Past exam of the particle physics course of the University of Oxford / 2019 / 3

Use the orthonormal basis $|u\bar{u}⟩,|d\bar{d}⟩,|s\bar{s}⟩$. The normalized flavour singlet isThe SU(3) invariant contraction pairs each fundamental flavour with its conjugate. Applying any supplied raising operator to the singlet gives zero because its quark and antiquark contributions cancel.

To fix the neutral isospin triplet phase consistently with the source's charged states, apply $I_{+}$ to $-|d\bar{u}⟩$:Thus one missing member of the meson octet isIndeed $I_{+}|T_0⟩=\sqrt{2}|u\bar{d}⟩$, the correct ladder operator strength for the $I=1$ triplet.

The other missing state is orthogonal to both $|1⟩$ and $|T_0⟩$. For coefficients $(a,b,c)$, orthogonality requires $a=b$ and $a+b+c=0$. Choose the convenient overall phase $c>0$ and normalize:It can also be obtained by adding the supplied ladder operations $V_{+}[-|s\bar{u}⟩]=|s\bar{s}⟩-|u\bar{u}⟩$ and $U_{+}[-|s\bar{d}⟩]=|s\bar{s}⟩-|d\bar{d}⟩$. The three displayed states are mutually orthonormal. The meson octet here is the traceless adjoint component of $\bar{3}\otimes 3$; the source's word “antisymmetric” does not impose antisymmetry on all its quark-antiquark flavour wavefunctions.

Only the singlet $|1⟩$ and isoscalar octet member $|8_0⟩$ can mix without violating isospin: they have the same $I=I_3=S=0$ and, for the vector multiplet, the same $J^{PC}=1^{--}$. The neutral triplet member has $I=1$ and is excluded in the isospin limit.

Set $|{\psi }_a⟩=|1⟩$ and $|{\psi }_b⟩=|8_0⟩$, with precisely the phases in a. Define $|n⟩=(|u\bar{u}⟩+|d\bar{d}⟩)/\sqrt{2}$. Taking overlaps givesTherefore the source's rotation convention is satisfied byThis yields $|\omega ⟩=|n⟩$ and $|\varphi ⟩=|s\bar{s}⟩$ exactly. Reversing the octet phase changes the quoted sign, with an irrelevant compensating phase on the physical phi state; exchanging the basis order gives the complementary angle. The physical statement of ideal vector-meson mixing is the pure strange and pure nonstrange content, rather than a convention-independent sign for the angle.

The observed dominance of kaons supports a predominantly $s\bar{s}$ phi meson. In a connected strong decay, the original strange quark and antiquark survive into different kaons, while a gluon produces a light $q\bar{q}$ pair:This gives $K^{-}{K}^{+}$ or ${\bar{K}}^0{K}^0$ and is allowed by the OZI rule. A pionic final state contains no strange valence constituents, so the initial $s\bar{s}$ pair must annihilate. The strong annihilation of a colour-singlet $1^{--}$ state requires at least three gluons: one gluon cannot carry the singlet colour, and two have the wrong charge-conjugation quantum number. This disconnected flavour topology is suppressed by the OZI rule, although it need not vanish.

The leptonic mode proceeds through $s\bar{s}\to {\gamma }^{\ast }\to {\ell }^{+}{\ell }^{-}$ and is suppressed by electromagnetic couplings; its width is proportional to ${\alpha }^2$ times the vector-meson overlap. Unlike a strong hadronic decay, it cannot create its final particles through quark rearrangement. Hence a branching fraction of order $10^{-4}$ is qualitatively consistent with electromagnetic annihilation competing with strong modes.

Kaon dominance is especially suggestive because phase space is limited: using the supplied neutral-kaon and phi masses,The two spin-zero kaons have orbital angular momentum $L=1$ to match the vector parent, so their partial width has the additional near-threshold factor $p_K^3$. Pionic modes have more phase space yet remain subdominant because of their annihilation topology. The observations therefore support, but do not establish exact, ideal vector-meson mixing: a small nonstrange admixture, electromagnetic amplitudes and nonperturbative strong dynamics can all contribute to the pionic rate.

Quark mass differences break flavour symmetry. Isospin exchanges $u,d$, so its strong breaking is controlled by the small difference $m_d-m_u$. U-spin exchanges $d,s$, so its strong breaking involves the much larger $m_s-m_d$. Relative to the characteristic strong-interaction scale, the latter is much larger; this explains why U-spin is a substantially poorer symmetry.

Flavour-dependent interactions are a second source. Electromagnetism distinguishes $u$ from $d$ because $Q_u=2/3$ and $Q_d=-1/3$, and therefore breaks isospin. However, $d,s$ both have charge $-1/3$, so their electromagnetic couplings alone do not break U-spin in the equal-mass limit. If two breaking mechanisms for each symmetry are required, the second common mechanism is the weak interaction: the CKM matrix and weak flavour transitions distinguish the flavours and do not conserve either strong-interaction flavour symmetry. The dominant difference in the quality of the two approximate strong symmetries is still the mass splitting.