Consider the vector space $\mathbb{Z}_2^n$ over the finite field. For vectors $a,b\in \mathbb{Z}_2^n$, we define the bitwise inner product $a\cdot b$ as follows:
$$a \cdot b = \sum_{i=1}^{n} a_ib_i (\text{mod} 2),$$
where $a_i,b_i$ are the $i$-th bits of vectors $a$ and $b$, respectively. In other words, the bitwise inner product $a\cdot b$ represents the number of bits where both vectors $a$ and $b$ have a $1$.
My question is about the nature of a map $f$ to a Hilbert space: $$ f: a \mapsto \left|{f(a)}\right>,$$ with the following properties: $$\lvert \left\langle f(b) \middle| f(a) \right \rangle \rvert^2 = \begin{cases} 0& \text{if } a\cdot b = 1,\\ 1& \text{if } a\cdot b =0. \end{cases}$$
What would such a map be, if the dimension of the Hilbert space is finite and denoted by $D$? It's sufficient if the values of $\lvert \left\langle f(b) \middle| f(a)\right\rangle\rvert^2$ can be categorized based on the value of $a\cdot b$, with the difference in values being at least $1/\text{poly}(n)$.