particle physics lecture 7 hadron spectroscopyhadron spectroscopy isospin symmetry and multiplets...

Particle Physicslecture 7

Hadron Spectroscopy

Yazid Delenda

Departement des Sciences de la matiereFaculte des Sciences - UHLB

http://delenda.wordpress.com/teaching/particlephysics/

Batna, 16 Novembre 2014

(http://delenda.wordpress.com) Particle Physics - lecture 7 1 / 25

Hadron Spectroscopy

We would like to study the catalogue of observed hadrons in nature andtheir properties (masses, spins, lifetimes, etc...). We shall try tounderstand their features in terms of the simple quark model: Baryons(B = qiqjqk) and Mesons M = qiqj .For simplicity we consider states made of light quarks only (u, d, s andtheir anti-particles) only,so we consider no heavy quarks c, t, b(C = B = T = 0).

Hadron Spectroscopy

Hadron Spectroscopy Isospin symmetry and multiplets

Isospin symmetry and multiplets

Isospin multiplets are families of hadrons with almost equal masses towithin a few MeV/c2. Examples include:

Family Spin B S Q

p(938), n(940) 12 1 0 1,0

π−(140),π0(135), π+ (140) 0 0 0 −1,0, +1

(498),K−(494) 0 0 −1 0,−1

where masses are given in units of MeV/c2.Members of the same familymust have the same spin (thus the name isospin),and same baryon numberand strangeness.However members have different electric charges.These members reflect a symmetry between u and d quarks and is calledisospin symmetry.

Family Spin B S Q

p(938), n(940) 12 1 0 1,0

π−(140),π0(135), π+ (140) 0 0 0 −1,0, +1

(498),K−(494) 0 0 −1 0,−1

Family Spin B S Q

p(938), n(940) 12 1 0 1,0

π−(140),π0(135), π+ (140) 0 0 0 −1,0, +1

(498),K−(494) 0 0 −1 0,−1

Family Spin B S Q

p(938), n(940) 12 1 0 1,0

π−(140),π0(135), π+ (140) 0 0 0 −1,0, +1

(498),K−(494) 0 0 −1 0,−1

Family Spin B S Q

p(938), n(940) 12 1 0 1,0

π−(140),π0(135), π+ (140) 0 0 0 −1,0, +1

(498),K−(494) 0 0 −1 0,−1

Family Spin B S Q

p(938), n(940) 12 1 0 1,0

π−(140),π0(135), π+ (140) 0 0 0 −1,0, +1

(498),K−(494) 0 0 −1 0,−1

Consider the quark contents of these families:

p = uud, n = udd

π+ = ud, π0 = uu, dd, π− = du

= sd, K− = su

so members differ by

u→ d, d→ u, u→ d, d→ u

It is convenient to define the new quantum number I3:

2(Nu −Nd)

or equivalently:

I3 = Q− Y

where Y is the hypercharge defined by:(http://delenda.wordpress.com) Particle Physics - lecture 7 4 / 25

p = uud, n = udd

π+ = ud, π0 = uu, dd, π− = du

= sd, K− = su

u→ d, d→ u, u→ d, d→ u

2(Nu −Nd)

or equivalently:

I3 = Q− Y

p = uud, n = udd

π+ = ud, π0 = uu, dd, π− = du

= sd, K− = su

u→ d, d→ u, u→ d, d→ u

2(Nu −Nd)

or equivalently:

I3 = Q− Y

p = uud, n = udd

π+ = ud, π0 = uu, dd, π− = du

= sd, K− = su

u→ d, d→ u, u→ d, d→ u

2(Nu −Nd)

or equivalently:

I3 = Q− Y

p = uud, n = udd

π+ = ud, π0 = uu, dd, π− = du

= sd, K− = su

u→ d, d→ u, u→ d, d→ u

2(Nu −Nd)

or equivalently:

I3 = Q− Y

Y = B + S + C + B + T = B + S

for C = B = T = 0 in our study.The hyper charge quantum number is thesame for all members of the family.Note that changing u↔ d changes I3 and Q by ±1 and leaves Yconstant.In general for each multiplet we define the isospin I as:

I = (I3)max

Then we always have:

I3 = I, I − 1, I − 2, ...,−I

hence there are 2I + 1 members in the family.

Y = B + S + C + B + T = B + S

I = (I3)max

I3 = I, I − 1, I − 2, ...,−I

Y = B + S + C + B + T = B + S

I = (I3)max

I3 = I, I − 1, I − 2, ...,−I

Y = B + S + C + B + T = B + S

I = (I3)max

I3 = I, I − 1, I − 2, ...,−I

Y = B + S + C + B + T = B + S

I = (I3)max

I3 = I, I − 1, I − 2, ...,−I

Y = B + S + C + B + T = B + S

I = (I3)max

I3 = I, I − 1, I − 2, ...,−I

Example

Family I3 I

p = uud +12 1

2n = udd −12

π+ = ud +11π0 = uu, dd 0

π+ = du −1K

0= sd +1

2 12K− = su −1

they are respectively isodoublet, isotriplet and isodoublet.There are alsoisosinglets with no partners,for example Λ = uds has I3 = I = 0.Why do these families have the same mass?

Example

Family I3 I

p = uud +12 1

2n = udd −12

π+ = ud +11π0 = uu, dd 0

π+ = du −1K

0= sd +1

2 12K− = su −1

Example

Family I3 I

p = uud +12 1

2n = udd −12

π+ = ud +11π0 = uu, dd 0

π+ = du −1K

0= sd +1

2 12K− = su −1

Example

Family I3 I

p = uud +12 1

2n = udd −12

π+ = ud +11π0 = uu, dd 0

π+ = du −1K

0= sd +1

2 12K− = su −1

Example

Family I3 I

p = uud +12 1

2n = udd −12

π+ = ud +11π0 = uu, dd 0

π+ = du −1K

0= sd +1

2 12K− = su −1

If we had identical quark masses mu = md and identical forces on u and d(both EM and strong), then replacing an up quark with a down quarkwould leave the same quark wave-function for the state,which would leavethe ground state energy (and thus mass) unchanged (m = E/c2).Wewould then have Mp = Mn, M

K0 = MK− exactly.Isospin would then be

an exact symmetry.The strong forces on the u and d (as well as on s, c, t and b) are flavourindependent (identical).However isospin is not an exact symmetry because:

md −mu ≈ 3MeV/c2 � hadron masses

Electromagnetic forces on u and d are different because they havedifferent electric charges: Qd = −1/3e, Qu = +2/3e.However theEM force is much smaller than strong force, so the symmetry justslightly broken.

This is a familiar example of a broken symmetry,but it is still a goodapproximation which is very predictive .For example it implies that all

members of an isospin family, e.g. (p, n), (π+,π0, pi−), (K0, K−) have

the same strong interactions , and many other predictions.

Hadron Spectroscopy Isospin in SU(2)

Isospin in SU(2)

Recall that the addition of the spin of two spin-1/2 particles gives thetriplet eigenstates |s,m〉 of total angular momemtum squared j = 1 andits projection:

|1, 1〉 = | ↑↑〉|1, 0〉 =

(| ↑↓〉+ | ↓↑〉)

|1,−1〉 = | ↓↓〉

We can analogously think of the pion as being composed of two spin-halfquarks (up ↔↑ and down ↔↓).

Isospin in SU(2)

|1, 1〉 = | ↑↑〉|1, 0〉 =

(| ↑↓〉+ | ↓↑〉)

|1,−1〉 = | ↓↓〉

Isospin in SU(2)

|1, 1〉 = | ↑↑〉|1, 0〉 =

(| ↑↓〉+ | ↓↑〉)

|1,−1〉 = | ↓↓〉

Isospin in SU(2)

So we may write the eigenstates of isospin |I, I3〉:

|1, 1〉 = |ud〉 = |π+〉|1, 0〉 =

(|uu〉+ |dd〉) = |π0〉

|1,−1〉 = |du〉 = |π−〉

So the pions here are eigenstates of isospin operator I2 and I3.

Isospin in SU(2)

|1, 1〉 = |ud〉 = |π+〉|1, 0〉 =

(|uu〉+ |dd〉) = |π0〉

|1,−1〉 = |du〉 = |π−〉

Isospin in SU(2)

|1, 1〉 = |ud〉 = |π+〉|1, 0〉 =

(|uu〉+ |dd〉) = |π0〉

|1,−1〉 = |du〉 = |π−〉

Hadron Spectroscopy Resonances

Resonances

Resonances are unstable hadrons, which decay by strong interactions tolighter hadrons.Consider a given quark system, e.g. ud (π+(135) in the ground state),uud (p (938) in the ground state), uds (Λ(1119) in the ground state).These ground states of bound quark systems are usually long lived, so theydecay by electromagnetic or weak interactions, or are stable (e.g. proton).Because the ground state is composite we expect “excited states” whichdecay by strong interactions with lifetime of order τ ∼ 10−22 to 10−24

seconds.

Resonances

seconds.

Resonances

seconds.

Resonances

seconds.

Resonances

There is however a problem in detecting these excited states (orresonances).Consider the distance travelled during the lifetime of aresonance:

d = cτγ ∼ cτ E2

mc2∼ 3× 1015 E

where γ is the time dilation factor and we replaced τ ∼ 10−23 s.The best known detector resolution for particle tracks is of order0.5µm.Thus the distance travelled by these resonances is typically tooshort to be observed directly.The best we can do is to detect them withtheir decay products.

Resonances

mc2∼ 3× 1015 E

Resonances

mc2∼ 3× 1015 E

Resonances

mc2∼ 3× 1015 E

Resonances

mc2∼ 3× 1015 E

Example: ∆++in formation

Let us first define what we mean by formation:When a particle X isproduced as the only particle (a+ b→ X) we say that this process isformation of X.When a particle X is produced along with other particles(a+ b→ X + ...) we say this process is production of X.The ∆++ particle was first detected in the formation process:

π+ + p→ ∆++,

which is followed by the (strong) decay

∆++ → π+ + p

to give the observed reaction:

π+ + p→ π+ + p

π+ + p→ ∆++,

∆++ → π+ + p

π+ + p→ π+ + p

π+ + p→ ∆++,

∆++ → π+ + p

π+ + p→ π+ + p

π+ + p→ ∆++,

∆++ → π+ + p

π+ + p→ π+ + p

π+ + p→ ∆++,

∆++ → π+ + p

π+ + p→ π+ + p

π+ + p→ ∆++,

∆++ → π+ + p

π+ + p→ π+ + p

via the process:

π+ π+

In the centre of mass frame the ∆ particle is produced at rest,hence thereaction occurs via resonance formation only when ECM = M∆.In otherframes the invariant mass of the π+p pair is the mass of the ∆ particleW = M∆.

via the process:

π+ π+

via the process:

π+ π+

via the process:

π+ π+

Hence we plot the event rate as a function of the centre of mass energy.

We see a distinct peak at ECM = M∆ = 1232 MeV.The observed peakwidth is Γ = decay rate = 120 MeV. For a particle at rest E = m andΓ = ~/τ = 1/τ in natural units, where τ is the lifetime of the particle.

τ =~γ

=6× 10−25GeV.s

120MeV= 5× 10−24s

which is a typical strong interaction lifetime.

Event rate

mp +mπ

τ =~γ

=6× 10−25GeV.s

120MeV= 5× 10−24s

which is a typical strong interaction lifetime.(http://delenda.wordpress.com) Particle Physics - lecture 7 15 / 25

Event rate

mp +mπ

τ =~γ

=6× 10−25GeV.s

120MeV= 5× 10−24s

Event rate

mp +mπ

τ =~γ

=6× 10−25GeV.s

120MeV= 5× 10−24s

Event rate

mp +mπ

τ =~γ

=6× 10−25GeV.s

120MeV= 5× 10−24s

Event rate

mp +mπ

τ =~γ

=6× 10−25GeV.s

120MeV= 5× 10−24s

∆++ in production

If the interpretation of the peak is correct we should see ∆++ inproduction with the same invariant mass and width (i.e. lifetime), forexample in production reactions like:

p+ p → ∆++ + n

p+ p → ∆++ + π− + p

π+ + p → ∆++ + π0

etc ...always followed by ∆++ → π+ + p.So for the first reaction we have:

p+ p → ∆++ + n|→ π+ + p

So the observed reaction is:

p+ p→ π+ + p+ n

∆++ in production

p+ p → ∆++ + n

p+ p → ∆++ + π− + p

π+ + p → ∆++ + π0

p+ p → ∆++ + n|→ π+ + p

p+ p→ π+ + p+ n

∆++ in production

p+ p → ∆++ + n

p+ p → ∆++ + π− + p

π+ + p → ∆++ + π0

p+ p → ∆++ + n|→ π+ + p

p+ p→ π+ + p+ n

∆++ in production

p+ p → ∆++ + n

p+ p → ∆++ + π− + p

π+ + p → ∆++ + π0

p+ p → ∆++ + n|→ π+ + p

p+ p→ π+ + p+ n

∆++ in production

p+ p → ∆++ + n

p+ p → ∆++ + π− + p

π+ + p → ∆++ + π0

p+ p → ∆++ + n|→ π+ + p

p+ p→ π+ + p+ n

∆++ in production

p+ p → ∆++ + n

p+ p → ∆++ + π− + p

π+ + p → ∆++ + π0

p+ p → ∆++ + n|→ π+ + p

p+ p→ π+ + p+ n

∆++ in production

with the mechanism:

∆ ++

To detect the ∆++ we use the fact that the mass of the unstable particleis equal to the invariant mass (W ) of its decay products.

∆++ in production

with the mechanism:

∆ ++

To detect the ∆++ we use the fact that the mass of the unstable particleis equal to the invariant mass (W ) of its decay products.

∆++ in production

Doing the experiment and plotting the event rate as a function of theinvariant mass of π+p pair.

1230MeV

Event rate

We see a peak with M∆ = 1232 MeV, Γ = 120 MeV independently of thebeam energy and angle of production.The same observation is achieved inother reactions.

∆++ in production

1230MeV

Event rate

∆++ in production

1230MeV

Event rate

∆++ isospin consideration

We have the strong decay:

∆++ → p+ π+

From the decay products (since strong interaction we have S conserved)wededuce that B∆ = 1, S∆ = 0, Q∆ = +2.Hence we deduce that ∆++ = uuu state.Thus computing the thirdcomponent of isospin I3 we have:

2(Nu −Nd) =

Now since ∆++ has I3 = 3/2, we must have I ≥ 3/2, and we must havethe following partners:

∆++ ∆+ ∆0 ∆−