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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:

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

Can anybody help?

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The quaternions are represented faithfully in two dimensions by the unit matrix and the Pauli matrices multiplied by the imaginary unit: $i = \sqrt{-1} X$, $j = \sqrt{-1} Y$ and $k = \sqrt{-1} Z$ respectively, thus you only need to write $H$ and $P$ the linear combinations: $H = -\frac{\sqrt{-1} }{\sqrt{2}}(i+k)$ and $P = \frac{1+\sqrt{-1}}{2}(1-k)$

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.

Clarifications

@Knot Log, Sorry I have misled you on two points:

1) The quaternions imaginary units should be represented as $i = \sqrt{-1} X$, $j = \sqrt{-1} Y$, $k = \sqrt{-1} Z$, as they have to square to $-1$ (I have corrected that in the main text) .

2) I forgot to mention that we need to work with the quaternion algebra over the complex field, i.e., we need to distinguish between the complex imaginary unit $\sqrt{-1}$ and the quaternion $i$, (I hope you took care of that in your analysis – in any case I have added the explicit expressions of $H$ and $P$ to the main text. In addition, it is correct that global phases are not important when you use the elements as quantum gates, and you correctly took equivalence classes.

However, Please see the article by Michel Planat , where in section 2.2. he mentions that the group generated by $H$ and $P$ should be of order 192, such that only when you remove a $\mathbb{Z}_8$ center you reach the 24 element Clifford group (I haven't done the work myself).

Moreover, it is possible to generate the $24$ element group directly without additional phases (Please see the lecture notes by Michel Devoret if you start with generators of unit determinant (for example, $R_x(\frac{\pi}{2})$ and $R_z(\frac{\pi}{2})$), because the Clifford group is geometric, as it is isomorphic to the octahedral group, the group of symmetries of the cube or the octahedron and all of its elements are rotations, i.e., with a unit determinant.

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  • $\begingroup$ Thank you for your clarification on the Clifford group / Clifford algebra / quaternions / quaternion group. You stated my question as "What can be said that the Clifford group is isomorphic to a subgroup of invertible elements of the quaternion algebra." Do you have an idea of how to determine these quaternions? $\endgroup$ – Knot Log Dec 11 '18 at 17:04
  • $\begingroup$ Yes, for example, you can write, by inspection: $H = \frac{1}{\sqrt{2}}(X+Z)$, then you can use the quaternion notation if you wish and write $H = \frac{1}{\sqrt{2}}(i+k)$, you can do a similar exercise and express $P$ as a linear combination of $1$ and $Z$. $\endgroup$ – David Bar Moshe Dec 11 '18 at 17:31
  • $\begingroup$ I think this is not correct. If I write $H=\frac{1}{\sqrt{2}}(i+k)$ and $P=\frac{1+i}{2}+\frac{1-i}{2}k$, these two elements 'generate' 48 different quaternions, which does not correspond to the 24 elements of the Clifford group. $\endgroup$ – Knot Log Dec 11 '18 at 21:53
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    $\begingroup$ You are correct. If after obtaining these 48 quaternions, you consider equivalence classes disregarding the minus sign, then there is an isomorphism with the 24 elements of the Clifford group. Thank you! $\endgroup$ – Knot Log Dec 11 '18 at 22:03
  • $\begingroup$ @Knot Log, Sorry I have misled you on two points, I have added a clarification in an update. $\endgroup$ – David Bar Moshe Dec 12 '18 at 7:47

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