~~I'm not aware of any proof that the Clifford group + any non-Clifford element gives a universal set of quantum gates. The closest related result that I know is that the Clifford group + any non-Clifford element densely generates an infinite group. (This is theorem 6.5 of Nebe et al. http://arxiv.org/abs/math/0001038v2). But this falls short of proving that the Clifford group + any non-Clifford element is universal, because it is possible you could densely generate a continuous group on $n$ qubits that is not all of $U(2^n)$. So I believe that proving this result remains open.
Note that in the case of the Clifford group on qubits, one can easily show that the Clifford group + any non-Clifford ~~**one-qubit** gate is universal, using the representation theory of $SU(2)$. But the more general statement about adding arbitrary elements to the Clifford group is open.

**Update**: The answer to the first problem is known. In particular, the statement that the Clifford group + any non-Clifford element yields a universal gate set follows from two theorems of Nebe, Rains and Sloane. First, Theorem 6.5 of Nebe et al (http://arxiv.org/abs/math/0001038v2) shows that adding any element to the Clifford group renders it infinite. Second, Corollary 6.8.2 of Nebe et al. (Self-Dual Codes and Invariant Theory (Springer 2006)) states that any infinite group which contains the Clifford group contains all of $SU(d^n)$. Combining these shows that the Clifford group + any non-Clifford element is universal, in any local dimension $d$ (prime or composite). This was previously discussed on stackexchange here: Universal sets of gates for SU(3)?

With regards to rotations by irrational multiples of pi: it's not necessary for your gate set to contain a rotation by an irrational multiple of pi to be universal - for example the gate set $CNOT, H,T$ is universal where $T$ is a Z-rotation by $\pi/8$. All that's required for universality is that the gate set densely generates all rotations. A priori it's possible that even as you take longer and longer products of your gate elements, all the resulting rotations are by rational multiples of pi, but the gate is still universal as the rational rotations are dense in the reals. So I don't believe one can argue that "the gate added to the Clifford group must generate an irrational rotation to be universal". We only know that "the gate added to the Clifford group must *densely* generate and irrational rotation to be universal".

This post imported from StackExchange at 2016-06-20 15:13 (UTC), posted by SE-user Adam Bouland