## 物理代写|热力学代写thermodynamics代考|Robustness of Noisy Channels

We now consider the effects of noise on the coupling strengths, known as offdiagonal noise, causing $$J_i \rightarrow J_i+J_i \tilde{\delta}i(t), \quad i=1, \ldots, N$$ Here $\tilde{\delta}_i(t)$ is a uniformly distributed random variable in the interval $\left[-\epsilon_J, \epsilon_J\right]$, $\epsilon_J>0$ characterizing the noise or disorder strength. When $\tilde{\delta}_i$ is time-dependent, the noise is fluctuating, whereas fixed $\tilde{\delta}_i$ corresponds to static noise. Such noises affect the bath-energy levels, while the central eigenvalue $\omega_z$ remains invariant. We may compare the performance of the boundary-couplings control solutions $g_p(t)$ under these types of noise: (i) Static noise. Static control can suppress such noise, but dynamical boundary control with $p=2$ can render the channel even more robust because it filters out the bath energies that damage the transfer. This is shown in Figure 14.4(b) in the strong-coupling regime for $g{\mathrm{M}p}=g{\mathrm{M}p}^{\text {opt }}$ where the advantage of the dynamical control with $p=2$ compared to the static control case $p=0$ is evident, at the expense of increasing the transfer time by only a factor of $2, t_2 / t_0 \approx 2$. In the weak-coupling regime, if we choose $g{\mathrm{M}{\mathrm{p}}}$ such that the transfer fidelity is similar for control with $p=0$ and with $p=2$, then the modulated control with $p=2$ yields transfer that is an order of magnitude faster. Remarkably, due to disorderinduced localization, the fidelity under static noise cannot be improved beyond the bound $$1-\bar{f} \propto N \epsilon_J^2 \quad\left(\epsilon_J \ll 1\right),$$ regardless of how small $g_0$ is. (ii) Markovian noise. The worst case for quantum state transfer is the absence of an energy gap around $\omega_z$. This case corresponds to Markovian noise characterized by $\left\langle\tilde{\delta}_i(t) \tilde{\delta}_i(t+\tau)\right\rangle=\delta(\tau)$, where the angle brackets denote the noise ensemble average. In this case, there is an analytical solution for the optimal modulation that can be approximated by $$g(t) \approx a g{\mathrm{M}}+b \sin ^q\left(t \pi / t_{\mathrm{f}}\right),$$
where
$$q \sim 3.5, b / a \sim 1 / 3, a \sim 0.84$$

## 物理代写|热力学代写thermodynamics代考|The Lindblad Theorem and Markovian Entropy Change

The evolution of a quantum system interacting with another quantum system [here a bath (B)] can be shown to conform to a completely positive (CP), trace-preserving map from the initial state of the system, $\rho(0)$, to the evolving state $\rho(t)$, provided that the system and the bath are initially uncorrelated, that is, they are described by a product state $\rho(0) \rho_{\mathrm{B}}, \rho_{\mathrm{B}}$ being the bath state. If, however, the evolution involves measurements (Chs. 9, 10), the map may decrease the trace. In what follows, we consider non-trace-increasing (NTI) maps $\mathcal{M}$, satisfying $\operatorname{Tr}(\mathcal{M} \rho) \leq 1$, which include both above types of evolution as special cases.
The general form of a CP-NTI map $\mathcal{M}$ is
$$\mathcal{M} \rho=\sum_j K_j^{\dagger} \rho K_j$$
where $K_j$ are known as the Kraus operators. The CP map $\mathcal{M}$ is trace preserving provided that $\sum_j K_j K_j^{\dagger}=I$, but more generally, for a CP-NTI map, $\sum_j K_j$ $K_j^{\dagger} \leq I$.
For any CP-NTI map $\mathcal{M}$, Lindblad’s $H$-theorem is the inequality
$$\mathcal{S}(\mathcal{M} \rho | \mathcal{M} \tilde{\rho}) \leq \mathcal{S}(\rho | \tilde{\rho})$$
for the relative entropy of arbitrary states $\rho$ and $\tilde{\rho}$, defined as
$$\mathcal{S}(\rho | \tilde{\rho})=k_{\mathrm{B}} \operatorname{Tr}(\rho \ln \rho-\rho \ln \tilde{\rho}),$$
where $k_{\mathrm{B}}$ is the Boltzmann constant.
Lindblad’s theorem (15.2) is the most general statement of the second law for the evolution of quantum open systems if we assume that any proper quantum evolution (including measurements) must conform to a CP-NTI map. It is known as the statement of the monotonicity of the quantum relative entropy. This theorem implies that, if a CP-NTI map $\mathcal{M}$ acting on a state $\rho$ has a steady (invariant) state $\rho_{\mathrm{ss}}$, such that
$$\mathcal{M} \rho_{\mathrm{ss}}=\rho_{\mathrm{ss}}$$
then the relative entropy with respect to the steady state $\rho_{\mathrm{ss}}$ does not increase under $\mathcal{M}$ :
$$\mathcal{S}\left(\mathcal{M} \rho | \rho_{\mathrm{ss}}\right) \leq \mathcal{S}\left(\rho | \rho_{\mathrm{ss}}\right)$$

## 物理代写|热力学代写热力学代考|噪声通道的鲁棒性

，其中
$$q \sim 3.5, b / a \sim 1 / 3, a \sim 0.84$$

## 物理代写|热力学代写热力学代考|林德布莱德定理和马氏熵变

CP-NTI映射的一般形式 $\mathcal{M}$
$$\mathcal{M} \rho=\sum_j K_j^{\dagger} \rho K_j$$
where $K_j$ 被称为克劳斯操纵者CP地图 $\mathcal{M}$ 跟踪保存是这样提供的吗 $\sum_j K_j K_j^{\dagger}=I$，但更一般地，对于CP-NTI地图， $\sum_j K_j$ $K_j^{\dagger} \leq I$.

$$\mathcal{S}(\mathcal{M} \rho | \mathcal{M} \tilde{\rho}) \leq \mathcal{S}(\rho | \tilde{\rho})$$

$$\mathcal{S}(\rho | \tilde{\rho})=k_{\mathrm{B}} \operatorname{Tr}(\rho \ln \rho-\rho \ln \tilde{\rho}),$$
where $k_{\mathrm{B}}$ 是玻尔兹曼常数。如果我们假设任何适当的量子演化(包括测量)必须符合CP-NTI映射，林德布莱德定理(15.2)是关于量子开放系统演化第二定律的最一般的陈述。它被称为量子相对熵单调性的表述。这个定理表明，如果CP-NTI映射 $\mathcal{M}$ 对一个国家起作用 $\rho$ 是否具有稳定(不变)状态 $\rho_{\mathrm{ss}}$，使
$$\mathcal{M} \rho_{\mathrm{ss}}=\rho_{\mathrm{ss}}$$

$$\mathcal{S}\left(\mathcal{M} \rho | \rho_{\mathrm{ss}}\right) \leq \mathcal{S}\left(\rho | \rho_{\mathrm{ss}}\right)$$

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