1
$\begingroup$

I would like to know what would be the approximation error of a QAOA circuit.

Suppose we have time-dependent Hamiltonian $H(t) = (1 - s(t))H_{init} + s(t)H_{prob}$ where $H_{init}$ in an initial Hamiltonian whose ground is $|\psi_0\rangle$, $H_{prob}$ is a Hamiltonian that encodes an optimization problem and $s(t)$ is an adiabatic schedule with $t \in [0,T]$. $H_{init}$ and $H_{prob}$ do not commute.

The solution to the Schrodinger equation $|\psi(t)\rangle$ is given by a unitary $U(t,0)$ such that $$|\psi(t)\rangle = U(t,0)|\psi_0\rangle$$ and \begin{align} U(t,0) = \lim_{p \rightarrow \infty} &\exp \left \{ -\frac{i \Delta t}{\hbar} H(p \Delta t) \right \} \exp \left \{ -\frac{i \Delta t}{\hbar} H((p-1) \Delta t) \right \} \cdots \\ &\exp \left \{ -\frac{i \Delta t}{\hbar} H(0) \right \}. \end{align} In the equation above $\Delta t = t/p$. We can approximate $U(t,0)$ with the QAOA circuit $\hat{U}(t,0)$ by setting $p$ to be finite and applying the first order Trotter approximation with error $O(\Delta t^2)$.

I would like to know if someone could provide analysis or refs on the error of $$||U(t,0) - \hat{U}(t,0)|| \leq \ ???$$ or alternatively $$U(t,0) = \hat{U}(t,0) + O(\textrm{???}) \textrm{ as} \ p \rightarrow \infty.$$

$\endgroup$

0

Your Answer

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