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## 物理代写|热力学代写thermodynamics代考|Applications to One- and Two-Qubit Gates

We next analyze quantum computations by two qubits that undergo random dephasing. For single-qubit gates on each qubit, the average fidelity is obtained as
$$\bar{f}\left(t_{\mathrm{f}}\right)=1-\frac{5}{12}\left[J_{11}^{(1)}\left(t_{\mathrm{f}}\right)+J_{22}^{(1)}\left(t_{\mathrm{f}}\right)\right],$$
where
$$J_{j k}^{(q)}(t)=\int_0^t d t^{\prime} \int_0^{t^{\prime}} d t^{\prime \prime} \Phi_{j k}\left(t^{\prime}-t^{\prime \prime}\right) \epsilon_j^{(q)}\left(t^{\prime}\right) \epsilon_k^{(q) *}\left(t^{\prime \prime}\right),$$
which implies
$$\operatorname{Re} J_{j k}^{(q)}(t)=\pi \int_{-\infty}^{\infty} d \omega G_{j k}(\omega) \epsilon_{j, t}^{(q)}(\omega) \epsilon_{k, t}^{(q) *}(\omega) .$$
Here, $J_{j k}^{(q)}(t)$ is the dephasing function modified by the fields with Rabi frequencies $\Omega_{j, k}^{(q)}\left(q=1,2_\psi, 2_{\Phi}\right), G_{j k}(\omega)=(2 \pi)^{-1} \int_{-\infty}^{\infty} d t \Phi_{j k}(t) e^{i \omega t}$ is the dephasing spectrum, and $\epsilon_{j, t}^{(q)}(\omega)=(2 \pi)^{-1 / 2} \int_0^t d t^{\prime} \epsilon_j^{(q)}\left(t^{\prime}\right) e^{i \omega t^{\prime}}$ is the finite-time Fourier transform of the modulation.

Equations (13.15)-(13.17) show that the fidelity is maximized by reducing the spectral overlap of the dephasing and modulation (control) spectra. Single-qubit gate fields do not cause cross-dephasing. Since (13.15) depends only on singlequbit dephasing, $\Phi_{j j}(t)$, because of the averaging over all initial qubits. For each initial entangled state that “suffers” from cross-dephasing (e.g., triplet, $\left|\Phi_{-}\right\rangle$), there is another entangled state that “benefits” from cross-dephasing (e.g., the singlet,

$\left.\left|\Psi_{-}\right\rangle\right)$. Equation (13.15) also shows that if one applies a gate field on one qubit, one can still benefit from applying a control field on the other, stored, qubit.
For two-qubit gate operations, the average fidelity at $t_{\mathrm{f}}$ is found to be:
$$\bar{f}\left(t_{\mathrm{f}}\right)=1-\frac{5}{24} \sum_{j, k=1,2}\left[J_{j k}^{(2) \Phi}\left(t_{\mathrm{f}}\right)+(-1)^{j+k} J_{j k}^{(2) \psi}\left(t_{\mathrm{f}}\right)\right]$$
In this expression, cross-dephasing corresponding to $j \neq k$ does not cancel out despite averaging. Yet, the cross-terms have opposite signs for the different twogate fields acting on the $\Psi$ and $\Phi$ Bell states. Thus, remarkably, the SWAP-gate fidelity may benefit from cross-dephasing.

It is noteworthy that by applying together both two-qubit gate fields, one can reduce dephasing, even if only one field is needed for the actual gate operation. For example, a two-qubit storage-gate field, with $\phi_{1,2}^{(2) \Phi}\left(t_{\mathrm{f}}\right)=2 \pi M(M=1,2, \ldots)$, concurrently applied with, for example, a SWAP gate, can reduce dephasing. This approach may result in longer gate durations, as the maximal peak power in the gate fields may be limited. Nevertheless, according to (13.15) and (13.18), this approach may be beneficial if the reduction in the dephasing due to the applied fields outweighs the increase in dephasing due to longer gate duration.

## 物理代写|热力学代写thermodynamics代考|Optimal Gate Protection

The protection of a given quantum operation from decoherence is most effective under bath-optimized task-oriented control (BOTOC), expounded in Chapter $12 .$ Here, we consider the implementation of a quantum-gate unitary operation within a given “gate time” $t$ for a pure input state $|\Psi\rangle$. In the interaction picture with respect to the desired gate operation, the projector $\hat{P}=\hat{\varrho}(0)=|\Psi\rangle\langle\Psi|$ is then used as the gradient operator, so that (12.22) is satisfied. Then, (12.24) yields the fidelity change as the score
$$P=\langle\Psi|\Delta \hat{\varrho}| \Psi\rangle=-t^2\left\langle\left\langle\Psi\left|\hat{H}^2\right| \Psi\right\rangle-\langle\Psi|\hat{H}| \Psi\rangle^2\right\rangle_{\mathrm{B}} .$$
To eliminate the dependence on $|\Psi\rangle$, we uniformly average over all $|\Psi\rangle$, whereby for any two operators $\hat{A}$ and $\hat{B}$ :
$$\overline{\langle\Psi|\hat{A}| \Psi\rangle\langle\Psi|\hat{B}| \Psi\rangle}=\frac{\operatorname{Tr} \hat{A} \hat{B}+\operatorname{Tr} \hat{A} \operatorname{Tr} \hat{B}}{d(d+1)},$$
$d$ being the Hilbert-space dimensionality of the system. The average score is then
$$\bar{P}=-t^2 \frac{d}{d+1}\left\langle\hat{H}^2\right\rangle_{\mathrm{id}},$$
where [in the notation of (12.6)] $\langle\ldots\rangle_{\text {id }}=\operatorname{Tr}\left(d^{-1} \hat{I} \otimes \hat{\varrho}B \ldots\right)$. Here we have used $\operatorname{Tr}{\mathrm{S}} \hat{H}=0$, corresponding to $\operatorname{Tr} \hat{S}j=0$. On account of $(12.21),\langle\hat{H}\rangle{\mathrm{B}}=\left\langle\hat{H}{\mathrm{I}}\right\rangle{\mathrm{B}}=0$,

$\langle\hat{H}\rangle_{\text {id }}=0$, so that $(13.21)$ is proportional to the variance of the Hamiltonian: $\operatorname{Var}(\hat{H})=\left\langle\hat{H}^2\right\rangle_{\mathrm{id}}-\langle\hat{H}\rangle_{\mathrm{id}}^2$. The gate error $\mathcal{E}$, which is the average fidelity decline (or the infidelity), then satisfies
$$\mathcal{E} \equiv-\bar{P}=t^2 \frac{d}{d+1} \operatorname{Var}(\hat{H}) .$$
The average over the initial states in the matrix $\Xi$, defined in (12.31), yields $\bar{\Xi}=$ $-\frac{d}{d+1} \boldsymbol{I}$, upon using $\operatorname{Tr}\left(\hat{S}j \hat{S}_k\right)=d \delta{j k}$ and $\operatorname{Tr} \hat{S}j=0$. Hence $$\mathcal{E}=\frac{d}{d+1} \int{-\infty}^{\infty} d \omega \operatorname{Tr}\left[\boldsymbol{\epsilon}t(\omega) \boldsymbol{\epsilon}_t^{\dagger}(\omega) \boldsymbol{G}(\omega)\right]$$ where $\boldsymbol{G}(\omega)$ and $\epsilon_r(\omega)$ are given by (12.32) and (12.33), respectively. Because of the requirement that $\mathcal{E} \geq 0, \boldsymbol{G}(\omega)$ must be a positive semidefinite matrix for any $\omega$. BOTOC then aims at finding the evolution operator of the system, $\hat{U}(t)$ (as in the control examples in Sec. 12.2), that minimizes $\mathcal{E}$, subject to the condition that the desired gate is executed over time interval $t{\mathrm{f}}$.

We may therefore conclude that, whereas each gate operation should be as fast as possible, the optimal overall pulse sequence may take longer than the gate time because of the storage control duration. This general principle is unparalleled by other approaches.

# 热力学代写

## 物理代写|热力学代写热力学代考|一个和两个qubit Gates的应用

.

$$\bar{f}\left(t_{\mathrm{f}}\right)=1-\frac{5}{12}\left[J_{11}^{(1)}\left(t_{\mathrm{f}}\right)+J_{22}^{(1)}\left(t_{\mathrm{f}}\right)\right],$$

$$J_{j k}^{(q)}(t)=\int_0^t d t^{\prime} \int_0^{t^{\prime}} d t^{\prime \prime} \Phi_{j k}\left(t^{\prime}-t^{\prime \prime}\right) \epsilon_j^{(q)}\left(t^{\prime}\right) \epsilon_k^{(q) *}\left(t^{\prime \prime}\right),$$
，这意味着
$$\operatorname{Re} J_{j k}^{(q)}(t)=\pi \int_{-\infty}^{\infty} d \omega G_{j k}(\omega) \epsilon_{j, t}^{(q)}(\omega) \epsilon_{k, t}^{(q) *}(\omega) .$$

$\left.\left|\Psi_{-}\right\rangle\right)$。式(13.15)还表明，如果一个人对一个量子位施加一个门场，他仍然可以从对另一个被存储的量子位施加一个控制场中获益。对于双量子比特门运算，$t_{\mathrm{f}}$处的平均保真度为:
$$\bar{f}\left(t_{\mathrm{f}}\right)=1-\frac{5}{24} \sum_{j, k=1,2}\left[J_{j k}^{(2) \Phi}\left(t_{\mathrm{f}}\right)+(-1)^{j+k} J_{j k}^{(2) \psi}\left(t_{\mathrm{f}}\right)\right]$$

## 物理代写|热力学代写热力学代考|最佳门保护

$$P=\langle\Psi|\Delta \hat{\varrho}| \Psi\rangle=-t^2\left\langle\left\langle\Psi\left|\hat{H}^2\right| \Psi\right\rangle-\langle\Psi|\hat{H}| \Psi\rangle^2\right\rangle_{\mathrm{B}} .$$

$$\overline{\langle\Psi|\hat{A}| \Psi\rangle\langle\Psi|\hat{B}| \Psi\rangle}=\frac{\operatorname{Tr} \hat{A} \hat{B}+\operatorname{Tr} \hat{A} \operatorname{Tr} \hat{B}}{d(d+1)},$$
$d$是系统的希尔伯特空间维数。那么平均分数是
$$\bar{P}=-t^2 \frac{d}{d+1}\left\langle\hat{H}^2\right\rangle_{\mathrm{id}},$$

$\langle\hat{H}\rangle_{\text {id }}=0$，使$(13.21)$与哈密顿量的方差成正比:$\operatorname{Var}(\hat{H})=\left\langle\hat{H}^2\right\rangle_{\mathrm{id}}-\langle\hat{H}\rangle_{\mathrm{id}}^2$。门误差$\mathcal{E}$，即平均保真度下降(或不忠)，然后满足
$$\mathcal{E} \equiv-\bar{P}=t^2 \frac{d}{d+1} \operatorname{Var}(\hat{H}) .$$
。矩阵$\Xi$(在(12.31)中定义)中初始状态的平均值在使用$\operatorname{Tr}\left(\hat{S}j \hat{S}_k\right)=d \delta{j k}$和$\operatorname{Tr} \hat{S}j=0$时得到$\bar{\Xi}=$$-\frac{d}{d+1} \boldsymbol{I}。因此，$$ \mathcal{E}=\frac{d}{d+1} \int{-\infty}^{\infty} d \omega \operatorname{Tr}\left[\boldsymbol{\epsilon}t(\omega) \boldsymbol{\epsilon}_t^{\dagger}(\omega) \boldsymbol{G}(\omega)\right]$$中$\boldsymbol{G}(\omega)$和$\epsilon_r(\omega)$分别由(12.32)和(12.33)给出。因为对于任何$\omega$,$\mathcal{E} \geq 0, \boldsymbol{G}(\omega)$必须是一个正半正定矩阵。BOTOC的目标是找到系统的进化算符$\hat{U}(t)$(如第12.2节的控制示例所示)，它使$\mathcal{E}$最小化，条件是期望的门在时间间隔$t{\mathrm{f}}\$上执行。

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## MATLAB代写

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