Super mixed Hodge structures? It's common in subjects that have some version of the "yoga of weights" that you have a functor called "Tate twist" and that the most natural version of it seems like it should be the square of something which obviously doesn't exist.  For example, in the category of mixed Hodge structures, there's the natural Tate Hodge structure $\mathbb{Q}(k)$ (which is, for example, the Hodge structure on the top degree of a smooth $k$-dimensional projective variety).   You'd like there to be a Hodge structure $\mathbb{Q}(\frac{1}{2})$ such that $\mathbb{Q}(\frac{1}{2})^{\otimes 2k}=\mathbb{Q}(k)$, but there's no way that such a thing could exist in the standard definition of mixed Hodge structures, because Hodge structures of pure odd weight must be even dimensional.
A way of fixing this is suggested by Beilinson, Ginzburg and Soergel - Koszul Duality Patterns in Representation Theory on page 514: in essence, you consider two copies of the category of mixed Hodge structures with one of them formally Tate twisted by $(\frac{1}{2})$.
I was writing a paper that used this idea, and thought of an explanation of this I like quite a bit better, and now am wondering if I'm really the first to have thought of it.
The idea of this:  you think of the two vector spaces in MHS as the even and odd parts of a super vector space (i.e. a $\mathbb{Z}/2$-graded vector space).  The definition of a Hodge structure on an even vector space is the usual one; the definition of a Hodge structure on a purely odd vector space is the usual one, but now the Hodge filtration is indexed by elements of $\mathbb{Z}+\frac{1}{2}$.
So, for example, $\mathbb{Q}(\frac{1}{2})$ is pure of weight 1, with $F_{-1/2}=\mathbb{C}$ and $F_{1/2}=0$ (with the purity condition being exactly that these guys are transverse).

Has anyone encountered this approach before?  Once I saw it, it seemed like obviously the right thing to do.

 A: I am not sure this is relevant to you, and you may know this already, but in the theory of monodromic ("exponential") mixed Hodge structures, there is a very natural (even) square root of the Tate motive. See Kontsevich-Soibelman's Cohomological Hall algebra, exponential Hodge structures and motivic Donaldson–Thomas invariants, Section 3.4: Square root of Tate motive. Most papers on DT theory will use this, see e.g. Davison-Meinhardt, Cohomological Donaldson–Thomas theory of a quiver with potential and quantum enveloping algebras, Section 2.1 for a summary.
A: If you introduce super-structure, it doesn't matter whether $\mathbb{C}(1/2)$ is even or odd. The reason is that your category of motives has a symmetric monoidal automorphism which acts by $-1$ on $\mathbb{C}(1/2);$ this implies that after introducing super-structure, there is a symmetric monoidal auto-equivalence which takes $\mathbb{C}(1/2)$ to the supershift $\Pi \mathbb{C}(1/2).$
More informally, I would suspect that's not the right thing to do. The reason is that there's nothing "super" about cohomology of varieties: the tensor product on motives compatible with cohomology is the usual one, modified by cohomological degree in the usual way. It sounds like Beilinson's $\mathbb{C}(1/2)$ is a cohomological degree-zero object, so it should behave commutatively. Another reason I think this isn't the right thing to do is that in the world of super-schemes, while motives will tend to "kill" the super-structure, you might still want to look at things like the Hodge-to-de Rham degeneration which will have genuinely "super" pieces. The fact that
$$\mathbb{A}^{0\mid 1}\times \mathbb{A}^{0\mid 1} \cong \mathbb{A}^{0\mid 2}$$
should intuitively mean that these super pieces correspond to a new "purely odd" piece of the cohomology rather than a square root of the Tate motive.
