Past exam of the particle physics course of the University of Oxford / 2016 / 3

Strictly, it is the infinitesimal generators, not every unitary matrix with unit determinant, that are traceless. For a continuous one-parameter family $U(ϵ)=\exp (iϵT)$, unitarity requires the generator $T$ to be Hermitian. The determinant condition givesDifferentiating at $ϵ=0$ yields $\mathrm{tr}T=0$. A Hermitian matrix has $n$ real diagonal parameters and $n(n-1)$ real off-diagonal parameters; one real trace constraint leaves $n^2-1$ independent generators. ThusThe identity matrix supplies a concrete counterexample to interpreting the claim as one about arbitrary group elements: $I_n\in SU(n)$ but $\mathrm{tr}{I}_n=n$. Also, $e^{i\mathrm{tr}T}=1$ at a single parameter value only fixes the trace modulo $2\pi$; the continuous neighbourhood is essential.

Embed the two off-diagonal Pauli matrices in each of the $(u,d)$, $(u,s)$ and $(d,s)$ two-dimensional subspaces. Together with two independent embedded diagonal Pauli matrices, this produces the Gell-Mann matrices:To construct the last matrix explicitly, the $(u,s)$ diagonal embedding is $D_{us}=\mathrm{diag}(1,0,-1)$ and the $(d,s)$ embedding is $D_{ds}=\mathrm{diag}(0,1,-1)$. Their difference is ${\lambda }_3$, and their sum divided by $\sqrt{3}$ is ${\lambda }_8$. There are only two independent diagonal generators.

Each matrix is Hermitian and traceless. Their inner products satisfy $\mathrm{tr}({\lambda }_a{\lambda }_b)=2{\delta }_{ab}$, proving linear independence. With $T_a={\lambda }_a/2$, arbitrary transformations near the identity have $U=\exp (i{\theta }_a{T}_a)$. The eight matrices span the generators; they are not a vector-space basis of group elements themselves.

Writing $E_{ij}=|i⟩⟨j|$, the ladder operators areThe ordered pair on each right-hand side gives the plus and minus operators respectively. Thus $I_{+}d=u$, $V_{+}s=u$ and $U_{+}s=d$, with reverse actions for the minus operators; their other actions on the fundamental triplet vanish.

The remaining Flavor SU(3) Cartan generators can be chosen as $I_3={\lambda }_3/2$ and $Y={\lambda }_8/\sqrt{3}$. Their eigenvalues label isospin and flavor hypercharge:These diagonal operators multiply a flavor eigenstate by an eigenvalue; exponentiating them changes relative phases rather than converting one quark flavour into another. The weight diagram makes the distinction visible: the ladder operators connect different points, while the diagonal eigenvalues supply the axes.

Represent a quark-antiquark state by a matrix $F$ with $q_i{\bar{q}}_j\leftrightarrow E_{ij}$. Under flavor symmetry, $F\mapsto UF{U}^{†}$, so an infinitesimal generator acts by $[T,F]$. The minus sign in this commutator is the antiquark transformation law; acting on the quark alone would miss states.

Start from ${\pi }^{+}=E_{ud}$. Three useful first steps areFurther actions give $[I_{-},E_{us}]=E_{ds}=K^0$, $[I_{-},E_{sd}]=-E_{su}=-K^{-}$ and $[I_{-},{\pi }^0]=\sqrt{2}{E}_{du}=\sqrt{2}{\pi }^{-}$. Overall phases are conventional. FinallyThus the second neutral octet state is obtained by removing the known ${\pi }^0$ component. Equivalently it is the normalized diagonal state orthogonal to both ${\pi }^0$ and the singlet. The full meson octet isTheir flavour matrices are traceless; the independent singlet is ${\eta }_1=(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$. The source uses ${\eta }^{\prime }$ for this ideal singlet. That notation is a basis convention here: the physical eta and eta prime mesons is not an exactly pure singlet because physical neutral pseudoscalar mesons mix.^[ref]

The neutral vector mesons with isospin zero have the same conserved quantum numbers, so an off-diagonal mass matrix can mix the singlet $V_1=(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$ and octet $V_8=(u\bar{u}+d\bar{d}-2s\bar{s})/\sqrt{6}$. Choose phases such thatAt the ideal mixing angle $\tan \theta =1/\sqrt{2}$, or $\theta \simeq 35.3^{\circ }$, these becomeThe overall sign of $\varphi$ is unobservable. The omega meson near $782\mathrm{MeV}$ decays predominantly to three pions, whereas the phi meson near $1020\mathrm{MeV}$ strongly favours kaon-antikaon channels when they are open. Suppression of its nonstrange channels agrees with the OZI rule and a predominantly strange state. The omega meson is below the two-kaon threshold, so its absence of that channel alone would not establish the mixing angle; masses and the broader decay pattern provide the evidence together.^[ref]