Lecture 2: Quarks and - Texas A&M Universitypeople.physics.tamu.edu/kamon/teaching/phys627/slide/... · 2019. 2. 2. · 2 Fermions and Bosons a) Fermions …spin ½ ; Bosons …spin

Lecture 2: Quarks and

“Symmetry in Quarks”

http://faculty.physics.tamu.edu/kamon/teaching/phys627/1

Fermions & Bosons, Wave Functions

Symmetry: Rotation, Parity (P), Charge Conjugate (C), SU(3),Isospin (I), Gauge symmetry

Conservation Laws

2

Fermions and Bosonsa) Fermions … spin ½ ; Bosons … spin 0, 1, 2

b) Spin-statistics theorem principal of quantum field theory

c) Wave function 𝝍 𝝍 𝟐 = probability

d) Pauli’s Exclusion Principle

Two or more identical fermions cannot exist in the same quantum

state. (No restriction for bosons Bose-Einstein condensation)

𝝍𝟏,𝟐𝟐= 𝝍𝟐,𝟏

𝟐

2 identical boson: 𝝍𝟏,𝟐 → +𝝍𝟐,𝟏

2 identical fermion: 𝝍𝟏,𝟐 → −𝝍𝟐,𝟏

𝑬𝜷

𝑬𝜶

𝜶 ≠ 𝜷

×

× ×

×1

2

2

1

[Q] is the above wavefunction for fermion or boson?

3

Symmetry in Two S=1/2 Particle System

S = 1Symmetric

S = 0Asymmetric

Symmetry = (-1)S+1

𝑆−| ۧ↑ = | ۧ↓

Consider spin of system of two spin-1/2 particles : 0, 1

| ۧ↑↑| ۧ↑↓ + | ۧ↓↑

| ۧ↓↓| ۧ↑↓ − | ۧ↓↑

ssystem = 0 or 1

4

PDG 2010

Quantum Numbers of Quarks

had

rons

Gell-Mann-Nishijima Formula

𝑸 = 𝑰𝒛 +ℬ + 𝑺 + 𝑪 + 𝑩 + 𝑻

𝟐

5

Quark Description of Light HadronsIso

spin

tri

ple

t st

ates

Iso

spin

tri

plet

stat

es

Iso

spin

dou

ble

t st

ates

We review “Isospin” symmetry.

6

Isospin Symmetry: 𝒖 and 𝒅

c) 𝑸 = 𝑰𝒛 +ℬ+𝑺+𝑪+𝑩+𝑻

𝟐

d) Isospin is conserved in strong interaction (same nuclear force 𝑉𝑝𝑝 =𝑉𝑛𝑛 = 𝑉𝑝𝑛). Not in e.m. interaction.

e) Where does the isospin symmetry come from?

i) ቊ𝑝 = (𝑢𝒖𝑑)𝑛 = (𝑢𝒅𝑑)

ii) 𝑚𝑢 ≈ 𝑚𝑑 (accidental near equality) … a sole reason for the isospin symmetry

a) Isospin symmetry: consider a proton and a neutron as different “charge” sub-states of one particle:

i) 𝑠 =1

2for p and n

ii) 𝑚𝑝 ≈ 𝑚𝑛

b) Isospin 𝐼 (angular momentum in the

isospin space). 𝐼 =1

2→ 𝑔𝐼 = 2𝐼 + 1 = 2

(2 states for p and n):

𝐼𝑧 =+1

2for 𝑝 = (𝑢𝑢𝑑)

−1

2for 𝑛 = (𝑢𝑑𝑑)

𝑝 = 10

, 𝑛 = 01

𝐼−| ۧ𝑢 = | ۧ𝑑

𝐽+ | ۧ𝑗, 𝑚 = (𝑗 𝑗 + 1 − 𝑚(𝑚 + 1) | ۧ𝑗,𝑚 + 1

𝐽− | ۧ𝑗, 𝑚 = (𝑗 𝑗 + 1 −𝑚(𝑚 − 1) | ۧ𝑗,𝑚 − 1

t Matrices

7

𝑝 = 10

, 𝑛 = 01

𝐼−| ۧ𝑢 = | ۧ𝑑

1

8

Isospin Symmetry: 𝒖 and 𝒅

c) 𝑸 = 𝑰𝒛 +ℬ+𝑺+𝑪+𝑩+𝑻

𝟐

d) Isospin is conserved in strong interaction.

e) Where does the isospin symmetry come from?

i) ൞

𝜋+ = (𝑢 ҧ𝑑)

𝜋0 = (𝑢ത𝑢 + 𝑑 ҧ𝑑)𝜋− = (𝑑ത𝑢)

ii) 𝑚𝑢 ≈ 𝑚𝑑 (accidental near equality) … a sole reason for the isospin symmetry

a) Isospin symmetry: consider three 𝜋mesons (𝜋+, 𝜋0, 𝜋−) as different “charge” sub-states of one particle:

i) 𝑠 = 0ii) 𝑚 ≈ 140 MeV

b) Isospin 𝐼 =1

1→ 𝑔𝐼 = 2𝐼 + 1 = 3 (3

states or triplet):

𝐽+ | ۧ𝑗, 𝑚 = (𝑗 𝑗 + 1 − 𝑚(𝑚 + 1) | ۧ𝑗,𝑚 + 1

𝐽− | ۧ𝑗, 𝑚 = (𝑗 𝑗 + 1 −𝑚(𝑚 − 1) | ۧ𝑗,𝑚 − 1

𝜋+ 𝑢 ҧ𝑑

𝜋0 (𝑢ത𝑢 + 𝑑 ҧ𝑑)/ 2 → ( Q singlet? )

𝜋− ത𝑢𝑑

Triplet states

Vector Meson Decay Width

9

Why ΤΓ(𝜌0 → 𝑒+𝑒−) Γ(𝜛0 → 𝑒+𝑒−) ≠ 1?

Isospin Triplet for Spin-1/2 𝚺 Baryons

10

𝝅+ 𝒖ഥ𝒅

𝝅𝟎 (𝒖ഥ𝒖 + 𝒅ഥ𝒅)/ 𝟐 → (𝐬𝐢𝐧𝐠𝐥𝐞𝐭? )

𝝅− ഥ𝒖𝒅

𝛴 baryons (𝑠 =1

2) are isospin triplet states of (𝑞𝑞𝑠), where 𝑞 = (𝑢 or 𝑑).

a) 𝛴+= uusb) 𝛴0= uds : 𝐼−Σ+ = 𝐼−𝑢𝑢𝑠 = ( 2𝑑𝑢𝑠 + 2𝑢𝑑𝑠)/2

c) 𝛴−= dds : 𝐼−Σ0 = Τ𝐼− 𝑑𝑢𝑠 + 𝑢𝑑𝑠 2 = 𝑑𝑑𝑠

Triplet states

→ (𝐬𝐢𝐧𝐠𝐥𝐞𝐭? )

𝐽+ | ۧ𝑗, 𝑚 = (𝑗 𝑗 + 1 − 𝑚(𝑚 + 1) | ۧ𝑗,𝑚 + 1

𝐽− | ۧ𝑗, 𝑚 = (𝑗 𝑗 + 1 −𝑚(𝑚 − 1) | ۧ𝑗,𝑚 − 1

Testing Isospin for 𝚺(𝒒𝒒𝒔)Branching fractions can be explained by isospin states.

11

Parity Operation

2

Spatial Inversion of Coordinates

1

[Q] Angular orbital momentum 𝐿parity

?

𝑷proton = 𝑷𝒖𝑷𝒖𝑷𝒅 = +𝟏

𝑷𝐟𝐞𝐫𝐦𝐢𝐨𝐧 = −𝑷𝐚𝐧𝐭𝐢−𝐟𝐞𝐫𝐦𝐢𝐨𝐧

𝑷𝐟𝐞𝐫𝐦𝐢𝐨𝐧 = +𝑷𝐚𝐧𝐭𝐢−𝐛𝐨𝐬𝐨𝐧

12

Classification of Physical QuantitiesBased on their rank and parity properties

We will look into the current forms with Dirac equation, later.13

Parity of Two Photon SystemBased on their rank and parity properties

Consider two-photon system in the decay of "spin 0" → 𝛾𝛾: .

Photon – E.M. wave (𝐸, 𝐵 )Possible expression of the matrix element with scalar nature is:

ℳ𝛾𝛾~𝑎𝑠 𝐸1 ∙ 𝐸2 + 𝑎𝑝𝑠 (𝐸1 × 𝐸2) ∙ Ԧ𝑝

Scalar Pseudoscalar

14

Parity of Two-Particle System

Exercise 1: Find the parity of a system of two spin-0 particles.

3

4

15

𝑷 𝒇ത𝒇 = 𝑷𝒇𝑷ത𝒇(−𝟏)ℓ

16

Parity of 𝑱/𝝍(𝒄ത𝒄)

2003y(3S)

y(2S)

J/y(1S)

∴ 𝐽𝑃=1− [Q] 𝐽𝑃(𝜋0)=0?

17

Exercise 2

Parity of Three-Particle System

5

𝒒𝟏

𝒒𝟐

𝒒𝟑

ℓ

ℓ′

Consider orbital angular momentum for three-quark state (𝑞1, 𝑞2, 𝑞3)

Note: we defined the nucleon parity to be +1. (𝑃proton=𝑃𝑢𝑃𝑢𝑃𝑑=+1)

∴ 𝑃𝑞1 = 𝑃𝑞2 = 𝑃𝑞3 = +1

(−𝟏)ℓ+ℓ′

18

Parity Conserving Interaction

𝒂 + 𝒃 →𝑯𝒄 + 𝒅

𝝍𝒕𝒐𝒕𝒂𝒍𝒊

𝝍𝒕𝒐𝒕𝒂𝒍𝒇

𝑷𝑯 = 𝑯 (invariance under parity operation)

𝝍𝒕𝒐𝒕𝒂𝒍𝒇

(𝒓) 𝑯 𝝍𝒕𝒐𝒕𝒂𝒍𝒊 (𝒓) = 𝝍𝒕𝒐𝒕𝒂𝒍

𝒇(−𝒓) 𝑯 𝝍𝒕𝒐𝒕𝒂𝒍

𝒊 (−𝒓)

= 𝑷𝒇𝑷𝒊 𝝍𝒕𝒐𝒕𝒂𝒍𝒇

(𝒓) 𝑯 𝝍𝒕𝒐𝒕𝒂𝒍𝒊 (𝒓)

∴ 𝑃𝑖 = 𝑃𝑓

Conservation of parity if 𝐻 is invariant under parity operation.Strong & e.m. interactions are parity-conserving interactions.

𝑃 𝐻 = 𝐻

𝑃𝜓𝑡𝑜𝑡𝑎𝑙(Ԧ𝑟) = 𝜓𝑡𝑜𝑡𝑎𝑙(−Ԧ𝑟)

19

Parity of 𝝅−

(i) Find 𝑃𝑖 of 𝜓𝑡𝑜𝑡𝑎𝑙𝑖

1

2

𝝅− + 𝒅+ → 𝒏+ 𝒏

Slow negative pion, captured

by deuterium (𝑝𝑛) in an

atomic S-state (ℓ = 0)

20

Parity of Final State

(ii) Find 𝑃𝑓 of 𝜓𝑡𝑜𝑡𝑎𝑙𝑓

= 𝜓𝑠𝑝𝑎𝑐𝑒𝜓𝑠𝑝𝑖𝑛

S = 1Symmetric

S = 0Asymmetric

Symmetry = (-1)S+1

| ۧ↑↑| ۧ↑↓ + | ۧ↓↑

| ۧ↓↓| ۧ↑↓ − | ۧ↓↑

ssystem = 0 or 1 −𝟏 = −𝟏 ℓ −𝟏 𝒔+𝟏

∴ ℓ + 𝒔 = "𝒆𝒗𝒆𝒏"

"n + 𝑛" is a system of two identical spin ½ particles. Thus, 𝜓𝑡𝑜𝑡𝑎𝑙𝑓 should

be anti-symmetric under interchange of two. Total symmetry is −1. “Space” symmetry = (−1)ℓ

“Spin” symmetry = (−1)𝑠+1

Spin Conservation−𝟏 = −𝟏 ℓ −𝟏 𝒔+𝟏

∴ ℓ + 𝒔 = "𝒆𝒗𝒆𝒏"

21

ℓ = 𝟏; 𝒔 = 𝟏

22

Parity Conservation & Parity of 𝝅−

𝑱𝑷=𝟎−

(iii) Impose parity conservation: 𝑃𝑖 = 𝑃𝑓

𝑷𝒇 = 𝑷𝒏𝒏 = −𝟏 ℓ(𝑷𝒏)𝟐= −𝟏 𝟏(+𝟏)𝟐= −𝟏

ℓ = 𝟏; 𝒔 = 𝟏

3

[Q] Is the result consistent with 𝑃 𝑞ത𝑞 = 𝑃𝑞𝑃ത𝑞(−1)ℓ?

23

Parity Violation

Parity Violation in Weak Interaction

1956

ParityOperaton

24

Parity of Omega (W) Baryon

V. E. Barnes; et al., "Observation of a Hyperon with Strangeness Minus Three". PRL 12 (8): (1964) 204. doi:10.1103/PhysRevLett.12.204

Bubble chamber trace of the first observed Ω baryon event at Brookhaven National Laboratory

https://en.wikipedia.org/wiki/Omega_baryon

𝒒𝟏

𝒒𝟐

𝒒𝟑

ℓ

ℓ′ 𝑷 = 𝑷𝒒𝟏𝑷𝒒𝟐𝑷𝒒𝟑(−𝟏)ℓ+ℓ′

[Q] Parity of Ω−?

25

Parity-Conserving Interaction

𝒆−

𝒆−A B

𝒑

𝒉′𝐬

Parity Operation

spin

Parity-conserving Interaction𝝈𝑨 = 𝝈𝑩

Charge Conjugation [1]

C parity is conserved with strong and electromagnetic interaction :

but not with the weak interaction (will be discussed later). We will discuss only C-parity conserved cases.

26

The charge conjugation መ𝐶 is the operation which replaces all particles by their anti-particles in the same state.

The charged particles cannot be 𝑪eigenstate.

For particles 𝒂 (= 𝜋+, 𝐾+, 𝑝) which have distinctive anti-particles, wavefunctions are changed with መ𝐶operation as :

For particles 𝜶 (= 𝜋0, g,…) which do not have distinctive anti-particles are eigenstates of the መ𝐶 operation as a single particle :

𝑪|𝜶 ۧ𝝍 = 𝑪𝜶|𝜶 ۧ𝝍

In this case one can get a quantum number C, called “C parity”.

𝑪| ۧ𝒂𝝍 = | ۧഥ𝒂𝝍

መ𝐶, 𝐻 = 0

መ𝐶2 = 1, therefore 𝐶𝜶 = ±1

𝑪 ۧ𝒆+ = ۧ𝒆− ≠ | ۧ𝒆+

𝑪 ۧ𝝅+ = ۧ𝝅− ≠ | ۧ𝝅+

Clearly the charge conjugation operation satisfies: 𝑪| ۧ𝜸 = 𝑪𝜸|𝜶𝜸 ۧ𝝍

| ۧ𝜸 ≡ | ൿ𝜶𝜸𝝍𝜸

In case of multi-particle system, the C-parity is a multiplicativenumber:

If the multi-particle system are symmetric with the Ĉ operation, the system is an eigenstate of the Ĉ operation and has definite C-parity, for example:

Charge Conjugation [2]

27

𝑪| ۧ𝜸𝜸 = 𝑪𝜸𝑪𝜸| ۧ𝜸𝜸 = −𝟏 𝟐 | ۧ𝜸𝜸

𝑪| ۧ𝑝 ҧ𝑝 = −𝟏 𝓵+𝒔| ۧ𝑝 ҧ𝑝

𝑪| ۧp+p− = −𝟏 𝓵| ۧp+p−

In general case which contains same number of fermions 𝑓 and anti-fermion ҧ𝑓, it is: 𝐶 ൿ𝑓 ҧ𝑓; 𝐽, 𝐿, 𝑆 = −1 𝐿+𝑆 ൿ𝑓 ҧ𝑓; 𝐽, 𝐿, 𝑆 ,

| ۧ𝜸 ≡ | ൿ𝜶𝜸𝝍𝜸

28

C-parity of g

ExerciseMaxwell’s Equations

C-parity of photon? It can be inferred from the behavior of the classical electromagnetic field under the charge conjugation.

Electric field and potential changes its sign with the charge conjugate:

Since 𝐸 = −𝛻𝜙 −𝜕 Ԧ𝐴

𝜕𝑡, we found 𝐶𝛾 = −1 𝑱𝑷𝑪(𝜸) = 𝟏−−

C-parity of p0

29

“Conserved” -> (Initial state C parity) = (Final state C parity)

C−parity of mulri−particle system is a multiplicative number.Therefore, 𝐶 | ۧ𝛾𝛾 = (−1)(−1)| ۧ𝛾𝛾 . ∴ 𝐶𝜋0 = +1

This agrees with 𝐶 ൿ𝑓 ҧ𝑓; 𝐽, 𝐿, 𝑆 = −1 𝐿+𝑆 ൿ𝑓 ҧ𝑓; 𝐽, 𝐿, 𝑆 , where 𝐿 + 𝑆 = 0.

𝜋0 dominantly decays into 2 photons with EM interaction: 𝜋0 → 𝛾𝛾 Feynman diagram:

𝑱𝑷𝑪(𝝅𝟎) = 𝟎−+

𝑱𝑷𝑪(𝜸) = 𝟏−−

30

Exercise: For a case of 𝝅+𝝅− system (𝒔𝝅± = 𝟎) with orbital angular momentum L, the C parity is 𝐶 ۧ𝝅+𝝅−; 𝐿 = −1 𝐿+𝑆 ۧ𝝅+𝝅−; 𝐿 .

The charge conjugation መ𝐶 is the operation which replaces all particles by their anti-particles in the same state.

31

S = 2Symmetric

S = 0Symmetric

S = 1Asymmetric

Symmetry = (-1)S

Exercise: For a case of 𝝅+𝝅− system (𝒔𝝅± = 𝟏) with orbital angular momentum L, the C parity is:

Particle-exchange Symmetry (total symmetry = +1 b/c Bose statistics)

𝑺𝝅+𝝅− = 𝟎, 𝟏, 𝐨𝐫 𝟐

𝝅+ 𝝅−

𝝅− 𝝅+ መ𝐶 ۧ𝝅+𝝅−; 𝐿 = 𝐶𝛼 ۧ𝝅−𝝅+; 𝐿 .

𝝅− 𝝅+

Charge conjugate operation

𝐶𝛼 | ۧ𝝅−𝝅+; 𝐿 →(−𝟏)𝑳(−𝟏)𝑺𝐶𝛼| ۧ𝝅+𝝅−; 𝐿

+𝟏 = (−𝟏)𝑳(−𝟏)𝑺𝐶𝛼∴ 𝐶𝛼 = (−𝟏)𝑳(−𝟏)𝑺= (−𝟏)𝑳+𝑺

𝑳

32

Exercise: Particle-exchange Symmetry in Two spin-1 Particles

Spin of system of two spin-1 particles : 0, 1, and 2

𝑱+ 𝟏 ۧ, 𝟏 = 𝟏 𝟏 + 𝟏 − 𝟏 𝟏 + 𝟏 ۧ𝟏, 𝟏 + 𝟏 =0

𝑱− 𝟏 ۧ, 𝟏 = 𝟏 𝟏 + 𝟏 − 𝟏 𝟏 − 𝟏 ۧ𝟏, 𝟏 − 𝟏 = 𝟐 | ۧ𝟏, 𝟎

𝑱− 𝟏 ۧ, 𝟎 = 𝟏 𝟏 + 𝟏 − 𝟎 𝟎 − 𝟏 ۧ𝟏, 𝟎 − 𝟏 = 𝟐 | ۧ𝟏,−𝟏

| ۧ2,2 = | ۧ𝟏, 1 | ۧ𝟏, 1

| ۧ2,1 =1

2ۧ𝟏, 0 ۧ𝟏, 1 +

1

2ۧ𝟏, 1 ۧ𝟏, 0

| ۧ2,0 =1

6ۧ𝟏,−1 ۧ𝟏, 1 +

1

6ۧ𝟏, 1 ۧ𝟏,−1 +

2

6ۧ𝟏, 0 ۧ𝟏, 0

| ۧ2, −1 =1

2ۧ𝟏, 0 ۧ𝟏,−1 +

1

2ۧ𝟏, −1 ۧ𝟏, 0

| ۧ2,−2 = | ۧ𝟏,−1 | ۧ𝟏,−1

−− −

| ۧ1,1 =1

2ۧ𝟏, 0 ۧ𝟏, 1 −

1

2ۧ𝟏, 1 ۧ𝟏, 0

| ۧ1,0 =1

2ۧ𝟏,−1 ۧ𝟏, 1 −

1

2ۧ𝟏, 1 ۧ𝟏,−1

| ۧ1,−1 =1

2ۧ𝟏,−1 ۧ𝟏, 0 −

1

2ۧ𝟏, 0 ۧ𝟏,−1

−−−

−−−−−−

| ۧ0,0 =1

3ۧ𝟏,−1 ۧ𝟏, 1 +

1

3ۧ𝟏, 1 ۧ𝟏,−1 −

1

3ۧ𝟏, 0 ۧ𝟏, 0

−− −−−−

S = 2Symmetric

S = 0Symmetric

S = 1Asymmetric

Symmetry = (-1)S

33

Particle-exchange Symmetry(total symmetry = -1)

𝒑 ഥ𝒑

ഥ𝒑 𝒑 መ𝐶 ۧ𝑝 ҧ𝑝; 𝐿 = 𝐶𝛼 ۧҧ𝑝𝑝; 𝐿 .

𝒑 ഥ𝒑

Charge conjugate operation

𝐶𝛼 | ۧҧ𝑝𝑝; 𝐿 →(−𝟏)𝑳(−𝟏)𝑺+𝟏𝐶𝛼|𝑝 ҧ𝑝 ۧ; 𝐿

−𝟏 = (−𝟏)𝑳(−𝟏)𝑺+𝟏𝐶𝛼∴ 𝐶𝛼 = (−𝟏)𝑳(−𝟏)𝑺= (−𝟏)𝑳+𝑺

𝑳

Exercise: For a case of 𝒑ഥ𝒑 system with orbital angular momentum L, the C parity is:

𝑺𝒑ഥ𝒑 = 𝟎 𝐨𝐫 𝟏

[Q] For a case of 𝐽/𝜓 (𝑐 ҧ𝑐) system with orbital angular momentum ℓ = 0, the C parity is ?

[Q] Why?

34

𝝅𝟎(𝒖ഥ𝒖) → 𝜸𝜸𝑱𝑷𝑪(𝝅𝟎) = 𝟎− +

𝑱𝑷𝑪(𝜸) = 𝟏− −

𝐶| ۧ𝛾𝛾 = 𝐶𝛾𝐶𝛾 = (−1)(−1)=+1

All vertices are allowed in QED.

𝑪 = (−𝟏)ℓ+𝒔

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iv.o

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709.3

371v1

Τ𝑱 𝝍 𝒄ത𝒄 → 𝜸𝜸 is C-parity violating process?

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Phys.Rev.D76:117101,2007

𝐶| ۧ𝛾𝛾 = 𝐶𝛾𝐶𝛾 = (−1)(−1)=+1𝑪 = (−𝟏)ℓ+𝒔 𝑪 Τ𝑱 𝝍 = (−𝟏)?+?

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Review an experimental technique:To measure Br(J/y gg), one needs to prepare a sample of J/y events and count the number of events in the gg final state.

Phys.Rev.D76:117101,2007 (Results)

J/y(1S), y(2S), y(3S), …

PDG 2016

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↑↓

↑↑↑↓ ↑↑↑↓

↑↑

𝝅+𝝅−

Τ𝑱 𝝍 𝜸

𝑪 𝒇ത𝒇 = (−𝟏)ℓ+𝒔

𝑷 𝒇ത𝒇 = 𝑷𝒇𝑷ത𝒇(−𝟏)ℓ

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𝝌𝒄𝟎(𝒄ത𝒄) Bound States

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𝒉𝒄(𝒄ത𝒄) Bound States

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C-parity of n

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CP of 𝝅+ → 𝝁+𝝂𝝁

Executive Summary of Lecture 02[1] Feynman’s words on discoveries of law of nature (from Lecture 01)o Rhythm and Patterno Character of discoveryo Confidence on one law ⇒ Avalanche of another discoveries

[2] Symmetries

o Fermions & Bosons, Wave Functions

o Pattern in quark description for hadron … underlying symmetry

o Symmetry: Isospin (I), Parity (P), Charge Conjugate (C)[3] Conservation Laws … Allowed or forbidden decayso Check type of interaction(s) - valid decay vertices (Feynman rules)o Check charge, spin conservationso Check baryon and lepton number conservationso Check P, C., …

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HW

Supplemental Materials

[1] Discovery of CP Violation

[2] Isospin in pN System

[3]

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Discovery of

CP Violation

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CP Violation in Neutral Kaons

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CP Violation in Neutral Kaons

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CP eigenstates:

2-pion system is CP = +13-pion system is CP = -1

KL is CP = -1

Thus, KL -> p+p-p0

Quiz - CP Violation Experiment

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Isospin in

pN System

Isospin in Two-particle System

Exercise 1

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https://en.wikipedia.org/wiki/Table_of_C

lebsch%E2%80%93Gordan_coefficients

m2j m1

m1

I- { |p+ > |p > }

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Z in 1987W in 198254

Summary

1

Z in 1987W in 198255

1

3

6

p- p interaction creates status of |3/2, -1/2> or |1/2, -1/2>.8

Z in 1987W in 1982

13

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13

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~200 mb

~60 mb

~20 mb

elastictotal

inelasticelastictotal

Why different?

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