C001 derivation

The claimed state is $$|\Psi^-\rangle = \tfrac{1}{\sqrt{2}}\left(|{\uparrow}{\downarrow}\rangle - |{\downarrow}{\uparrow}\rangle\right).$$

Normalization. The two product kets are orthonormal, so $$\langle\Psi^-|\Psi^-\rangle = \tfrac{1}{2}\left(\langle{\uparrow}{\downarrow}|{\uparrow}{\downarrow}\rangle + \langle{\downarrow}{\uparrow}|{\downarrow}{\uparrow}\rangle\right) = \tfrac{1}{2}(1+1) = 1.$$

Antisymmetry. Under the exchange (SWAP) operator $P|a,b\rangle = |b,a\rangle$, $$P|\Psi^-\rangle = \tfrac{1}{\sqrt2}\left(|{\downarrow}{\uparrow}\rangle - |{\uparrow}{\downarrow}\rangle\right) = -|\Psi^-\rangle,$$ so it is the antisymmetric combination, distinct from the three symmetric triplet states $|{\uparrow}{\uparrow}\rangle,\,|{\downarrow}{\downarrow}\rangle,\,(|{\uparrow}{\downarrow}\rangle+|{\downarrow}{\uparrow}\rangle)/\sqrt2$.

Maximal entanglement. Writing $\rho = |\Psi^-\rangle\langle\Psi^-|$ and tracing out ion 2, $$\rho_1 = \mathrm{Tr}_2\,\rho = \tfrac{1}{2}\left(|{\uparrow}\rangle\langle{\uparrow}| + |{\downarrow}\rangle\langle{\downarrow}|\right) = \tfrac{1}{2}\mathbb{1}.$$ The reduced state is maximally mixed, with von Neumann entropy $S(\rho_1) = \ln 2$ (one ebit), the maximum for a qubit; hence the state is maximally entangled.

Total-spin-0 singlet. With $S_k = \tfrac12(\sigma_k\otimes\mathbb1 + \mathbb1\otimes\sigma_k)$, one finds $S_x|\Psi^-\rangle = S_y|\Psi^-\rangle = S_z|\Psi^-\rangle = 0$, so $\mathbf S^2|\Psi^-\rangle = 0$: total angular momentum zero, the singlet.

All four properties hold, so the definition is the standard singlet Bell state.