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

Chiral DynamicsChiral DynamicsHowHowss and Why and Whyss

2nd lecture: Goldstone bosons

Martin Mojžiš, Comenius University23rd Students’ Workshop, Bosen, 3-8.IX.2006

a brief history of strong interactions

23rd Students’ Workshop, Bosen, 3-8.IX.2006 Martin Mojžiš, Comenius University

|exp T| S 4int i}xdH{-iffi

pre-QCD

< < ff | | ...HHintint... | i >| i >

| f , i >| f , i >known hadronic states

HHint int

unknown Hamiltonian(in terms of hadrons)

QCD

< < ff | | ...HHintint... | i >| i >

HHintint

known Hamiltonian

| f , i >| f , i >unknown hadronic states

(in terms of quarks)

ChPT

< < ff | | ...HHintint... | i >| i >

| f , i >| f , i >known hadronic states

HHintint

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


SU(2) isospin symmetry

... qMDiqQCDL

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the symmetry

Why linear and unitary? So it happened.


Nothing to do with the superposition principle or probability conservation.

classical conservation laws


(Noether’s theorem)

)( )()( xxx

symmetry conservation law current or charge

δ = 0 μ jμ(x) = 0 jμ = δφ /(μφ )

δ = ε μ μ(x) μ jμ(x) = 0 jμ = δφ /(μφ ) – μ

δL = 0 dt Q(t) = 0 Q = d3x δφ /(0φ )

δL = ε dt (t) dt Q(t) = 0 Q = d3x δφ /(0φ ) –

δS = 0 μ Iμ(x) = 0 Iμ not known explicitly

the generators


• linear field transformation φi Tij φj

• infinitesimal transformation φi φi – i εa (ta)ij φj

δaφi = – i (ta)ij φj

• Lie algebra [ta ,tb] = i fabc tc

• SU(2) as a special case ta = ½ τa (Pauli matrices)

• conserved charges Qa = – i d3x 0φi (τa)ij φj

the generators – another incarnation


• quantization: linear operators φi H Qa

• quantum conservation laws [H , Qa] = 0

[Pμ , Qa] = 0

• Qk’s form the Lie algebra [Qa , Qb] = i fabc Qc

• realization of the original symmetry in the Fock space

• in the QCD the knowledge of Qk is not sufficient for the knowledge of hadronic states transformations

transformations of states


• Qa known explicitly in terms of quark fields

• |h(adron) not known explicitly in terms of quark fields

• Qa | h explicitly unknown, with the same energy

• eiαaQa representation of the symmetry group

• eiαaQa | h multiplet of states with the same energy

• observed in the hadronic spectrum

questions and comments


• 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 symmetriesLie groups with pertinent irreducible representationswere postulated as the symmetries of strong interactions

SU(2) SU(2) chiral symmetry

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RRLL USUUSU )1()2()1()2(

the symmetry


)1 ( 521

, LR

L

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chirality

the generators


• formally: isospin + L,R tLa, tR

a

• the Lie algebra [tLa ,tL

b] = i f abc tL

c

[tRa ,tR

b] = i f abctRc

[tLa ,tR

b] = 0

• useful combinations tVa =tR

a+tLa tA

a=tRa –tL

a

• the Lie algebra [tVa ,tV

b] = i f abc tV

c

[tAa ,tA

b] = i f abctVc

[tVa ,tA

b] = i f abc tA

c

the generators – another incarnation


• charges QVa , QV , QA

a , QA

• conservation laws [H , Q] = 0 , [Pμ , Q] = 0

• the Lie algebra [QVa

, , QVb] = i f abc

QVc 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


• even without the explicit knowledge of Qa | 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! [QAa

,QAb] = i f abcQV

c

isospin generators would also annihilate those states

spontaneous symmetry breakdown


• Nambu (60’s)• if Qa | h = 0 does not help

what about trying Qa | 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


• are there really any massless hadrons?pions are almost massless, with right quantum numbersthe 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 bosonsbut we are not able to write this down explicitlyreason: not only | h but also | 0 became complicated

even some more


• 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|j0|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


• we have identified the symmetries of the QCDas 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


• 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 operatorsi.e. transformations do not change number of particles

• this is inconvenient for the axial generators Qa

they should change the number of Goldstone bosons

how do the GB fields transform?


• 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?


• 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 3rd lecture

Documents

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