Let $\{A_n\}$ and $\{B_n\}$ be two bases for an $N$-dimensional Hilbert space. Does there exist a unit vector $V$ such that:
$$(V\cdot A_j)\;(A_j\cdot V) = (V\cdot B_j)\;(B_j\cdot V) = 1/N\;\;\; \ \text{for all} \ 1\le j\le N?$$
Notes and application:
That the $\{A_n\}$ and $\{B_n\}$ are bases means that
$$(A_j\cdot A_k) =\left\{\begin{array}{cl}
1&\;\text{if }j=k,\\
0&\;\text{otherwise}.\end{array}\right.$$
In the physics notation, one might write $V\cdot A_j = \langle V\,|\,A_j\rangle$. In quantum mechanics, $P_{jk} = |\langle A_j|B_k\rangle|^2$ is the "transition probability" between the states $A_j$ and $B_k$. "Unbiased" means that there is no preference in the transition probabilities. A subject much studied in quantum information theory is "mutually unbiased bases" or MUBs. Two mutually unbiased bases satisfy
$|\langle A_j|B_k\rangle|^2 = 1/N\;\;$ for all $j,k$.
If it is true that the vector $V$ always exists, then one can multiply the rows and columns of any unitary matrix by complex phases so as to obtain a unitary matrix where each row and column individually sums to one.
If true, then $U(n)$ can be written as follows:
$$U(n) = \exp(i\alpha)
\begin{pmatrix}1&0&0&0…\\0&e^{i\beta_1}&0&0…\\0&0&e^{i\beta_2}&0…\end{pmatrix}
M
\begin{pmatrix}1&0&0&0…\\0&e^{i\gamma_1}&0&0…\\0&0&e^{i\gamma_2}&0…\end{pmatrix}$$
where the Greek letters give complex phases and where $M$ is a "magic" unitary matrix, that is, $M$ has all rows and columns individually sum to 1.
And $M$ can be written as $M=\exp(im)$ where $m$ is Hermitian and has all rows and columns sum to 0. What's significant about this is that the $m$ form a Lie algebra. Thus unitary matrices can be thought of as complex phases, plus a Lie algebra. This is a new decomposition of unitary matrices.
Since $m$ is Hermitian and has all rows and columns sum to 0, it is equivalent to an $(n-1)\times(n-1)$ Hermitian matrix with no restriction on the row and column sums. And this shows that $U(n)$ is equivalent to complex phases added to an object (the $M$ matrices) that is equivalent to $U(n-1)$. This gives a recursive definition of unitary matrices entirely in terms of complex phases.
Best Answer
I believe the answer to be yes, and it follows by some symplectic geometry of Lagrangian intersections.
Let $U$ be the unitary matrix so that $B_j = U A_j$. Without loss of generality, we will also assume that $A_j = e_j$. This means that $B_j = U e_j$.
We will identify $\mathbb C^N = \mathbb R^{2N}$.
Then, the first condition on the vector $V$ is that: $$ |(V, e_j)|^2 = \frac{1}{N}, j=1, \dots, N $$ This is equivalent to saying that $V = \frac{1}{\sqrt{N}} \sum \mathrm e^{i \theta_j} e_j$, or, in other words, that $V$ lies in the Lagrangian torus in $\mathbb R^{2N}$ with the standard symplectic structure $\sum dx_j \wedge dy_j$, defined by $\{ |x_j|^2 + |y_j|^2 = \frac{1}{\sqrt{N}} \}$.
The second condition on $V$ is that $$ |(V, U e_j)|^2 = \frac{1}{N}. $$ Thus, $U^* V$ also should lie in the torus $L$. Thus, the vector $V$ exists if and only if $L \cap UL$ is non-empty.
(Note the first condition gives automatically that $V$ is a unit vector.)
Right now, I don't see how to take advantage of the linearity in this problem, so I will use an incredibly high powered theory (Floer theory). If I think of a better solution, I will update.
Notice that the action of $U$ on $\mathbb C^N$ induces a map on $\mathbb CP^{N-1}$. Furthermore, if we write $U=\mathrm{e}^{iH}$ for a Hermitian $H$, then $U$ is the time-1 map of the Hamiltonian flow generated by the Hamiltonian $$h(v) = \frac{1}{2} \Re (v, Hv).$$
Finally, we note that $L$ projects to the Clifford torus $L'$ in $\mathbb CP^{N-1}$. It is known for Floer theoretic reasons (not sure who first proved it... there are now many proofs in the literature) that the Clifford torus is not Hamiltonian displaceable, so there must always exist an intersection point. After normalizing, this lifts to an intersection point in $\mathbb C^N$, as desired.
Note that the Floer homology argument is a very powerful tool. I suspect that a much simpler proof can be found, since this doesn't use the linear structure.
EDIT: Apparently my use of the term "Clifford torus" is non-standard. Here is what I mean by it: Consider $\mathbb CP^{N-1}$ as the quotient of the unit sphere in $\mathbb C^{N}$ by the $S^1$ action by multiplication by a unit complex number (as we have defined here). In the unit sphere there is a real $N$ dimensional torus given by $|z_1| = |z_2| = \dots = |z_N| = \frac{1}{\sqrt{N}}$. The image of this $N$ dimensional torus by the quotient map is an $N-1$ dimensional torus in $\mathbb CP^{N-1}$. Equivalently, it is the torus given in homogeneous coordinates on $\mathbb CP^{N-1}$ by $[e^{i \theta_1}, e^{i \theta_2}, \dots, e^{i \theta_{N-1}}, 1]$.