Hamiltonian simulation for 2 qubit gates

Question

I have been reading this paper, and at the end they have given exact decomposition of Hamiltonian simulation step, where they have decomposed a matrix $A$ into pauli matrices and done the operation $e^{iAt}$.

The decomposition of A is as follows: $A=\frac{1}{4}(I\otimes I +9Z\otimes X+5X\otimes Z+3 Y\otimes Y)$, where I, X, Y and Z are Pauli matrices.

The hamiltonian simulation is $e^{iAt}$ and it's circuit is:

They have given the following description of the circuit:

Another observation is that the commuting terms are the stabilizers of the 2-qubit cluster state [5]. So we attempt to use controlled phase gates to get the correct terms. We can rotate the first qubit about the x -axis by an angle $5\theta$ and the second qubit about the x-axis by angle $9\theta$. The structure of $e^{3i\theta X\otimes X} $ is a x-rotation on the computational basis states {$|00\rangle$, $|11\rangle$} and another on {$|01\rangle$, $|10\rangle$}. A CNOT gate converts these bases into single qubit bases, controlled off the target qubit. Since both implement the same rotation but controlled off opposite values, we can remove the control. The overall circuit is shown in figure 3a, which can be further simplified by combining the two controlled gates at the end as in figure 3b.

Can anyone explain the logic behind this?

Answer 1

So, you want to implement a time evolution $$ e^{iA t}. $$The first thing to note is that $$ e^{iAt}=U^{\dagger}e^{iUAU^\dagger t}U. $$ So, we can choose a $U$ that diagonalises $A$. This works particularly well here because all the terms commute and are the stabilizers of the two-qubit cluster state. For those that know the cluster state, you'll know that controlled-phase is critical in its creation. So, let's try $U$ being controlled phase $$ UAU^\dagger=\frac14(I+9I\otimes X+5X\otimes I+3X\otimes X) $$

Next, because each of the terms commute, $$ e^{it\frac14(I+9I\otimes X+5X\otimes I+3X\otimes X)}=e^{it/4}e^{9it I\otimes X/4}e^{5it X\otimes I/4}e^{3it X\otimes X/4}. $$ So, now, if you look at (b) from your figure, you'll see controlled-phase, X rotation on first qubit by $5\theta$ ($e^{5it X\otimes /4I}$ provided $\theta=t/2$), X rotation on second qubit by $9\theta$ ($e^{9it I\otimes X/4}$), some other gates, and a controlled phase. The "some other gates" must correspond to $e^{3it X\otimes X}$, which you can probably analyse now. I might note that there appears to be an extra factor of 2 floating around somewhere!