]{}Suppose $U(t) = cH$ for some scalar $c$ in ${{\mathbb C}}$ and some $H$ in ${{\mathbb C}}^{n \times n}$ such that each entry of $H$ is algebraic. By a well-known result from linear algebra, we know that for the complex matrix $itA$ we have $$\det(e^{itA}) = e^{\operatorname{tr}(itA)} = 1.$$ If we assume that $U(t) = cH$, then this implies that $$c^n \det(H) = 1.$$ Since $H$ has algebraic entries, it follows that $\det(H)$ is algebraic. This implies that $c$ is algebraic, which further implies that all of the entries of U(t) are algebraic.[  ]{}

Complex Hadamards and SRGs {#srg_section}
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Let $X$ be a strongly regular graph with parameters $(n,k,{\lambda},\mu)$ and let $A$ be its adjacency matrix, with spectral decomposition $$A = kE_k + \theta E_\theta + \tau E_\tau,$$ where $k > \lambda \geq \tau$.

The following result of Chan [@chan08] is instrumental in our classification of strongly regular graphs that admit uniform mixing.

\[thm:chan\] Let $A$ be the adjacency matrix of a primitive strongly regular graph $X$ with eigenvalues $k$, ${\theta}$, $\tau$, and $W = I + xA +y\bar{A}$. If $W$ is a complex Hadamard matrix, then $X$ or $\bar{X}$ has one of the following parameter sets $(n,k,{\lambda},\mu)$:

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