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# 物理代写|量子力学代写Quantum mechanics代考|The Density Operator on the Bloch Sphere

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## 物理代写|量子力学代写Quantum mechanics代考|The Density Operator on the Bloch Sphere

Consider that the following pure qubit state
$$|\psi\rangle \equiv \cos (\theta / 2)|0\rangle+e^{i \varphi} \sin (\theta / 2)|1\rangle$$
has the following density operator representation:
\begin{aligned} |\psi\rangle\langle\psi|= & \left(\cos (\theta / 2)|0\rangle+e^{i \varphi} \sin (\theta / 2)|1\rangle\right)\left(\cos (\theta / 2)\langle 0|+e^{-i \varphi} \sin (\theta / 2)\langle 1|\right) \ = & \cos ^2(\theta / 2)|0\rangle\left\langle 0\left|+e^{-i \varphi} \sin (\theta / 2) \cos (\theta / 2)\right| 0\right\rangle\langle 1| \ & \quad+e^{i \varphi} \sin (\theta / 2) \cos (\theta / 2)|1\rangle\left\langle 0\left|+\sin ^2(\theta / 2)\right| 1\right\rangle\langle 1| . \end{aligned}
The matrix representation, or density matrix, of this density operator with respect to the computational basis is as follows:
$$\left[\begin{array}{cc} \cos ^2(\theta / 2) & e^{-i \varphi} \sin (\theta / 2) \cos (\theta / 2) \ e^{i \varphi} \sin (\theta / 2) \cos (\theta / 2) & \sin ^2(\theta / 2) \end{array}\right] .$$
Using trigonometric identities, it follows that the density matrix is equal to the following matrix:
$$\frac{1}{2}\left[\begin{array}{cc} 1+\cos (\theta) & \sin (\theta)(\cos (\varphi)-i \sin (\varphi)) \ \sin (\theta)(\cos (\varphi)+i \sin (\varphi)) & 1-\cos (\theta) \end{array}\right]$$

## 物理代写|量子力学代写Quantum mechanics代考|An Ensemble of Ensembles

The most general ensemble that we can construct is an ensemble of ensembles, i.e., an ensemble $\mathcal{F}$ of density operators where
$$\mathcal{F} \equiv\left{p_X(x), \rho_x\right}$$
The ensemble $\mathcal{F}$ essentially has two layers of randomization. The first layer is from the distribution $p_X(x)$. Each density operator $\rho_x$ in $\mathcal{F}$ arises from an ensemble $\left{p_{Y \mid X}(y \mid x),\left|\psi_{x, y}\right\rangle\right}$. The conditional distribution $p_{Y \mid X}(y \mid x)$ represents the second layer of randomization. Each $\rho_x$ is a density operator with respect to the above ensemble:
$$\rho_x \equiv \sum_y p_{Y \mid X}(y \mid x)\left|\psi_{x, y}\right\rangle\left\langle\psi_{x, y}\right|$$
The ensemble $\mathcal{F}$ has its own density operator $\rho$ where
$$\rho \equiv \sum_{x, y} p_{Y \mid X}(y \mid x) p_X(x)\left|\psi_{x, y}\right\rangle\left\langle\psi_{x, y}\right|=\sum_x p_X(x) \rho_x .$$
The density operator $\rho$ is the density operator from the perspective of someone who does not possess $x$. Figure 4.1 displays the process by which we can select the ensemble $\mathcal{F}$.

## 物理代写|量子力学代写Quantum mechanics代考|The Density Operator on the Bloch Sphere

$$|\psi\rangle \equiv \cos (\theta / 2)|0\rangle+e^{i \varphi} \sin (\theta / 2)|1\rangle$$

\begin{aligned} |\psi\rangle\langle\psi|= & \left(\cos (\theta / 2)|0\rangle+e^{i \varphi} \sin (\theta / 2)|1\rangle\right)\left(\cos (\theta / 2)\langle 0|+e^{-i \varphi} \sin (\theta / 2)\langle 1|\right) \ = & \cos ^2(\theta / 2)|0\rangle\left\langle 0\left|+e^{-i \varphi} \sin (\theta / 2) \cos (\theta / 2)\right| 0\right\rangle\langle 1| \ & \quad+e^{i \varphi} \sin (\theta / 2) \cos (\theta / 2)|1\rangle\left\langle 0\left|+\sin ^2(\theta / 2)\right| 1\right\rangle\langle 1| . \end{aligned}

$$\left[\begin{array}{cc} \cos ^2(\theta / 2) & e^{-i \varphi} \sin (\theta / 2) \cos (\theta / 2) \ e^{i \varphi} \sin (\theta / 2) \cos (\theta / 2) & \sin ^2(\theta / 2) \end{array}\right] .$$

$$\frac{1}{2}\left[\begin{array}{cc} 1+\cos (\theta) & \sin (\theta)(\cos (\varphi)-i \sin (\varphi)) \ \sin (\theta)(\cos (\varphi)+i \sin (\varphi)) & 1-\cos (\theta) \end{array}\right]$$

## 物理代写|量子力学代写Quantum mechanics代考|An Ensemble of Ensembles

$$\mathcal{F} \equiv\left{p_X(x), \rho_x\right}$$

$$\rho_x \equiv \sum_y p_{Y \mid X}(y \mid x)\left|\psi_{x, y}\right\rangle\left\langle\psi_{x, y}\right|$$

$$\rho \equiv \sum_{x, y} p_{Y \mid X}(y \mid x) p_X(x)\left|\psi_{x, y}\right\rangle\left\langle\psi_{x, y}\right|=\sum_x p_X(x) \rho_x .$$

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