Strong interactions Flashcards
Isospin symmetry of the strong interaction
What explains the almost identical mass of the proton and the neutron?
Any linear combination of proton and neutron states would be an acceptable state of the nucleon, which would still look the same to strong interactions, and this 2D space is called an internal space with corresponding internal quantum numbers.
If all the states look the same to strong interactions, then strong interactions are invariant under a general unitary transformation of the nucleon state: this means that strong interactions are symmetric under SU(2) transformations of the nucleon state which is called isospin symmetry with isospin as the quantum number.
Isospin symmetry of the strong interaction
How can we describe isospin symmetry?
While entirely unrelated to spin in physical origin, isospin displays exactly the same mathematical structure, U is a rep of SU(2) that is completely reducable.
- assuming that SU(2) is an exact symmetry: strong int. will be unchanged by SU(2) trafos, so H(strong) = invariant or [U,H(strong)] = 0
- for isospin trafos: U(α) = e^(iαI) = e^(iασ/2)
- [I,H(strong)] = 0 also, so strong int. conserve isospin
- isospin commutes with translations, rotations —» I commutes with E,p,J as well: I^2 and I3 can be diagonalized together with them
Consequences of isospin symmetry:
- the spectrum of the theory is organised in degenerate isospin multiplets, which form the
bases of irreducible representations of SU(2) - isospin is conserved in dynamical hadronic processes, i.e. in decay and scattering processes
Isospin symmetry of the strong interaction
What are mass multiplets?
Consequence of isospin symmetry: the existence of multiplets of hadrons with nearly-degenerate masses. There are then 2I + 1 set of linearly independent particle states |m, ~p; s, sz ; I, I3〉 with the same mass and spin, as well as p and s3, forming degenerate particle multiplets. As soon as isospin is broken, as it actually is in nature, these states are expected to display slightly different masses as well as further quantum numbers, which makes them different particles in all respects.
- protons, neutrons: isodoublet, I = 1/2, I3(p, n) = (+1/2 , − 1/2)
- antiprotons, antineutrons: isodoublet, I = 1/2, I3(¯n, ¯p) = (+1/2 , − 1/2)
- pions: isotriplet, I = 1, I3(π+, π0, π−) = (+1, 0, −1)
- Σ baryons: isotriplet, I = 1, I3(Σ+, Σ0, Σ−) = (+1, 0, −1)
- kaons: two isodoublets, I = 1/2, I3(K+, K0) = (+1/2 , −1/2), I3( ¯K0, K−) = (+1/2 , −1/2), distinguished by strangeness
- ∆ resonances: isoquartet, I = 3/2, I3(∆++, ∆+, ∆0, ∆−) = (+3/2, +1/2, –1/2, –3/2)
- η meson, Λ baryon: isosinglets
Isospin symmetry of the strong interaction
How to obtain the isospin multiplets from quarks?
The quark states have to be composed. We can trace isospin symmetry back to the fact that strong interactions are blind to the flavour of quarks, except for the difference in their masses. If quark masses were identical, then we could “rotate” flavours into each other without any physical effect. The Hamiltonian of strong interactions is invariant under unitary rotations in (u, d) space, up to a small symmetry breaking term proportional to (m(d) − m(u))/Λ.
Mesons: singlets or triplets and doublets
- q¯q : 1/2 ⊗ 1/2 = 0 ⊕ 1
- s¯q : 0 ⊗ 1/2 = 1/2
- q¯s : 1/2 ⊗ 0 = 1/2
Baryons:
- non-strange: doublets, quartets, 1/2 ⊗ 1/2 ⊗ 1/2 = 1/2 ⊕ 1/2 ⊕3/2
- singly-strange: singlets, triplets, 0 ⊗ 1/2 ⊗ 1/2 = 0 ⊕ 1
- doubly strange: doublets, 0 ⊗ 0 ⊗ 1/2 = 1/2
- triply-strange: singlets, 0 ⊗ 0 ⊗ 0 = 0
In general, all elementary particles different from u and d are assigned I = 0.
Isospin symmetry of the strong interaction
What’s the Gell-Mann–Nishijima formula?
The empirical relation between isospin and charge: Q = I3 + (1/2)(B + S) = I3 + Y/2.
- Y: hypercharge, also conserved
This comes from empirical observations: no two states with the same electric charge belonged to the same multiplet and no gaps in electric charge where observed in multiplets, therefore leading to a linear relation between Q and I3.
If the charges Q, B, and S of the three quarks u, d, s, satisfy it, then any of the plethora of hadrons built out of them and their antiquarks will automatically satisfy it. Assigning charges that fulfill the Gell-Mann–Nishijima formula is indeed possible (e.g.: q(u) = 2/3, b(u) = b(d) = b(s) = 1/3).
Isospin symmetry of the strong interaction
How does isospin invariance effect dynamical processes?
Decay processes: forbids some processes from happening
- G-parity can be defined: G ≡ CR2, R2|I I3〉 = (−1)^(I−I3) |I −I3〉
Scattering processes: relations between scattering amplitudes can be derived
- [I, S] = 0, so the initial and final state must have the same eigenvalues of I^2 and I3 for the matrix element not to vanish
- e.g.: for proton-pion+ elastic scattering there’s a peak at the mass of the ∆++ resonance for which I = 3/2, so M(3/2) will dominate over M(1/2)
Quark model
How can hadrons be classified? What’s the explanation for this?
The isospin quantum numbers I, I3, the baryon number B, and the strangeness S allow a full classification of the known hadrons, entirely characterising their states together with four-momentum and spin. In the (I3, Y) plane:
- eight lightest 1/2-spin baryions, eight lightest pseudoscalar mesons: hexagonal array
- nine 3/2-spin resonances: triangular array
These fit into isospin multiplets with a 150 MeV mass splitting within each multiplet: with a decrease in S comes an extra 150 MeV of mass.
Explanation: existence of an approximate symmetry extending the SU(2)I isospin symmetry and the U(1)Y symmetry associated with hypercharge: SU(3) ⊃ SU(2)I × U(1)Y × U(1)B, the last one being an extra factor (SU(3) doesn’t mix states with diff baryon numbers)
- an eight-dimensional rep with the desired decomposition into isospin-hypercharge multiplet
Quark model
What is SU(3)?
The group of unitary unimodular 3×3 complex matrices, i.e., SU(3) = {U ∈ M(3,3)(C) | U †U = 1 , det U = 1}. It’s an 8D Lie group, with corresponding Lie algebra the space of Hermitian traceless matrices equipped with the matrix commutator.
- U = e^(iA) can be written where A = A† and tr(A) = 0, A = ∑(a)α(a)t(a), {t(a)}, a = 1,…,8 as basis elements (A make an 8D real vector space with real coefficients)
Quark model
What’s the algebra of SU(3)?
The basis matrices obey the commutation relation: [t(a), t(b)] = if(abc)t(c) with tr[t(a)t(b)] = 1/2 δ(ab) as the normalization.
- structure constants: cyclic, antisymmeric under exchange of any pair of indices
Choosing the generators, they should break the symmetry to the smaller groups: t(a) = 1/2 λ(a), λ(a) being the Gell-Mann matrices.
- t1,2,3: generators I of the SU(2)I isospin subgroup, t(a) = I(a) , a = 1, 2, 3
- t8: generator Y of the U(1)Y subgroup, t8 = √3/2 Y
- t4,5 and 1/2 t3 + √3/2 t8: SU(2)V subgroup, t4,5 = V1,2, 1/2 t3 + √3/2 t8 = V3 = 1/2 I3 + 3/4 Y
- t6,7 and –1/2 t3 + √3/2 t8: SU(2)W subgroup, t6,7 = W1,2, –1/2 t3 + √3/2 t8 = W3 = –1/2 I3 + 3/4 Y
In full analogy with the SU(2) case, we now define the ladder operators of our SU(2) subgroups, i.e., I± = I1 ± iI2, V± = V1 ± iV2 and W± = W1 ± iW2 with commutation relations also analogous to SU(2).
SU(3) is rank 2: for any choice of basis no more than two generators commute and can be diagonalised simultaneously
- we choose to diagonalize I3 and Y because these are conserved quantities used to label particles irl
The V± and W± are meant to connect hadron states with hypercharge differing by one in the same multiplet, and so their effect on the simultaneous eigenvector of I3 and Y has to be an increase or decrease of Y by one unit.
Quark model
What are the commutation relations of the ladder operators of SU(3)?
The commutation relations of the algebra are a set of simultaneous eigenvalue equations for certain linear operators acting on the algebra of the group:** [H, E^(j)±] = ±α^(j)E^(j)±**.
- α^(j): root vectors that make up the root system of SU(3)
- eight eigenvalues in total: ±α^(j) and H1 = I3, H2 = √3/2 Y
- eight eigenvectors: E^(j)±, H that form a complete basis for the algebra
- from the Jacobi identity: [E^(i)(s) , E^(j)(t)] ∝ E^(k)(u), s,t,u = ±1, k = 0,1,2,3, otherwise the commutators are zero
Quark model
How to define representations of SU(3)?
Since we are interested in unitary representations of the group, we look for Hermitian representations of the algebra. We can take as a basis of our representation space a complete set of simultaneous eigenvectors of the representatives of I3 and Y. Reps are then characterised by the corresponding pairs of eigenvalues, called weights or weight vectors, and can be represented graphically as diagrams in the (I3, Y ) plane called weight diagrams.
- we want to find diagrams that match the baryon octet or any other multiplet of hadrons
Quark model
What are the simplest representations of SU(3)?
Trivial rep: 1D, denoted as 1
- D(T)(g) = 1 for all g
- all the elements of the Lie algebra of the group are represented by zero
- weights: (I3, Y) = (0, 0), singlet
Fundamental rep: 3D, denoted as 3
- D(F)(U (α)) = U(α) = e^(iα·t)
- generators: T^(F)(a) = t(a)
- weights: (I3, Y) = (1/2 , 1/3), (−1/2 , 1/3); (0, −2/3), isodoublet + isosinglet
Complex-conjugate rep: 3D, denoted as ¯3
- D(C)(U(α)) = U(α)∗ = e^(iα·(−t∗))
- generators: T(C)(a) = (−t(a))∗
- weights: (I3, Y) = (–1/2 , –1/3), (1/2 , –1/3); (0, 2/3), isodoublet + isosinglet
Adjoint rep: 8D, denoted as 8
- a linear op. from the algebra to itself
- equal to its complex conjugate
- AdU A ≡ U AU † ≈ A + iε(a)[t(a), A] = A + iε(a) ad(t(a))A
- Ad(U) and ad(t(a)) do provide us with representative matrices for the elements of the group and of the algebra, respectively
- weights: (I3, Y) = (1/2 , 1),(− 1/2 , 1); (1, 0),(0, 0),(−1, 0); (0, 0); (1/2 , −1),(−1/2 , −1), doublet + triplet + singlet + doublet
- corresponds precisely to the baryon octet
Quark model
What’s the decuplet representation of SU(3)?
The 10 rep that accomodates the baryonic resonances as well, prefectly reproducing their quantum numbers.
- the stated can be obtained by using the I–, V–, W– ladder operators, starting from the state that the raising operators annihilate
- since a rep should also provide reps of the subgroups: applying the SU(2)I/V/W rotations of e^(iπ I1/V1/W1) on an eigenvector of I3/V3/W3 and Y/Y(V)/Y(W) gives a new one with opposite eigenvalue
- so the weight diagram should be symmetric under reflections through the I3/V3/W3 = 0 axis
Quark model
What’s the eightfold way?
If SU(3) symmetry really is the explanation behind the observed hadronic multiplets, then all the hadronic multiplets should fit into irreducible representations, not just the lightest baryons. To complete this list:
- η meson for the meson octet: I = Y = 0 (discovered experimentally, GM-N formula)
- Ω− for the baryon decuplet: s = 3/2, I = 0, S = -3, Q = -1 (discovered from the mass splitting between particles with diff strangeness)
So the eightfold way is the classification of hadron multiplets in terms of irreducible representations of SU(3) .
Quark model
What are the irreps of SU(3)? What’s the goal? How to obtain them?
We want to explain why nature shows only a small subset of irreps and not others. The solution is to assume that hadrons are bound states of more fundamental particles and their antiparticles (reducing the number of elementary particles). Assumption: the constituents make a triplet, transforming in the fundamental rep of SU(3).
Method: all irreducible reps can be obtained by reducing tensor products of fundamental (3) and complex conjugate (¯3) reps using weight diagrams (drawing them on top of each other and adding the weights)
- ¯3 ⊗ 3 = 8 ⊕ 1 (outer triangle: octet + middle: singlet)
- 3 ⊗ 3 = 6 ⊕ ¯3 (inner triangle: antifund. rep + remaining six non-deg. states: 6)
- 6 ⊗ 3 = 10 ⊕ 8 (using symmetry properties, using descent ops)
Altogether, the reps appearing in baryon multiplet + one singlet that doesn’t appear in nature: 3 ⊗ 3 ⊗ 3 = (6 ⊕ ¯3) ⊗ 3 = 10 ⊕ 8 ⊕ 8 ⊕ 1
- baryons = bound states of 3 quarks
- mesons = bound state of quark + antiquark
Quark model
How to determine and what are the quantum numbers of quarks? Why do this in the first place?
Quarks never show up alone in experiments, we have to assume permanent confinement in hadrons. This can be solved by assigning them quantum numbers, making them dynamical. By construction, quarks correspond to the simultaneous eigenstates of I3 and Y in the fundamental representation, so the quark content of each hadron automatically follows from the corresponding values of I3 and Y, which are fixed by rep. Three types of flavours: up, down, strange + antiquarks. For each hadron: I3 = 1/2 (n(u) − n(d)) , Y = 1/3 (n(u) + n(d) − 2n(s)), and the n(u,d,s) can be expressed in terms of Y and I3 with
- n(u) + n(d) + n(s) = 3 for baryons,
- n(u) + n(d) + n(s) = 0 for mesons.
From the quark compositions, since Q, B and S are additive quantities, knowing the quantum numbers of the hadrons, the quark quantum numbers can be obtained: s = 1/2, Q(u) = 2/3, Q(d,s) = –1/3, S(s) = 1, S(u,d) = 0.
Quark model
What shows the explicit breaking of the SU(3) symmetry?
The strong int. doesn’t differentiate between quarks if they have the same masses, which would be the case if SU(3) was an exact symmetry (it rotates quark flavours into each other). Assigning the masses based on negligible binding energy compared to the quark masses and the linear growth of hadron masses with strangeness from the masses of baryons we find that:
- constituent masses: m(u, d) ~ 300 MeV, m(s) ~ 450 MeV
that are different from the smaller current masses. Explanation: most of the mass of a hadron does not come from the quark masses, but rather from the interaction energy between quarks, as mediated by gluons.
All in all, to explain the differences in baryon masses it is required to introduce an explicit breaking of the symmetry.
Quark model
What is colour and what’s the purpose of introducing it?
Purpose: it should be possible to assign a wave function to each baryon consistent with Fermi-Dirac statistics that is also antisymmetric under the exchange of quarks but this doesn’t happen if the wave function looks like: ψ = ψ(space)ψ(spin)ψ(flavour).
- ψ(space): naturally symmetric at the lowest level (l = 0)
- ψ(flavour): the ladder ops don’t change symmetry properties (the starting ∆++ has uuu content for which the flavour w.f. is symmetric)
- only based on FD statistics we’d get and octet of s = 1/2 and a singlet of s = 3/2 baryons (which is not the exp. result)
Solution: color, an extra degree of freedom to which an extra SU(Nc) symmetry is associated
Quark model
How does the introduction of color solve the problem with Fermi-Dirac statistics for baryons?
Since there are no further degeneracies among hadron masses, then not only the colour wave function for baryons must be antisymmetric under exchange, but in general it should also be a singlet of SU(Nc), for all hadrons.
- δ(i1i2) ψ^(1)(i1) ψ^(2)(i2)∗: allows a singlet for a fund./c.c. pair, so mesons
- ε(i1…i(Nc)) ψ^(1)(i1) . . . ψ^(Nc)(i(Nc)): antisymmetric, allows a singlet out of Nc fund. objects, so for baryons
If Nc = 3 is assumed:
- explanation of why baryons have 3 quarks
- allows a w.f. with symmetric spin-flavour parts (which gives correct results for both the octet and decuplet states)
- three quarks can’t make a totally antisymm. spin w.f., so no use for the flavour singlet for baryons (rep puzzle solved)
Quark model
What’s the Gell-Mann–Okubo mass formula?
It explains how the mass splittings can be determined for the hadronic multiplets: The strong int. Hamiltonian: H = [2m(ud) + m(s)]/3 1 + [m(ud) − m(s)]/√3 λ8 = H0 + H8, where H8 is the symmetry breaking part that transforms as the 8th component of the adjoint rep.
- the breaking has to preserve isospin and strangeness, so it has to be an isosinglet with I = Y = 0 (the smallest rep with such an element is the 8)
- assumption: H8 is a small perturbation: m(B) = m(0) + ∆m(B) —» ∆m(B) = 〈B(0)|H8|B(0)〉, where B(0) is the unperturbed state and this has to diagonalised within the degenate subspace
The result for this for each baryon:
mB = m(0)(R) + 〈B(0)|H8|B(0)〉 = ˜m(0)(R) + δ˜m1(R) Y(B) + δ˜m2(R){I(B)[I(B) + 1] − 1/4 Y(B)^2}
- there are two symm. breaking terms because the 8 rep appears twice at most in the decomposition in irreps of ¯R ⊗ R and the δ˜m terms come from the coefficients of the decomposition (the final coefficients were redefined)
- applying for mesons the mixing of the flavour singlet and octet have to be taken into account, the perturbation is not small here (spontanious breaking of chiral symmetry, the left and right parts can be treated separately)
Quark model
What is the confinement of quarks and gluons?
Quarks and gluons have to be bound in colorless states within hadrons, without the possibility of them being liberated by pulling them apart.
- color: conserved quantity (at every vertex in QCD), zero color in, zero color out for physical processes due to the physical states being singlets under color trafos
- the potential between quarks and antiquarks looks similar to a Coulomb potential with an extra linear term, so for large distances between the quarks it becomes flat
- string breaking: at sufficient energy stored in the system, a quark-antiquark pair can be created from the vacuum, the new particles binding to the old ones, preventing their liberation
