Explicit example of a linear map $\alpha: \ell^2 \to B(\ell^2)$ satisfying the canonical anti-commutation relations? I've been trying to understand the CAR algebra - in particular, I'm trying to make sense of how one can view the CAR algebra as the $2^\infty$ UHF algebra.
I'd like to have an explicit example of a function $\alpha: \ell^2 \to B(\ell^2)$ to play around with. (Since I know sometimes notation can vary in textbooks, here $\ell^2$ refers to sequences $x := (x_n)$ such that $|| x ||_2 = \sum_n |x_n|^2 < \infty$. Further, $B(\ell^2)$ refers to the bounded linear maps on $\ell^2$, i.e. the infinite square matrices $A$ such that $\sum_n |(Ax)_n|^2 < C||x||_2$ for some $C > 0 \in \mathbb{R}$.).
So, is there a standard / typical / straightforward example of a linear map $\alpha: \ell^2 \to B(\ell^2)$ such that for all $x,y \in \ell^2$,

*

*$\alpha(x)\alpha(y)+\alpha(y)\alpha(x)=0$, and

*$\alpha(x)^*\alpha(y) + \alpha(y)\alpha(x)^* = \langle y,x\rangle I$?

Thanks!
 A: This is essentially the only one I've seen.
This one can be found in chapter 5.2 of Bratteli & Robinsons's Operator algebras and quantum statistical mechanics, volume 2, although I'll use my own notation (I don't think they use wedges, but using them is a huge help to me) . They essentially take the "anti-symmetric Fock space", which can be viewed as the infinite direct sum of "wedge products" of Hilbert spaces:
$$\mathcal{F}_-(\mathcal{H}) = \bigoplus_{n=0}^{\infty} \mathcal{H}^{\wedge n}. $$
I only use the wedge here since I think it aligns with the wedge product in abstract algebra - I won't get into the details, its essentially an "anti-symmetric tensor product". $\mathcal{H}^{\wedge n}$ is the closed linear span of "wedges" $x_1 \wedge \cdots \wedge x_n$ which lie in $\mathcal{H}^{\otimes n}$. Explicitly, one can view these as
$$x_1 \wedge \cdots \wedge x_n = \frac{1}{\sqrt{n!}}\sum_{\sigma \in \mathfrak{S}_n} \text{sgn}(\sigma)x_{\sigma(1)} \otimes \cdots \otimes x_{\sigma(n)}, $$ where $\mathfrak{S}_n$ is the premutation group on a set of $n$ elements. I won't go further into it (besides the inner product at the end).
If you start with a separable infinite-dimensional $\mathcal{H}$, one can take $\alpha: \mathcal{H} \to B(\mathcal{F}_-(\mathcal{H})) \simeq \mathcal{B}(\mathcal{H})$ to be the creation operator: if $h \in \mathcal{H}$, take
$$ \alpha(h)(x_1 \wedge\cdots \wedge x_n) = h \wedge x_1 \wedge \cdots \wedge x_n; $$
we then extend by linearity and continuity to a see that $\alpha(h)$ can be defined on all of $\mathcal{F}_-(\mathcal{H})$.
Using the explicit form of the wedges, one can go through a difficult computation to show that $\alpha$ satisfies the canonical anti-commutation relations (I guess after passing through any isomorphism $B(\mathcal{F}_-(\mathcal{H})) \simeq \mathcal{B}(\mathcal{H})$).
Just a remark on what the inner product is on these wedge products (this is around in some functional analysis books that I can't remember, but I don't think its too hard to prove once you've got the right determinant formula): for two n-wedges,
$$ \langle x_1 \wedge \cdots \wedge x_n, y_1 \wedge \cdots \wedge y_n \rangle = \det\left((\langle x_i,y_j\rangle)_{i,j}\right). $$
