= Solution
Use <flavor hypercharge> $Y=B+S$, where $B$ is <baryon number> and $S$ is <strangeness>. The <isospin> coordinate is $I_3$. The <baryon octet> has coordinates
$$
\begin{array}{c|rrrrrrrr}
\text{state}&p&n&\Sigma^+&\Sigma^0&\Sigma^-&\Lambda^0&\Xi^0&\Xi^-\\\hline
I_3&\tfrac12&-\tfrac12&1&0&-1&0&\tfrac12&-\tfrac12\\
Y&1&1&0&0&0&0&-1&-1
\end{array}
$$
Here the <nucleons> have $S=0$, the <Sigma baryons> and <Lambda baryon> have $S=-1$, and the <Xi baryons> have $S=-2$. The central <Sigma baryon> belongs to an <isospin> triplet while the central <Lambda baryon> is an <isospin> singlet; equal coordinates do not identify the states.
The pseudoscalar <meson octet> is
$$
\begin{array}{c|rrrrrrrr}
\text{state}&K^+&K^0&\pi^+&\pi^0&\pi^-&\eta_8&\bar K^0&K^-\\\hline
I_3&\tfrac12&-\tfrac12&1&0&-1&0&\tfrac12&-\tfrac12\\
Y&1&1&0&0&0&0&-1&-1
\end{array}
$$
The <pions> have $S=0$, the upper <kaons> have $S=+1$, and their lower antiparticles have $S=-1$. The two central states are the neutral <pion> and the <Eta octet state>. This is the octet basis of <flavor symmetry>; the physical eta can also mix with the flavor-singlet state.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-44-flavor-octets.png]
{title=Flavor SU(3) baryon and pseudoscalar meson octets in isospin and strong hypercharge coordinates}
{height=500}
For the <flavor SU(3) Cartan generators>, choose the Hermitian physics convention and the <inner product> $\langle A,B\rangle=\operatorname{tr}(AB)$. Then an orthonormal diagonal basis is
$$
\boxed{h_1=\frac1{\sqrt2}\operatorname{diag}(1,-1,0),\qquad h_2=\frac1{\sqrt6}\operatorname{diag}(1,1,-2),\qquad\operatorname{tr}(h_ih_j)=\delta_{ij}.}
$$
Strictly, $ih_1,ih_2$ are elements of the anti-Hermitian <SU(3) Lie algebra>; $h_1,h_2$ are the corresponding Hermitian observables. In the quark basis $(u,d,s)$, the up and down <quarks> form an <isospin> doublet and the strange <quark> is a singlet. Their $I_3$ values are $(1/2,-1/2,0)$. Each <quark> has <baryon number> $1/3$, and their <strangeness> values are $(0,0,-1)$. It follows that
$$
\boxed{I_3=\frac{h_1}{\sqrt2},\qquad Y=\sqrt{\frac23}\,h_2=\operatorname{diag}\left(\frac13,\frac13,-\frac23\right).}
$$
These <flavor hypercharge> conventions differ from the <electroweak hypercharge> convention. If instead the inner product is $2\operatorname{tr}(AB)$, the orthonormal basis is $t_3=h_1/\sqrt2$, $t_8=h_2/\sqrt2$, and the same operators are $I_3=t_3$, $Y=2t_8/\sqrt3$.
In the ordinary <quark model>, the <proton> has valence content $uud$ and charge $+1$, while the <neutron> has content $udd$ and charge zero. Additivity of <electric charge> gives $2q_u+q_d=1$ and $q_u+2q_d=0$, hence $q_u=2/3$, $q_d=-1/3$. The <Sigma baryon> $\Sigma^-$ has content $dds$ and charge $-1$, giving $2q_d+q_s=-1$. Thus the <quark> triplet's <electric charges>, in units of the positive elementary charge, are
$$
\boxed{(q_u,q_d,q_s)=\left(\frac23,-\frac13,-\frac13\right).}
$$
The <Gell-Mann--Nishijima formula> is consequently
$$
\boxed{Q=I_3+\frac Y2=\frac{h_1}{\sqrt2}+\frac{h_2}{\sqrt6}=\operatorname{diag}\left(\frac23,-\frac13,-\frac13\right).}
$$
It also reproduces every <baryon octet> charge from the first diagram. On <antiquarks> the additive quantum numbers reverse sign, and combining a <quark> with an <antiquark> reproduces the <meson octet> charges.
Because the down and strange <quarks> have identical <electric charge>, $Q$ commutes with the <U-spin> generators
$$
U_1=\frac{E_{23}+E_{32}}2,\qquad U_2=\frac{E_{23}-E_{32}}{2i},\qquad U_3=\frac12\operatorname{diag}(0,1,-1).
$$
They satisfy $[U_a,U_b]=i\epsilon_{abc}U_c$; equivalently the anti-Hermitian matrices $iU_a$ span an $\mathfrak{su}(2)$ subalgebra. The entries on the $d,s$ block of $Q$ are equal, so \b[$[Q,U_a]=0$ for all three <U-spin> generators]. The <electric charge> is therefore constant within each irreducible <U-spin> multiplet. For example, <U-spin> relates $\pi^+$ and $K^+$, and relates $p$ and $\Sigma^+$, without changing their charge. It does not imply exact mass degeneracy: unequal down- and strange-quark masses break <U-spin>.
For <pion-nucleon octet channels>, assume the collision is governed by the <strong interaction>. The initial <baryon number> is one and <strangeness> is zero, so an outgoing <meson>-<baryon> pair must preserve $B=1$, $S=0$, and <electric charge>. Thus its total <flavor hypercharge> is $Y=1$. The allowed types are
$$
\boxed{\pi N,\qquad\eta_8N,\qquad K\Lambda,\qquad K\Sigma.}
$$
A <kaon> of $S=+1$ can accompany a <Lambda baryon> or <Sigma baryon> of $S=-1$. An antikaon cannot balance the nonpositive <strangeness> of an octet <baryon>. A <Xi baryon> would require a meson of $S=+2$, which the <meson octet> does not contain.
Resolving these types by <electric charge> gives all possible pairs:
$$
\begin{array}{c|l|l}
Q&\text{incoming}&\text{outgoing pairs allowed by additive charges}\\\hline
2&\pi^+p&\pi^+p,\ K^+\Sigma^+\\
1&\pi^0p,\ \pi^+n&\pi^+n,\ \pi^0p,\ \eta_8p,\ K^+\Lambda^0,\ K^+\Sigma^0,\ K^0\Sigma^+\\
0&\pi^-p,\ \pi^0n&\pi^-p,\ \pi^0n,\ \eta_8n,\ K^0\Lambda^0,\ K^+\Sigma^-,\ K^0\Sigma^0\\
-1&\pi^-n&\pi^-n,\ K^0\Sigma^-
\end{array}
$$
In the <isospin>-symmetric approximation, total <isospin> is conserved as well: the incoming $1\otimes\tfrac12$ contains $I=\tfrac12,\tfrac32$. The $\pi N$ and $K\Sigma$ channels contain both values, while $\eta_8N$ and $K\Lambda$ contain only $I=\tfrac12$. The extreme-charge initial states are pure $I=\tfrac32$, consistently excluding $\eta_8N$ and $K\Lambda$. <Clebsch-Gordan coefficients> relate amplitudes in different charge channels; the table establishes permission, not equal probabilities. Electromagnetism and unequal up- and down-quark masses introduce small violations of <isospin> symmetry.
Finally, <energy> and <momentum conservation> require $\sqrt{s}\geq m_M+m_B$ for a particular pair, where $s$ is the squared total <four-momentum>. Only channels above their own threshold can occur. Total <angular momentum> and <parity symmetry in quantum field theory> constrain the <partial waves> of the <meson-baryon scattering>: a pseudoscalar <meson> and a positive-parity spin-one-half <baryon> have pair parity $-(-1)^L$ and total <angular momentum> $J=L\pm\tfrac12$ (only $J=\tfrac12$ for $L=0$). Initial and final <partial waves> must have matching $J$ and parity. Sufficient energy can open a channel, but cannot remove the <electric charge conservation>, <baryon number>, or <strangeness> constraints.
Back to article page
