Given a Hilbert space: $\mathcal{h}$
Consider representations of the CAR-algebra: $\mathcal{A}_\text{CAR}^{(\prime)}(\mathcal{h})$
In Bratelli & Robinson it is stated the uniqueness: $$\Phi:\mathcal{A}_\text{CAR}(\mathcal{h})\to\mathcal{A}_\text{CAR}'(\mathcal{h})$$
My guess would be: $\Phi[a(f)]:=a'(f)$
Extended to the algebra via: $$\Phi[A+B]:=\Phi[A]+\Phi[B],\,\Phi[\lambda A]:=\lambda\Phi[A]$$ $$\Phi[AB]:=\Phi[A]\Phi[B],\,\Phi[A^*]:=\Phi[A]^*,\Phi[1]:=1$$ Especially, it is an isometry then: $\|\Phi[A]\|=\|A\|$
Similarly for the Weyl algebra: $\mathcal{W}^{(\prime)}(\mathcal{h})$
Why do these construction work?
Ok, I think I got it now...
Both work perfectly fine as they are always nondegenerate: $$\mathcal{A}_\text{CAR}:\quad a(f\neq0)\neq0$$ $$\mathcal{W}:\quad W(f)\neq0$$
A counterexample is provided by the angular momentum algebra: $$\mathcal{J}:\quad [J_i,J_j]=\imath\varepsilon_{ijk}J_k$$ There one has one trivial representation: $J_x=J_y=J_z=0$