1
$\begingroup$

I have a few questions regarding dynamics rescaling for zero noise extrapolation. In the paper Error mitigation for short-depth quantum circuits, in equation (30), they write

We redefine $T \rightarrow T^{\prime}=c T$ as well as $J_\alpha(t) \rightarrow J_\alpha^{\prime}(t)=c^{-1} J_\alpha\left(c^{-1} t\right) $ from which also $ \rho(t) \rightarrow \rho^{\prime}(t)=\rho\left(c^{-1} t\right) (30)$

We claim that this rescaling maps $\rho_\lambda^{\prime}\left(T^{\prime}\right)=\rho_{c \lambda}(T)$ if the noise operator $\mathcal{L}$ does not depend on the Hamiltonian couplings $J_\alpha(t)$ and is constant in time.

Context: $$ \rho_\lambda(T)=\rho(0)-i \int_0^T[K(t), \rho(t)] d t+\lambda \int_0^T \mathcal{L}(\rho(t)) d t . $$

We can now choose a re-parametrization of the evolution $c^{-1} J_\alpha\left(c^{-1} t\right)$ and an increased runtime $c T$, and write $$ \rho_\lambda^{\prime}\left(T^{\prime}\right)=\rho(0)-i \int_0^{c T}\left[K^{\prime}(t), \rho^{\prime}(t)\right] d t+\lambda \int_0^{c T} \mathcal{L}\left(\rho^{\prime}(t)\right) d t $$ with $K^{\prime}(t)=\sum_\alpha c^{-1} J_\alpha\left(c^{-1} t\right) P_\alpha$. If we now substitute the integration variable according to $t=c t^{\prime}$, we have that $d t=c d t^{\prime}$, which leads to $$ \begin{aligned} \rho_\lambda^{\prime}\left(T^{\prime}\right) & =\rho(0)-i \int_0^T \sum_\alpha c^{-1} J_\alpha\left(t^{\prime}\right)\left[P_\alpha, \rho\left(t^{\prime}\right)\right] c d t^{\prime}+\lambda \int_0^T \mathcal{L}\left(\rho^{\prime}(t)\right) c d t^{\prime} \\ & =\rho(0)-i \int_0^T \sum_\alpha\left[K\left(t^{\prime}\right), \rho\left(t^{\prime}\right)\right] d t^{\prime}+\lambda c \int_0^T \mathcal{L}\left(\rho\left(t^{\prime}\right)\right) d t^{\prime} \\ & =\rho_{c \lambda}(T) \end{aligned} $$

My questions:

  1. Why is equation (30) justified? How does rescaling T and J accomplish density matrix rescaling (equation 30) when both commutator and dissipator terms are present? Their claim will only be right if in the integral representation, one makes the substitution for $\rho'(t)=\rho(t/c)$.

If I just have local hamiltonian (unitary evolution, assuming hamiltonians at different times commute), then one can show that $\rho(t)$ is also rescaled as $\rho(t/c)$.

$U(T,0)*\rho(0)*U^{\dagger}(T,0)=\rho(T)$

$U'(T,0)*\rho(0)*U'^{\dagger}(T,0)=\rho'(T)=\rho(T/c)$ where

$U(T,0)=exp(-i\int_0^T H(t)dt)$ and $U'(T,0)=exp(-i\int_0^T (1/c)*H(t/c)dt)=U(T/c,0)$

But here I have both Hamiltonian and noise term.

  1. What if I don't have a local hamiltonian evolution and have just the dissipator term, can I still make this claim?
$\endgroup$

0

Your Answer

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