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This question can be formulated in two ways. Let there be two $d$-dimensional orthonormal bases $B_{1}$ and $B_{2}$. I refer to the elements of $B_{1}$ by $\lvert\nu_{i}\rangle$ and to the elements of $B_{2}$ by $\lvert\omega_{j}\rangle$. These two bases are mutually unbiased namely: $\lvert\langle\omega_{j}\lvert\nu_{i}\rangle\lvert^{2}=1/d$ for all $\lvert\omega_{j}\rangle$ and $\lvert\nu_{i}\rangle$. We define some vector $\lvert\tau\rangle$ such that $\lvert\tau\rangle$ is orthonormal to one element of $B_{2}$, say $\lvert\omega_{m}\rangle$ and it is mutually unbiased with respect to all the elements in $B_{1}$: $\lvert\langle\tau\lvert\nu_{i}\rangle\lvert^{2}=1/d$ for all $\lvert\nu_{i}\rangle$. Can we conclude that $\lvert\tau\rangle$ is equal to a member of $B_{2}$ say $\lvert\omega_{p}\rangle$ for some $p\neq m$ up to a phase difference?

Another way to make the same conclusion is to ask this: Let there be three orthonormal bases $B_{1}$, $B_{2}$ and $B_{3}$. $B_{1}$ and $B_{2}$ are mutually unbiased, so are $B_{2}$ and $B_{3}$. Can we conclude that $B_{1}$ and $B_{3}$ are either mutually unbiased or have equivalent Hadamard matrices? This question for me is particularly important since in the literature it is always talked about sets of pairwise mutually unbiased bases. But there's no proof given for the fact that three mutually unbiased bases should necessarily be pairwise MUBs.

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The answer to your first question is no.

Consider the three-dimensional Hilbert space $\mathbb C^4$, and let $B_1$ be the canonical basis and $$ B_2=\left\{ \frac{1}{2}\begin{pmatrix}1\\1\\1\\1\end{pmatrix}, \frac{1}{2}\begin{pmatrix}1\\ i\\-1\\-i\end{pmatrix}, \frac{1}{2}\begin{pmatrix}1 \\ -1\\1\\-1\end{pmatrix}, \frac{1}{2}\begin{pmatrix}1\\ -i\\-1\\i\end{pmatrix} \right\}. $$ Then the vector $| \tau \rangle=\tfrac12(1,i,-i,-1)^{T}$ is unbiased with respect to $B_1$, orthogonal to the first vector of $B_2$, and not present in $B_2$ even up to a phase.

This same trick can be used to produce a third basis, $$ B_3=\left\{ \frac{1}{2}\begin{pmatrix}1\\1\\1\\1\end{pmatrix}, \frac{1}{2}\begin{pmatrix}1\\ i\\-i\\-1\end{pmatrix}, \frac{1}{2}\begin{pmatrix}1 \\ -1\\-1\\1\end{pmatrix}, \frac{1}{2}\begin{pmatrix}1\\ -i\\i\\-1\end{pmatrix} \right\}, $$ which is mutually unbiased with respect to $B_1$, but has no such relation with $B_2$.

On the other hand, this third basis is obviously related to $B_2$ by a simple permutation. I'm not sure what sort of equivalences you're allowing for the Hadamard matrices, which I presume you mean to be $H=\sum_j| \omega_j \rangle\langle \nu_j |$. If you require that $H'\sim h$ if and only if $H'=U^\dagger H U$, then I am not sure but I suspect the Hadamard matrices of $B_2$ with $B_1$ and $B_3$ with $B_1$ are not equivalent. If you allow relations of the form $H'=U H$ then you're in dangerous territory as any basis is reachable in this way.

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