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# 物理代写|量子力学代写Quantum mechanics代考|Separable States

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## 物理代写|量子力学代写Quantum mechanics代考|Separable States

Let us now consider two systems $A$ and $B$ whose corresponding ensembles are correlated in a classical way. We describe this correlated ensemble as the joint ensemble
$$\left{p_X(x),\left|\psi_x\right\rangle \otimes\left|\phi_x\right\rangle\right}$$
It is straightforward to verify that the density operator of this correlated ensemble has the following form:
$$\mathbb{E}_X\left{\left(\left|\psi_X\right\rangle \otimes\left|\phi_X\right\rangle\right)\left(\left\langle\psi_X\right| \otimes\left\langle\phi_X\right|\right)\right}=\sum_x p_X(x)\left|\psi_x\right\rangle\left\langle\psi_x|\otimes| \phi_x\right\rangle\left\langle\phi_x\right|$$
By ignoring Bob’s system, Alice’s local density operator is of the form
$$\mathbb{E}_X\left{\left|\psi_X\right\rangle\left\langle\psi_X\right|\right}=\sum_x p_X(x)\left|\psi_x\right\rangle\left\langle\psi_x\right|$$
and similarly, Bob’s local density operator is
$$\mathbb{E}_X\left{\left|\phi_X\right\rangle\left\langle\phi_X\right|\right}=\sum_x p_X(x)\left|\phi_x\right\rangle\left\langle\phi_x\right|$$
States of the form in (4.110) can be generated by a classical procedure. A third party generates a symbol $x$ according to the probability distribution $p_X(x)$ and sends the symbol $x$ to both Alice and Bob. Alice then prepares the state $\left|\psi_x\right\rangle$ and Bob prepares the state $\left|\phi_x\right\rangle$. If they then discard the symbol $x$, the state of their systems is given by $(4.110)$.

## 物理代写|量子力学代写Quantum mechanics代考|Separable States and the CHSH Game

One motivation for Definitions 4.3 .2 and 4.3 .3 was already given above: for a separable state, there is a classical procedure that can be used to prepare it. Thus, for an entangled state, there is no such procedure. That is, a non-classical (quantum) interaction between the systems is necessary to prepare an entangled state.

Another related motivation is that separable states admit an explanation in terms of a classical strategy for the $\mathrm{CHSH}$ game, discussed in Section 3.6.2. Recall from (3.163) that classical strategies $p_{A B \mid X Y}(a, b \mid x, y)$ are of the following form:
$$p_{A B \mid X Y}(a, b \mid x, y)=\int d \lambda p_{\Lambda}(\lambda) p_{A \mid \Lambda X}(a \mid \lambda, x) p_{B \mid \Lambda Y}(b \mid \lambda, y)$$

If we allow for a continuous index $\lambda$ for a separable state, then we can write such a state as follows:
$$\sigma_{A B}=\int d \lambda p_{\Lambda}(\lambda)\left|\psi_\lambda\right\rangle\left\langle\left.\psi_\lambda\right|A \otimes \mid \phi\lambda\right\rangle\left\langle\left.\phi_\lambda\right|B\right.$$ Recall that in a general quantum strategy, there are measurements $\left{\Pi_a^{(x)}\right}$ and $\left{\Pi_b^{(y)}\right}$, giving output bits $a$ and $b$ based on the input bits $x$ and $y$ and leading to the following strategy: \begin{aligned} & p{A B \mid X Y}(a, b \mid x, y) \ & =\operatorname{Tr}\left{\left(\Pi_a^{(x)} \otimes \Pi_b^{(y)}\right) \sigma_{A B}\right} \ & =\operatorname{Tr}\left{\left(\Pi_a^{(x)} \otimes \Pi_b^{(y)}\right)\left(\int d \lambda p_{\Lambda}(\lambda)\left|\psi_\lambda\right\rangle\left\langle\left.\psi_\lambda\right|A \otimes \mid \phi\lambda\right\rangle\left\langle\left.\phi_\lambda\right|B\right)\right}\right. \ & =\int d \lambda p{\Lambda}(\lambda) \operatorname{Tr}\left{\Pi_a^{(x)}\left|\psi_\lambda\right\rangle\left\langle\left.\psi_\lambda\right|A \otimes \Pi_b^{(y)} \mid \phi\lambda\right\rangle\left\langle\left.\phi_\lambda\right|B\right}\right. \ & =\int d \lambda p{\Lambda}(\lambda)\left\langle\left.\psi_\lambda\right|A \Pi_a^{(x)} \mid \psi\lambda\right\rangle_A\left\langle\left.\phi_\lambda\right|B \Pi_b^{(y)} \mid \phi\lambda\right\rangle_B . \end{aligned}

## 物理代写|量子力学代写Quantum mechanics代考|Separable States

$$\left{p_X(x),\left|\psi_x\right\rangle \otimes\left|\phi_x\right\rangle\right}$$

$$\mathbb{E}_X\left{\left(\left|\psi_X\right\rangle \otimes\left|\phi_X\right\rangle\right)\left(\left\langle\psi_X\right| \otimes\left\langle\phi_X\right|\right)\right}=\sum_x p_X(x)\left|\psi_x\right\rangle\left\langle\psi_x|\otimes| \phi_x\right\rangle\left\langle\phi_x\right|$$

$$\mathbb{E}_X\left{\left|\psi_X\right\rangle\left\langle\psi_X\right|\right}=\sum_x p_X(x)\left|\psi_x\right\rangle\left\langle\psi_x\right|$$

$$\mathbb{E}_X\left{\left|\phi_X\right\rangle\left\langle\phi_X\right|\right}=\sum_x p_X(x)\left|\phi_x\right\rangle\left\langle\phi_x\right|$$
(4.110)中形式的状态可以通过经典程序生成。第三方根据概率分布$p_X(x)$生成符号$x$，并将符号$x$发送给Alice和Bob。然后Alice准备状态$\left|\psi_x\right\rangle$, Bob准备状态$\left|\phi_x\right\rangle$。如果他们放弃符号$x$，他们的系统的状态由$(4.110)$给出。

## 物理代写|量子力学代写Quantum mechanics代考|Separable States and the CHSH Game

$$p_{A B \mid X Y}(a, b \mid x, y)=\int d \lambda p_{\Lambda}(\lambda) p_{A \mid \Lambda X}(a \mid \lambda, x) p_{B \mid \Lambda Y}(b \mid \lambda, y)$$

$$\sigma_{A B}=\int d \lambda p_{\Lambda}(\lambda)\left|\psi_\lambda\right\rangle\left\langle\left.\psi_\lambda\right|A \otimes \mid \phi\lambda\right\rangle\left\langle\left.\phi_\lambda\right|B\right.$$回想一下，在一般的量子策略中，有测量$\left{\Pi_a^{(x)}\right}$和$\left{\Pi_b^{(y)}\right}$，根据输入位$x$和$y$给出输出位$a$和$b$，并导致以下策略: \begin{aligned} & p{A B \mid X Y}(a, b \mid x, y) \ & =\operatorname{Tr}\left{\left(\Pi_a^{(x)} \otimes \Pi_b^{(y)}\right) \sigma_{A B}\right} \ & =\operatorname{Tr}\left{\left(\Pi_a^{(x)} \otimes \Pi_b^{(y)}\right)\left(\int d \lambda p_{\Lambda}(\lambda)\left|\psi_\lambda\right\rangle\left\langle\left.\psi_\lambda\right|A \otimes \mid \phi\lambda\right\rangle\left\langle\left.\phi_\lambda\right|B\right)\right}\right. \ & =\int d \lambda p{\Lambda}(\lambda) \operatorname{Tr}\left{\Pi_a^{(x)}\left|\psi_\lambda\right\rangle\left\langle\left.\psi_\lambda\right|A \otimes \Pi_b^{(y)} \mid \phi\lambda\right\rangle\left\langle\left.\phi_\lambda\right|B\right}\right. \ & =\int d \lambda p{\Lambda}(\lambda)\left\langle\left.\psi_\lambda\right|A \Pi_a^{(x)} \mid \psi\lambda\right\rangle_A\left\langle\left.\phi_\lambda\right|B \Pi_b^{(y)} \mid \phi\lambda\right\rangle_B . \end{aligned}

## MATLAB代写

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