$t$, uniform mixing occurs on $X$ if and only if it occurs on the complement $\overline{X}$.

[*Proof.* ]{}Note that $$\begin{gathered}
    e^{i\bar{A}t} = e^{i(J-I-A)t} = e^{i(J-I)t}e^{-iAt} \\ =        
        \left(e^{i(n-1)t}\left(\frac{1}{n}J\right) +    
        e^{-it}\left(I-\frac{1}{n}J\right)\right)e^{-iAt}.\end{gathered}$$ If $t$ denote an integer multiple of $2 \pi/n$, then the above equation reduces to $$e^{i\bar{A}t} = e^{-it}e^{iAt},$$ which implies that the complement of $X$ has uniform mixing if and only if $X$ does.[  ]{}

In Section \[srg\_section\], we see that the previous lemma applies to all cases of strongly regular graphs that admit uniform mixing.

Recall that an $n \times n$ matrix $H$ is a complex Hadamard matrix if and only if

(i) [$H H^{*} = nI.$]{}

(ii) [$H \circ {{\mkern2mu\overline{\mkern-2muH}}} = J.$]{}

With this terminology we see that uniform mixing occurs on $X$ at time $t$ if and only if $\sqrt{n} U(t)$ is a complex Hadamard matrix.

For the known examples of uniform mixing, the corresponding entries of $\sqrt{n}U(t)$ are roots of unity. Such matrices are called *Butson-type*. They live in a more general family of complex Hadamard matrices whose entries are all algebraic numbers. In general, we see that if $U_X(t)$ has all algebraic entries, then the ratio of any two eigenvalues of $X$ must be transcendental. 