Let $\{\lvert a_j\rangle\}$ and $\{\lvert b_k\rangle\}$ be a pair of mutually unbiased bases (MUBs): $$\lvert\langle a_i\rvert b_j\rangle\rvert=1/\sqrt N.\tag A $$

Let us consider the unitary operators $A$ and $B$ defined as the cyclic operators whose eigenvectors are the elements of the bases: $$A\lvert a_j\rangle=\gamma_N^j \lvert a_j\rangle, \quad B\lvert b_k\rangle=\gamma_N^k \lvert b_k\rangle,$$ where $\gamma_N^j\equiv \exp(2\pi ij/N)$, and thus $A^N=B^N=I$.

It turns out (see for example this review at (1.3), pag. 7) that (A) is equivalent to: $$\operatorname{tr}(A^m B^n)=N \delta_{m0}\delta_{n0}.\tag B$$

By explicitly writing the trace we get $$\operatorname{tr}(A^m B^n)=\sum_{jk} \gamma_N^{jm+kn}\lvert\langle a_j\rvert b_k\rangle\rvert^2,$$ from which it follows that (A) implies (B), using $\sum_j \gamma_N^{jm}=N\delta_{m,0}$.

What is an easy way to show the opposite implication, that is, that (B) implies (A)?


1 Answer 1


If we multiply (B) by $\gamma_N^{-ms-nt}$ we get: $$\sum_{jk}\gamma_N^{jm+kn-ms-nt}\lvert\langle a_j\rvert b_k\rangle\rvert^2=N \gamma_N^{-ms-nt} \delta_{m0}\delta_{n0}.$$ Summing over $m$ and $n$ we get on the LHS $$\sum_{jkmn}\gamma_N^{jm+kn-ms-nt}\lvert\langle a_j\rvert b_k\rangle\rvert^2= \sum_{jk}\left(\sum_{mn}\gamma_N^{m(j-s)+n(k-t)}\right)\lvert\langle a_j\rvert b_k\rangle\rvert^2 =N^2 \lvert\langle a_s\rvert b_t\rangle\rvert^2,$$ while the RHS simply gives $N$. This directly implies $\lvert\langle a_s\rvert b_t\rangle\rvert=1/\sqrt N$ for all $s,t$.


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