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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
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).
(http://delenda.wordpress.com) Particle Physics - lecture 7 2 / 25
Hadron Spectroscopy
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).
(http://delenda.wordpress.com) Particle Physics - lecture 7 2 / 25
Hadron Spectroscopy
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).
(http://delenda.wordpress.com) Particle Physics - lecture 7 2 / 25
Hadron Spectroscopy
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).
(http://delenda.wordpress.com) Particle Physics - lecture 7 2 / 25
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
K0
(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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 3 / 25
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
K0
(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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 3 / 25
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
K0
(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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 3 / 25
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
K0
(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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 3 / 25
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
K0
(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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 3 / 25
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
K0
(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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 3 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
Consider the quark contents of these families:
p = uud, n = udd
π+ = ud, π0 = uu, dd, π− = du
K0
= 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:
I3 =1
2(Nu −Nd)
or equivalently:
I3 = Q− Y
2
where Y is the hypercharge defined by:(http://delenda.wordpress.com) Particle Physics - lecture 7 4 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
Consider the quark contents of these families:
p = uud, n = udd
π+ = ud, π0 = uu, dd, π− = du
K0
= 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:
I3 =1
2(Nu −Nd)
or equivalently:
I3 = Q− Y
2
where Y is the hypercharge defined by:(http://delenda.wordpress.com) Particle Physics - lecture 7 4 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
Consider the quark contents of these families:
p = uud, n = udd
π+ = ud, π0 = uu, dd, π− = du
K0
= 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:
I3 =1
2(Nu −Nd)
or equivalently:
I3 = Q− Y
2
where Y is the hypercharge defined by:(http://delenda.wordpress.com) Particle Physics - lecture 7 4 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
Consider the quark contents of these families:
p = uud, n = udd
π+ = ud, π0 = uu, dd, π− = du
K0
= 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:
I3 =1
2(Nu −Nd)
or equivalently:
I3 = Q− Y
2
where Y is the hypercharge defined by:(http://delenda.wordpress.com) Particle Physics - lecture 7 4 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
Consider the quark contents of these families:
p = uud, n = udd
π+ = ud, π0 = uu, dd, π− = du
K0
= 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:
I3 =1
2(Nu −Nd)
or equivalently:
I3 = Q− Y
2
where Y is the hypercharge defined by:(http://delenda.wordpress.com) Particle Physics - lecture 7 4 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 5 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 5 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 5 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 5 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 5 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 5 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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
2
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?
(http://delenda.wordpress.com) Particle Physics - lecture 7 6 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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
2
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?
(http://delenda.wordpress.com) Particle Physics - lecture 7 6 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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
2
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?
(http://delenda.wordpress.com) Particle Physics - lecture 7 6 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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
2
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?
(http://delenda.wordpress.com) Particle Physics - lecture 7 6 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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
2
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?
(http://delenda.wordpress.com) Particle Physics - lecture 7 6 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 7 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 7 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 7 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 7 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 7 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 7 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 7 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 7 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 7 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 8 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 8 / 25
Hadron Spectroscopy Isospin symmetry and multiplets
Isospin symmetry and multiplets
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 8 / 25
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√2
(| ↑↓〉+ | ↓↑〉)
|1,−1〉 = | ↓↓〉
We can analogously think of the pion as being composed of two spin-halfquarks (up ↔↑ and down ↔↓).
(http://delenda.wordpress.com) Particle Physics - lecture 7 9 / 25
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√2
(| ↑↓〉+ | ↓↑〉)
|1,−1〉 = | ↓↓〉
We can analogously think of the pion as being composed of two spin-halfquarks (up ↔↑ and down ↔↓).
(http://delenda.wordpress.com) Particle Physics - lecture 7 9 / 25
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√2
(| ↑↓〉+ | ↓↑〉)
|1,−1〉 = | ↓↓〉
We can analogously think of the pion as being composed of two spin-halfquarks (up ↔↑ and down ↔↓).
(http://delenda.wordpress.com) Particle Physics - lecture 7 9 / 25
Hadron Spectroscopy Isospin in SU(2)
Isospin in SU(2)
So we may write the eigenstates of isospin |I, I3〉:
|1, 1〉 = |ud〉 = |π+〉|1, 0〉 =
1√2
(|uu〉+ |dd〉) = |π0〉
|1,−1〉 = |du〉 = |π−〉
So the pions here are eigenstates of isospin operator I2 and I3.
(http://delenda.wordpress.com) Particle Physics - lecture 7 10 / 25
Hadron Spectroscopy Isospin in SU(2)
Isospin in SU(2)
So we may write the eigenstates of isospin |I, I3〉:
|1, 1〉 = |ud〉 = |π+〉|1, 0〉 =
1√2
(|uu〉+ |dd〉) = |π0〉
|1,−1〉 = |du〉 = |π−〉
So the pions here are eigenstates of isospin operator I2 and I3.
(http://delenda.wordpress.com) Particle Physics - lecture 7 10 / 25
Hadron Spectroscopy Isospin in SU(2)
Isospin in SU(2)
So we may write the eigenstates of isospin |I, I3〉:
|1, 1〉 = |ud〉 = |π+〉|1, 0〉 =
1√2
(|uu〉+ |dd〉) = |π0〉
|1,−1〉 = |du〉 = |π−〉
So the pions here are eigenstates of isospin operator I2 and I3.
(http://delenda.wordpress.com) Particle Physics - lecture 7 10 / 25
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 11 / 25
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 11 / 25
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 11 / 25
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 11 / 25
Hadron Spectroscopy Resonances
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
mc2
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 12 / 25
Hadron Spectroscopy Resonances
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
mc2
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 12 / 25
Hadron Spectroscopy Resonances
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
mc2
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 12 / 25
Hadron Spectroscopy Resonances
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
mc2
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 12 / 25
Hadron Spectroscopy Resonances
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
mc2
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 12 / 25
Hadron Spectroscopy Resonances
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
(http://delenda.wordpress.com) Particle Physics - lecture 7 13 / 25
Hadron Spectroscopy Resonances
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
(http://delenda.wordpress.com) Particle Physics - lecture 7 13 / 25
Hadron Spectroscopy Resonances
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
(http://delenda.wordpress.com) Particle Physics - lecture 7 13 / 25
Hadron Spectroscopy Resonances
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
(http://delenda.wordpress.com) Particle Physics - lecture 7 13 / 25
Hadron Spectroscopy Resonances
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
(http://delenda.wordpress.com) Particle Physics - lecture 7 13 / 25
Hadron Spectroscopy Resonances
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
(http://delenda.wordpress.com) Particle Physics - lecture 7 13 / 25
Hadron Spectroscopy Resonances
Example: ∆++in formation
via the process:
∆++
π+ π+
p p
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∆.
(http://delenda.wordpress.com) Particle Physics - lecture 7 14 / 25
Hadron Spectroscopy Resonances
Example: ∆++in formation
via the process:
∆++
π+ π+
p p
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∆.
(http://delenda.wordpress.com) Particle Physics - lecture 7 14 / 25
Hadron Spectroscopy Resonances
Example: ∆++in formation
via the process:
∆++
π+ π+
p p
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∆.
(http://delenda.wordpress.com) Particle Physics - lecture 7 14 / 25
Hadron Spectroscopy Resonances
Example: ∆++in formation
via the process:
∆++
π+ π+
p p
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∆.
(http://delenda.wordpress.com) Particle Physics - lecture 7 14 / 25
Hadron Spectroscopy Resonances
Example: ∆++in formation
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 15 / 25
Hadron Spectroscopy Resonances
Example: ∆++in formation
Hence we plot the event rate as a function of the centre of mass energy.
M∆
Γ∆
Event rate
ECM
mp +mπ
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.(http://delenda.wordpress.com) Particle Physics - lecture 7 15 / 25
Hadron Spectroscopy Resonances
Example: ∆++in formation
Hence we plot the event rate as a function of the centre of mass energy.
M∆
Γ∆
Event rate
ECM
mp +mπ
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.(http://delenda.wordpress.com) Particle Physics - lecture 7 15 / 25
Hadron Spectroscopy Resonances
Example: ∆++in formation
Hence we plot the event rate as a function of the centre of mass energy.
M∆
Γ∆
Event rate
ECM
mp +mπ
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.(http://delenda.wordpress.com) Particle Physics - lecture 7 15 / 25
Hadron Spectroscopy Resonances
Example: ∆++in formation
Hence we plot the event rate as a function of the centre of mass energy.
M∆
Γ∆
Event rate
ECM
mp +mπ
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.(http://delenda.wordpress.com) Particle Physics - lecture 7 15 / 25
Hadron Spectroscopy Resonances
Example: ∆++in formation
Hence we plot the event rate as a function of the centre of mass energy.
M∆
Γ∆
Event rate
ECM
mp +mπ
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.(http://delenda.wordpress.com) Particle Physics - lecture 7 15 / 25
Hadron Spectroscopy Resonances
∆++ 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
(http://delenda.wordpress.com) Particle Physics - lecture 7 16 / 25
Hadron Spectroscopy Resonances
∆++ 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
(http://delenda.wordpress.com) Particle Physics - lecture 7 16 / 25
Hadron Spectroscopy Resonances
∆++ 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
(http://delenda.wordpress.com) Particle Physics - lecture 7 16 / 25
Hadron Spectroscopy Resonances
∆++ 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
(http://delenda.wordpress.com) Particle Physics - lecture 7 16 / 25
Hadron Spectroscopy Resonances
∆++ 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
(http://delenda.wordpress.com) Particle Physics - lecture 7 16 / 25
Hadron Spectroscopy Resonances
∆++ 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
(http://delenda.wordpress.com) Particle Physics - lecture 7 16 / 25
Hadron Spectroscopy Resonances
∆++ in production
with the mechanism:
∆ ++
π+
pn
p
p
To detect the ∆++ we use the fact that the mass of the unstable particleis equal to the invariant mass (W ) of its decay products.
(http://delenda.wordpress.com) Particle Physics - lecture 7 17 / 25
Hadron Spectroscopy Resonances
∆++ in production
with the mechanism:
∆ ++
π+
pn
p
p
To detect the ∆++ we use the fact that the mass of the unstable particleis equal to the invariant mass (W ) of its decay products.
(http://delenda.wordpress.com) Particle Physics - lecture 7 17 / 25
Hadron Spectroscopy Resonances
∆++ in production
Doing the experiment and plotting the event rate as a function of theinvariant mass of π+p pair.
1230MeV
Event rate
Wπ+p
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 18 / 25
Hadron Spectroscopy Resonances
∆++ in production
Doing the experiment and plotting the event rate as a function of theinvariant mass of π+p pair.
1230MeV
Event rate
Wπ+p
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 18 / 25
Hadron Spectroscopy Resonances
∆++ in production
Doing the experiment and plotting the event rate as a function of theinvariant mass of π+p pair.
1230MeV
Event rate
Wπ+p
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.
(http://delenda.wordpress.com) Particle Physics - lecture 7 18 / 25
Hadron Spectroscopy Resonances
∆++ 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:
I3 =1
2(Nu −Nd) =
3
2
Now since ∆++ has I3 = 3/2, we must have I ≥ 3/2, and we must havethe following partners:
∆++ ∆+ ∆0 ∆−
uuu uud udd dddI3 = 3/2 1/2 −1/2 −3/2
(http://delenda.wordpress.com) Particle Physics - lecture 7 19 / 25
Hadron Spectroscopy Resonances
∆++ 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:
I3 =1
2(Nu −Nd) =
3
2
Now since ∆++ has I3 = 3/2, we must have I ≥ 3/2, and we must havethe following partners:
∆++ ∆+ ∆0 ∆−
uuu uud udd dddI3 = 3/2 1/2 −1/2 −3/2
(http://delenda.wordpress.com) Particle Physics - lecture 7 19 / 25
Hadron Spectroscopy Resonances
∆++ 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:
I3 =1
2(Nu −Nd) =
3
2
Now since ∆++ has I3 = 3/2, we must have I ≥ 3/2, and we must havethe following partners:
∆++ ∆+ ∆0 ∆−
uuu uud udd dddI3 = 3/2 1/2 −1/2 −3/2
(http://delenda.wordpress.com) Particle Physics - lecture 7 19 / 25
Hadron Spectroscopy Resonances
∆++ 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:
I3 =1
2(Nu −Nd) =
3
2
Now since ∆++ has I3 = 3/2, we must have I ≥ 3/2, and we must havethe following partners:
∆++ ∆+ ∆0 ∆−
uuu uud udd dddI3 = 3/2 1/2 −1/2 −3/2
(http://delenda.wordpress.com) Particle Physics - lecture 7 19 / 25
Hadron Spectroscopy Resonances
∆++ 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:
I3 =1
2(Nu −Nd) =
3
2
Now since ∆++ has I3 = 3/2, we must have I ≥ 3/2, and we must havethe following partners:
∆++ ∆+ ∆0 ∆−
uuu uud udd dddI3 = 3/2 1/2 −1/2 −3/2
(http://delenda.wordpress.com) Particle Physics - lecture 7 19 / 25
Hadron Spectroscopy Resonances
∆++ 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:
I3 =1
2(Nu −Nd) =
3
2
Now since ∆++ has I3 = 3/2, we must have I ≥ 3/2, and we must havethe following partners:
∆++ ∆+ ∆0 ∆−
uuu uud udd dddI3 = 3/2 1/2 −1/2 −3/2
(http://delenda.wordpress.com) Particle Physics - lecture 7 19 / 25
Hadron Spectroscopy Resonances
∆++ isospin consideration
all with M∆ ≈ 1230 MeV and Γ ≈ 120 MeV since they decay by stronginteractions (which does not care about whether u or d are involved in thedecay).Historically the ∆++ was discovered first. The existence of ∆+, ∆0 and∆− was predicted and subsequently confirmed in formation reactions like:
π− + p→ ∆0 → π− + p
and production reactions like:
p+ π− → π+ + ∆−
|→ n+ π−
where (in the last reaction) a peak is observed in the plot of the decay rateas a function of the invariant mass of the (nπ−) pair around 1230 MeV.
(http://delenda.wordpress.com) Particle Physics - lecture 7 20 / 25
Hadron Spectroscopy Resonances
∆++ isospin consideration
all with M∆ ≈ 1230 MeV and Γ ≈ 120 MeV since they decay by stronginteractions (which does not care about whether u or d are involved in thedecay).Historically the ∆++ was discovered first. The existence of ∆+, ∆0 and∆− was predicted and subsequently confirmed in formation reactions like:
π− + p→ ∆0 → π− + p
and production reactions like:
p+ π− → π+ + ∆−
|→ n+ π−
where (in the last reaction) a peak is observed in the plot of the decay rateas a function of the invariant mass of the (nπ−) pair around 1230 MeV.
(http://delenda.wordpress.com) Particle Physics - lecture 7 20 / 25
Hadron Spectroscopy Resonances
∆++ isospin consideration
all with M∆ ≈ 1230 MeV and Γ ≈ 120 MeV since they decay by stronginteractions (which does not care about whether u or d are involved in thedecay).Historically the ∆++ was discovered first. The existence of ∆+, ∆0 and∆− was predicted and subsequently confirmed in formation reactions like:
π− + p→ ∆0 → π− + p
and production reactions like:
p+ π− → π+ + ∆−
|→ n+ π−
where (in the last reaction) a peak is observed in the plot of the decay rateas a function of the invariant mass of the (nπ−) pair around 1230 MeV.
(http://delenda.wordpress.com) Particle Physics - lecture 7 20 / 25
Hadron Spectroscopy Resonances
∆++ isospin consideration
all with M∆ ≈ 1230 MeV and Γ ≈ 120 MeV since they decay by stronginteractions (which does not care about whether u or d are involved in thedecay).Historically the ∆++ was discovered first. The existence of ∆+, ∆0 and∆− was predicted and subsequently confirmed in formation reactions like:
π− + p→ ∆0 → π− + p
and production reactions like:
p+ π− → π+ + ∆−
|→ n+ π−
where (in the last reaction) a peak is observed in the plot of the decay rateas a function of the invariant mass of the (nπ−) pair around 1230 MeV.
(http://delenda.wordpress.com) Particle Physics - lecture 7 20 / 25
Hadron Spectroscopy Resonances
∆++ isospin consideration
all with M∆ ≈ 1230 MeV and Γ ≈ 120 MeV since they decay by stronginteractions (which does not care about whether u or d are involved in thedecay).Historically the ∆++ was discovered first. The existence of ∆+, ∆0 and∆− was predicted and subsequently confirmed in formation reactions like:
π− + p→ ∆0 → π− + p
and production reactions like:
p+ π− → π+ + ∆−
|→ n+ π−
where (in the last reaction) a peak is observed in the plot of the decay rateas a function of the invariant mass of the (nπ−) pair around 1230 MeV.
(http://delenda.wordpress.com) Particle Physics - lecture 7 20 / 25
Hadron Spectroscopy Resonances
∆++ isospin consideration
all with M∆ ≈ 1230 MeV and Γ ≈ 120 MeV since they decay by stronginteractions (which does not care about whether u or d are involved in thedecay).Historically the ∆++ was discovered first. The existence of ∆+, ∆0 and∆− was predicted and subsequently confirmed in formation reactions like:
π− + p→ ∆0 → π− + p
and production reactions like:
p+ π− → π+ + ∆−
|→ n+ π−
where (in the last reaction) a peak is observed in the plot of the decay rateas a function of the invariant mass of the (nπ−) pair around 1230 MeV.
(http://delenda.wordpress.com) Particle Physics - lecture 7 20 / 25
Hadron Spectroscopy Resonances
Quark diagrams
Consider the decay:∆++ → π+ + p
which effectively is:uuu→ uud+ ud
i.e. a dd pair is created.One can represent this process by the diagram:
(http://delenda.wordpress.com) Particle Physics - lecture 7 21 / 25
Hadron Spectroscopy Resonances
Quark diagrams
Consider the decay:∆++ → π+ + p
which effectively is:uuu→ uud+ ud
i.e. a dd pair is created.One can represent this process by the diagram:
(http://delenda.wordpress.com) Particle Physics - lecture 7 21 / 25
Hadron Spectroscopy Resonances
Quark diagrams
Consider the decay:∆++ → π+ + p
which effectively is:uuu→ uud+ ud
i.e. a dd pair is created.One can represent this process by the diagram:
(http://delenda.wordpress.com) Particle Physics - lecture 7 21 / 25
Hadron Spectroscopy Resonances
Quark diagrams
Consider the decay:∆++ → π+ + p
which effectively is:uuu→ uud+ ud
i.e. a dd pair is created.One can represent this process by the diagram:
uuu
duu
du
∆++{
} π+
}p
(http://delenda.wordpress.com) Particle Physics - lecture 7 21 / 25
Hadron Spectroscopy Resonances
Quark diagrams
Similarly for the reaction:
π+ + p→ ∆++ → π+ + p
We have the following diagram:
duu
du
∆++ = uuu
} π+
}p
d u u
du
π + {
p{
Formation of ∆++
via dd annihilation
Decay of ∆++
via dd creation
(http://delenda.wordpress.com) Particle Physics - lecture 7 22 / 25
Hadron Spectroscopy Resonances
Quark diagrams
Similarly for the reaction:
π+ + p→ ∆++ → π+ + p
We have the following diagram:
duu
du
∆++ = uuu
} π+
}p
d u u
du
π + {
p{
Formation of ∆++
via dd annihilation
Decay of ∆++
via dd creation
(http://delenda.wordpress.com) Particle Physics - lecture 7 22 / 25
Hadron Spectroscopy Other Examples:
Other Examples:
Consider the reaction:
π− + p → n+X0
|→ π+ + π−
giving the overall observed reaction:
π− + p→ π+ + π− + n
with the following mechanism:
} π−
}π +
d u u
ud
π − {
p{
d u
} nd
X0 = uu
u
ud
d
(http://delenda.wordpress.com) Particle Physics - lecture 7 23 / 25
Hadron Spectroscopy Other Examples:
Other Examples:
Consider the reaction:
π− + p → n+X0
|→ π+ + π−
giving the overall observed reaction:
π− + p→ π+ + π− + n
with the following mechanism:
} π−
}π +
d u u
ud
π − {
p{
d u
} nd
X0 = uu
u
ud
d
(http://delenda.wordpress.com) Particle Physics - lecture 7 23 / 25
Hadron Spectroscopy Other Examples:
Other Examples:
Consider the reaction:
π− + p → n+X0
|→ π+ + π−
giving the overall observed reaction:
π− + p→ π+ + π− + n
with the following mechanism:
} π−
}π +
d u u
ud
π − {
p{
d u
} nd
X0 = uu
u
ud
d
(http://delenda.wordpress.com) Particle Physics - lecture 7 23 / 25
Hadron Spectroscopy Other Examples:
Other Examples:
Consider the reaction:
π− + p → n+X0
|→ π+ + π−
giving the overall observed reaction:
π− + p→ π+ + π− + n
with the following mechanism:
} π−
}π +
d u u
ud
π − {
p{
d u
} nd
X0 = uu
u
ud
d
(http://delenda.wordpress.com) Particle Physics - lecture 7 23 / 25
Hadron Spectroscopy Other Examples:
Other Examples:
The reaction can also go by the following mechanism:
} π+
}π −d
uu
ud
π − {
p{
du
} nd
X0 = dd
d
du
u
so X0 can also be formed by dd just like π0.Peaks are observed atWπ+π− = mX if X0 exists.
(http://delenda.wordpress.com) Particle Physics - lecture 7 24 / 25
Hadron Spectroscopy Other Examples:
Other Examples:
The reaction can also go by the following mechanism:
} π+
}π −d
uu
ud
π − {
p{
du
} nd
X0 = dd
d
du
u
so X0 can also be formed by dd just like π0.Peaks are observed atWπ+π− = mX if X0 exists.
(http://delenda.wordpress.com) Particle Physics - lecture 7 24 / 25
Hadron Spectroscopy Other Examples:
Other Examples:
The reaction can also go by the following mechanism:
} π+
}π −d
uu
ud
π − {
p{
du
} nd
X0 = dd
d
du
u
so X0 can also be formed by dd just like π0.Peaks are observed atWπ+π− = mX if X0 exists.
(http://delenda.wordpress.com) Particle Physics - lecture 7 24 / 25
Hadron Spectroscopy Other Examples:
Other Examples:
Data: Three peaks are observed:
Event rate
Wπ+p
ρ0(770) Γ = 150MeV
f0(1275) Γ = 180MeV
ρ(1700) Γ = 200MeV
(http://delenda.wordpress.com) Particle Physics - lecture 7 25 / 25
Hadron Spectroscopy Other Examples:
Other Examples:
Data: Three peaks are observed:
Event rate
Wπ+p
ρ0(770) Γ = 150MeV
f0(1275) Γ = 180MeV
ρ(1700) Γ = 200MeV
(http://delenda.wordpress.com) Particle Physics - lecture 7 25 / 25
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