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Properties conserved in Properties conserved in Strong and EM interactionsStrong and EM interactions

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OutlineOutline• Multiplets• Strangeness• Charge Conjugation (C)• Parity (P)• Time Reversal (T)• T Inv. and Weak Decays: CPT Inv.• Isospin Concept• Isospin Calculation• Isospin Conservation• Application

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MultipletsMultiplets• Hadrons can be arranged in small groups with all members of the group

having nearly the same mass.• The members of a group are called by the same name (pions, kaons, nu

cleons, antikaons, etc).• The groups are called multiplets and the number of particles in a group

is called the multiplicity of that group.• B=Baryon number• S=Strangeness• Y=Hypercharge

=Twice of Qave

• S=Y-BThis formula must be altered for particles that have Charm.

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NoteNote• K- is antiparticle of K+.• But Σ- not antiparticle of Σ+ .• How about π? Why they can stay in same multiplet?

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ReminderReminder• Universal conservation

– conservation of energy– conservation of baryon number – conservation of lepton number– conservation of charge– conservation of momentum

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StrangenessStrangeness• Decay time

– Strong ~10-23 sec– EM ~ 10-21 sec– Weak ~ 10-10

sec

• Nine example for weak decay which violate the conservation of strangeness

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Charge Conjugation (Charge Conjugation (CC))• Charge Conjugation (C) is the operation of changing

a particle into its antiparticle.• The strong and EM interactions are said to be invaria

nt under charge conjugation.• If, in the first interaction, A is

scattered through a certain angle with a certain probability, then, in the second interaction, A is scattered through the same angle with the same probability provided, of course, the experimental conditions (initial energy, momentum, and spin) are the same.

time decay same the with

C under Invariance

CBACBA

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CC Non-Invariance : Weak Non-Invariance : Weak

• The electron from μ- the tend to leave the decay site preferentially in the direction opposite to the spin direction of the μ- while the positrons from μ+ tend to leave preferentially in the same direction as the spin of the μ+. The decay is not invariant under charge conjugation.

e

e

e

e

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Parity (Parity (PP))• The P Operator (P), works on x, p, L, S. • P reverses the components of all true vectors. • For example, the parity operation on radius and mom

entum vectors produces:

• Note that angular momentum is the product of two vectors so it does not get reversed by the parity operator:

• Spin is a form of angular momentum so it too does not change sign under the P.

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Inv. Under PP : Strong & EM, not Weak.• Inv. case: if A + B → C + D,

and P(A) = A’; P(B) = B’; P(C) = C’ then A’ + B’ → C’ + D’

• Non-Inv. case:eepn

• Non-Inv. case: neutrino itself

P( )=

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CPCP Invariance Invariance• Neutrino itself

http://universe-review.ca/I15-25-CPviolation1.jpg

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CPCP Violation Violation• K0 can transform into their antiparticles (in which each quark is r

eplaced with the other's antiquark) and vice versa, but such transformation does not occur with exactly the same probability in both directions

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Time Reversal (Time Reversal (TT))• The T makes time run backwards.• This reverses both momentum and spin

and changes ingoing particles to outgoing and outgoing to incoming.

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CPT CPT Invariance Invariance• Many physicists believe that all interacti

ons must be CPT invariant.

• If this is true and it is also true that CP is violated in weak interactions, then T must also be violated in just the right way to make weak interactions CPT invariant.

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Isospin ConceptIsospin Concept• Multiplet Splitting and Charge.• Examples of Hadron multiplets are:

a doublet of nucleons and a triplet of pions.

• The members of a multiplet have the same strangeness, hypercharge, spin, baryon number, electron family number, and muon family number, but differ in charge and differ slightly in mass (about 5MeV differences in mass).

• It is believed that if electromagnetic interactions were turned off, the members of a multiplet would be identical.

• This hypothesis is born out by the study of nucleon-nucleon scattering where neutron-neutron scattering can be deduced from proton-proton scattering by “turning off the proton's charge" in the scattering equations.

• The algebra of isospin is exactly the same as for spin, which accounts for its name. It should be emphasized at the outset, however, that isospin is not spin. Spin exists in real space but isospin exists only in an abstract mathematical isospin space.

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Isospin CalculationsIsospin Calculations• Each multiplet is assigned a isospin number I.• Positive integer or half an odd positive integer.• Isospin may be considered to be a vector with magnitude

(I(I + 1))1/2 (compare with the spin vector of magnitude ћ (s(s + 1))1/2 .)

• To emphasize that isospin is not a vector in coordinate space, we will not call the axes x, y, z but rather 1, 2, 3.

• The 3 component, I3, may take on any one of the values I, I-1, I-2, …, -I in a fashion similar to the values of the z component of angular momentum.

• Each of these values corresponds to a different member of the multiplet with I3.

• I3 = I assigned to the particle with the most positive charge and I3= -I assigned to the particle with the most negative charge.

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Isospin CalculationsIsospin Calculations• If a particle is in a state with definite I3 , the uncertainty principle

prohibits it from having definite values of I1 or I2 so we ignore those components.

• Multiplicity. If the isospin of a multiplet is I, there are 2I + 1 particles in the multiplet.

• This result follows from counting the possible values of I3.

• Thus singlets have I = 0, doublets have I = 1/2, and triplets have I = 1. Later we shall run into a quartet (I = 3/2).

• Note that an antiparticle has the same value of I as the corresponding particle but its I3 has the opposite sign.

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Isospin CalculationsIsospin Calculations• The magnitude of the total isospin is (I(I + 1))1/2 where I is a

positive integer or half an odd positive integer and, for a given collection of particles, the maximum value I can have is the sum of the individual particle's I's:

• The value of I3 for a collection of particles is the sum of the individual particle's I3's:

• I must be greater than or equal to | I3 |. Thus I can have any one of the values Imax, Imax-1, Imax-2, …, | I3 |.

• For example if one considers π+p scattering:Iπ+ =1; Ip =1/2; Imax =3/2; I3π+ =1; I3

p =1/2; I3 =3/2; so I=3/2

• For π-p scattering: Iπ- =1; Ip =1/2; Imax =3/2; I3π- =-1; I3

p =1/2; I3 =-1/2; so I=3/2 or 1/2

i

iII )(33

i

iII )(max

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Isospin ConservationIsospin Conservation• Relationship: Q, S, B, I3; Y, I3.

• Note that the signs of S, B, I3, and q are opposite for a particle and its antiparticle.

• Both I and I3 are conserved during strong interactions and decays. Since q, S, and B are conserved there, the conservation of I3 is not a new conservation law.

• For weak interactions neither S, I3, nor I need be conserved although q and B are universally conserved.

• For electromagnetic interactions, I need not be conserved.

33 22I

YI

BSq

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Deducing Interactions from Decay ProductsDeducing Interactions from Decay Products• If particles appear for which isospin is not defined (γ, leptons), th

en the interaction is not strong.

• If neutrinos appear, then the interaction must be weak.

• If photons appear, the interaction must be include electromagnetic.

• It is sometimes impossible to distinguish between weak and electromagnetic interactions on the basis of product particles alone. In particular, when C, P, or T violation occurs, the products may not show the violation (neutrinos, of course, do).

• What you are really doing is finding the dominant part of the interaction for the determination of lifetime.

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More ExampleMore Example• K+,Λ 0,Ξ0,Ω- only decay by weak • K+ is the lightest strange particle and a combination of the energy and st

rangeness conservation laws forces its decay to be weak.• The Λ 0 is the lightest strange baryon and conservation of energy, strang

eness, and baryon number precludes its decay via the strong or electromagnetic interactions.

• The Ξ0, if it decayed strongly or electromagnetically, must decay to particles with a net strangeness of -2.

– Clearly, ΣΣ, ΛΛ, ΛΣ, ΣK meet energy conservation requirements.– Two K's have less total mass and conserve strangeness and charge.– However the products must include a baryon and all baryon have to

o much mass to join the kaons as decay products.– The Ξ0, must decay weakly. In fact, the most probable decay is

• The Ω- also decays weakly can be predicted by invoking conservation of strangeness, energy and baryon number.

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More ExampleMore Example• As an example, consider the strong or electromagnetic decay of the Σ+.• To conserve strangeness, which strong and electromagnetic decays require,

the Σ+ must decay either to a Λ 0 or a kaon, the only strange particles with less mass.

• Consider first the decay to a Λ 0 . – The other decay products must have net strangeness 0, total mass less than 73.8

MeV, and charge +1. – These conditions can be met only if a positron appears among the products.– Other products must have net charge 0, and electron family number +1.– Only the electron's neutrino will do.– Since a neutrino is required, parity invariance is violated and the decay

cannot be strong or electromagnetic.• Now consider the possibility of decay to a K0.

– Strangeness is conserved. Other products must have net charge +1, mass less than 695.6 MeV, and strangeness 0. In addition, at least one baryon must be involved.

– There is no baryon with mass less than 695.6 eV. Thus this decay cannot occur via the strong or electromagnetic interactions.

• The decay most often observed is

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More ExampleMore Example• This interaction conserves charge and baryon number.• It can conserve angular momentum, energy, and mom

entum. It conserves strangeness.

• We conclude that isotopic spin could be conserved. So, barring violation of C, P, or T, the interaction is strong.

S 0 0 1 -1

I 1 ½ ½ 0

I3 -1 ½ -½ 0

I 3/2 or ½

½

I3 -½

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Special case : sigma decay Special case : sigma decay ΣΣ00→→ΛΛ 00++γγ• The decay of the Σ0.• The products must include a single baryon (conservation of energy

prohibits 2 baryons and an antibaryon). • If the decay is to be strong the product particles must have

strangeness =-1. This requirement leads to the conclusion that one of the products is either Λ 0 or one of the anti-kaons. A kaon and any baryon together are too massive so the decay must be to a Λ 0.

• Any other particles that could be produced must have total mass less than 76.9 MeV: i.e., photons, electrons, or neutrinos.

• We have determined that the decay products of Σ0 must be the Λ 0 plus photons or leptons.

• Can this decay take place via the strong interaction? • The Σ0 has isotopic spin 1 while the Λ 0 has isotopic spin 0. • All the other possible products (leptons and photons) have isotopic spin 0.

• We conclude that the decay cannot take place via the strong interaction. • However all the conservation laws for the electromagnetic interaction do

hold in fact the decay does not violate any conservation law we have discussed except isospin. The decay takes place via the electromagnetic interaction.

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