Quantum numbers II

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Quantum numbers II

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M. Cobal, PIF 2006/7

Where does isospin comes from?

In quark model: proton = uud, neutron = udd

One is obtained by the other by changing a u for a d

®The closeness in mass of p and n reflects the fact that M_u and M_d must be very similar

p and n have finite size: rms of charge distribution of order≅1fm.

The Coulomb energy term (needed to put charge on p) is then ≅

Since M_n-M_p = 1.3 MeV → M_d>M_uby 2-3 MeV

The costituent mass of each quark is of order 300 MeV (one third of the nucleon mass), M_u and M_d are equal within 1%

This near equality in the mass of the lightest quarks is the reason for isospin simmetry

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Lo spin isotopico

Dà corpo all’idea che adroni aventi masse prossime, e gli stessi

spin e parità intrinseca ma cariche elettriche diverse, sono la stessa cosa per quanto riguarda l’interazione forte (indipendenza dalla

carica).

Questo fu formalizzato assegnando a ciascuno dei gruppi di adroni di cui sopra uno stesso numero quantico I, analogo a quello di spin (isospin o spin isotopico).

Come per lo spin, la terza componente dell’isospin si supponeva quantizzata, e i suoi diversi valori distinguevano fra i membri di uno stesso gruppo (multipletto).

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Sarebbero la stessa cosa, per quanto riguarda le interazioni forti, per cominciare, il protone e il neutrone. Essi divengono quindi

stati di una stessa particella. Lo stato di un nucleone può allora essere scritto come una matrice colonna a due elementi, che danno rispettivamente l’ampiezza di probabilità che si tratti di un protone o di un neutrone.

Gli stati di protone e neutrone si scriveranno allora come:

e

Essi sono autostati dell’operatore matriciale corrispondenti agli autovalori 1 e -1

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Introdotti anche gli operatori matriciali e

si verifica che le tre matrici soddisfano alle regole di commutazione:

e Inoltre

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Una generica trasformazione nello spazio degli stati si scrive:

La trasformazione deve trasformare una base ortonormale in una base ortonormale, la matrice U deve essere unitaria.

Le matrici unitarie hanno determinante di modulo 1.

Se il determinante vale 1, si chiamano unimodulari.

Le matrici in questione formano gruppo: il gruppo SU(2), gruppo delle trasformazioni unitarie unimodulari.

Le individuano l’algebra di Lie del gruppo.

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L’algebra dei generatori può essere generalizzata in termini di operatori astratti:

Le matrici 2x2 date sopra che soddisfano a quest’algebra

agiscono sulla rappresentazione bi-dimensionale (fondamentale)

del gruppo. Si possono trovare via via matrici NxN che soddisfano all’algebra. I multipletti che sono trasformati da esse costituiscono rappresentazioni N-dimensionali di SU(2). La rappresentazione

tridimensionale è nota come la regolare. Essa descrive particelle di isospin 1, dunque un tripletto (es.: i pioni, nei loro tre stati di carica).

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L’isospin si supponeva rigorosamente conservato nelle interazioni forti. Le piccole differenze di massa all’interno di un multipletto si attribuivano alle interazioni elettromagnetiche che distinguevano fra di esse (esempio: protone e neutrone, doppietto di isospin ½.)

Ora, l’indipendenza dalla carica si può tradurre nell’invarianza dell’interazione forte per rotazioni nello spazio dell’isospin, cui corrisponde la conservazione dello spin isotopico.

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Isospin I

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Nuovi numeri quantici: la stranezza

Negli anni 50, i kaoni e le Λ erano noti come particelle

strane. La stranezza del loro comportamento stava nel fatto che erano prodotte con sezioni d’urto tipiche delle interazioni forti, ma decadevano con vite medie proprie delle deboli.

Nel 1952 Abraham Pais avanzò la sua ipotesi della “produzione associata”: un kaone e una Λ interagivano “forte” soli in coppie, e come tali potevano essere prodotti da interazioni forti; ma,

lasciate a se stesse, potevano decadere solo via interazione debole.

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Particles with I=0 are isosinglets :

Λ(116)=uds, I(J)^P = 0(1/2)⁺

-By introducing isospin, new criteria for non-exotic particles:

In all observed interactions these criteria are satisfied, confirming once again the quark model

This allows predictions of possible multiplet members

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I due nucleoni non si distinguerebbero nel caso di una simmetria

esatta. L’esatta simmetria è rotta dall’interazione elettromagnetica.

Anche i barioni ordinari, nucleoni,e iperoni, per un totale di 8, si

assomigliano abbastanza. Si può pensare a una simmetria rotta, che di per sé li porrebbe in un unico multipletto.

D’altra parte, le interazioni forti possiedono altre leggi di conservazione oltre a quella dello spin isotopico: quella del numero barionico e quella della stranezza.

La simmetria, una volta rotta, dovrebbe lasciare inviolate la

conservazione dell’isospin, del numero barionico e della stranezza.

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SU(3). Ritrovare un ordine.

Furono Murray Gell-Mann e Yuval Ne’eman (1961) a rendersi

conto, indipendentemente, che il gruppo di simmetria che garantiva tutto ciò era la più immediata generalizzazione di SU(2), cioè il

gruppo SU(3) delle matrici unitarie unimodulari 3x3.

M.Gell-Mann, Y. Ne’eman, The Eightfold Way, Benjamin, 1964

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Sembra presentarsi subito una difficoltà: la rappresentazione fondamentale è tridimensionale, può quindi albergare tre

particelle e non otto. Bisognerà dunque saltarla, questa

rapresentazione fondamentale, e partire dalla regolare, che otto-dimensionale. Va osservato che il contenuto in multipletti di isospin e l’attribuzione ad essi di un valore definito della stranezza è prescritto.

La figura mostra l’ottetto barionico.

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Ma non è proibita l’esistenza di altri ottetti – non barionici:

l’ottetto dei mesoni pseudoscalari

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e quello dei mesoni vettoriali

N.B.: c’è qualche problema di

sovrabbondanza, per il quale rimando a Gell-Mann e

Ne’eman, op. cit.

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Si passa poi alle rappresentazioni di maggiore dimensionalità. La prima è il decupletto. Vi trovano posto stati eccitati degli iperoni, nonché la Δ, ex “risonanza 33”. Ma il vertice, la Ω, è mancante …

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Deve trattarsi di un singoletto isotopico con spin 3/2, parità +, massa di circa of 1.680 MeV, carica negativa, numero barionico +1, stranezza = -3, stabile rispetto a decadimenti forti.

Un barione con queste caratteristiche fu scoperto nel 1964 da un gruppo di fisici di Brookhaven e delle Università di Rochester e Syracuse, condotto da N. Samios, usando la camera a bolle da 80 pollici.

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Strangeness S

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-Suppose we observe production of the Σ+ baryon in a strong interaction:

Which then decays weakly:

It follows that Σ⁺ baryon quantum numbers are: B=1, Q=1, S=-1, and hence Y=0 and I₃=1

Since I₃> 0 ⇒there are more multiplet members!

If a baryon has I₃=1, the only possibility for isospin is I=1 and we have a triplet: Σ⁺, Σ⁰, Σ^-

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Λ+γ π- + n Masses and quark composition of Σ-baryons:

Σ⁺(1189)=uus, Σ⁰(1193)=uds, Σ^-(1197) = dds

It clearly indicates that d-quark is heavier than u-quark Combining different methods, it has been found:

2≤ m_d-m_u ≤ 4 (MeV/c²)

which is negligible comparing to hadron masses

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Isospin in two-nucleon

Isospin of two-nucleons, each with I = ½ : (in analogy with combination of two spin ½)

triplet with I=1

symmetric under label inter-charge

singlet with I= 0

antysimmetric for label inter-charge

Other notation:

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• Example: In each case: final state has I = 1

p+p → d + π⁺ p+n → d + π⁰

I = 1 0 1 I= 0 o 1 0 1 This is a pure I= 1 In this case: 50%

I=0

reaction and 50% I=1

Conservation of isospin → reactions can proceed ONLY through the I=1 channel.

• Important application of I conservation: in strong interactions of non-identical particles (mixture of different isospin states)

e.g.: pion-nucleon scattering I_π = 1, I_N = ½ → I_ToT= ½ or

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If strong interactions depend only on I and not on I₃, the 3x2 π-n scattering processes can be described in terms of two isospin

amplitudes

These 2 decays have I₃ = ± 3/2:

described by I= 3/2 amplitude These 4 decays have I₃ = ± 1/2:

described by I= 3/2 or ½ amplitudes

The weights of the two amplitudes in the mixture are given by the Clebsh-Gordan coefficients (see Appendix C1 of Perkins)

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• For pion-nucleon scattering:

I = 3/2 I = 1/2

Pion Nucleon I3 = 3/2 1/2 -1/2 -3/2 1/2 -1/2

π⁺ p 1

π⁺ n

π⁰ p -

π⁰ n

π^- p -

π^- n 1

• Calculate cross-section for:

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Where H is an isospin operator: H₁ if it operates on initial/final states with I = ½, H₃ for states with I = 3/2

By conservation of isospin: no operator connecting I = ½ with I = 3/2.

• Reaction ( π⁺ p → π⁺ p) is a pure I = 3/2, I₃ = +3/2→

• For reaction ( π^- p → π^. p) we may write:

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• For reaction ( π^- p → π⁰ n) we may write:

• Then:

• Limiting situations are:

Experimentally: (σ_π+p/ σ_π−p)_tot = 3, proving that the I = 3/2

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Finally, there are 2 discrete symmetries associated with reversing the direction of some quantity. These are:

1. Charge conjugation – particles into anti-particles.

2. Time reversal – changing the direction of time.

Interesting because it is not obvious whether the laws of nature should look the same for any of these changes, and the answer was surprising when these symmetries were first tested.

Example of a neutron and its decay to illustrate each of the two symmetries. Neutrons have spin angular momentum of ½ and

decay as:

n→p e ν

Other symmetries

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Charge conjugation

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Let’s denote particle which have distinct anti-particles by “a”, and otherwise by “α”

Then:

That is, the final state acquires a phase factor Otherwise:

That is, from the particle in the initial state, we arrive to the antiparticle in the final state

The 2^nd transformation turns antiparticles back to particles, and hence:

For multiparticle states the transformation is:

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-Other eigenstates can be constructed from particle-antiparticle pairs:

For a state of definite L, interchanging between particle and antiparticle reverses their relative position vector. For example:

For fermion-antifermion pairs theory predicts:

This implies that π⁰, being a ¹S₀ state of and must have C-parity of 1.

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C_γ can be inferred from classical field theory:

and since all electric charges swap, electric field and scalar potential also change sign:

Which upon substitution into: gives C_γ = -1

Tests of C-invariance

• Another confirmation of C-invariance comes from observation of η-meson decays:

These are em decays, and first two clearly indicate that C =1.

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Charge conjugation (C) simply means to change each particle into its anti-

particle. This changes the sign of each of the charge-like numbers. The neutron is neutral, nonetheless it has charge-like quantum numbers. It is made of three quarks, and charge conjugation change them into three anti-quarks. Charge

conjugation leaves spin and momentum unchanged.

The interesting question is, does a world composed completely of anti-matter have the same behavior. For example, in neutron decay, there is a correlation between the spin of the neutron and the direction of the electron that is

emitted when the neutron decays. The electron spin is also directed opposite to its direction of motion.

← → momentum direction n ν p e

⇐ ⇐ ⇒ ⇐ spin direction

Charge conjugated: ← → momentum direction n ν p e

This is not what an anti-neutron decay looks like! The laws of physics responsible for neutron decay are not invariant with respect to charge conjugation.

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The parity operation (P) changes the direction (sign) of each of the spatial

coordinates. Hence, it changes the sign of momentum. Since spin is like angular momentum (the cross product of a vector direction and a vector momentum, both of which change sign under the parity operation), spin does not change direction under the parity operation.

⇐ ⇐ ⇒ ⇐ spin direction Parity operation:

→ ← momentum direction n ν p e

The world would look different under the parity operation, since now the electron’s spin would be in the same direction as its momentum.

The world is not symmetric under the parity operation!

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The lack of symmetry under the parity operation was

discovered in the fifties following the suggestion of Lee and Yang that this symmetry was not well tested experimentally. It is now known that parity is violated in the weak interaction, but not in strong and electromagnetic interactions.

The situation with charge conjugation symmetry is similar; the lack of symmetry under charge conjugation exists only in the weak interaction.

The Standard Model incorporates parity violation and charge conjugation symmetry violation in the structure of the weak interaction properties of the quarks and leptons and in the form of the weak interaction itself.

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In weak interactions C invariance is broken:

LH neutrino ν → LH antineutrino

(which dos not exists)

However under combined CP:

LH neutrino ν → RH antineutrino

Weak interactions are eigenstates of CP This statement is not completely true:

CP violation in weak interactions does occur at the 10^-4 level

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Time reversal means to reverse the direction of time. Here we need to be a bit more careful. There are a number of ways in which we can consider time

reversal. For example, if we look at collisions on a billiard table when the cue ball strikes the colored balls on the break, it would clearly violate our sense of how things work if time were reversed. It is very unlikely that we would have a set of billiard balls moving in just the directions and speeds necessary for

them to collect and form a perfect triangle at rest, with the cue ball moving away. However, if we look at any individual collision, reversing time results in a perfectly normal looking collision (if we ignore the small loss in kinetic energy due to inelasticity in the collision). The former lack of time reversal invariance has to do with the laws of thermodynamics; we here are interested in

individual processes for which the laws of thermodynamics are not important.

Time reversal reverses momenta and also spin, since the latter is the

cross product of a momentum (which changes sign) and a coordinate, which does not.

Time Reversal

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Now let’s consider what happens when we apply time reversal (T) to the case of the neutron decay.

⇐ ⇐ ⇒ ⇐ spin direction Time reversal:

→ ← momentum direction n ν p e

⇒ ⇒ ⇐ ⇒ spin direction

This looks just fine, the electron spin is opposite to its momentum and the electron direction is opposite to the neutron’s spin.

So, at least for neutron decay, the laws of physics appear to be symmetric under time reversal invariance.

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Now let’s consider what happens when we apply all three symmetry operations to the case of neutron decay.

Time reversal: → ← momentum direction n ν p e

Parity plus time reversal: ← → momentum direction n ν p e

T, P and Charge conjugation: ← → momentum direction n ν p e

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The result that applying C, P, and T leaves the physical laws unchanged is not

surprising. Since CP leaves things unchanged (for neutron decay) and T also does, applying all three should also work fine.

In fact, there is a theorem that says that under rather general conditions, any set of physical laws that can be described by a field theory will be unchanged under the CPT operation. There are many consequences to this theorem, for example that the total lifetime and mass of a particle is identical to that of its anti-particle.

There are some considerations of conditions under which physical laws are not invariant under CPT, but sensitive experimental tests of CPT invariance have not shown any evidence for its breakdown.

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Could there be evidence of violation of one or more of these symmetries in neutrons?

Consider the case if neutrons had an electric dipole moment (edm). The neutron has no charge, but it does have charged quarks inside it. If the charge is distributed such that the negative and positive charge is separated by some distance (within the neutron), then it would have a dipole moment. The value of the dipole moment is the value of the positive charge times the distance between the positive and negative charges. The direction of the dipole moment must be aligned with the spin. Assume the neutron dipole moment points in the same direction as the spin:

. T P C CP CPT

← ← → → ← ← dipole moment direction n n n n n n

⇐ ⇒ ⇐ ⇐ ⇐ ⇒ spin direction

Charge conjugation correctly turns a neutron into an anti-neutron, with the spin and electric dipole moment in opposite directions. CPT does the same. However, both T and CP produce non-physical particles, with the relative direction of spin and edm incorrect. The existence of a neutron edm explicitly violates CP and T symmetries.

No such evidence for a neutron edm is seen.

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There is evidence for violations of CP symmetry and hence of T symmetry. Until recently, that evidence existed solely in the decays of neutral kaons. A neutral kaon is a meson consisting of a strange quark and a down quark. The physical particles with definite mass and lifetime are combinations of a kaon and an anti-kaon, much the way that circularly polarized light is a combination of vertically and horizontally polarized light.

Now, one combination of K⁰and K⁰ that makes a physical particle with definite

mass and lifetime is mostly a CP eigenstate with eigenvalue +1 (K⁰_S) and another combination is a CP eigenstate with eigenvalue –1 (K⁰_L) . It was a surprising result found in 1964 that the K⁰_Ldecayed into a pair of pions that were in a CP eigenstate with eigenvalue +1. This implied violation of CP symmetry in kaon decays.

The manifestation of CP violation is restricted to the weak interaction. Hence

processes that involve only the electromagnetic and strong interactions appear to be CP conserving.

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• Until 1964, it was believed that CP simmetry was respected. That year Christensons discovered that:

Normally: (CP = -1)

But also: (CP= 1, prob = 2x10^-3)

• Origin of this CP violation is not known, but it is a necessary pre- requisite to the development of a baryon-antibaryon asimmetry

• CP implies T violation through the CPT theorem

• In weak interactions limits found on T-violating contributions are below the 10^-3 level.

• In strong interactions, studied with forward-backward reactions:

if the interactions are invariant under T and P → matrix elements for the forward and backward reactions must be the same (< 5x10^-4)

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• CP violation will be mentioned, linked to violation under time reversal T

• CPT Theorem: All interactions are invariant under the successive operation of C(=charge conjugation) P(=parity) and T=time reversal) taken in any order

Consequences of the CPT theorem that may be tested with

experiments relate to properties of particles and antiparticles:

Must have:

- same mass - same lifetime

- equal and opposite electric charge and magnetic moments

CPT conservation

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Summary

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Quantity Under C Under T Under P

position r r -r

momentum p -p -p

Spin σ −σ σ

E field E E -E

B field B -B B

Magnetic dipole momentum

σ⋅B σ⋅B σ⋅B

Electric dipole momentum

σ ⋅ E −σ ⋅ E −σ ⋅ E

Long. polarisation σ ⋅ p σ ⋅ p −σ ⋅ p

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Hadron Quantum Numbers

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• Characteristics of a hadron:

1) Mass

2) Quantum numbers arising from simmetries: J,P,C. Common notation:

- J^P (e.g. for proton: ½⁺), or

- J^PC if a particle is also an eigenstate of C (e.g. For π⁰: 0⁺)

3) Internal quantum number’s: Q and B (always conserved), S, C, , T (conserved in em and strong interactions)

How it is known the quantum numbers of a new hadron?

Hadron quantum numbers

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Some a priori knowledge is needed:

Considering the lightest 3 quarks (u,d,s), the possible 3- and 2- quarks states will be

Mesons with S = -1 and Q =1 are forbidden

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• Particles which fall out of above restrictions are called exotic

particles ( )

• From observations of strong interactions processes, quantum numbers of any particles can be deduced.

•Observation of pions

confirm these predictions ensuring that pions are non- exotic

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Assuming that K^- is a strange meson, one can predict quantum numbers of Λ-baryon:

And further, for K+ meson:

 All of the more than 200 observed hadrons satisfy this kind of predictions, and NO exotic particles have been found so far,

 Confirms validity of the quark model which suggests that only quark-antiquark and 3-quarks (or antiquark) states exist

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Even more quantum numbers..

It is convenient to introduce another quantum number which are conserved in strong and em interactions:

-Sum of all internal quantum numbers, except of Q hypercharge Y = B+S+C+ +T

- Instead of Q: I₃ = Q – Y/2

which is to be treated as a projection of a new vector Isospin I = (I₃)_max

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-Hypercharge Y, isospin I and its projection I3 are additive quantum numbers, so that the quantum numbers for hadrons can be deduced from those of quarks:

-Proton and neutron both have isospin of ½, and also very close masses. They are said to belong to an isospin doublet

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Isospin ‘up’ en isospin ‘down’

•

• Heisemberg (1932): neutron and proton might be treated as different charge sub-states of one particle: the neutron

• For every nucleon: new quantum number – isospin- with value I = ½

• There are 2 substates, with I_z, or I₃ = ± ½

• Charge is given by Q/e = ½ + I₃if we assign I₃ = + ½ to the proton and I₃ = - ½ to the neutron

• Complete analogy with the description of a particle of ordinary

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Other examples:

• Useful concept, because it is quantum number conserved in strong interactions ⇒ these depends on I, not on I^3.

• In strong interactions we do not distinguish between n and p:

they are degenerate states

• Em interactions do not conserve I.

They couple to electric charge Q and then “single-out” the I³-axis in the isospin space

• Isospin simmetry apply between all baryons and mesons which transform one to another by change of u and d quarks.

I₃ = +1 I₃ = -1