the special case of $U(\Lambda) = U(L(p))$ acting on ${\ensuremath{|k,s\rangle}}$ the Wigner transformation is $W(L(p), k) = [L^{-1}(L(p)k)]L(p)L(k) = L^{-1}(p)L(p)I
= I$, which consistently produces no rotation in spin space, as Eq.(\[e:sp\]) demands.

Explicit generators of the unitary representation
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One-particle states are defined through the action of creation operator $a^\dag({\ensuremath{\textbf p}},s)$ on the vacuum as ${\ensuremath{|{\ensuremath{\textbf p}},s\rangle}} = a^\dag({\ensuremath{\textbf p}},s){\ensuremath{|0\rangle}}$. It follows from Eq.(\[e:psfin\]) that under a Lorentz transformation annihilation and creation operators in quantum field theory behave as $$\begin{aligned}
 \label{e:adrule}
 {{a'}^\dag}({\ensuremath{\textbf p}},s) &=& U(\Lambda){a^\dag}({\ensuremath{\textbf p}},s){U^\dag}(\Lambda) \\
&\equiv&\int d^3{\ensuremath{\textbf p}}\,\,[{{U}^*}_{({\ensuremath{\textbf p}},s)({\ensuremath{\textbf p}}',s')}(\Lambda)]{a^\dag}({\ensuremath{\textbf p}}',s')\\
&=& \sqrt{\frac{(\Lambda p)^0}{p^0}}D_{s' s}\bigl( W(\Lambda,p) \bigr)
a^\dag({\ensuremath{\textbf p}}_\Lambda, s'),\end{aligned}$$ with $p^i_\Lambda \equiv \Lambda^i\,_\mu{p^\mu}$; and by taking the adjoint, $$\begin{aligned}
\label{e:arule}
 a'({\ensuremath{\textbf p}},s) &=& U(\Lambda)a({\ensuremath{\textbf p}},s)U^\dag(\Lambda) \\
 &\equiv& \int d^3{\ensuremath{\textbf p}}\,\,[{U}_{({\ensuremath{\textbf p}},s)({\ensuremath{\textbf p}}',s')}(\Lambda)]a({\ensuremath{\textbf p}},s') \\
 &=& \sqrt{\frac{(\Lambda p)^0}{p^0}}D_{s s'}\bigl( W^{-1}(\Lambda,p)
 \bigr)a({\ensuremath{\textbf p}}_\Lambda,s').\label{e:43}\end{aligned}$$ We may hence deduce from the last equation that $${U}_{({\ensuremath{\textbf p}},s)({\ensuremath{\textbf p}}',s')}(\Lambda) = \sqrt{\frac{(\Lambda p)^0}{p^0}}D_{s s'}\bigl(
W^{-1}(\Lambda,p)\bigr)\delta({\ensuremath{\textbf p}}'-{\ensuremath{\textbf p}}_\Lambda),$$ and verify by direct calculation, using $D^\dag(W)= [D(W)]^{-1}= D(W^{-1})$, that the transformation $U(\Lambda)$ is indeed unitary i.e. $$\label{e:45}
\int
d^3{\ensuremath{\textbf p}}''\,\,{U}_{({\ensuremath{\textbf p}},s)({\ensuremath{\textbf p}}'',s'')}(\Lambda){U}^\dag_{({\ensuremath{\textbf p}}'',s'')({\ensuremath{\textbf p}}',s')}(\Lambda)
= \delta({\ensuremath{\textbf p}}-{\ensuremath{\textbf p}}')\delta_{s s'}.$$ But with the formula of the infinitesimal Wigner rotation of Eq.(\[e:wrg\]) at hand, it is possible 