Can we write CNOT gate in terms of Quaternion?

A quaternion can be represented as $$a + bi + cj + dk$$ where a, b, c, d, are real numbers, and i, j, k, are symbols that can be interpreted as unit-vectors pointing along the three spatial axes. . I am wondering what is the representation of CNOT gate in terms of quaternions. Since it is controlled-X gate, it should be something with $$b$$. $$cj$$ and $$dk$$ should be 0. What about a?

• Quaternion lies in $\mathbb{R}^4$, which is isomorphic to $\mathbb{C}^2$. They can describe the information of a single 2-level system. To convey information about 2 qubits, you will need 4 Quaternions. Operations on two qubits lie in the operator space of $\mathbb{C}^4$, $\mathcal{B}(\mathbb{C}^4)$, which is $\mathbb{C}^{16}$, which is isomorphic to $\mathbb{R^{32}}$. So technically, I think it should be possible to convey information about operations on 4-level systems with 8 Quaternions, using some arbitrary mathematical model of description. Commented Sep 28, 2023 at 22:10
• But it seems very inelegant. Commented Sep 28, 2023 at 22:12

Quaternion lies in $$\mathbb{R}^4$$, which is isomorphic to $$\mathbb{C}^2$$. They can describe the information of a single 2-level system. To convey information about 2 qubits, you will need 4 Quaternions. Operations on two qubits lie in the operator space of $$\mathbb{C}^4$$, $$\mathcal{B}(\mathbb{C}^4) = (\mathbb{C}^4)^{\otimes2}$$, which is $$\mathbb{C}^{16}$$, which is isomorphic to $$\mathbb{R^{32}}$$.
For describing 3 qubits ($$\mathbb{C}^8 \cong \mathbb{R}^{16}$$), you will need 4 quaternions and operations on 3 qubits, i.e., 3 qubit gates ($$\mathbb{C}^{64} \cong \mathbb{R}^{128}$$) will require 32 quaternions.