Generator of two-qubit quantum gate

Question

I would like to know how to derive the explicit form of the GENERATOR of a general two-qubit gate (also here), e.g., controlled-rotation Y.

From the definition: $$\exp(-i\theta G) \ ,$$

I see it is: $$G=\begin{pmatrix}0&0&0&0\\0&0&0&0\\0&0&0&-i\\0&0&i&0\end{pmatrix} \ ,$$

but, I would like to know how to write it in the form of Pauli Y operator.

Many thanks.

[Update] The question is about if a unitary two-qubit gate is given, how to derive its generator explicitly. For instance, if we look at a Molmer Sorenson gate, we will immediately know its generator is XX or YY or ZZ. For Controlled rotation gate, if we do the calculation $-\log[U]/(i\theta)$, we have the G in the main text. The problem is how to write it in form of Pauli-Y and identity matrix (with tensor product or other operation).

Answer 1

I'm not sure I fully understood your question, but the thing you exponentiate to get a controlled $Y$ gate, $$ \mathsf{CY} = \frac{1}{2} \left( \mathbb{1} \otimes \mathbb{1} + Z \otimes \mathbb{1} + \mathbb{1} \otimes Y - Z \otimes Y \right)$$ is simply $$ G = \frac{1}{4} \left( \mathbb{1} - Z \right) \otimes \left( \mathbb{1} - Y \right),$$ which is the projector onto the $-1$ eigenstates of $Z$ and $Y$ for the respective qubits, so that $$ \mathsf{CY} = e^{- i \pi G}.$$ The matrix representation of this differs from what you wrote in your question, but will generally depend on the choice of basis for the two-qubit system.

To answer your more general question, I suggest looking at the two answers to this question on the QC stack exchange. Any controlled-something gate will certainly involve the Pauli $Z$ on the first qubit.