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数学计算|MATH3076/MATH3976 Mathematical Computing代写 Sydney代写

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这是一份Sydney悉尼大学MATH3076/MATH3976 的成功案例

数学计算|MATH3076/MATH3976 Mathematical Computing代写 Sydney代写


Disjoint union axiom): Let $X_{1}$ and $X_{2}$ be compact oriented surfaces with boundaries whose labels are $\ell_{1}$ and $\ell_{2}$ respectively.
$$
F\left(X_{1} \text { I } X_{2}, \ell_{1} \text { Ш } \ell_{2}\right)=F\left(X_{1}, \ell_{1}\right) \otimes F\left(X_{2}, \ell_{2}\right) .
$$
(Gluing axiom): Let $\tilde{X}$ be a compact oriented surface obtained by sluing together a pair of boundary circles with dual labels in $X$.
$$
F(\tilde{X}, \ell)=\bigoplus_{x \in L} F(X, \ell \cup{x, \hat{x}})
$$
Duality axiom): Let $X^{}$ stnds for $X$ with the reversed orientation and labels applied $.$ $$ F\left(X^{}\right)=(F(X))^{} \text {, where we mean that } A^{}={ }^{t} \bar{A} .
$$





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MATH3076/MATH3976 COURSE NOTES ：


Let $X$ be a topological space. For any $p, q \in X$ if there is a continuous $\operatorname{map} f:[0,1] \rightarrow X$ such that $f(0)=p, f(1)=q$, then $X$ is called arcwise connected and such a map is called a path in $X$. In what follows, we always consider an arcwise connected topological space and so we refer it simply “a space” hereafter.

Let $X$ be a space, and for $p, q, r \in X$ let $f$ be a path connecting $p$ with $q$ and $g$ a path connecting $q$ with $r$. Then we define a product of paths, $f \cdot g$, as follows:
$$
f \cdot g(s)= \begin{cases}f(2 s) & (0 \leq s \leq 1 / 2) \ g(2 s-1) & (1 / 2 \leq s \leq 1)\end{cases}
$$
The $f \cdot g$ is a new path in $X$ connecting $p$ with $r$.
When there is a path $l$ such that the starting point and the end point coincide, i.e., a continuous map
$$
l:[0,1] \rightarrow X, l(0)=p, l(1)=p
$$
is called a loop in $X$ with a base point $p$.
















数学计算|Mathematical Computing代写 MATH 397C

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这是一份umass麻省大学 MATH 397C作业代写的成功案例

数学计算|Mathematical Computing代写 MATH 397C
问题 1.

A quantum circuit which models the phase-flip channel is shown in Fig. 6. Let $\rho_{S}$ be the first qubit input state while $(1-p)|0\rangle\langle 0|+p| 1\rangle\langle 1|$ be the second qubit input state. The circuit is the inverted controlled $-\sigma_{z}$ gate
$$
V=I \otimes|0\rangle\left\langle 0\left|+\sigma_{z} \otimes\right| 1\right\rangle\langle 1| .
$$
The output of this circuit is
$$
\begin{aligned}
&V\left(\rho_{S} \otimes[(1-p)|0\rangle\langle 0|+p| 1\rangle\langle 1|]\right) V^{\dagger} \
&=(1-p) \rho_{S} \otimes|0\rangle\left\langle 0\left|+p \sigma_{z} \rho_{S} \sigma_{z} \otimes\right| 1\right\rangle\langle 1|
\end{aligned}
$$

证明 .

from which we obtain
$$
\mathcal{E}\left(\rho_{S}\right)=(1-p) \rho_{S}+p \sigma_{z} \rho_{S} \sigma_{z}
$$
The second qubit input state may be a pure state
$$
\left|\psi_{E}\right\rangle=\sqrt{1-p}|0\rangle+\sqrt{p}|1\rangle,
$$
for example. Then we find
$$
\mathcal{E}\left(\rho_{S}\right)=\operatorname{tr}{E}\left[V \rho{S} \otimes\left|\psi_{E}\right\rangle\left\langle\psi_{E}\right| V^{\dagger}\right]=E_{0} \rho_{S} E_{0}^{\dagger}+E_{1} \rho_{S} E_{1}^{\dagger}
$$
where the Kraus operators are
$$
E_{0}=\left\langle 0|V| \psi_{E}\right\rangle=\sqrt{1-p} I, E_{1}=\left\langle 1|V| \psi_{E}\right\rangle=\sqrt{p} \sigma_{z}
$$


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MMATH 397C COURSE NOTES ：

(1) For any $a, b, c \in G$ we have
$$
\mu(a, \mu(b, c))=\mu(\mu(a, b), c) .
$$
(2) For any $a \in G$ there exists an element $e \in G$ such that
$$
\mu(a, e)=\mu(e, a)=a,
$$
where $e$ is called a unit element of $G$.
(3) For any $a \in G$ there exists an element $a^{\prime} \in G$ such that
$$
\mu\left(a, a^{\prime}\right)=\mu\left(a^{\prime}, a\right)=e,
$$
where $a^{\prime}$ is called an inverse element of $a$ and we write $a^{-1} \in G$.