We are given a set of $d$ normalized vectors on a $d$-dimensional complex vector space: $e_1$, $e_2$... $e_d$, where $$\langle e_j,e_j\rangle=1$$ for all $j$. These are not necessarily mutually orthogonal or linearly independent.
The question is: Can we always find $(d-1)$ unitary matrices, call them $U_1\dots U_{d-1}$, that together with $U_0:=1$ (the identity) have the following property: $$\langle U_ke_j,U_le_j\rangle=\delta_{kl}$$ for all $j,k,l$?
There are specific examples where I know it is possible:
- If all the $e_j$'s are the same, say $e_j=e$: let $f_1,f_2\dots f_d$ be an orthonormal basis spanning the space, with $f_1=e$. Then just construct unitary matrices $U_j$ such that $U_{j-1}f_1=f_j$.
- If the $e_j$'s are orthonormal, then just construct $U_1$ with the property that $U_1e_j=e_{j+1}$ for all $j$, and then use $$U_j=U_1^j.$$ Here $j+1$ is meant modulo $d$.
- If $d=2$: I don't want to go into the details here, but there's a pretty straightforward construction that works for any $d=2$ case.
Sanity check: Each $d$-dimensional unitary matrix has $d^2$ real degrees of freedom. Therefore, in this case, the total number of degrees of freedom is $d^2(d-1)$. The total number of real constraint equations, on the other hand, is (the number of pairs $(k,l)$)$\times$(the number of $j$'s for each pair)$\times2=\left(\frac{d(d-1)}2\right)\times d\times2=d^2(d-1)$.
This naive constrint-counting does not rule out the possibility that the answer to my question is "yes", but it does make it look quite unlikely, because if the answer is yes, then the satisfying unitaries are completely determined!
But I don't know the answer in general. If you're wondering why I have this question, it occurs in the course of my research in quantum information theory. If the question is answered in the affirmative, then there is a cute little result that I can include in my upcoming paper. If I get an answer here, I will acknowledge this group's help in my paper! :)
Thanks!
I believe the answer to your question is negative when no further restrictions on $\{e_j\}$ are imposed.
A necessary condition for $\langle U_ke_j, U_le_j \rangle = \delta_{k,l}$ is that there exist $d$ subspaces $S_j \subset \mathbb{C}^d$, $j=1,\cdots,d$, that are mutually orthogonal, i.e. $\mathcal{R}(S_k)\perp\mathcal{R}(S_l)$, $k \neq l$, where $\mathcal{R}(S)$ is the range of (sub)space $S$. Clearly this can only be possible if $\{S_j\}$ are one-dimensional, so the question becomes whether the elements of a one-dimensional subspace, say, $\mathcal{S}_1$, can be written as $Ue$, where $U$ is unitary and $e$ arbitrary. This is impossible since $U$ represents a basis for the $d$-dimensional space $\mathbb{C}^d$ and, therefore, there exist $e$ such that $Ue \notin \mathcal{S}_1$.