the states, (and ignoring for the moment normalization factors and species labels), it can be checked that the quadruplet is simply[@wootters] $$\begin{split}
  {\ensuremath{|B^0\rangle}}&\propto~~~~{\ensuremath{|+, 1; -,2\rangle}} - {\ensuremath{|-,1; +, 2\rangle}}  \\
  {\ensuremath{|B^1\rangle}}&\propto~~i({\ensuremath{|+, 1; +,2\rangle}} - {\ensuremath{|-,1; -, 2\rangle}})  \\
  {\ensuremath{|B^2\rangle}}&\propto~~~~{\ensuremath{|+, 1; +,2\rangle}} + {\ensuremath{|-,1; -, 2\rangle}}  \\
  {\ensuremath{|B^3\rangle}}&\propto -i({\ensuremath{|+, 1; -,2\rangle}} + {\ensuremath{|-,1; +, 2\rangle}}).
  \end{split}$$ For ease of comparison we have also simplified the momenta indices ${\ensuremath{\textbf p}}_{1,2}$ to $1,2$ respectively. Therefore, apart from multiplicative constants, these are, respectively, the familiar “singlet" (${\ensuremath{|B^0\rangle}}$) and “triplet" $({\ensuremath{|B^{1,2,3}\rangle}})$ Bell states encountered in non-relativistic quantum information science. The conventional assignment of the four Bell states[@nielsen] is related to the present one by ${\ensuremath{|\beta_{11}\rangle}}={\ensuremath{|B^0\rangle}}, {\ensuremath{|\beta_{10}\rangle}} = -i{\ensuremath{|B^1\rangle}}, {\ensuremath{|\beta_{00}\rangle}} =
{\ensuremath{|B^2\rangle}},{\ensuremath{|\beta_{01}\rangle}} = i{\ensuremath{|B^3\rangle}}$. But we would like to advocate the convention in Eq.(\[e:con\]) because it is the particular combination of $({\ensuremath{\tilde\sigma}}^\mu\sigma^2)_{s s'}a^\dag({\ensuremath{\textbf p}}_1,s; n_1)a^\dag({\ensuremath{\textbf p}}_2,s';n_2){\ensuremath{|0\rangle}}$ which provides us with complete and concise description of the behaviour of Bell states under [*arbitrary*]{} Lorentz transformations. The upshot is that under generic $\Lambda$, these Bell states undergo a rotation among themselves, and transform as $$\label{e:tyt}
{\ensuremath{|B^\mu({\ensuremath{\textbf p}}_1,{\ensuremath{\textbf p}}_2)\rangle}}' = U(\Lambda){\ensuremath{|B^\mu({\ensuremath{\textbf p}}_1,{\ensuremath{\textbf p}}_2)\rangle}} =
\sqrt{\frac{(\Lambda p_1)^0(\Lambda p_2)^0}{p_1^0p^0_2}}\,
R_\nu\,^\mu(\Lambda,p_1,p_2){\ensuremath{|B^\nu({{\ensuremath{\textbf p}}_1}_\Lambda, {{\ensuremath{\textbf p}}_2}_{\Lambda})\rangle}},$$ for which $R_\nu\,^\mu(\Lambda,p_1,p_2) \in SO(4)$ is explicitly listed 