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# 物理代写|量子力学代写Quantum mechanics代考|Depolarizing Channels

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## 物理代写|量子力学代写Quantum mechanics代考|Depolarizing Channels

The depolarizing channel is a “worst-case scenario” channel. It assumes that we completely lose the input qubit with some probability, i.e., it replaces the lost qubit with the maximally mixed state. The map for the depolarizing channel is
$$\rho \rightarrow(1-p) \rho+p \pi$$
where $\pi$ is the maximally mixed state: $\pi=I / 2$.
Most of the time, this channel is too pessimistic. Usually, we can learn something about the physical nature of the channel by some estimation process. We should only consider using the depolarizing channel as a model if we have little to no information about the actual physical channel.

EXercise 4.7.3 (Pauli Twirl) Show that randomly applying the Pauli operators $I, X, Y, Z$ with uniform probability to any density operator gives the maximally mixed state:
$$\frac{1}{4} \rho+\frac{1}{4} X \rho X+\frac{1}{4} Y \rho Y+\frac{1}{4} Z \rho Z=\pi$$
(Hint: Represent the density operator as $\rho=\left(I+r_x X+r_y Y+r_z Z\right) / 2$ and apply the commutation rules of the Pauli operators.) This is known as the “twirling” operation.

EXERCISE 4.7.4 Show that we can rewrite the depolarizing channel as the following Pauli channel:
$$\rho \rightarrow(1-3 p / 4) \rho+p\left(\frac{1}{4} X \rho X+\frac{1}{4} Y \rho Y+\frac{1}{4} Z \rho Z\right) .$$
EXERCISE 4.7.5 Show that the action of a depolarizing channel on the Bloch vector is
$$\left(r_x, r_y, r_z\right) \rightarrow\left((1-p) r_x,(1-p) r_y,(1-p) r_z\right)$$

## 物理代写|量子力学代写Quantum mechanics代考|Amplitude Damping Channels

The amplitude damping channel is an approximation to a noisy evolution that occurs in many physical systems ranging from optical systems to chains of spin$1 / 2$ particles to spontaneous emission of a photon from an atom.

In order to motivate this channel, we give a physical interpretation to our computational basis states. Let us think of the $|0\rangle$ state as the ground state of a two-level atom and let us think of the state $|1\rangle$ as the excited state of the atom. Spontaneous emission is a process that tends to decay the atom from its excited state to its ground state, even if the atom is in a superposition of the ground and excited states. Let the parameter $\gamma$ denote the probability of decay so that $0 \leq \gamma \leq 1$. One Kraus operator that captures the decaying behavior is
$$A_0=\sqrt{\gamma}|0\rangle\langle 1|$$
The operator $A_0$ annihilates the ground state:
$$A_0|0\rangle\langle 0| A_0^{\dagger}=0$$
and it decays the excited state to the ground state:
$$A_0|1\rangle\left\langle 1\left|A_0^{\dagger}=\gamma\right| 0\right\rangle\langle 0| .$$
The Kraus operator $A_0$ alone does not specify a physical map because $A_0^{\dagger} A_0=$ $\gamma|1\rangle\langle 1|$ (recall that the Kraus operators of any channel should satisfy the condition $\left.\sum_k A_k^{\dagger} A_k=I\right)$. We can satisfy this condition by choosing another operator $A_1$ such that
$$A_1^{\dagger} A_1=I-A_0^{\dagger} A_0=|0\rangle\langle 0|+(1-\gamma)| 1\rangle\langle 1|$$

## 物理代写|量子力学代写Quantum mechanics代考|Depolarizing Channels

$$\rho \rightarrow(1-p) \rho+p \pi$$

$$\frac{1}{4} \rho+\frac{1}{4} X \rho X+\frac{1}{4} Y \rho Y+\frac{1}{4} Z \rho Z=\pi$$
(提示:将密度算子表示为$\rho=\left(I+r_x X+r_y Y+r_z Z\right) / 2$，并应用泡利算子的交换规则。)这就是所谓的“旋转”操作。

$$\rho \rightarrow(1-3 p / 4) \rho+p\left(\frac{1}{4} X \rho X+\frac{1}{4} Y \rho Y+\frac{1}{4} Z \rho Z\right) .$$

$$\left(r_x, r_y, r_z\right) \rightarrow\left((1-p) r_x,(1-p) r_y,(1-p) r_z\right)$$

## 物理代写|量子力学代写Quantum mechanics代考|Amplitude Damping Channels

$$A_0=\sqrt{\gamma}|0\rangle\langle 1|$$

$$A_0|0\rangle\langle 0| A_0^{\dagger}=0$$

$$A_0|1\rangle\left\langle 1\left|A_0^{\dagger}=\gamma\right| 0\right\rangle\langle 0| .$$
Kraus操作符$A_0$单独不指定物理映射，因为$A_0^{\dagger} A_0=$$\gamma|1\rangle\langle 1|(回想一下，任何通道的Kraus操作符都应该满足条件\left.\sum_k A_k^{\dagger} A_k=I\right))。我们可以通过选择另一个算子A_1来满足这个条件$$ A_1^{\dagger} A_1=I-A_0^{\dagger} A_0=|0\rangle\langle 0|+(1-\gamma)| 1\rangle\langle 1|$\$

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