Anybody have example of two-qubit non-Pauli and non-Clifford quantum gate? A lot of known quantum gates are in the Pauli group (I,X,Z,Y) or in the Clifford group (H,P,Cnot). I need examples of the quantum gates that aren't in this groups. Also, are there are matlab functions to check if a quantum gate (2x2 or 4x4) is in any of these groups? Or maybe there are matlab routines that generate quantum gates outside of this groups?
Thanks..
 A: Any gate of the form diag$(1,1,1,\exp(i\phi))$ is not in $C_n$ for any $n$ unless $\phi = 2\pi k/2^n$ for some integers $k$ and $n$.  This can be proven by induction using the similar result for single-qubit gates.  I'm not sure if this is included in any published paper.
We don't have a good characterization of gates in $C_n$ for $n > 2$, so there is no known more general method of generating them, or even checking if a gate has this property.
A: The Pauli and Clifford groups only contain finitely many elements, so almost any unitary is not going to be in them.
Just ask matlab to make you a random unitary.  For example, almost any one qubit phase gate is not in these groups.
I'm not aware of a matlab function that checks for membership in these groups.  However, you could write a simple code for the small gate sizes you mentioned.  Since elements $U$ of the Clifford group satisfy $U (Pauli) U^\dagger = (Pauli)'$ you could run through all Pauli operators and make sure they are mapped to each other e.g. by computing the overlap of operators using something like the matrix inner product $(M,N) = tr(M^\dagger N)$ since one has $(\sigma^a, \sigma^b) = tr(\sigma^a \sigma^b) = 2 \delta^{ab}$.
There is probably a better way, but this silly algorithm should work if you only care about 2x2 and 4x4 gates.
A: Precisely because the Clifford group is generated by the operators
$$ S = \begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix} \qquad H = \tfrac{1}{\sqrt 2} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \qquad \mathbf{cnot} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix} $$
and tensor products with the identity, it follows that every Clifford operator is of the form $2^{-n/2} M$, where $M$ is a matrix over the Gaussian integers (i.e. complex numbers where the real and imaginary parts are both integers). Any unitary which is not of this form is therefore not a Clifford group operator.
