is therefore a stochastic process.

In a continuous-time quantum walk, instead of a probability distribution $p(t)$ we use a quantum state ${|\psi(t)\rangle}$ defined on a state space with orthonormal basis $\{{|1\rangle},
\ldots, {|n\rangle} \}$. Instead of a stochastic process, the walk is governed by a unitary evolution. From Schrödinger’s equation for the evolution of a quantum system $$i \frac{\mathrm{d}{|\psi(t)\rangle}}{\mathrm{d}t } = L {|\psi(t)\rangle},$$ we have the solution $${|\psi(t)\rangle} = e^{iLt}{|\psi(0)\rangle}.$$ Since the Laplacian $L$ is Hermitian, $e^{iLt}$ is unitary. Assume our quantum walk begins at some vertex $j$, so ${|\psi(0)\rangle} = {|j\rangle}$. If at some time $t$ we measure the state ${|\psi(t)\rangle}$ in the vertex basis $\{{|1\rangle}, \ldots, {|n\rangle} \}$, then the probability of outcome ${|k\rangle}$ is $${\left|{\langlek|\psi(t)\rangle}\right|}^2 = {\left|{\langlek|}e^{iLt}{|j\rangle}\right|}^2 =  {\left|(e^{iLt})_{kj}\right|}^2.$$ If this probability distribution is uniform over all vertices $k$, and this is true for all initial states ${|j\rangle}$, then the graph is said to have uniform mixing.

In fact we will define the unitary evolution using the adjacency matrix rather than the Laplacian. When $X$ is regular, the resulting states differ only by an overall phase factor, and this does not affect uniform mixing.

In this paper we focus on the question of which graphs have uniform 