Quark bound states.

7. Quark bound states.#

Learning objectives:

know the quantum numbers of particles in the lowest lying multiplets;

7.1. Particles and resonances#

In this unit 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 $10^{34}$ 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 $c^{2}$ ) and the uncertainty principle tells us that

$Δ E Δ t \geq ℏ$

Which can be used to calculate the uncertainty on the particle energy (and hence its mass). The uncertainty $Δ E$ is called the “width” of a particle and is indicated by the letter $Γ$ . We saw how the measurement of the width of the $Z^{0}$ 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 with axes labelled as (1,2,3) rather than ( $x, y, z)$ . The strong interaction will treat particles based on their isospin, and will not distinguish states with a different charge. For instance the interaction of protons and neutrons inside a nucleus interact is mediated by the strong force which is independent on their charges.

Isospin can have values that are either integers or half-integer and follows the same algebra as the spin of a particle. Due to the Heisenberg principle, we can only define the magnitude of the isospin and one of its components. By convention we choose the component 3 as the independent component that can be defined and we call that $I_{3}$ .

Nucleons ( $p, n)$ , pi mesons $(π^{+}, π^{0}, π^{-})$ and the baryons known as $Δ (Δ^{+ +}, Δ^{+}, Δ^{0}, Δ^{-})$ are three examples of groups of similar mass particles differing in charge by one unit. These group of particles are isospin multiplets with the same isospin magnitude and different values of the component $I_{3}$ . The charge $Q$ in each case is due to the orientation of an “isospin vector” in our hypothetical space, therefore $Q$ depends on the third component $I_{3}$ . Thus the nucleons belong to an isospin doublet with $I = 1 / 2$

p = | I, I_{3} ⟩ = | \frac{1}{2}, \frac{1}{2} ⟩; n = | I, I_{3} ⟩ = | \frac{1}{2}, - \frac{1}{2} ⟩

Similarly the pions form an isospin triplet with $I = 1$

π^{+} = | I, I_{3} ⟩ = | 1, 1 ⟩; π^{0} = | I, I_{3} ⟩ = | 1, 0 ⟩; π^{-} = | I, I_{3} ⟩ = | 1, - 1 ⟩

The $Δ$ ‘s forms a quadruplet with $I = 3 / 2$ .

$Δ^{+ +} = | I, I_{3} ⟩ = | \frac{3}{2}, \frac{3}{2} ⟩; Δ^{+} = | I, I_{3} ⟩ = | \frac{3}{2}, \frac{1}{2} ⟩; Δ^{0} = | I, I_{3} ⟩ = | \frac{3}{2}, - \frac{1}{2} ⟩; Δ^{-} = | I, I_{3} ⟩ = | \frac{3}{2}, - \frac{3}{2} ⟩$

The rule for electric charge can then be written $Q = e (\frac{1}{2} B + I_{3})$ , where $B$ is the baryon number which is 1 for nucleons and the $Δ^{'}$ s and 0 for mesons such as the $π^{'}$ s.

In terms of quarks, the $u$ and $d$ form an isospin doublet with $B = 1 / 3$

u = | \frac{1}{2}, \frac{1}{2} ⟩; d = | \frac{1}{2}, - \frac{1}{2} ⟩

Following the rules of angular momentum composition, three quarks with $I = 1 / 2$ can combine to form $I_{tot} = 1 / 2$ or $3 / 2$ . $I_{tot} = 1 / 2$ gives the nucleons while $I_{tot} = 3 / 2$ 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 = ψ_{spin} ψ_{space} ψ_{isospin} ψ_{colour}

And must be antisymmetric overall under interchange of two quarks (as quarks are fermions). $ψ_{colour}$ is always antisymmetric (as hadrons are colourless); the symmetry of $ψ_{space}$ is given by $(- 1)^{'}$ and $/$ (the orbital angular momentum) is zero for all the long-lived hadrons we consider, so $ψ_{space}$ is symmetric. Thus the product

ψ_{spin} ψ_{isospin}

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 $I_{3}$ ) 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#

7.3.1. Deuteron#

After defining the isospin and as an example, we look at the combination of two objects with spin $1 / 2$ and isospin $1 / 2$ . 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 $s_{2} = 1 / 2$ by $↑$ and $s_{2} = - 1 / 2$ by $↓$ , and $l_{3} = 1 / 2$ by p (proton) and $l_{3} = - 1 / 2$ 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:

p p, \frac{1}{\sqrt{2}} (p n + n p), n n, \frac{1}{\sqrt{2}} (p n - n p)

The first 3 are symmetric, corresponding respectively to $(I, I_{3}) = (1, 1), (1, 0), (1, - 1)$ . The last is antisymmetric, $(I, I_{3}) = (0, 0)$ . Similarly the spin states are

↑↑, \frac{1}{\sqrt{2}} (↑↓ + ↓↑), ↓↓, \frac{1}{\sqrt{2}} (↑↓ - ↓↑)

corresponding to $(s, s_{2}) = (1, 1), (1, 0), (1, - 1), (0, 0)$ respectively. (The $1 / \sqrt{2}$ factors are simply the normalisations.)

The allowed states are those where spin and isospin have opposite symmetry; the antisymmetric isospin state and symmetric spin states $(s = 1)$ give the bound deuteron, while the complementary states are only allowed for free pairs of protons and neutrons. (This is discussed further in the Nuclear Physics course next semester.)

Example 7.1 Nucleon interaction cross sections

Consider the two reactions: i) $p p \to d π^{+}$

ii) $p n \to d π^{0}$

Use isospin symmetry to show that the cross sections ratio is $σ_{i} / σ_{i i} = 2$

7.3.2. $Δ$ spin and isospin configurations#

We now return to the quarks, denoting $I_{3} = 1 / 2$ by $u$ and $I_{3} = - 1 / 2$ by $d$ . Once again, we assume that the strong interactions sees the $u$ - and $d$ -quarks as isospin states of the same particle. Since the colour part of the wavefunction is always antisymmetric, the combined

ψ_{spin} ψ_{isospin}

must be symmetric under the exchange of two quarks. First we will consider symmetric isospin combinations of the three quarks.

uuu is $(I, l_{3}) = (3 / 2, 3 / 2)$ so must be the $Δ^{+ +}$ .
uud is $I_{3} = 1 / 2$ ; forming combinations which are symmetric under 2-quark

permutations we get $1 / \sqrt{3} (u u d + u d u + d u u)$ . This must be $(I, I_{3}) = (3 / 2, 1 / 2)$ so the $Δ^{+}$ .

Similarly $1 / \sqrt{3} (u u d + u d u + d u u)$ is $(I, I_{3}) = (3 / 2, - 1 / 2)$ so the $Δ^{0}$ .
ddd is $(I, I_{3}) = (3 / 2, - 3 / 2)$ so the $Δ^{-}$ .

Note: The first and last state must be $I = 3 / 2$ . There must therefore be symmetric states with $I = 3 / 2$ and intermediate $I_{3}$ values, and the above combinations are the only possibilities! The spin states for the $Δ$ must also be symmetric, and we can write these down in an analogous way:

$| S, S_{z} ⟩ = (3 / 2, 3 / 2)$ is $↑↑↑$ .
$| S, S_{z} ⟩ = (3 / 2, 1 / 2)$ is $1 / \sqrt{3} (↑↑↓ + ↑↓↑ + ↓↑↑)$ .
$| S, S_{z} ⟩ = (3 / 2, - 1 / 2)$ is $1 / \sqrt{3} (↑↓↓ + ↓↑↓ + ↓↓↑)$
$| S, S_{z} ⟩ = (3 / 2, - 3 / 2)$ is $↓↓↓$ .

It was implied earlier that the nucleon states were antisymmetric in isospin and antisymmetric in spin. In fact, it is impossible to write down 3-quark states that are antisymmetric under the interchange of all pairs! (Try it!) However, it is possible to use “partially antisymmetric” isospin and spin states as a basis to find the required states which are symmetric overall.

First consider one pair of quarks. Ignoring the normalisation factors, $(u ↑ d ↓ - d ↑ u ↓)$ is antisymmetric under exchange of isospin labels; $(u ↑ d ↓ - u ↓ d ↑$ ) is antisymmetric under exchange of spin labels; so ( $u ↑ d ↓ - u ↓ d ↑ - d ↑ u ↓ + d ↓ u ↑$ ) is antisymmetric under exchange of isospin or spin labels but symmetric overall (i.e. when both the spin and isospin properties of a pair of particles are swapped). This combination has $I = 0$ and $s = 0$ .

We can now combine this state with a third quark ( $u ↑$ ) and symmetrise to find the combined state representing a proton $(/_{3} = 1 / 2)$ with spin $s_{z} = 1 / 2$ . Simply adding a $u ↑$ on the left gives:

u ↑ u ↑ d ↓ - u ↑ u ↓ d ↑ - u ↑ d ↑ u ↓ + u ↑ d ↓ u ↑

This must be made symmetric under interchange of pairs of quarks. Permuting first and second adds on

u ↑ u ↑ d ↓ - u ↓ u ↑ d ↑ - d ↑ u ↑ u ↓ + d ↓ u ↑ u ↑

while permuting first and third adds

d ↓ u ↑ u ↑ - d ↑ u ↓ u ↑ - u ↓ d ↑ u ↑ + u ↑ d ↓ u ↑

(Note that interchange of $2^{nd}$ and $3^{rd}$ quarks was already considered above).

Gathering terms together, and including the normalisation factor (equal as usual to the reciprocal of the square-root of the sum of the squares of the coefficients) gives

\begin{array}{r} \begin{matrix} p ↑= \frac{1}{\sqrt{18}} (2 u ↑ u ↑ d ↓ - u ↑ u ↓ d ↑ - u ↓ u ↑ d ↑ + \\ + 2 u ↑ d ↓ u ↑ - u ↑ d ↑ u ↓ - u ↓ d ↑ u ↑ + \\ + 2 d ↓ u ↑ u ↑ - d ↑ u ↓ u ↑ - d ↑ u ↑ u ↓) \end{matrix} \end{array}

The spin-down state of the proton and the spin states of the neutron can be found in the same way.

7.4. Strangeness#

It was observed that some unstable particles produced in strong interactions had a long lifetime. This unusual stability for strongly interacting particles led to the term of strangeness. Such particles are always produced in pairs (associated production), and the quantum number of strangeness, $S$ , was introduced, which is conserved in strong interactions. Thus in the interaction $π^{-} p \to Λ^{0} K^{0}$ , the $Λ^{0}$ is assigned $S = - 1$ and the $K^{0}$ has $S = + 1$ . These strange particles can only decay by the weak interaction, which does not conserve strangeness. Process when a change of one unit of strangeness are allowed if mediated by the charge weak interaction.

The total or mean isospin $I_{3}$ for the particles in a multiplet must be zero, so we need to modify the formula for the electric charge to include the strangeness quantum number

Q = e (I_{3} + \frac{1}{2} B + \frac{1}{2} S) = e (I_{3} + \frac{1}{2} Y)

where $Y = B + S$ is known as the hypercharge. (This formula is known as the Gell-Mann Nishijima relation).

Families of particles with similar properties (e.g. same spin and parity) can be plotted in terms of $Y$ versus $I_{3}$ , and form regular geometrical patterns (see figures 7.1 to 7.4 and Table 7.1):

mesons with spin-parity $(J^{P}) = 0^{-}$ form an octet;
mesons with $J^{p} = 1^{-}$ form a nonet;
baryons with $J^{p} = 1 / 2^{+}$ form an octet;
baryons with $J^{P} = 3 / 2^{+} +$ form a decuplet.

The difference in shape (i.e. the allowed combinations of quarks) between the baryon octet and decuplet is entirely a consequence of symmetry constraints.

Isospin multiplet			B	$S$	I	$< Q > / e$	$Y = B + S$
$Δ^{+ +}$	$π^{+}$	$π^{0}$	0	0	1	0	0
	p	n	1	0	$1 / 2$	$1 / 2$	1
	$Δ^{+}$	$Δ^{0}$	1	0	$3 / 2$	$1 / 2$	1
		$Λ$	1	-1	0	0	0
	$Σ^{+}$	$Σ^{0} Σ$	1	-1	1	0	0
	$K^{+}$	$K^{0}$	0	1	$1 / 2$	1/2	1
		$\overset{―}{K^{0}}$	0	-1	$1 / 2$	$- 1 / 2$	-1
		$Ξ^{0}$	1	-2	$1 / 2$	$- 1 / 2$	-1
			1	-3	0	-1	-2

Table 7.1 A selection of strange and non-strange baryon and meson multiplets.

In terms of quarks we can introduce a new flavour of quark, the strange quark s. This has charge $- 1 / 3$ and baryon number $1 / 3$ (like a d quark) but $I = 0$ and $S = - 1$ . It is also somewhat heavier than the $u$ and $d$ quarks. Since baryons consist of (qqq), it is clear why no positive baryons exist with $| S | > 1$ , while negative baryons are found with $S = - 2$ or -3 .

Fig. 7.1 The lowest mass pseudoscalar-meson states $(ρ^{p} = 0^{-})$ , with quark assignments indicated. (The states at the origin are displaced slightly for clarity.)

Fig. 7.2 The vector-meson nonet $(J^{P} = 1^{-})$ . (Quark assignments are the same as above.)

Fig. 7.3 The baryon octet of spin-parity $J^{P} = {\frac{1}{2}}^{+}$ .

Fig. 7.4 The baryon decuplet with spin-parity $J^{P} = {\frac{3}{2}}^{+}$

Example 7.2: Strange hadrons spin

Consider the case of hadrons with two light ( $u$ or d) quarks and one $s$ quark, or two $s$ quarks and one light quark. Can you explain why the $Σ$ and $Ξ$ exist in both the spin $1 / 2$ and spin $3 / 2$ multiplets, while the $Λ$ only exists with spin $1 / 2$ ?

Solution

The $Σ$ states

The $Σ$ is a three-quark bound states, where one of the quarks is a strange and the other two are up and/or down quarks. All the three quarks have spin $1 / 2$ , while only the up and down have isospin and should be considered identical, the strange quark has isospin zero and different quantum numbers from the up and down, so it is “different” from the others.

We approach this in two steps. First we combine the two identical light-quarks and identify the possible quantum states of this combination, second we combine these states with the additional strange quark.

Step 1

The $ψ_{spin} \times ψ_{isospin}$ wavefunction has to be symmetric for exchange of identical particles, e.g. it has to be symmetrical if we exchange the two light quarks. There are two ways to achieve this:

The $ψ_{spin}$ and $ψ_{isospin}$ are both symmetric for particle exchange. This is the case for states with spin=1 and isospin=1, and there are three separate states with three different $I_{3}$ values.
The $ψ_{spin}$ and $ψ_{isospin}$ are both antisymmetric for particle exchange. This is the case for states with spin $= 0$ and isospin=0, and there is only one such state.

Step 2

The two states defined above are then combined with a strange quark. This can lead to

For the case where $ψ_{spin}$ and $ψ_{isospin}$ are both symmetric for particle exchange. Since the strange quark has no isospin, the isospin value is given by the combination of lightquarks defined abobe and we have a three-quark state with isospin $I = 1$ , and three states

u u s, \frac{1}{\sqrt{2}} (u d s + d u s), d d s

These are the $Σ^{+}, Σ^{0}, Σ^{-}$ , respectively. Note that the $Σ^{0}$ is a linear combination of two states, which is symmetric under exchange of the first two particles.

When treating the spin, we need to combine states with spin 1 , with the strange quark with spin $1 / 2$ . The rule for combination of angular momenta suggest that such combination leads to states with either spin $1 / 2$ or spin $3 / 2$

The $Ξ$ states

In the case of the $Ξ$ , we use a similar argument. However in this case the particle has strangeness -2 , which means that the particles have two identical strange quarks and one lightquark ( $u$ or d). Therefore there are two states that can exist

ssu,ssd

Which correspond to the $Ξ^{0}$ and $Ξ^{-}$ respectively. Since the strange quarks have isospin=0, the isospin of the state is given by the isospin of the light-quarks and is $1 / 2$ . Therefore this is a doublet of particles, and we don’t have to worry about the $u$ and d-quarks being identical under particle exchange, as there is no exchange.

However we should make sure that the wavefunction is symmetric under exchange of the two identical s-quarks. Since the only relevant quantum number is the spin, the only way the two $s$ -quarks can combine is into a state with spin=1, which is symmetric under particle exchange.

Finally, we need to combine the (ss) state with a light quark to obtain a three-quark bound state. We follow the usual angular momentum combination rules, and realise that this could lead to states with spin=1/2 and spin=3/2

Bonus (The $Λ^{0}$ state)

When discussing the $Σ$ states in the first part of the example, we neglected the case when $ψ_{spin}$ and $ψ_{isospin}$ are both antisymmetric for particle exchange. In this case, when we combine this with a the s-quark, we have a state with isospin $= 0$ , and total $spin = 1 / 2$ . This state is composed by

\frac{1}{\sqrt{2}} (u d s - u s d)

e.g. totally antisymmetric if we exchange the $u$ and $d$ quarks. Has a total charge of 0 , and spin $1 / 2$ . This is the $Λ^{0}$ in the diagram above.

7.5. Further quarks#

Other, still heavier quarks also exist. The charm quark, $c$ , has a charge of $2 / 3$ , like the $u$ , and can be considered as a partner to the s. In 3 dimensions (see figure 10.5) particles containing c quarks can be plotted, and again show regular patterns.

o-Mesons

(a)

(c)

(b)

${\frac{3}{2}}^{+}$ Baryons

(d)

Fig. 7.5 Multiplets of hadrons containing up, down, strange and charm quarks. The slices through these figures where charm $= 0$ correspond to the plane figures already shown in the previous diagrams, though containing new particles in the case of the mesons composed of $c \bar{c}$ .

We thus have 2 doublets or generations of quarks $- (d, u)$ and $(s, c)$ . Since there are 3 doublets of leptons, there are theoretical reasons for expecting a third doublet of quarks too. Particles containing $b$ quarks (bottom or beauty) were discovered in 1977. The $b$ is an even heavier version of the d. Its partner, the $\underset{―}{t}$ (top or truth) was first seen in 1994, and it has the greatest mass of any known fundamental particle at $174 GeV / c^{2}$ .

Flavour	Charge/e	$B$	$I$	$l_{3}$	$S$	$c$	$b$	$t$	Mass $(GeV / c^{2})$
d	$- 1 / 3$	$1 / 3$	$1 / 2$	$- 1 / 2$	0	0	0	0	0.005
u	$+ 2 / 3$	$1 / 3$	$1 / 2$	$+ 1 / 2$	0	0	0	0	0.002
s	$- 1 / 3$	$1 / 3$	0	0	-1	0	0	0	0.095
c	$+ 2 / 3$	$1 / 3$	0	0	0	+1	0	0	1.3
b	$- 1 / 3$	$1 / 3$	0	0	0	0	-1	0	4.2
t	$+ 2 / 3$	$1 / 3$	0	0	0	0	0	+1	174

Table 7.2 Quark quantum numbers and masses. Note the convention that quarks with a negative electric charge carry a negative flavour quantum number.

The masses quoted are “bare masses” - when bound in a hadron the effective masses differ, especially for the lightest quarks. (Binding and kinetic energies mean that the $u$ and $d$ quarks can be treated as effectively equal in mass.)

7.6. Strange particles and Cabibbo angle#

As we have already seen, the strong and electromagnetic interactions conserve quark flavour, whereas the weak interaction may change it. In many weak decays, the changes are within a generation, e.g. in the beta decay the W couples a $u$ to a $d$ quark, such as in the decay

π^{-} \to u \bar{d}

Or to a $c$ and $s$ quark as in the decay

D^{+} \to \overset{―}{K_{0}} π^{+}

However, this is not always the case, and strangeness-changing decays are observed as already anticipated last week. One example is the decay

K^{-} \to π^{0} e^{-} \overset{―}{v_{e}}

Where the W couples an s to a $u$ quark. The observation is that such strangeness-changing decays are slightly weaker than strangeness-conserving weak decays.

Cabibbo explained this by proposing that the eigenstates of the weak interaction are different from those of the strong interaction. The strong interaction eigenstates are the $u, d, s, c, b$ and $t$ quarks, with well-defined isospin, strangeness etc. The eigenstates of the weak interaction, which does not conserve $I, S$ etc., are said to be those of “weak isospin” $T$ . For simplicity, let us start by considering the first 2 generations alone. The weak eigenstates are the leptons and orthogonal linear combinations of the familiar quarks

(\binom{v_{e}}{e}) (\binom{v_{μ}}{μ}) (\binom{u}{d_{c}}) (\binom{c}{s_{c}})

With

\begin{array}{r} \begin{aligned} d_{c} = α d + β s \\ s_{c} = - β d + α s \end{aligned} \end{array}

Where the mixing of the $d$ and $s$ states is done in such a way to preserve normalisation

α^{2} + β^{2} = 1

The simplest way to ensure that the sum of the square of two numbers is 1 is to indicate them as $α = \sin θ_{C}$ and $β = \cos θ_{C}$ , where $θ_{C}$ is the Cabibbo angle. A value of $\sin θ_{C} = 0.25$ is consistent with the observed apparent variation of weak coupling constant with reaction type. The relationship between weak and strong eigenstates in 2 generations can also be expressed as

\begin{array}{r} (\binom{d_{c}}{s_{c}}) = (\begin{array}{cc} \cos θ_{C} & \sin θ_{C} \\ - \sin θ_{C} & \cos θ_{C} \end{array}) (\binom{d}{s}) \end{array}

In other words we can say that we transform the quark eigenstates for the strong interaction into the eigenstates for the weak interaction by performing a “rotation” with an angle $θ_{C}$ , which is experimentally measured.

7.7. The CKM matrix#

The same formalism can be used for 3 generations, and the mixing matrix, known as the Cabibbo-Kabayashi-Maskawa or CKM matrix, can be parameterised in a number of ways.

\begin{array}{r} (\begin{array}{l} d^{'} \\ s^{'} \\ b^{'} \end{array}) = M (\begin{array}{l} d \\ s \\ b \end{array}) \end{array}

The magnitudes of the matrix elements have been determined experimentally, and are given by

\begin{array}{r} M = (\begin{array}{ccc} 0.9745 \pm 0.001 & 0.2245 \pm 0.0004 & 0.036 \pm 0.0001 \\ 0.2244 \pm 0.0004 & 0.9736 \pm 0.0001 & 0.0421 \pm 0.0008 \\ 0.0090 \pm 0.0002 & 0.0413 \pm 0.007 & 0.99911 \pm 0.0003 \end{array}) \end{array}

Note that the values along the leading diagonal are quite close to one, those adjacent to it are significantly smaller, and the elements in the top-right and bottom-left corners are much smaller. This means that the mixing results in states which contain a small admixture of the quark from the next generation, while mixing between 1st and 3rd generation quarks is extremely small.

Flavour-changing weak interactions always occur via the charged current. That is, they always involve transitions between the two members of the same weak isospin doublet, e.g. between c and $s^{'}$ (in either direction), or between u and $d^{'}$ . The mixing of the negative quarks plays a role for both initial and final state quarks. For example, the decay of a c quark is always to an s’ weak eigenstate, which will be bound in a hadron as one of the strong eigenstates of which it can be considered a mixture. On the other hand, when a hadron containing an s quark decays, the s must be considered a mixture of $d^{'}, s^{'}$ and $b^{'}$ weak eigenstates, and these decay to $u, c$ and $t$ respectively.

Physically, the relative probability of producing hadrons containing the respective quarks in a weak decay is determined by the elements of the CKM matrix. For example, when a top quark decays it produces a $b^{'}$ quark. This is bound in a hadron by the strong interaction, so must be revealed as a strong eigenstate. The $b^{'}$ is most likely to result in a particle containing a $b$ quark, with a smaller probability of an s quark and almost negligible likelihood of producing a d quark. Therefore, the near-diagonal structure of the CKM matrix means that weak decays are most likely to be within a generation if allowed by conservation of energy (a particle cannot decay into one that is heavier) or to the next generation below if this is not allowed. The most likely overall decay chain of a b quark is therefore $b \to c \to s \to u$ .

When the charged-current weak decays are considered along with binding into strong eigenstates in the hadrons, the elements of $M$ can be interpreted as giving the effective transition strengths between quarks as follows:

\begin{array}{r} M = (\begin{array}{lll} u \leftrightarrow d & u \leftrightarrow s & u \leftrightarrow b \\ c \leftrightarrow d & c \leftrightarrow s & c \leftrightarrow b \\ t \leftrightarrow d & t \leftrightarrow s & t \leftrightarrow b \end{array}) \end{array}

(Again, physical decays must always be to the lighter quark.)

For two generations, one parameter was required to describe the mixing. This was the Cabibbo angle. With three generations, 4 independent parameters are needed to define a general unitary matrix, and the individual matrix elements may have imaginary parts. One possible parameterisation of the CKM matrix is given below. Note that the following material is provided for completeness only, and is not examinable! (Further details are provided in the text books.)

\begin{array}{r} M = (\begin{array}{ccc} c_{12} c_{13} & s_{12} c_{13} & s_{13} e^{- i δ} \\ - s_{12} c_{23} - c_{12} s_{23} s_{13} e^{i δ} & c_{12} c_{23} - s_{12} s_{23} s_{13} e^{i δ} & s_{23} c_{13} \\ s_{12} s_{23} - c_{12} c_{23} s_{13} e^{i δ} & - c_{12} s_{23} - s_{12} c_{23} s_{13} e^{i δ} & c_{23} c_{13} \end{array}) \end{array}

where $c_{i j} = \cos θ_{i j}$ and $s_{i j} = \sin θ_{i j} s$ with $i$ and $j$ being generation labels ${i, j = 1, 2, 3}$ . In the limit $θ_{23} = θ_{13} = 0$ the third generation decouples, and the situation reduces to Cabibbo mixing of the first two generations, with $θ_{12}$ identified with the Cabibbo angle. The real angles $θ_{12}, θ_{13}, θ_{23}$ can all be chosen to lie in the first quadrant. $c_{13}$ is known to differ from unity only in the sixth decimal place.

If the parameter $δ$ is non-zero, then the matrix is complex, and the small degree of CP violation present in the weak interaction can be explained naturally. It has not yet been conclusively proven that this is the explanation for all the observed CP violation!

[The above parameterisation and values are taken from the Particle Physics Data Booklet, from “Review of Particle Physics”, Phys. Rev. D98, 03001 January 2018, by the Particle Data Group.]

7.8. Why doesn’t the Weak Neutral Current allow Decays like $s \to$ d?#

You may have wondered why all the discussion of quark decays via the weak interaction has concentrated on the charged current (resulting in decays between charge $2 / 3$ quarks and charge $- 1 / 3$ quarks or vice versa). This is because interactions involving the neutral current cannot change quark flavour (an explanation known as the GIM mechanism). It might be expected that, since the strong quark eigenstates are mixed in the weak interaction, a decay such as $s \to d$ would be allowed The following argument should show why this is not the case. The strong eigenstates can be represented in terms of weak eigenstates using

s = s_{C} \cos θ + d_{C} \sin θ

The weak interaction always couples states in the same weak doublet. For the neutral current, this means coupling $s_{C}$ to $s_{C}$ and $d_{C}$ to $d_{C}$ . The produced quarks will eventually be bound in hadrons by the strong interaction, so we need to consider how the weak states are represented in terms of strong states:

\begin{array}{r} \begin{aligned} s_{C} = s \cos θ - d \sin θ \\ d_{c} = d \cos θ + s \sin θ \end{aligned} \end{array}

Overall, therefore, the s coupling, via the weak neutral current, would be:

s \to (s \cos θ - d \sin θ) \cos θ + (d \cos θ + s \sin θ) \sin θ = s

In other words, the s quark only couples via the neutral current to the s quark; there

7.9. The $K^{0}$ System and Strangeness Regeneration#

We have seen that the strong and weak eigenstates, when expressed in terms of quarks, are different. We now look at evidence for different eigenstates in the observed, free particles themselves. This is seen in the neutral kaon system.

The $K^{0}$ ( $d \bar{s})$ has strangeness 1 , and can be produced in $π p$ collisions by the strong interaction in association with a $Λ$ hyperon (i.e. strange baryon). In contrast, the $\overset{―}{K^{0}} (\bar{d} s)$ has strangeness -1 , and as there are no baryons of positive strangeness, this can only be formed in higher energy collisions. At low energies, a pure sample of $K^{0}$ can therefore be produced.

The weak interaction does not conserve strangeness. It therefore does not distinguish between $K^{0}$ and $\overset{―}{K^{0}}$ . The eigenstates of the weak interactions are not those of strangeness but (approximately) those of CP and are written as

\begin{array}{r} \begin{array}{ll} K_{1} = \frac{1}{\sqrt{2}} (| K^{0} ⟩ + | \overset{―}{K^{0}} ⟩) & with CP eigenvalue + 1 \\ K_{2} = \frac{1}{\sqrt{2}} (| K^{0} ⟩ - | \overset{―}{K^{0}} ⟩) & with CP eigenvalue - 1 \end{array} \end{array}

The $K_{1}$ decays to two pions, while the $K_{2}$ must decay to a 3 -pion final state. As there is much more phase space available for the $K_{1}$ decay, its decay rate is about 600 times greater than that for the $K_{2}$ .

Consider a beam of pions striking a thin solid target, mounted in a vacuum. Strong interactions will occur, resulting in the production of $K^{0}$ (with no $\overset{―}{K^{0}}$ ). These then travel on and decay through the weak interaction. The $K^{0}$ is equivalent to an equal mixture of $K_{1}$ and $K_{2}$ , and the $K_{1}$ component rapidly decays away. If a further solid target (known as a regenerator) is introduced some distance downstream, the remaining kaons will interact with it via the strong interaction. Here the surviving $K_{2}$ component must be seen as an equal mixture of $K^{0}$ and $\overset{―}{K^{0}}$ , so $S = + 1$ and -1 states will be produced, even though only $K^{0}$ (with $S$ $= + 1$ ) was present initially! In fact, the $\overset{―}{K^{0}}$ state will be preferentially absorbed, with only some of the $K^{0}$ emerging from the regenerator. Here, weak decays will again occur, and an equal mixture of $K_{1}$ and $K_{2}$ will again be observed, even though previously the entire $K_{1}$ component had decayed away!

Quantum mechanically, the situation is exactly equivalent to a series of Stern-Gerlach experiments, with crossed magnetic field gradients. If an unpolarised beam of neutral, spin $1 / 2$ atoms travelling along the $z$ -axis encounters a region with a field gradient in the $x$ -direction, it will split into two equal divergent beams with $s_{x} = \pm 1 / 2$ . If one of these is blocked and the other allowed into a second region where the field gradient is in the $y$ -direction, it will split again into equal divergent beams with $s_{y} = \pm 1 / 2$ . If now one of these is selected and encounters a region with a field gradient again in the $x$ -direction, it will split into two equal beams with $s_{x} = \pm 1 / 2$ even though one of these components had previously been eliminated.

This mixture of states would lead to the identification of two states that can be differentiated using the lifetime as an observable. In other words we have two states that we call “long” and “short” with the lifetimes of the two particles being different by more than 3 orders of magnitude:

Lifetime of $K_{S}^{0}$ is $0.89 \times 10^{- 10} s$
Lifetime of $K_{L}^{0}$ is $0.52 \times 10^{- 7} s$

However it was observed in 1964 that the decay

K_{L}^{0} \to π^{+} π^{-}

Is also observed. Which implies that $K_{L}^{0}$ is not a CP eigenstate, but a combination of the two states

| K_{L}^{0} ⟩ = \frac{1}{{(1 + | ϵ |^{2})}^{1 / 2}} [ϵ | K_{1}^{0} ⟩ + | K_{2}^{0} ⟩]

Experimentally, we measure the value of $| ϵ | = 2.2 \times 10^{- 3}$ .

One interesting consequence of this phenomenon is that the $K_{L}^{0}$ is not a symmetric mixture of the two strangeness states $K^{0}$ and $\overset{―}{K^{0}}$ , so the two decays:

\begin{array}{r} \begin{aligned} K^{0} \to π^{-} e^{+} v_{e} \\ \overset{―}{K^{0}} \to π^{+} e^{-} \overset{―}{v_{e}} \end{aligned} \end{array}

Are observed as decays of the $K_{L}^{0}$ and have different probability of occurring. In other words this results into an asymmetry between matter and antimatter. There are several discussions on how this can be used to check if an alien civilization is made of matter of antimatter before getting into direct contact with them. (Or avoid in case they are made of antimatter!)

7.10. Further reading#

The concept of resonances and quarkonium states are discussed in [Perkins] Section 4.1 and 4.2. We did not cover these in the course this year
A description of the isospin and its role in strong interaction is given in [Martin] Section 5.1
Quarks in hadrons to form multiplets is covered in [Perkins] Sections 4.3 and 4.4
Chapters 7 and 8 of “Ideas of particle physics” give a simple description of isospin and how $u, d$ , s quarks combine to make hadrons