# Zero noise extrapolation for error mitigation: Meaning of rescaled density matrix, specifically when there is no local hamiltonian evolution

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?