Past exam of the physics A course of the University of Cambridge / 2010 / Paper 2 / Section B / B9 / b / Solution

The lowest singlet state isFor the lowest triplet states take $\lambda =\pm 1$ and $m_s=-1,0,1$:orthogonality of $u_0$ and $u_{\lambda }$ verifies wavefunction normalization of the antisymmetric spatial factor. At this same energy there are also two excited singlet states, obtained by replacing its minus sign with a plus sign and the spin factor by $|0,0⟩$. Together they reproduce the eight states counted in part a.