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Particle Physics WS 2012/13 (30.11.2012)

Stephanie Hansmann-Menzemer Physikalisches Institut, INF 226, 3.101 Content of Today

Symmetries in Particle Physics  Emmy Noether Theorem: any symmetry result in an conserved observable

 Symmetry of isospin SU(2): „u and d quark are not distinguishable in strong IA“ [this is only an approximative symmetry e.g. m(u)~m(d)] conservation of isospin in strong IA

 Flavour symmetry SU(3): „ u, d and s quark are not distinguishable in strong IA“ [this is an even more approximative symmetry: m(s) ~ 150 MeV, m(u)~ few MeV] conservation of hypercharge in strong IA

 Colour symmetry SU(3): „quarks of different colour are not distinguishable among each other“ this is an exact symmetry conservation of colour in all IA

Experimental evidence for Colour 2 Reminder: Emmy-Noether Theorem

Suppose physics is invariant under (linear) transformation: ψ → ψ‘ = U ψ † T* U = U † † † Ensure normalization: ψ ψ 푑4푥 = 1 ψ 푈 푈ψ 푑4푥 = 1

† 푈 푈 = 1 U has to be unitary

To ensure physics is invariant under transformation:

† † † ψ † 퐻 ψ 푑4푥 = ψ′ 퐻 ψ′푑4푥 = ψ 푈 퐻 푈 ψ 푑4푥

H = U† H U UH = HU [H, U] = 0

Now consider infinitesimal transformation: U = 1 + i ε G G: generator of transformation

U†U = (1 – iεG†)(1+iεG) = 1 + iε(G-G†) + O(ε2)

† † 푈 푈 = 1 G = G G is hermitian, thus coresponds to an observable quantity g [H,U] = 0 [H,G] = 0 g is a conserved observable quantity!

For each infinite symmetry transformation, there is an conserved observable quantity! 3 Reminder: Emmy-Noether Theorem

Example: infinite 1D spatial translation: x → x + δx

휕ψ(푥) 휕 Ψ(x) → ψ‘(x) = ψ(x + δx) = ψ(x) + δx = (1+ δx ) ψ(x) 휕푥 휕푥

휕 U P = -i ψ‘(x) = (1-i δx P )ψ(x) x 휕푥 x U x component of momentum is conserved!

In general the symmetry operation may depend on more than one parameters:

U = 1 + i ε 퐺

in our example: 푟 → 푟 + δ푥 U = 1 – i δ푥 푃

A finite transformation (α) can be expressed as a series of infinitesimal transformations:

푛 α U(α) = lim 1 + 푖 퐺 = 푒푖α퐺 푛→∞ 푛

Emmy Noether Theorem: For each symmetry transformation, there is an conserved quantity4! Symmetries in Particle Physics: Isospin Proposed by Heisenberg (1932) :

Proton and neutron are same particles with respect to strong IA!

Indications: m(p) ~ m(n) and nuclear binding energies (if one ignores Coulomb IA) are similar 1 0 Proton and neutron are two states of single particle: the nucleon p= ; n= ; 0 1 In analogy to , call this isospin. Expect physics to be invariant under rotation in this space

Later extended idea to u, d quark: m(u) ~ m(d)

1 0 푢′ 푢 푢 푢 푢 u = , d = , = U = 11 12 0 1 푑′ 푑 푢21 푢22 푑

rotation in 2D u-d quark space

Strong IA is invariant to rotation in isospin space, U: 2x2 unitary matrix 4 complex matrix elements 8 real parameters + 4 unitary conditions 4 parameters

In group theory these matrices form U(2) group

unitary matrix 2D 5 Reminder: Definition of a Group

* is operation which connects elements of a group

There is a neutral element e, that for all group elements a: e*a = a = a*e Completeness: a, b element of the group → a*b as well element of the group For each element a there is an inverse element a-1, so that a*a-1 = a-1*a = e

Example 1: entire numbers, “+” is group operation, “0” is neutral element

Example 2: rational numbers without 0, “*” is group operation, “1” is neutral element

Example 3: element a: rotation around 60o, “*” is group operation, complete list of group elements: {a,a2,a3,a4,a5,a6=e} → finite group, dimension 6

Subgroup: element a: rotation around 120° {a2,a4,a6=e} → finite group, dimension 3

6 Isospin 1 0 1 0 U = 푒푖φ; U †= 푒−푖φ; U U †=1 Ψ‘ = U1ψ for all wave functions 1 0 1 1 0 1 1 1 not distinbuishable from original system

U1 ϵ U(2), however U1 has no physical relevance: absolute phases cannot be measured Left with three independent matrices which form SU(2) subgroup of U(2)

special All unitary matrices have |det(U)| = 1, all elements of SU(2) have det(U)=1.

1 + 푖ε퐺11 푖ε퐺12 2 U = 1 + iεG det(U) = det = 1 + iε(G11+G22) + O(ε ) 푖ε퐺21 1 + 푖ε퐺22

G11+G22 = Trace(G) = 0 Generators of rotation in isospin space are traceless 2x2 matrices

A set of linear independent traceless 2x2 matrices are e.g. the

0 1 0 −푖 1 0 σ = ; σ = ; σ = Note: Isospin has NOTHING 1 1 0 2 푖 0 3 0 −1 to do with spin – JUST SAME MATH

1 2D represenation of generator of Isospin symmetry group: 퐼 = 퐺 = σ 2 7 Isospin SU(2) – 2D representation

1 0 states: |u> = |d> = 0 1 symmetry transformation: U = 푒푖ασ

commutator relations: [σi, σj] = 2iεijkσk generators: σ1, σ2, σ3

σ σ isospin operators : 푇 = ; 푇 = 푖 i=1,2,3 2 푖 2 1 1 0 1 1 1 1 0 0 1 3. component: T |u> = = |푢 > T |d> = = − |푑 > 3 2 0 −1 0 2 3 2 0 −1 1 2

1 1 3 casimir : 푇2 |푢 > = (σ 2 + σ 2 + σ 2) = |푢 > 4 1 2 3 0 4 1 0 3 푇2 |푑 > = (σ 2 + σ 2 + σ 2) = |푑 > 4 1 2 3 1 4

2 Commutation properties: [푇 , Ti] for i=1,2,3 = 0 [Tj, Ti] ≠ 0 for i≠j only two observables simultanously measurable

2 Choose common eigenstages of 푇 and T3 : |I,I3>

2 푇 퐼, 퐼3 > = 퐼 퐼 + 1 퐼, 퐼3 > T3 퐼, 퐼3 > = 퐼3 퐼, 퐼3 > 8 Isospin SU(2) – 2D representation I I I I 3 1 3 0 states: |u> = |1/2, 1/2> = |d> = |1/2, -1/2> = 0 1

0 1 0 0 ladder operator: T = T + i T = T = T - i T = + 1 2 0 0 - 1 2 1 0

T+|u> = 0 T+|d> = |u> T_|u> = |d> T_|d> = 0

T+

T_

d u

-1/2 +1/2 I3

Isospin is generator of symmetry of strong IA, thus isopin (I, I3) are conserved in strong IA!

9 General Properties of Isospin Operators

To describe a set of total isospin = I, need (2I+1) dimensional representation of SU(2)

For any represenation of SU(2), following relation are valid:

[Ti, Tj] = 2iεijkIk

T3|I,I3> = I3|I,I3>

2 푇 퐼, 퐼3 > = 퐼 퐼 + 1 퐼, 퐼3 >

T± |I,I3> = 퐼 ∓ 퐼3 퐼 ± 퐼3 + 1 |퐼, 퐼3 ± 1 >

2 [T3, 푇 ] = 0

10 Isospin SU(2) – 3D representation 1 0 0 states: |1, 1> = 0 |1, 0> = 1 |1,-1> = 0 e.g. π+ , π0, π- 0 0 1

0 1 0 0 −푖 0 1 0 0 1 1 generators: J = 1 0 1 J = 푖 0 −푖 J = 0 0 0 1 2 2 2 3 0 1 0 0 푖 0 0 0 −1

commutator relations: [Ji, Jj] = 2iεijkJk

isospin operators : 푇 = 퐽; 푇3 = 퐽3

3. component: T3|1,1> = 1,1 > T3|1,0> = 0 T3|1,−1> = − 1, −1 >

1 0 0 2 2 2 2 casimir operator: 푇 = 퐽1 + 퐽2 + 퐽3 = 2 0 1 0 0 0 1 0 1 0 0 0 0 ladder operator: T+ = T1 + i T2 = 2 0 0 1 T- = T1 - i T2 = 2 1 0 0 0 0 0 0 1 0 11

Combining (Iso-)spins (goal: get wave function of proton) general rule (again identically to ): I3 components are added, I ≥ I3

|uu> = |1/2,1/2> |1/2, 1/2> = |1,+1> |dd> = |1/2,-1/2> |1/2, -1/2> = |1,-1> T _ |1,1> = 2 |1,0 > |푢 > How does |1,0> look like? T_ |uu> = (T_|u>) + |u>(T_ |u>) = |d>|u> + |u>|d> 1 |1,0> = (|du> + |ud>) 2

Instead of studying ladder operators consider symmetry arguments: |1,0> is linear combination of |1,1> and |1,-1> ; |1,1> and |1,-1> are symmetric |1,0> must be symmetric

1 |0,0> orthogonal to |1,0> |0,0> = (|du> - |ud> 2

triplett symmetrisch under singlett antisymmetrisch under exchange of quarks exchange of quarks |1,1> = |uu> 1 1 |1,0> = (|ud> + |du>) |0,0> = (|du> - |ud>) 2 2 |1,-1> = |dd> 12 Combining (Iso)spins From four possible combinations of isospin doublets obtain a triplet of isospin 1 states and a singlet isospin 0 state: 2 ⨂ 2 = 3 ⨁ 1 1 dd (ud+du) uu 1 2 (du-du) I3 ⨁ 2

-1 0 +1 0 I3 T± T± Can move around within multiplets using ladder operators.

States with different total isospin are physically different: the isospin 1 triplet is symmetric Under interchange of quarks where as singlet is anti-symmetric.

Now add additional up or down quark

ddu uud 1 1 1 1 ddd (ud+du)d (ud+du)u uuu (du-du)d (du - du) u 2 2 2 2 ⨁ I3 I -3/2 -1/2 +1/2 +3/2 -1/2 +1/2 3 6 2

Use ladder operators and orthogonality to group the 6 states into isospin multiplets. 13 Combining (Iso)Spins

T± |I,I3> = 푰 ∓ 푰ퟑ 푰 ± 푰ퟑ + ퟏ |푰, 푰ퟑ ± ퟏ >

14 Baryon Isospin Wave Function

2 ⨂ 2 ⨂ 2 = (3 ⨁1) ⨂ 2 = 4 ⨁ 2 ⨁ 2

Different multiplets have different symmetry properties:

|3/2,3/2> = |uuu> 1 S: Sym quartett |3/2,1/2> = (|uud> + |udu> + |duu>) 3 Symmetric under interchange 1 |3/2,-1/2> = (|udd> + |dud> + |ddu>) Of any quark 3 |3/2,-3/2> = |ddd>

1 |1/2,1/2> = (2|uud> - |udu> - |duu>) 6 Ms: Mixed sym. 1 |1/2,-1/2> = (-2|ddu> +|udd> + |dud>) sym for 1↔ 2 6

1 |1/2,1/2> = (|udu> - |duu>) 2 MA: Mixed sym. 1 |1/2,-1/2> = (|udd> - |ddu>) Anti-sym for 1↔ 2 2

Mixed symmetriy states have no definite sym. Under interchange of 1↔ 3 15 Baryon Spin Wave Function Can apply exact same mathematics to derive possible combinations of wave functions for three Spin ½ particles |3/2,3/2> = |↑↑↑> 1 |3/2,1/2>= (|↑↑↓> + |↑↓↑> + |↓↑↑>) 3 S: Sym quartett 1 |3/2,-1/2> = (|↑↓↓> + |↓↑↓> + |↓↓↑>) 3 |3/2,-3/2> = |↓↓↓>

1 |1/2,1/2> = (2|↑↑↓> - |↑↓↑> - |↓↑↑>) M : Mixed sym. 16 S |1/2,-1/2> = (-2|↓↓↑> +|↑↓↓> + |↓↑↓>) sym for 1↔ 2 6

1 |1/2,1/2> = (| ↑↓↑> −| ↓↑↑>) 2 MA: Mixed sym. 1 |1/2,-1/2> = (|↑↓↓> - |↓↓↑>) Anti-sym for 1↔ 2 2

Can now form total wave function of proton ….

16 Total Baryon Wave Function

(see later in lecture)

17 Total Baryon Wave Function

Wave function e.g. helpfull to determine magneticP moment of proton/neutron (see homeworks18 ). |퐽, 푀 > = |퐽 , 푚 > 퐽 , 푚2 >< 퐽 , 푚 , 퐽 , 푚 퐽, 푀 > M = m +m 푚1,푚2 1 1 2 1 1 2 2 1 2 I ϵ [ |I1-I2|, I1+I2] Glebsch Gordan Coefficients

19 Examples for Application of Glebsch-Gordan-Coefficients

|퐽, 푀 > = |퐽 , 푚 > 퐽 , 푚2 >< 퐽 , 푚 , 퐽 , 푚 퐽, 푀 > M = m +m 푚1,푚2 1 1 2 1 1 2 2 1 2 J ϵ [ |J1-J2|, J1+J2] C J How does M

|J,M> = |3/2, ½> composition look like?

Possible contributions: m1 m2

|3/2,1/2> = C1|1,0>|1/2,1/2> + C2 |1,1>|1/2,-1/2>

C1 : J=3/2, M=1/2, m1 = 0, m2 = 1/2 C1 = 2/3

C2 : J=3/2, M=1/2, m1 = 1, m2 = -1/2 C2 = 1/3

Glebsch-Gordan-Coefficients can be derived by applying ladder operators …. 20 Flavour SU(3)

Ansatz: m(s) ~150 MeV – small on nucleon scale (~ 1 GeV) ↔ u,d,s are approximately degenerated

Physics is invariant under rotation in 3D flavour space:

1 0 0 푢′ 푢 u = 0 ; d = 1 ; s = ; 0 푑′ = U 푑 ; 0 0 1 푠′ 푠

U is 3x3 unitary matrix 9 complex entries 18 real parameters 9 fixed due to unitarity condition UU†=1 One matrix is (as for 2D case) just multiplication with phase 8 remaining free parameters describe subgroup SU(3) of group of unitarity matrices U(3)

Generators (in 3D representation) of SU(3) are 8 hermitian matrices λi i=1,…,8 (Gell-Mann matrices)

Define 푇 = λ/2 U = 푒푖α푇 : group of translation in flavour space 21 Gell-Mann-Matrices λi

T± = 1/2(λ1 ±iλ2)

T3 = λ3/2 (Isospin)

V± = 1/2(λ4 ±iλ5)

U± = 1/2(λ6 ±iλ7)

Hypercharge Y = λ8/3; Y = B + S

Gell-Mann Matrices are baryon number strangeness generators of SU(3) flavour symmetry group 22 Hypercharge Y

Hypercharge Y = λ8/3; Y = B + S

baryon number strangeness

I3 el. Q = I3 + Y/2

B S Y Q I3

u 1/3 0 1/3 2/3 1/2 d 1/3 0 1/3 -1/3 -1/2 s 1/3 -1 -2/3 -1/3 0 푢 -1/3 0 -1/3 -2/3 -1/2 I 3 푑 -1/3 0 -1/3 +1/3 1/2 푠 -1/3 +1 2/3 +1/3 0

23 Combining Quark+Antiquark

What is composition of Y=0, I3=0 state? Exploit ladder operators

2 independent states, third state not part of multiplet 24 Combining Quark+Antiquark

Charged and neutral pions are member of same isospin doublet, thus should be in same flavour multiplet as well.

singulet state must be symmetric in flavour

You can apply any ladder operator on singulet and it yields 0.

Exploiting orthogonality:

25 Example: Pseudo-Scalars and Vector-Mesons

Similar combinations can be done for baryons. Particles of same multiplett have similar Masses. Historically very important to establish quark model!

26 Historical Relevance

In history the order was different than our approach up to now:

Physicis started with the „particle zoo“ of more several 100 observed „light“ particles and started to sort them by their mass and properties (branching ratios, …). (early days only strong and elm IA was observed)

they then concluded on the existance of partons/quarks

they made predictions about not yet observed particles in multipletts

27