7 Quark bound states.
7.1 Particles and resonances
This week we will talk about particles as bound states of quarks. Most particles have a short
lifetime and decay into lower mass states. A few particles and bound states are stable, as
there are no states with lower mass they can decay to. One such example is the proton, which
is a bound state of three quarks. There are no allowed decays for the proton, so the proton is
stable, and we know its lifetime to be longer than 1034 years.
For an unstable particle, the mass cannot be exactly determined because of the Heisenberg
uncertainty principle. In the particle rest frame the total energy of the particle is equal to its
mass (multiplied by c2) and the uncertainty principle tells us that
Δ𝐸𝐸 Δ𝑡𝑡 ≥ ℏ
Which can be used to calculate the uncertainty on the particle energy (and hence its mass).
The uncertainty Δ𝐸𝐸 is called the “width” of a particle and is indicated by the letter Γ. We saw
how the measurement of the width of the 𝑍𝑍 0 boson can be used to constrain the number of
neutrino families.
In a collision experiment the cross section of a process increases when the process goes
through an unstable particle as intermediate state. This sharp increase is called a “resonance”
and is how new particles were observed a few decades ago.
7.2 Isospin
It was noticed that many groupings of particles of similar mass and properties fitted in to
common patterns. One way to characterise these is using isotopic spin or isospin, I. This
quantity has nothing to do with the real spin of the particle, but obeys the same addition laws
as the quantum mechanical rules for adding angular momentum or spin. When the orientation
of an isospin vector is considered, it is in some hypothetical space, not in terms of the x, y and
z axes of normal co-ordinates.
Nucleons (p, n), pi mesons (𝜋𝜋 + , 𝜋𝜋 0 , 𝜋𝜋 − ) and the baryons known as Δ (Δ ++ , Δ+ , Δ0 , Δ− ) are
three examples of groups of similar mass particles differing in charge by one unit. The charge
Q in each case can be considered as due to the orientation of an “isospin vector” in some
hypothetical space, such that Q depends on the third component I3. Thus the nucleons belong
to an isospin doublet with 𝐼𝐼 = 1⁄2
1 1 1 1
𝑝𝑝 = |𝐼𝐼, 𝐼𝐼3 ⟩ = � , � ; 𝑛𝑛 = |𝐼𝐼, 𝐼𝐼3 ⟩ = � , − �
2 2 2 2
Similarly the pions form an isospin triplet with 𝐼𝐼 = 1
𝜋𝜋 + = |𝐼𝐼, 𝐼𝐼3 ⟩ = |1,1⟩; 𝜋𝜋 0 = |𝐼𝐼, 𝐼𝐼3 ⟩ = |1,0⟩; 𝜋𝜋 − = |𝐼𝐼, 𝐼𝐼3 ⟩ = |1, −1⟩
The Δ’s forms a quadruplet with 𝐼𝐼 = 3⁄2.
3 3 3 1 3 1 3 3
Δ++ = |𝐼𝐼, 𝐼𝐼3 ⟩ = � , � ; Δ+ = |𝐼𝐼, 𝐼𝐼3 ⟩ = � , � ; Δ0 = |𝐼𝐼, 𝐼𝐼3 ⟩ = � , − � ; Δ− = |𝐼𝐼, 𝐼𝐼3 ⟩ = � , − �
2 2 2 2 2 2 2 2
57
, 1
The rule for electric charge can then be written 𝑄𝑄 = 𝑒𝑒 � 𝐵𝐵 + 𝐼𝐼3 �, where B is the baryon
2
number which is 1 for nucleons and the Δ’s and 0 for mesons such as the 𝜋𝜋’s. In terms of
quarks, the u and d form an isospin doublet with B=1/3
1 1 1 1
𝑢𝑢 = � , � ; 𝑑𝑑 = � , − �
2 2 2 2
Three quarks with I = ½ can combine to form Itot = ½ or 3/2. Itot = ½ gives the nucleons while
Itot = 3/2 forms the Δ’s.
It is useful to consider the symmetry of the quarks inside these baryons. The internal
wavefunction can be written as a product of terms,
Ψ = 𝜓𝜓𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜓𝜓𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜓𝜓𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝜓𝜓𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
And must be antisymmetric overall under interchange of two quarks (as quarks are fermions).
𝜓𝜓𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 is always antisymmetric (as hadrons are colourless); the symmetry of 𝜓𝜓𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 is given
by (–1)l and l (the orbital angular momentum) is zero for all the long-lived hadrons we consider,
so 𝜓𝜓𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 is symmetric. Thus the product
𝜓𝜓𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜓𝜓𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖
must be symmetric.
This explains the correlation between allowed spin and isospin states for the baryons: the Δ’s
have I = 3/2 and s = 3/2 (both symmetric), while the nucleons have I = 1/2 and s = 1/2 (both
antisymmetric).
In strong interactions, the total isospin vector (as well as I3) is conserved. This is not true in
electromagnetic or weak interactions. The conservation of isospin has observable effects on
the relative rates of strong interactions.
7.3 Baryons and Quark Symmetry
Deuteron
We will look at the combination of two objects with spin ½ and isospin ½. Since two quarks do
not form a bound state, it is helpful to consider combinations of two nucleons as these have
similar quantum numbers. In other words the strong interaction sees the proton and the
neutron as two different isospin states of the same particle.
In this case, since there is no colour part to the total wavefunction, the product of spin and
isospin states must be antisymmetric. We will denote sz = ½ by ↑ and sz = –½ by ↓, and I3 = ½
by p (proton) and I3 = –½ by n (neutron).
First consider just the isospin. There are 4 combinations of two nucleons: pp, pn, np, nn. The
first and last are obviously symmetric under interchange; the other 2 do not have a defined
symmetry. We can produce the following combinations:
1 1
𝑝𝑝𝑝𝑝, (𝑝𝑝𝑝𝑝 + 𝑛𝑛𝑛𝑛), 𝑛𝑛𝑛𝑛, (𝑝𝑝𝑝𝑝 − 𝑛𝑛𝑛𝑛).
√2 √2
The first 3 are symmetric, corresponding respectively to (I, I3) = (1, 1), (1, 0), (1, –1). The last is
antisymmetric, (I, I3) = (0, 0). Similarly the spin states are
58
7.1 Particles and resonances
This week we will talk about particles as bound states of quarks. Most particles have a short
lifetime and decay into lower mass states. A few particles and bound states are stable, as
there are no states with lower mass they can decay to. One such example is the proton, which
is a bound state of three quarks. There are no allowed decays for the proton, so the proton is
stable, and we know its lifetime to be longer than 1034 years.
For an unstable particle, the mass cannot be exactly determined because of the Heisenberg
uncertainty principle. In the particle rest frame the total energy of the particle is equal to its
mass (multiplied by c2) and the uncertainty principle tells us that
Δ𝐸𝐸 Δ𝑡𝑡 ≥ ℏ
Which can be used to calculate the uncertainty on the particle energy (and hence its mass).
The uncertainty Δ𝐸𝐸 is called the “width” of a particle and is indicated by the letter Γ. We saw
how the measurement of the width of the 𝑍𝑍 0 boson can be used to constrain the number of
neutrino families.
In a collision experiment the cross section of a process increases when the process goes
through an unstable particle as intermediate state. This sharp increase is called a “resonance”
and is how new particles were observed a few decades ago.
7.2 Isospin
It was noticed that many groupings of particles of similar mass and properties fitted in to
common patterns. One way to characterise these is using isotopic spin or isospin, I. This
quantity has nothing to do with the real spin of the particle, but obeys the same addition laws
as the quantum mechanical rules for adding angular momentum or spin. When the orientation
of an isospin vector is considered, it is in some hypothetical space, not in terms of the x, y and
z axes of normal co-ordinates.
Nucleons (p, n), pi mesons (𝜋𝜋 + , 𝜋𝜋 0 , 𝜋𝜋 − ) and the baryons known as Δ (Δ ++ , Δ+ , Δ0 , Δ− ) are
three examples of groups of similar mass particles differing in charge by one unit. The charge
Q in each case can be considered as due to the orientation of an “isospin vector” in some
hypothetical space, such that Q depends on the third component I3. Thus the nucleons belong
to an isospin doublet with 𝐼𝐼 = 1⁄2
1 1 1 1
𝑝𝑝 = |𝐼𝐼, 𝐼𝐼3 ⟩ = � , � ; 𝑛𝑛 = |𝐼𝐼, 𝐼𝐼3 ⟩ = � , − �
2 2 2 2
Similarly the pions form an isospin triplet with 𝐼𝐼 = 1
𝜋𝜋 + = |𝐼𝐼, 𝐼𝐼3 ⟩ = |1,1⟩; 𝜋𝜋 0 = |𝐼𝐼, 𝐼𝐼3 ⟩ = |1,0⟩; 𝜋𝜋 − = |𝐼𝐼, 𝐼𝐼3 ⟩ = |1, −1⟩
The Δ’s forms a quadruplet with 𝐼𝐼 = 3⁄2.
3 3 3 1 3 1 3 3
Δ++ = |𝐼𝐼, 𝐼𝐼3 ⟩ = � , � ; Δ+ = |𝐼𝐼, 𝐼𝐼3 ⟩ = � , � ; Δ0 = |𝐼𝐼, 𝐼𝐼3 ⟩ = � , − � ; Δ− = |𝐼𝐼, 𝐼𝐼3 ⟩ = � , − �
2 2 2 2 2 2 2 2
57
, 1
The rule for electric charge can then be written 𝑄𝑄 = 𝑒𝑒 � 𝐵𝐵 + 𝐼𝐼3 �, where B is the baryon
2
number which is 1 for nucleons and the Δ’s and 0 for mesons such as the 𝜋𝜋’s. In terms of
quarks, the u and d form an isospin doublet with B=1/3
1 1 1 1
𝑢𝑢 = � , � ; 𝑑𝑑 = � , − �
2 2 2 2
Three quarks with I = ½ can combine to form Itot = ½ or 3/2. Itot = ½ gives the nucleons while
Itot = 3/2 forms the Δ’s.
It is useful to consider the symmetry of the quarks inside these baryons. The internal
wavefunction can be written as a product of terms,
Ψ = 𝜓𝜓𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜓𝜓𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜓𝜓𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝜓𝜓𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
And must be antisymmetric overall under interchange of two quarks (as quarks are fermions).
𝜓𝜓𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 is always antisymmetric (as hadrons are colourless); the symmetry of 𝜓𝜓𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 is given
by (–1)l and l (the orbital angular momentum) is zero for all the long-lived hadrons we consider,
so 𝜓𝜓𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 is symmetric. Thus the product
𝜓𝜓𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜓𝜓𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖
must be symmetric.
This explains the correlation between allowed spin and isospin states for the baryons: the Δ’s
have I = 3/2 and s = 3/2 (both symmetric), while the nucleons have I = 1/2 and s = 1/2 (both
antisymmetric).
In strong interactions, the total isospin vector (as well as I3) is conserved. This is not true in
electromagnetic or weak interactions. The conservation of isospin has observable effects on
the relative rates of strong interactions.
7.3 Baryons and Quark Symmetry
Deuteron
We will look at the combination of two objects with spin ½ and isospin ½. Since two quarks do
not form a bound state, it is helpful to consider combinations of two nucleons as these have
similar quantum numbers. In other words the strong interaction sees the proton and the
neutron as two different isospin states of the same particle.
In this case, since there is no colour part to the total wavefunction, the product of spin and
isospin states must be antisymmetric. We will denote sz = ½ by ↑ and sz = –½ by ↓, and I3 = ½
by p (proton) and I3 = –½ by n (neutron).
First consider just the isospin. There are 4 combinations of two nucleons: pp, pn, np, nn. The
first and last are obviously symmetric under interchange; the other 2 do not have a defined
symmetry. We can produce the following combinations:
1 1
𝑝𝑝𝑝𝑝, (𝑝𝑝𝑝𝑝 + 𝑛𝑛𝑛𝑛), 𝑛𝑛𝑛𝑛, (𝑝𝑝𝑝𝑝 − 𝑛𝑛𝑛𝑛).
√2 √2
The first 3 are symmetric, corresponding respectively to (I, I3) = (1, 1), (1, 0), (1, –1). The last is
antisymmetric, (I, I3) = (0, 0). Similarly the spin states are
58
