Context and definitons

I want to study the evolution of a three qubit Heisenberg Hamiltonian: $$H=(I\otimes X\otimes X)+(X\otimes X\otimes I)+(I\otimes Y\otimes Y)+(Y\otimes Y\otimes I)+(I\otimes Z\otimes Z)+(Z\otimes Z\otimes I)$$ where $X,Y$ and $Z$ are the Pauli gates.

For that I trotterize the evolution operator as $$U=\exp(-\mathrm i H t)\approx[\exp(-\mathrm i H t/n)]^n\approx[\exp(-\mathrm i H^{(1,2)} t/n)\exp(-\mathrm i H^{(0,1)} t/n)]^n $$ for a large $n$, where $H^{(i,j)}$ means separating the Hamiltonian into operators that only apply to the qubits $i$ and $j$, such that $H=H^{(0,1)}+H^{(0,2)}$.

Let me define the two qubit parametrized gate $$G^{(i,j)}(t_x,t_y,t_z)=\exp[-\mathrm i (t_x X^{(i)}X^{(j)}+t_y Y^{(i)}Y^{(j)}+t_z Z^{(i)}Z^{(j)})],$$ where $t_x,t_y,t_z$ are parameters.

The trotterization I want consists of the application of $n$ times $G^{(0,1)}(t,t,t)G^{(1,2)}(t,t,t)$.

What I am looking for

In order to reduce my circuit, I wish to manipule each couple of steps in the trotterization. According to this paper (if I am reading correctly) https://link.aps.org/doi/10.1103/PhysRevA.106.012412 (arxiv), there would be a way to reflect a circuit with three qubits such that

$$G^{(0,1)}(t_0,t_0,t_0)G^{(1,2)}(t_0,t_0,t_0)G^{(0,1)}(t_0,t_0,t_0)G^{(1,2)}(t_0,t_0,t_0)$$ $$=G^{(1,2)}(t_{1,x},t_{1,y},t_0)G^{(0,1)}(t_{2,x},t_{2,y},t_0)G^{(1,2)}(t_{3,x},t_{3,y},t_0)G^{(1,2)}(t_0,t_0,t_0)$$

known as the Yang-Baxter equivalency where $t_{k,\ell}$ are the new parameters I am looking for ( $k\in\{1,2,3\}$ and $\ell\in\{x,y\}$) and $t_0=t/n$ is known.

The article gives a way to obtain the new parameters $t_{k,\ell}$ for a more general case. I am trying to solve that for the example given above, where initially all $G$ gates start with the same parameter $t_0$.

Could you help me obtain the new $t_{k,\ell}$? I do not have an example on how to solve the equations. This techniques allows to reduce the depth of the circuit. An example of how can this be useful is found here: https://github.com/rajat709/Quantum-Simulation/blob/main/Quantum_Simulation.ipynb

  • $\begingroup$ @unknown Imagine it is a circuit with three qubits, I just gave them labels 0,1,2. $\endgroup$
    – Mauricio
    Aug 3, 2022 at 8:22
  • $\begingroup$ I believe that if there is an equality between parametrized quantum circuits $PQC_1(x)=PQC_2(y)$ their angles must be linearly related $x_i=\sum A_{ij} y$. You can get such a relation for your circuits simply by expanding $e^{-it(XX+YY+ZZ)}=1-it(XX+YY+ZZ)+O(t^2)$ and solving equations at the linear order in $t$. It would help if you were to write exactly what equations are to be solved. $\endgroup$ Aug 3, 2022 at 11:12
  • $\begingroup$ @NikitaNemkov yes these are parametrized circuits. Good idea a linear expansion may help. The equation is $G(t,0,1)G(t,1,2)G(t,0,1)=G(t1,1,2)G(t2,0,1)G(t3,1,2)$ $\endgroup$
    – Mauricio
    Aug 3, 2022 at 11:39
  • $\begingroup$ I can't get your equation to work literally. Are you sure all three operators $XX, YY, ZZ$ must have the same coefficients within each $G$?. $\endgroup$ Aug 3, 2022 at 11:48
  • $\begingroup$ @NikitaNemkov maybe I am misreading the utility of the identity. The inspiration is this github.com/rajat709/Quantum-Simulation/blob/main/… I wanted to work out a single fold $\endgroup$
    – Mauricio
    Aug 3, 2022 at 22:35


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