Exponential representation of single-qubit gates

For arbitrary single-qubit gate $U_{\mathbf{1}}$, we can express it in such a form

$$ U_{\mathbf{1}} = e^{i\alpha}\exp[-i\frac{\theta}{2}\vec{v}\cdot\vec{\sigma}]= e^{i\alpha}\exp[-i\frac{\theta}{2}(xX+yY+zZ)], $$

where $\alpha$ is the global phase, $\theta$ is the rotation angle of $U$ on the Bloch sphere, $\vec{v}=(x,y,z)$ is a real 3-d vector with length equal to 1, and $\vec{\sigma}=(X, Y, Z)$ stand for Pauli matrix vector. This fact can be found in Exercise 4.8 from the Nielson and Chuang's textbook.

Generalization to two-qubit gates

Similarly, I am wondering if we have a conclusion for two-qubit gates like

$$ U_{\mathbf{2}} = e^{i\alpha}\exp[-i\frac{\theta}{2}\vec{v_{\mathbf{2}}}\cdot\vec{\sigma_{\mathbf{2}}}], $$

where I guess $\vec{v_{\mathbf{2}}}$ is a real 15-d vector (as we have 15 non-trivial two-qubit Pauli bases), and $\vec{\sigma_{\mathbf{2}}} = (IX, IY, IZ, XI, XX..., ZZ)$ is the Pauli matrix vector with 15 elements. Does there exist such a generalization? If the answer is no, can we still define rotation angle for arbitrary two-qubit gates?


1 Answer 1


I think what you want is the KAK decomposition. An available implementation is cirq.kak_decomposition.

The KAK decomposition decomposes an arbitrary two-qubit unitary gate $U_2$ into single qubit gates and same-basis Pauli coupling terms:

$$U_2 = (A \otimes B) \cdot (X \otimes X)^x \cdot (Y \otimes Y)^y \cdot (Z \otimes Z)^z \cdot (C \otimes D) e^{i\theta}$$

where $A, B, C, D \in \mathbb{SU}(2)$ and $x, y, z, \theta \in \mathbb{R}$.

For two-qubit Clifford gates the single qubit gates will be Cliffords and the coupling terms can be simplified to the presence or absence of a CZ and a SWAP. So two-qubit Cliffords gates can be broken down into ones that are CZ-like, SWAP-like, CZSWAP-like, and identity-like.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.