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How do I find an explicit isomorphism between the elements of the Clifford group and some 24 quaternions?

The easy part:
The multiplication of matrices should correspond to multiplication of quaternions.

The identity matrix $I$ should be mapped to the quaternion $1$.

The hard part:
To what should the other elements of the Clifford group be mapped? Since the following to elements generate the entire group, mapping these will be sufficient:

$$H=frac{1}{sqrt{2}}begin{bmatrix}1&1\1&-1end{bmatrix}text{ and }P=begin{bmatrix}1&0\0&iend{bmatrix}$$

Can anybody help?

The quaternions are represented faithfully in two dimensions by the unit matrix and the Pauli matrices $X$, $Y$ and $Z$ respectively, thus you only need to write $H$ and $P$ as linear combinations Pauli matrices, which I presume you know how to do.

However, this is not an isomorphism between the Clifford group and the quaternions, because here we use the quaternions as an algebra not a group. What can be said that the Clifford group is isomorphic to a subgroup of invertible elements of the quaternion algebra.

The term Quaternion group is reserved to another subgroup of invertible elements of the quaternion algebra consisting of $pm 1$, $pm (-iX = R_x(pi))$, $pm (-iY = R_y(pi))$ and $pm (-iZ = R_z(pi))$. This group is called the quaternion group. This group can be generated by $pi$ rotations around two major axes. However, the order-8 quaternion group is not isomorphic to the order-24 Clifford group, which can be generated by $frac{pi}{2}$ rotations around two major axes. The Clifford group is in a certain sense the square root of the quaternion group.