Past exam of the particle and nuclear physics course of the University of Cambridge / 2014 / Paper 1 / 3

Set $c=1$ and use metric $(+---)$. In the rest frame of B, $p_B=(m_B,0)$ and $p_C=p_B+p_t$. Since $p_t^2=0$,The denominator is $m_B$ because this is the B rest frame, not the C rest frame. In this frame C and $t$ have equal, parallel three-momenta.

The second decay has ${\mathbf{p}}_A=-{\mathbf{p}}_s$ in the B rest frame. Using $p_A^2=m_A^2$ and $p_s^2=0$ givesThis is an ordinary two-body decay in its parent's rest frame.

Let $q=(m_C^2-m_B^2)/(2m_B)$ and $p=(m_B^2-m_A^2)/(2m_B)$. In the B frame these are the energies of $t$ and $s$. If $\theta$ is their relative angle,The minimum has parallel massless particles; the maximum has antiparallel ones. The isotropic decay allows both limiting directions. The bounds are Lorentz invariant even though the B frame makes their calculation simple.

Since ${\mathbf{p}}_A=-{\mathbf{p}}_s$, the same angle gives $m_{At}^2=m_A^2+2q(E_A+p\cos \theta )$. ThereforeThe minimum now occurs when $s$ and $t$ are antiparallel. The eventwise identity $m_{At}^2+m_{st}^2=m_C^2+m_A^2-m_B^2$ independently checks the opposite ordering of the two endpoints.

Use $K^{-}=s\bar{u}$ and $p=uud$. In the first reaction a proton $u$ can join the incoming $\bar{u}$ to form the $u\bar{u}$ component of the neutral pion, leaving $uds$. In the second, a $u\bar{u}$ pair can annihilate into a gluon which creates $s\bar{s}$; the remaining $u$ joins $\bar{s}$ in the outgoing kaon, leaving $dss$. Thusbaryon number is one in both cases. Strong strangeness conservation provides a separate check: the first baryon has strangeness $-1$; the second has strangeness $-2$ because the outgoing $K^{+}$ carries $+1$. The Feynman diagrams show representative quark-line flavour rearrangement and pair-annihilation topologies. Bound hadrons require nonperturbative binding, so a single drawn gluon is not an exclusive cross-section calculation.

The radiative $X^0$ decay conserves strangeness and proceeds electromagnetically. A strong $\Lambda \pi$ decay is closed because even the lighter neutral-pion threshold exceeds the supplied $X^0$ mass. The $Y^{-}$ decay has about $65\mathrm{MeV}$ of available energy, but changes strangeness from $-2$ to $-1$, so it requires the weak interaction. Its lifetime is therefore ordinarily much longer than the electromagnetic lifetime of $X^0$; no numerical lifetime follows from these masses alone.

Both particles are baryons with half-integer spin. A radiative transition to spin $1/2$ can couple through photon multipoles $L\ge 1$, giving $J_X=L\pm 1/2$. Electromagnetic parity is conserved, but without the electric/magnetic multipolarity neither $J_X$ nor its parity is fixed. In the weak $Y^{-}$ decay, coupling orbital $L$ to the final $\Lambda$ spin gives $J_Y=L\pm 1/2$; parity conservation cannot be imposed on that weak amplitude. The supplied modes therefore do not determine unique spin-parity assignments.

The masses and flavours identify the familiar candidates as ${\Sigma }^0$ and ${\Xi }^{-}$. Their established ground-state assignments are both $1/2^{+}$, consistent with the lowest-orbital quark model, but this is additional empirical/model information rather than a consequence of the two decay channels alone.^[ref]