𝜇 𝑗 𝜇 = 0 Probability

density 𝜌 = 𝑗 0 = Ψ𝛾 0 Ψ = Ψ 𝛾 0 𝛾 0 Ψ = Ψ 𝟙Ψ = Ψ Ψ ⟹ 𝜌 = ∑ Ψ 𝛼 †

Ψ 𝛼 ≥ 0

4 𝛼=1 … positive definite 𝑗 𝜇 = Ψ𝛾 𝜇 Ψ = Ψ 𝛾 0 𝛾 𝜇 Ψ

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