is a $(2{\theta}+1)$-th root of unity. Since ${\left|{\mathrm{Re}}(x)\right|}$ is too small, this never occurs. Thus the only conference graph having uniform mixing is the $3 \times 3$ Latin square.

The combination of the work above proves Theorem \[srg\_theorem\].

Bipartite graphs {#bipartite_graphs}
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Next we consider bipartite graphs. Assume $X$ is bipartite on $n$ vertices with adjacency matrix $A$ such that $$A =\begin{pmatrix} 0&B\\ B^T&0 \end{pmatrix}$$ Then $$A^{2k} = {\begin{pmatrix} (BB^T)^k&0\\ 0&(B^TB)^k \end{pmatrix}},$$ $$\qquad A^{2k+1} = {\begin{pmatrix} 0&(BB^T)^kB\\ (B^TB)^kB^T&0 \end{pmatrix}},$$ and $$U(t) = e^{itA} = {\begin{pmatrix} \cos(tBB^T)& i\sin(tBB^T)B\\ i\sin(B^TB)B^T& \cos(tB^TB) \end{pmatrix}}.$$ To simplify notation we write $$U(t) = {\begin{pmatrix} C_1(t)& iK(t)\\ iK^T(t)&C_2(t) \end{pmatrix}}.$$ What we are seeing in this expression for $U(t)$ is a reflection of the fact that any normal matrix can be written as a sum $C+iK$, where $C$ and $K$ are commuting Hermitian matrices; this is relevant because $U(t)$ is normal. ($2N = (N+N^*) +i(-iN+iN^*)$)

We immediately rederive Kay’s result [@Kay11] concerning the phase factors for perfect state transfer in bipartite graphs.

If $X$ is a bipartite graph on $n > 2$ vertices that admits uniform mixing, then $n$ is divisible by four.

[*Proof.* ]{}Now suppose $U(\tau)$ is flat. Using the notation above, we see that all entries 