identity $$\label{e:sigmatrans}
   {\cal U}_1{\ensuremath{\tilde\sigma}}^\mu{\cal U}_2^{-1} = R_\nu^{~\mu} {\ensuremath{\tilde\sigma}}^\nu.$$ Returning to one-particle states, the Lorentz transformation for each party of the entangled pair is (disregarding, for the moment, normalization factors which are not relevant to the discussion below) $$\begin{aligned}
  {\ensuremath{|{\ensuremath{\textbf p}}_1,s,n_1\rangle}}  &\longmapsto& {\cal U}_1(\Lambda)_{s' s}{\ensuremath{|{{\ensuremath{\textbf p}}_1}_{\Lambda},s',n_1\rangle}} \\
  {\ensuremath{|{\ensuremath{\textbf p}}_2,s,n_2\rangle}} &\longmapsto& {\cal U}_2(\Lambda)_{s' s}{\ensuremath{|{{\ensuremath{\textbf p}}_2}_{\Lambda},s',n_2\rangle}}.
  \end{aligned}$$ The crucial observation is that the quadruplet ($\mu = 0,1,2,3)$ of states defined by $({\ensuremath{\tilde\sigma}}^\mu\sigma^2)_{s s'} {\ensuremath{|{\ensuremath{\textbf p}}_1, s, n_1; {\ensuremath{\textbf p}}_2, s';n_2\rangle}}$ transforms as $$\begin{aligned}
({\ensuremath{\tilde\sigma}}^\mu\sigma^2)_{s s'} {\ensuremath{|{\ensuremath{\textbf p}}_1,s,n_1;{\ensuremath{\textbf p}}_2,s',n_2\rangle}} &\longmapsto&
[{\cal U}_1({\ensuremath{\tilde\sigma}}^\mu\sigma^2){\cal U}_2^T]_{s s'}{\ensuremath{|{{\ensuremath{\textbf p}}_1}_{\Lambda},s,n_1;{{\ensuremath{\textbf p}}_2}_{\Lambda},
s',n_2 \rangle}}   \\
&=& [{\cal U}_1 {\ensuremath{\tilde\sigma}}^\mu{\cal U}^{-1}_2\sigma^2]_{s s'}{\ensuremath{|{{\ensuremath{\textbf p}}_1}_{\Lambda},s,n_1;
{{\ensuremath{\textbf p}}_2}_{\Lambda},s',n_2 \rangle}} \\
&=& R_\nu^{~\mu} ({\ensuremath{\tilde\sigma}}^\nu\sigma^2)_{s
s'}{\ensuremath{|{{\ensuremath{\textbf p}}_1}_{\Lambda},s,n_1;{{\ensuremath{\textbf p}}_2}_{\Lambda}, s',n_2 \rangle}}.
$$ In arriving at the last result we have used $\sigma^2 {\cal U}^T_2 ={\cal
U}^{-1}_2\sigma^2$ which follows from $\sigma^2\sigma^i\sigma^2=
-(\sigma^i)^T$; as well as the identity of Eq.(\[e:sigmatrans\]) in the final step. As a consequence, Eq.(\[e:tyt\]) is therefore valid. Furthermore Eq.(\[e:sigmatrans\]) also yields the relation $$R_\nu^{~\mu}(\Lambda) = \frac{1}{2}\eta_{\nu\alpha}Tr[{\cal U}_1(\Lambda){\ensuremath{\tilde\sigma}}^\mu
                 {\cal U}_2^{-1}(\Lambda) {\ensuremath{\tilde\sigma}}^\alpha].$$

In passing we mention two special cases:\
For Lorentz transformations which are pure rotations, $({\textbf{X}}= -{\textbf{Y}}= \bm\theta$ as $\bm\phi({\ensuremath{\textbf p}}_{1,2}) =0$); the explicit form of the $SO(4)$ matrix in Appendix B then yields $$R_\nu^{~\mu} = 