# Why are two bases mutually unbiased if and only if $\operatorname{tr}(A^m B^n)=N \delta_{m0}\delta_{n0}$?

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)?

• inverse Fourier transform? Commented Sep 24, 2018 at 17:05

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 $$s,t$$.