Practical implementation of Hamiltonian Evolution

Question

Following from this question, I tried to look at the cited article in order to simulate and solve that same problem... without success. Mainly, I still fail to understand how the authors managed to simulate the Hamiltonian evolution through the circuit shown at the bottom of Fig.4. Even exponentiating classically the matrix I do not get values of the gates shown in the Quirk circuit that @Blue linked along his question.

I tried to look at the paper in which the Group Leader Optimization algorithm is explained, but I still have troubles understanding how do they assign the rotation angles to the different gates.

Answer 1

I don't know why/how the authors of that paper do what they do. However, here's how I'd go about it for this special case (and it is a very special case):

You can write the Hamiltonian as a Pauli decomposition $$ A=15\mathbb{I}\otimes\mathbb{I}+9Z\otimes X+5X\otimes Z-3Y\otimes Y. $$ Update: It should be $+3Y\otimes Y$. But I don't want to redraw all my diagrams etc., so I'll leave the negative sign.

Now, it is interesting to note that every one of these terms commutes. So, that means that $$ e^{iA\theta}=e^{15i\theta}e^{9i\theta Z\otimes X}e^{5i\theta X\otimes Z}e^{-3i\theta Y\otimes Y}. $$ You could work out how to simulate each of these steps individually, but let me make one further observation first: these commuting terms are the stabilizers of the 2-qubit cluster state. That may or may not mean anything to you, but it tells me that a smart thing to do is apply a controlled-phase gate. $$ CP\cdot A\cdot CP=15\mathbb{I}\otimes\mathbb{I}+9\mathbb{I}\otimes X+5X\otimes \mathbb{I}-3X\otimes X. $$ (You may want to check the sign of the last term. I didn't compute it carefully.) So, if we start and end our sequence with controlled-phase gates, then 2 of the terms are easy to get right: we rotate the first qubit about the $x$ axis by an angle $5\theta$, and the second qubit about the $x$ axis by an angle $9\theta$.

The only thing we are left to get right is the $X\otimes X$ rotation. If you think about the structure of $e^{-3i\theta X\otimes X}$, this is like an $x$ rotation on the basis states $\{|00\rangle,|11\rangle\}$, and another one on $\{|01\rangle,|10\rangle\}$. A controlled-not converts these bases into the single-qubit bases controlled off the target qubit. But since both implement the same controlled-rotation but controlled off opposite values, we can just remove the control. Thus, the overall circuit is: This simplifies slightly by combining the two controlled-gates at the end: Note that I have not included the global phase term here because that's not the sensible way to do it. When you make controlled-($e^{iA\theta}$), you apply the "global phase" as a $z$ rotation on the control qubit.