Chiral Dynamics How s and Why s 2 nd lecture: Goldstone bosons Martin Mojžiš, Comenius University23 rd Students’ Workshop, Bosen, 3-8.IX.2006

a brief history of strong interactions 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University pre-QCD | f, i > known hadronic states H int unknown Hamiltonian (in terms of hadrons) QCD H int known Hamiltonian | f, i > unknown hadronic states (in terms of quarks) ChPT | f, i > known hadronic states H int effective Hamiltonian (in terms of hadrons)

from the  QCD to the  ChPT the ChPT not derived, but constructed the procedure based on the symmetries  ChPT shares all the symmetries of  QCD the symmetries were identified in the pre-QCD period then incorporated into and understood within the QCD the most prominent: the chiral symmetry 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University

SU(2) isospin symmetry the symmetry Why linear and unitary?So it happened. 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University Nothing to do with the superposition principle or probability conservation.

classical conservation laws 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University (Noether’s theorem) symmetryconservation lawcurrent or charge δ  = 0  μ j μ (x) = 0 j μ = δφ   /  (  μ φ ) δ  = ε  μ μ (x)  μ j μ (x) = 0 j μ = δφ   /  (  μ φ ) – μ δL = 0d t Q(t) = 0 Q =  d 3 x δφ   /  (  0 φ ) δL = ε d t  (t)d t Q(t) = 0 Q =  d 3 x δφ   /  (  0 φ ) –  δS = 0  μ I μ (x) = 0 I μ not known explicitly

the generators 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University linear field transformationφ i  T ij φ j infinitesimal transformationφ i  φ i – i ε a (t a ) ij φ j δ a φ i = – i (t a ) ij φ j Lie algebra[t a,t b ] = i f abc t c SU(2) as a special caset a = ½ τ a (Pauli matrices) conserved chargesQ a = – i  d 3 x  0 φ i (τ a ) ij φ j

the generators – another incarnation 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University quantization: linear operators φ i H Q a quantum conservation laws[H, Q a ] = 0 [P μ, Q a ] = 0 Q k ’s form the Lie algebra[Q a, Q b ] = i f abc Q c realization of the original symmetry in the Fock space in the QCD the knowledge of Q k is not sufficient for the knowledge of hadronic states transformations

transformations of states 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University Q a known explicitly in terms of quark fields |h (adron)  not known explicitly in terms of quark fields Q a | h  explicitly unknown, with the same energy e iα a Q a representation of the symmetry group e iα a Q a | h  multiplet of states with the same energy observed in the hadronic spectrum

questions and comments 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University why to bother with charges if they are of no explicit use? we are going to see some use of them shortly what about SU(3)? the same story with Pauli matrices  Gell-Mann ones what was the historical development? patterns in hadronic masses  approximate symmetries Lie groups with pertinent irreducible representations were postulated as the symmetries of strong interactions

SU(2)  SU(2) chiral symmetry the symmetry 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University chirality

the generators 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University formally: isospin + L,R t L a, t R a the Lie algebra[t L a,t L b ] = i f abc t L c [t R a,t R b ] = i f abc t R c [t L a,t R b ] = 0 useful combinationst V a =t R a +t L a t A a =t R a –t L a the Lie algebra[t V a,t V b ] = i f abc t V c [t A a,t A b ] = i f abc t V c [t V a,t A b ] = i f abc t A c

the generators – another incarnation 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University charges Q V a, Q V, Q A a, Q A conservation laws[H, Q] = 0, [P μ, Q] = 0 the Lie algebra[Q V a,, Q V b ] = i f abc Q V c etc. realization of the original symmetry in the Fock space again, the knowledge of charges is not sufficient for the knowledge of hadronic states transformations

transformations of states 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University even without the explicit knowledge of Q a | h  quite a lot can be said about the hadronic multiplets to each isospin multiplet there should be a mirror multiplet with same masses and opposite parity no trace of this in the particle spectrum! could be that axial generators just annihilate | h  states? NO! [Q A a,Q A b ] = i f abc Q V c isospin generators would also annihilate those states

spontaneous symmetry breakdown 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University Nambu (60’s) if Q a | h  = 0 does not help what about trying Q a | 0   0 (SSB) Goldstone (prior to Nambu): this leads to the existence of spinless massless particles in the theory (with the precisely given quantum numbers) this is (in a sense) observed in the hadronic spectrum the spectrum does not overrule the chiral symmetry it rather supports the symmetry (in the SSB form)

questions and comments 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University are there really any massless hadrons? pions are almost massless, with right quantum numbers the mass splitting within the flavor multiplets tells us that for strong interactions 150 MeV is a small number so the pions are close enough to masslessness what are the axial generators doing with states? they should in a sense create the Goldstone bosons but we are not able to write this down explicitly reason: not only | h  but also | 0  became complicated

even some more 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University what about U(1) V and U(1) A ? U(1) V happens to be the baryon number symmetry (easy) U(1) A happens not to survive in quantum worlds (difficult) what is the precise formulation of the Goldstone theorem? if for a Noether charge Q there is an operator A for which  0|[Q,A]|0   0 then there is a massless state |G  for which  0|j 0 |G  G|A|0   0 a simple choice of A leads to  0|[Q,A]|0  =  0|qq|0  famous condensate (here it pays off to know Q explicitly)

(almost) do it yourself  ChPT 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University we have identified the symmetries of the QCD as well as the lightest particles low-energy effective theory should start as a theory of fields of these particles sharing all the symmetries with the QCD one should start with the transformation properties of these fields and to construct the invariant  ChPT the transformation properties of Goldstone boson fields are not known explicitly!

transformations of fields 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University so far we have considered either transformations of fields or of states both were linear, for different reasons for the states the reason is the superposition principle for the quantum fields: creation operators transformed to linear combinations of the creation operators i.e. transformations do not change number of particles this is inconvenient for the axial generators Q a they should change the number of Goldstone bosons

how do the GB fields transform? 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University according to a non-linear realization of the group as to the unbroken isospin subgroup the pion triplet behaves quite ordinary so for the unbroken isospin subgroup the realization should become a linear representation there is an infinite number of such realizations becoming representation when restricted to the subgroup which one is the one?

which choice is the right one? 23 rd Students’ Workshop, Bosen, 3-8.IX.2006Martin Mojžiš, Comenius University any will do!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! they are all equivalent as to the S-matrix elements the off-shell Green functions do depend on a choice but the measurable quantities do not choose the most convenient realization of the symmetry start to build up the most general invariant Lagrangian this is going to be the topic of the 3 rd lecture