Practical Problem

Let $S \in \mathbb{R}_{+}^{n \times m}$ a column stochastic matrix (i.e. a matrix with non-negative components whose columns sum to one), and $R \in \mathbb{R}_{+}^{n \times m}$ be a row stochastic matrix. Show that$\|S\mathbf{v}\|_{1}\leq\|\mathbf{v}\|_{1},\forall\mathbf{v}\in\mathbb{C}^{m}$ and $\|R\mathbf{v}\|_{\infty}\leq\|\mathbf{v}\|_{\infty},\forall\mathbf{v}\in\mathbb{C} ^{m}$

Idea: We want to use the property that $\sum^n_{x=1}s_{xy}=1\text{ or }\sum^n_{y=1}r_{xy}=1$
Then we need to group $\sum_x$ with $s_{xy}$. Using triangular inequality
Grouping $\sum_yr_{xy}$ needs taking the maximum

Proof

For $S=(s_{xy}),||S\vec{v}||_{1}=\sum_{x=1}^{n}|(S\vec{v})_{x}|=\sum_{x=1}^{n}\left|\sum _{y=1}^{m}s_{xy}v_{y}\right|\leq\sum_{y}\underbrace{\sum_{x}s_{xy}}_{=1}|v_{y}|=\sum _{y}|v_{y}|=||\vec{v}||_{1}$

$\|R\vec{v}\|_{\infty}=\max_{x\in\left\lbrack n\right\rbrack}|\sum_{y=1}^{m}r_{xy} v_{y}|=\left|\sum_{y=1}^{m}r_{iy}v_{y}\right|\leq\sum_{y=1}^{m}r_{iy}\left|v_{y}\right |$ for some $i\in[n]$

Then $||R\vec{v}||\leq\sum_{y=1}^{m}r_{iy}\max|v_{y}|=\sum_{y=1}^{m}r_{iy}|v_{k}|$ for some $k\in [m]$

Then $\|R\vec{v}\|_{\infty}\leq|v_{k}|\leq\max_{j\in[n]}|v_{j}|=||\vec{v}||_{\infty}$ since the sum of entries in a row is $1$
Let $M : \tilde{A} \rightarrow B$ be a linear map and denote its transpose map by $M^T : B \rightarrow A$. Show that $I \otimes M|\Omega^{A\tilde{A}}\rangle = M^T \otimes I|\Omega^{B\tilde{B}}\rangle$ where $|\Omega^{A\tilde{A}}\rangle := \sum_{y=1}^{|A|} |yy\rangle^{A\tilde{A}}$ and $|\Omega^{B\tilde{B}}\rangle := \sum_{y=1}^{|B|} |yy\rangle^{B\tilde{B}}$

Proof

$I\otimes M^{\tilde{A}\rightarrow B}|\Omega^{A\tilde{A}}\rangle=I\otimes M^{\tilde{A}\rightarrow B}(\sum_{y=1}^{|A|}|yy\rangle^{A\tilde{A}})=\sum_{y=1}^{|A|}|y\rangle^{A}\otimes M^{\tilde{A}\rightarrow B}|y\rangle^{\tilde{A}}$ $=\sum_{y=1}^{|A|}|y\rangle^{A}\otimes\sum_{x=1}^{|B|}M_{xy}|x\rangle=\sum_{y=1}^{|A|} \sum_{x=1}^{|B|}M_{xy}|yx\rangle$

$M^{T}\otimes I|\Omega^{B\tilde{B}}\rangle=M^{T}\otimes I(\sum_{x=1}^{|B|}|xx\rangle ^{B\tilde{B}})=\sum_{x=1}^{|B|}M^{T}|x\rangle^{B}\otimes|x\rangle^{\tilde{B}}=\sum _{x=1}^{|B|}\sum_{y=1}^{|A|}M_{xy}|y\rangle\otimes|x\rangle$
Thus they are equal
Let $|\psi\rangle, |\phi\rangle \in \mathbb{C}^n$. Show that $\mathrm{Tr}[|\psi\rangle\langle\phi|] = \langle\phi|\psi\rangle.$

Proof

Assume $|\psi\rang =\begin{pmatrix} a_1\\a_2\\\vdots\\a_n \end{pmatrix}$ and $\lang \phi|=\begin{pmatrix} b_1&b_2&\cdots&b_n \end{pmatrix}$, then $|\psi\rangle\langle\phi|=\begin{pmatrix} a_1b_1&a_1b_2&\cdots&a_1b_n\\ a_2b_1&a_2b_2&\cdots&a_2b_n\\ \vdots&\vdots&\ddots&\vdots\\ a_nb_1&a_nb_2&\cdots&a_nb_n\\ \end{pmatrix}$

Then $\text{Tr}[|\psi\rangle\langle\phi|]=\sum_{i=1}^{n}a_{i}b_{i}$, also $\langle\phi|\psi\rangle=\sum_{i=1}^{n}a_{i}b_{i}=\text{Tr}[|\psi\rangle\langle\phi |]$
Show that for any Hermitian operators $M, \sigma \in \text{Herm}(A)$, with $\sigma \ge 0$, the following holds: $\|\sqrt{\sigma} M \sqrt{\sigma}\|_1 \le \|M\|_2 \|\sigma\|_2$

Idea: Diagonalize $M$ and use Cauchy Schwartz inequality and do some computation to show $\lang \phi_x|\sigma^2|\phi_x\rang\geq \lang \phi_x|\sigma|\phi_x\rang^2$(Matrix expansion)

Proof

Since $M$ is hermitian, then $M=\sum_{k}u_{k}|\phi_{k}\rangle\!\langle \phi_{k}|$
Then $||\sqrt{\sigma}M\sqrt{\sigma}||_{1}=||\sqrt{\sigma}(\sum_{k}\mu_{k}|\phi_{k}\rangle \!\langle \phi_{k}|)\sqrt{\sigma}||_{1}\leq \sum_{x}|u_{x}|\cdot||\sqrt{\sigma}|\phi _{x}\rang\lang \phi_{x}\|\sqrt{\sigma}||_{1}$ by triangle inequality
$\le\Big(\sum_{k}\mu_{k}^{2}\Big)^{1/2}\Big(\sum_{k}\text{Tr}[|\sqrt{\sigma}|\phi _{x}\rang\lang \phi_{x}|\sqrt{\sigma}|]^{2}\Big)^{1/2}=\|M\|_{2}\Big(\sum_{k}\langle \phi_{k}|\sigma|\phi_{k}\rangle^{2}\Big)^{1/2}$ by Cauchy–Schwarz inequality: $a_{k}:=|\mu_{k}|,\,b_{k}:=\|v_{k}\|^{2},\,\,\,\,\sum_{k} a_{k} b_{k} \le \left(\sum_{k} a_{k}^{2}\right)^{1/2}\left(\sum_{k} b_{k} ^{2}\right)^{1/2}$ and since $\sqrt{\sigma}|\phi _{x}\rang\lang \phi_{x}\|\sqrt{\sigma}$ is positive
We know $||\sigma||_2=\text{Tr}[|\sigma|^2]^\frac{1}{2}=\text{Tr}[\sigma^2]^\frac{1}{2}=(\sum_x\lang \phi_x|\sigma^2|\phi_x\rang)^\frac{1}{2}$ and $\lang\phi_x|\sigma^2|\phi_x\rang =\sum_y\lang \phi_x|\sigma\phi_y\rang\lang \phi_y|\sigma|\phi_x\rang\geq \lang \phi_x|\sigma|\phi_x\rang^2$
Hence $\|\sqrt{\sigma} M \sqrt{\sigma}\|_1 \le \|M\|_2\,\|\sigma\|_2$.
Let $A$ and $B$ be two Hilbert spaces, $M, N \in \text{Pos}(A)$, and $p, q \in [1, \infty)$ such that $\frac{1}{p} + \frac{1}{q} = 1$.
Use the Hölder inequality of the Schatten norm to show that $\text{Tr}[MN] \leq \frac{1}{p}\text{Tr}[M^{p}] + \frac{1}{q}\text{Tr}[N^{q}]$ with equality if and only if $M^p = N^q$.

Idea: First coping with the problem that $\text{Tr}[MN]\neq \text{Tr}[|MN|]=||MN||_1$(using $\frac{1}{2}=\frac{1}{2p}+\frac{1}{2q}$)
Take log on holder inequality, then using Jensen's inequality and log is increasing to get it.
Then we can also get two combined inequality by log. Remember two condition: $M^p=cN^q$ and $\text{Tr}[M^p]=\text{Tr}[N^q]$ to get the result.

Proof
1. We prove $\text{Tr}[MN]\leq\frac{1}{p}\text{Tr}[M^{p}]+\frac{1}{q}\text{Tr}[N^{q}]$
  
  $\operatorname{Tr}[MN]=\|M^{1/2}N^{1/2}\|_{2}^{2}\le \|M^{1/2}\|_{2p}^{2}\,\|N^{1/2} \|_{2q}^{2}\text{ where }\tfrac12=\tfrac{1}{2p}+\tfrac{1}{2q}\;(\Leftrightarrow\; \tfrac1p+\tfrac1q=1)$
  
  $=\Big(\operatorname{Tr}\!\big[(M^{1/2})^{2p}\big]\Big)^{1/p}\Big(\operatorname{Tr}\!\big[(N^{1/2})^{2q}\big]\Big)^{1/q} =\big(\operatorname{Tr}[M^{p}]\big)^{1/p}\big(\operatorname{Tr}[N^{q}]\big)^{1/q}$
  Thus we have $\text{Tr}[MN]\leq\text{Tr}[M^{p}]^{\frac{1}{p}}\cdot\text{Tr}[N^{q}]^{\frac{1}{q}}$ by Hölder inequality
  We take log on both sides $\Rightarrow \log(\text{Tr}[MN]) \leq \log(\text{Tr}[M^p]^{\frac{1}{p}}) + \log(\text{Tr}[N^q]^{\frac{1}{q}})$ $= \frac{1}{p}\log(\text{Tr}[M^{p}]) + \frac{1}{q}\log(\text{Tr}[N^{q}])$ $\leq \log \left(\frac{1}{p}\text{Tr}[M^{p}] + \frac{1}{q}\text{Tr}[N^{q}]\right)$ because log is concave on $(0, \infty)$ and $\frac{1}{p} + \frac{1}{q} = 1$, by Jensen's inequality
  
  Then we get $\text{Tr}[MN] \leq \frac{1}{p} \text{Tr}[M^p] + \frac{1}{q} \text{Tr}[N^q]$ since $\log$ is increasing 2. NTP: $\text{Tr}[MN] = \frac{1}{p}\text{Tr}[M^{p}] + \frac{1}{q}\text{Tr}[N^{q}]$ iff $M^p = N^q$
  
  $\Leftarrow)$ $\frac{1}{p}\text{Tr}[M^{p}]+\frac{1}{q}\text{Tr}[N^{q}]=(\frac{1}{p}+\frac{1}{q} )\text{Tr}[M^{p}]=\text{Tr}[M^{p}]$
  By Hölder inequality: $\text{Tr}[MN]\le\text{Tr}[M^{p}]^{\frac{1}{p}}\cdot\text{Tr}[N^{q}]^{\frac{1}{q}} =\text{Tr}[M^{p}]^{\frac{1}{p}+\frac{1}{q}}=\text{Tr}[M^{p}]$
  
  Then $\text{Tr}[MN] = \frac{1}{p}\text{Tr}[M^{p}] + \frac{1}{q}\text{Tr}[N^{q}]$ equality holds when $\text{Tr}[MN]=\text{Tr}[M^{p}]$
  Since $\frac{1}{p}+ \frac{1}{q}= 1 \implies \frac{1}{q}= 1 - \frac{1}{p}= \frac{p-1}{p} \implies q = \frac{p}{p-1}$
  $\text{Tr}[MN]=\text{Tr}[M\cdot M^{\frac{p}{q}}]=\text{Tr}[M^{\frac{p+q}{q}}]=\text{Tr} [M^{\frac{p+\frac{p}{p-1}}{\frac{p}{p-1}}}]=\text{Tr}[M^{p}]$
  
  $\Rightarrow$)
  
  We have $\text{Tr}[MN] = \frac{1}{p}\text{Tr}[M^{p}] + \frac{1}{q}\text{Tr}[N^{q}]$ and by Hölder inequality $\text{Tr}[MN]\le\text{Tr}[M^{p}]^{\frac{1}{p}}\cdot\text{Tr}[N^{q}]^{\frac{1}{q}}$
  
  Also in the beginning we have $\log\left(\text{Tr}[M^{p}]^{\frac{1}{p}}\cdot\text{Tr}[N^{q}]^{\frac{1}{q}}\right )\leq \log \left(\frac{1}{p}\text{Tr}[M^{p}] + \frac{1}{q}\text{Tr}[N^{q}]\right)$ by concavity
  
  Then $\text{Tr}[M^{p}]^{\frac{1}{p}}\cdot\text{Tr}[N^{q}]^{\frac{1}{q}}\leq\frac{1}{p} \text{Tr}[M^{p}]+\frac{1}{q}\text{Tr}[N^{q}]$ by concavity
  
  Thus we have $\text{Tr}[MN] \;\le\; \underbrace{\text{Tr}[M^{p}]^{\frac{1}{p}}\text{Tr}[N^{q}]^{\frac{1}{q}}} _{\text{Hölder bound}}\;\le\; \underbrace{\frac{1}{p}\text{Tr}[M^{p}] + \frac{1}{q}\text{Tr}[N^{q}]} _{\text{Young RHS}}$
  Since hypothesis, all three quantities must be equal.
  - Equality in the concavity step (strict concavity of log) forces $\text{Tr}[M^{p}] = \text{Tr}[N^{q}].$
  - Equality in Hölder for Schatten norms (with $M,N\in\text{Pos}(A)$) implies there exists $c\ge 0$ such that $M^{p}= c\,N^{q}$
  Taking traces of this relation gives $\text{Tr}[M^{p}] = c\,\text{Tr}[N^{q}]$. Using $\text{Tr}[M^{p}] = \text{Tr}[N^{q}]$ yields $c=1$.
  
  Hence $M^{p} = N^{q}.$
Let $\left|\Phi^{A B}\right\rangle:=\frac{1}{\sqrt{d}} \sum_{z \in[d]}|z z\rangle$ be a 2-qudit (normalized) maximally entangled state in $A B \cong \mathbb{C}^{d} \otimes \mathbb{C}^{d}$. Consider a family of $d^{2}$ states in $A B$ defined by $\left|\psi_{xy}^{AB}\right\rangle=T^{x}\otimes S^{y}\left|\Phi^{AB}\right\rangle ,\,x,y\in[d]$ where $T$ and $S$ are the phase and shift operators defined by $T|z\rangle=e^{i \frac{2 \pi z}{d}}|z\rangle$ and $S|z\rangle=|z \oplus 1\rangle$, where $\oplus$ is the plus modulo $d$, and $z \in[d]$.
1. Show that $\left\{\left|\psi_{x y}^{A B}\right\rangle\right\}_{x, y \in[d]}$ is an orthonormal basis of $A B$.
  
  Proof
  
  $|\psi_{xy}^{AB}\rangle= T^x\otimes S^y\,|\Phi^{AB}\rangle= \frac{1}{\sqrt d}\sum_{z\in[d]} T^x|z\rangle \otimes S^y|z\rangle= \frac{1}{\sqrt d}\sum_{z\in[d]} e^{i\frac{2\pi x z}{d}}\,|z\rangle\otimes|z\oplus y\rangle$
  
  Then $\langle \psi_{x'y'}|\psi_{xy}\rangle = \frac{1}{d}\sum_{z',z\in[d]}e^{-i\frac{2\pi x' z'}{d}}e^{i\frac{2\pi x z}{d}}\langle z'|\langle z'!\oplus! y'|\,\big(|z\rangle\otimes|z!\oplus! y\rangle\big) \ = \frac{1}{d}\sum_{z\in[d]}e^{i\frac{2\pi z (x-x')}{d}}\langle z!\oplus! y'|z!\oplus! y\rangle = \frac{1}{d}\,\delta_{y,y'}\sum_{z\in[d]}e^{i\frac{2\pi z (x-x')}{d}} = \frac{1}{d}\,\delta_{y,y'}\sum_{z}e^{i\frac{2\pi z (x-x')}{d}} $
  
  $= \frac{1}{d}\,\delta_{y,y'} \sum_{z} e^{i\frac{2\pi z (x-x')}{d}}\quad\begin{cases}\text{if } x'=x,\ \langle \psi_{xy}|\psi_{x'y'}\rangle=\delta_{y,y'} \\\text{if } x'\neq x,\ \langle \psi_{xy}|\psi_{x'y'}\rangle=0\end{cases}$ 2. Show that the reduced density matrix of $\left|\psi_{x y}^{A B}\right\rangle$ is the maximally mixed state for all $x, y \in[d]$.
  
  Proof
  
  $\operatorname{Tr}_B\!\left[\,|\psi_{xy}^{AB}\rangle\langle\psi_{xy}^{AB}|\,\right]= \operatorname{Tr}_B\!\left[\frac{1}{d}\sum_{z,z'\in[d]} e^{i\frac{2\pi x (z-z')}{d}}\big(|z\rangle\langle z'|\big)\otimes\big(|z\oplus y\rangle\langle z'\oplus y|\big)\right]$
  
  $= \frac{1}{d}\sum_{z,z'} e^{i\frac{2\pi x (z-z')}{d}} |z\rangle\langle z'|\,\operatorname{Tr}\!\left[|z\oplus y\rangle\langle z'\oplus y|\right]= \frac{1}{d}\sum_{z} |z\rangle\langle z|= \frac{1}{d}\,I$ 3. Find a protocol for faithful teleportation of a qudit from Alice's lab to Bob's lab. Assume that the joint measurement that Alice performs on her two qudits is a basis measurement in the basis $\left\{\left|\psi_{x y}^{A B}\right\rangle\right\}_{x, y \in[d]}$.
  What are the unitary operators performed by Bob? How many classical bits (cbits) Alice transmits to Bob?
  
  Idea: Discuss about initial state, measurement, probability, post-measurement state, then define unitary.
  
  Proof
  
  Initial state $|\psi\rangle^{A^{\prime}AB}=|\psi\rangle^{A^{\prime}}\otimes|\Phi\rangle^{AB}=|\psi \rangle^{A^{\prime}}\otimes\frac{1}{\sqrt{d}}\sum_{z^{\prime}\in[d]}|z^{\prime}z^{\prime} \rangle^{AB}.$
  
  Measurements $\Pi_{x,y}^{A^{\prime}A}=|\psi_{xy}^{A^{\prime}A}\rangle\langle\psi_{xy}^{A^{\prime}A} |,\sum_{x,y}\Pi_{x,y}^{A^{\prime}A}=I^{A^{\prime}A},\Pi_{x,y}\Pi_{x^{\prime},y^{\prime}} =\delta_{x,x^{\prime}}\delta_{y,y^{\prime}}\Pi_{x,y}.$
  
  Probability $P(x,y)=\Pr(x,y)=\langle\psi|^{A^{\prime}AB}\,\Pi_{x,y}^{A^{\prime}A}\otimes I^{B} \,|\psi\rangle^{A^{\prime}AB}=\frac{1}{d^{2}}$
  
  Post‑measurement state $|\varphi_{x,y}^{A^{\prime}AB}\rangle=\frac{1}{\sqrt{P(x,y)}}\,\Big(\Pi_{x,y}^{A^{\prime}A} \otimes I^{B}\Big)\,|\psi\rangle^{A^{\prime}AB}.$
  
  Assume $|\psi\rangle^{A'}=\sum_{t\in[d]}\alpha_{t}|t\rangle.$
  
  Then $|\varphi^{A'AB}_{x,y}\rangle= d\,|\psi^{A'A}_{x y}\rangle\otimes\Big(\langle \psi ^{A'A}_{x y}|\otimes I^{B}\Big)\,|\psi\rangle^{A'AB}.$
  
  $= d\,|\psi^{A'A}_{x y}\rangle\otimes \left(\frac{1}{d}\sum_{t,z'z\in[d]}\,\omega^{-x z}\,\alpha_{z}\lang z|t\rang\lang z\oplus y|z'\rang|z'\rang^{B}\right)$
  
  $=d|\psi_{xy}^{A^{\prime}A}\rangle\otimes\frac{1}{d}\sum_{z}\alpha_{z}\omega^{-xz} |z\oplus y\rangle^{B}= |\psi^{A'A}_{x y}\rangle \otimes \Big(S^{y}T^{-x}|\psi\rangle ^{B}\Big)$ where $|\psi\rangle=\sum_{z}\alpha_z|z\rangle$.
  
  Thus $U_{x,y}=(S^{y}T^{-x})^{-1}=T^{x}S^{-y}.$ Note: Reserve $|\psi_{xy}^{A^{\prime}A}\rangle$ can reduce the complexity of computation
  
  Alice sends $(x,y) \in [d]\times[d]$, which is $2 \log_2 d$ cbits.
  
  Since $4 \ \rightarrow\ 2\ \text{b.ts}$, $9 \ \rightarrow\ \log_2 9\ \text{b.ts}$ and $\log d^{2} \ \Rightarrow\ 2\log_2 d\ \text{b.ts}$
Let $\rho \in \mathcal{D}(A)$ be a density matrix.
1. Show that $\sqrt{\rho}\otimes I^{\tilde{A}}|\Omega^{A\tilde{A}}\rangle$ is a purification of $\rho$.
  
  Proof
  
  First $\sqrt{\rho}\otimes I|\Omega\rangle=(\sqrt{\rho}\otimes I)(\sum_{x=1}^{|A|}|x\rangle |x\rangle ))=\sum_{x=1}^{|A|}\sqrt{\rho}|x\rangle\otimes|x\rangle=|\psi^{A\tilde{A}}\rangle$
  
  Then $\text{Tr}_{\tilde{A}}(|\psi^{A\tilde{A}}\rangle\langle\psi^{A\tilde{A}}|) = \text{Tr}_{\tilde{A}} ((\sum_{x=1}^{|A|}\sqrt{\rho}| x\rangle \otimes |x\rangle) (\sum_{y=1}^{|A|}\langle y|\sqrt{\rho}\otimes \langle y |))$ since $\rho$ is hermitian
  $= \text{Tr}_{\tilde{A}}(\sum_{x,y}\sqrt{\rho}|x\rangle \langle y| \sqrt{\rho}|x\rangle \langle y|) = \sum_{x,y}^{|A|}\sqrt{\rho}|x\rangle \langle y| \sqrt{\rho}\,\text{Tr} ( |x\rangle \langle y|)$
  $= \sum_{x,y} \sqrt{\rho} |x\rangle \langle y| \sqrt{\rho} \otimes \delta_{x,y}$
  $= \sum_{x} \sqrt{\rho}|x\rangle \langle x| \sqrt{\rho}= \sqrt{\rho}\sum_{x} |x\rangle \langle x| \sqrt{\rho} = \rho$ 2. Show that $|\psi^{AB}\rangle \in AB$ and $|\phi^{AC}\rangle \in AC$ (assuming $|B| \leq |C|$ ) are two purifications of $\rho \in \mathcal{D}(A)$ if and only if there exists an isometry matrix $V: B \to C$ such that $|\phi^{AC}\rangle = I^{A}\otimes V^{B \to C}|\psi^{AB}\rangle.$
  
  Idea: $\Leftarrow$) Note $I^A\otimes V^{B\to C}$ is a whole(conjugate)
  
  $\Rightarrow$) Discuss that we can use schmidt decomposition in the same basis of same space. Then define $V$ and prove it's isometry($\lang Vv|Vv\rang =\lang v|v\rang$)
  
  Proof
  
  $\Rightarrow$)
  
  First we prove it's possible to decompose like this $|\psi^{AB}\rangle=\sum_{x=1}^{|A|}\sqrt{p_{x}}|u_{x}^{A}\rangle|v_{x}^{B}\rangle$ and $|\phi^{AC}\rangle=\sum_{x=1}^{|A|}\sqrt{q_{x}}|u_{x}^{A}\rangle|z_{x}^{C}\rangle$
  By schmidt decomposition: $|\psi^{AB}\rangle=\sum_{x=1}^{|A|}\sqrt{p_{x}}|u_{x}^{A}\rangle|v_{x}^{B}\rangle$, then $\rho^{A}=\text{Tr}_{B}[\psi^{AB}]=\sum_{x=1}^{|A|}p_{x}|u_{x}\rangle\langle u_{x} |^{A}$
  Then $|u_x\rang$ are eigenvectors of $\rho$ with eigenvalues $p_x$.
  Since the reduced density matrix of $|\psi\rang$ and $|\phi\rang$ is $\rho$, then the eigenvalue should be same.
  If all eigenvalues of $\rho$ are distinct, then the eigenbasis is unique since eigenbasis is orthonormal basis. Then the orthonomal basis in $A$ should be same
  If there are some repeated eigenvalues, the eigenbasis are not unique. Then we are allowed to choose the same one.
  
  Thus by schmidt decomposition: $|\phi^{AC}\rangle=\sum_{x=1}^{|A|}\sqrt{q_{x}}|u_{x}^{A}\rangle|z_{x}^{C}\rangle$ and $\rho^{A}=\text{Tr}_{C}[\phi^{AC}]=\sum_{x=1}^{|A|}q_{x}|u_{x}\rangle\langle u_{x} |^{A}$
  
  Then $\sum_{x=1}^{|A|}p_{x}|u_{x}\rangle\langle u_{x}|^{A}=\sum_{x=1}^{|A|}q_{x}|u_{x} \rangle\langle u_{x}|^{A}$, then $p_x=q_x$
  
  Then $|\phi^{AC}\rangle=\sum_{x=1}^{|A|}\sqrt{p_{x}}|u_{x}^{A}\rangle|z_{x}^{C}\rangle$
  
  Also $(I^{A}\otimes U)|\psi^{AB}\rangle=(I^{A}\otimes U)\sum_{x=1}^{|A|}\sqrt{p_{x}}|u _{x}^{A}\rangle|v_{x}^{B}\rangle=\sum_{x=1}^{|A|}\sqrt{p_{x}}|u_{x}^{A}\rangle\otimes U|v_{x}^{B}\rangle$
  
  We define $U:B\to C$ s.t. $U|v_{x}^{B}\rangle=|z_{x}^{C}\rangle$ for all $x$, let show this is a isometry.
  
  Since $\langle Uv|Uv\rangle=\langle\sum_{x}Uc_{x}v_{x}|\sum_{x}Uc_{x}v_{x}\rangle=\sum_{x} \langle Uc_{x}v_{x}|Uc_{x}v_{x}\rangle+\sum_{x\neq y}\langle Uc_{x}v_{x}|Uc_{y}v_{y} \rangle=\sum_{x}||c_x||^2\langle z_{x}|z_{x}\rangle+\sum_{x\neq y}\bar c_xc_y\langle z_{x}|z_{y}\rangle =\sum_x||c_x||^2+0=\sum_x||c_x||^2=\langle v|v\rangle$, thus $U$ is a isometry.
  
  Thus we complete the proof
  
  $\Leftarrow$) Assume $|\psi^{AB}\rangle$ is a purification of $\rho$, then $\text{Tr}_B[|\psi^{AB}\rang\lang \psi^{AB}|]=\rho$
  
  Since $|\phi^{AC}\rangle = I^{A}\otimes V^{B \to C}|\psi^{AB}\rangle$, then $\text{Tr}_{C}(|\phi^{AC}\rangle\langle\phi^{AC}|)=\text{Tr}_{C}\left(I^{A}\otimes V^{B\to C}\psi^{AB}\cdot\left(I^{A}\otimes V^{B\to C}\right)^{*}\right)=\text{Tr} _{B}\left(I^{A}\otimes I\psi^{AB}\right)=\text{Tr}_{B}(\psi^{AB})=\rho$ 3. Use Part 2 to provide alternative (simpler!) proof of the claim in Exercise 2.27.
  Exercise 2.27. Let $\left\{\left|\psi_{x}\right\rangle, p_{x}\right\}_{x \in[m]}$ and $\left\{\left|\phi_{y}\right\rangle, q_{y}\right\}_{y \in[n]}$ be two ensembles of quantum states in A with $m \geq n$. Show that they correspond to the same density matrix $\rho:=\sum_{x \in[m]}p_{x}\left|\psi_{x}\right\rangle \langle\psi_{x}|= \sum_{y \in[n]}q_{y}| \phi_{y}\rangle\left\langle\phi_{y}\right|$ if and only if there exists an $m \times n$ isometry matrix $V=\left(v_{x y}\right)$ (i.e. $V^{*} V=I_{n}$) such that $\sqrt{p_{x}}\left|\psi_{x}\right\rangle=\sum_{y \in[n]} v_{x y} \sqrt{q_{y}}\left|\phi_{y}\right\rangle \quad \forall x \in[m] .$
  
  Idea: Define $|\Psi\rang \text{ and }|\Phi\rang$(schmidt decomposition) and using (2)
  
  Proof
  
  Using Schmidt Decomposition
  
  Define purifications on orthonormal bases $\{|y\rangle\}_{y=1}^n$ and $\{|x\rangle\}_{x=1}^m$:
  $|\Psi^{AB}\rangle=\sum_{y=1}^n \sqrt{q_y}\,|\phi_y\rangle^A\otimes|y\rangle^B,\quad|\Phi^{AC}\rangle=\sum_{x=1}^m \sqrt{p_x}\,|\psi_x\rangle^A\otimes|x\rangle^C.$
  Then easily $\operatorname{Tr}_B|\Psi\rangle\langle\Psi|=\sum_{y}q_y|\phi_y\rangle\langle\phi_y|,\quad\operatorname{Tr}_C|\Phi\rangle\langle\Phi|=\sum_{x}p_x|\psi_x\rangle\langle\psi_x|.$
  
  $\Rightarrow$) Since $n=|B|\le |C|=m$, by Part 2 there exists an isometry $V:B\to C$ such that
  $|\Phi^{AC}\rangle=(I^A\otimes V^{B\to C})\,|\Psi^{AB}\rangle.$
  Since this, then $V^{B\to C}=\sum_{x\in[m]}\sum_{y\in [n]}v_{xy}|x\rang \lang y|$, then $V|y\rangle=\sum_{x=1}^m v_{xy}\,|x\rangle$.
  Then $\sum_{x=1}^{m} \sqrt{p_{x}}\,|\psi_{x}\rangle\otimes|x\rangle=\sum_{y=1}^{n} \sqrt{q_{y}} \,|\phi_{y}\rangle\otimes V|y\rangle=\sum_{x=1}^{m} \Big(\sum_{y=1}^{n} v_{xy}\sqrt{q_{y}} \,|\phi_{y}\rangle\Big)\otimes|x\rangle$
  Then $\sqrt{p_{x}}\,|\psi_x\rangle=\sum_{y=1}^n v_{xy}\sqrt{q_{y}}\,|\phi_y\rangle\quad\forall x\in[m]$
  
  $\Leftarrow$) If there exists an $m\times n$ isometry $V=(v_{xy})$ with the above relations, then
  $\sum_{x}p_{x} |\psi_{x}\rangle\langle\psi_{x}|= \sum_{x}\sum_{y,y'}v_{xy}v_{xy'} ^{*}\sqrt{q_{y}q_{y'}}\,|\phi_{y}\rangle\langle\phi_{y'}|= \sum_{y,y'}(V^{\ast} V )_{y'y}\sqrt{q_{y}q_{y'}}\,|\phi_{y}\rangle\langle\phi_{y'}|= \sum_{y}q_{y} |\phi _{y}\rangle\langle\phi_{y}|$ since $V^\ast V=I_n$.
  Hence the ensembles yield the same density matrix.
Let $A \cong \mathbb{C}^n$ be a Hilbert space $A$, and let $\text{Herm}(A)$ be the Hilbert space consisting of all Hermitian matrices on $A$. Give an example of a basis (not necessarily orthogonal) of $\text{Herm}(A)$ consisting of pure density matrices in $\mathfrak{D}(A)$.

Solution

Claim: $\text{Herm}(A) = \text{span} \left\{\left\{|x\rangle\langle x|\right\}_{x=1}^{n}\cup \left\{\left (|x\rangle+|y\rangle\right)\left(\langle x|+\langle y|\right)\right\}_{x \neq y}\cup \left\{\left(|x\rangle+i|y\rangle\right)\left(\langle x|-i\langle y|\right)\right \}_{x \neq y}\right\}$

Another Problem in Practical Exam

Give an example of an informationally complete POVM in $\mathrm{Herm}(A)$.

$A=\sum_{i}a_{i}|e_{i}\rangle\langle e_{i}|+\sum_{i<j}b_{ij}(|e_{i}\rangle\langle e_{j}|+|e_{j}\rangle\langle e_{i}|)+\sum_{i>j}c_{ij}\cdot i\cdot(|e_{i}\rangle\langle e_{j}|-|e_{j}\rangle\langle e_{i}|)$
Let $\Lambda_{\alpha_{1i}}=|e_{i}\rangle\langle e_{i}|,\Lambda_{\alpha_{2ij}}=\left(|e_{i}\rangle +|e_{j}\rangle\right)\left(\langle e_{j}|+\langle e_{i}|\right),\Lambda_{\alpha_{3ij}}= \left(i|e_{i}\rangle+|e_{j}\rangle\right)\left(\langle e_{j}|-i\langle e_{i}|\right )$
Then let $\Lambda=\sum_{i}\Lambda_{\alpha_{1i}}+\sum_{i<j}\Lambda_{\alpha_{2ij}}+\sum_{i>j} \Lambda_{\alpha_{3ij}}=\sum_\alpha\Lambda_\alpha$

We define $\tilde{\Lambda}_\alpha=\Lambda^{-\frac{1}{2}}\Lambda_\alpha\Lambda^{-\frac{1}{2}}$ is a IC-POVM since it spans the $\text{Herm}(A)$ and $\sum_\alpha\tilde{\Lambda}_\alpha=I$
Suppose Alice and Bob share the state $|\psi^{AB}\rangle = \frac{1}{2}|00\rangle + \frac{\sqrt{3}}{2}|11\rangle$. Show that there exists a 2-outcome (basis) measurement that Alice can perform, such that with some probability greater than zero, the state of Alice and Bob after the measurement becomes the maximally entangled state $|\Phi_{+}^{AB}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$.

Idea: Define $M_0$ and calculate $M_1$, then use it. Remember the probability formula for generalized measurement is $p_0=\lang \psi |M_0^*M_0\otimes I^B|\psi\rang$ and post-measurement state is $\frac{1}{\sqrt{p_0}}M_0\otimes I^B|\psi\rang$

Proof

We cannot perform element projection since $p_{x} = |\psi\rangle\langle\psi|$, then $p_{x}\otimes I(\frac{1}{2}|00\rangle + \frac{\sqrt{3}}{2}|11\rangle)=|\psi\rang\otimes(...)$ is not entangled

Alice performs a 2-outcome generalized measurement: $M_{0}= \begin{pmatrix} 1 & 0 \\[2 pt] 0 & \tfrac{1}{\sqrt{3}} \end{pmatrix},M_{1}=\sqrt{I-M_{0}^{*}M_{0}}= \begin{pmatrix} 0 & 0 \\[2 pt] 0 & \sqrt{\tfrac23} \end{pmatrix}$
This is a valid 2-outcome POVM since $M_{0}^{*} M_{0}+M_{1}^{*} M_{1}=I$.
The probability is $p_{0} \;=\; \|(M_{0}\otimes I)|\psi\rangle\|^{2}= \left(\tfrac{1}{2}\right)^{2} + \left(\tfrac{\sqrt{3}}{2}\cdot\tfrac{1}{\sqrt{3}}\right)^{2}= \tfrac{1}{4}+\tfrac {1}{4}=\tfrac{1}{2}$
The post-measurement state is $\frac{(M_{0}\otimes I)\,|\psi^{AB}\rangle}{\sqrt{p_{0}}}=\frac{|00\rangle+|11\rangle}{\sqrt{2}} =|\Phi_{+}^{AB}\rangle.$
Consider four qubit systems $A, B, C$, and $D$, in the double-singlet state $|\Psi_{-}^{AB}\rangle \otimes |\Psi_{-}^{CD}\rangle$.
1. Show that a joint Bell measurement on system $BC$ generates a maximally entangled state on system $AD$ (along with another maximally entangled state on system $BC$) for all four possible outcomes of the measurement.
  
  Idea: Express $|\Psi_{-}^{AB}\rangle \otimes |\Psi_{-}^{CD}\rangle$ and regroup in $AD\otimes BC$, then express it in bell basis and perform bell measurement
  
  Proof
  
  Let $|\Phi_{\pm}\rangle=\tfrac{1}{\sqrt{2}}(|00\rangle\pm|11\rangle),|\Psi_{\pm}\rangle =\tfrac{1}{\sqrt{2}}(|01\rangle\pm|10\rangle).$
  Then $|\Psi_-^{AB}\rangle|\Psi_-^{CD}\rangle=\tfrac{1}{2}\big(|0_A1_B0_C1_D\rangle-|0_A1_B1_C0_D\rangle-|1_A0_B0_C1_D\rangle+|1_A0_B1_C0_D\rangle\big).$
  $=\tfrac{1}{2}\big(|01\rangle_{AD}|10\rangle_{BC}-|00\rangle_{AD}|11\rangle_{BC}- |11\rangle_{AD}|00\rangle_{BC}+|10\rangle_{AD}|01\rangle_{BC}\big).$
  Bell basis: $|00\rangle=\tfrac{1}{\sqrt{2}}(|\Phi_{+}\rangle+|\Phi_{-}\rangle),|11\rangle=\tfrac {1}{\sqrt{2}}(|\Phi_{+}\rangle-|\Phi_{-}\rangle),|01\rangle=\tfrac{1}{\sqrt{2}}(| \Psi_{+}\rangle+|\Psi_{-}\rangle),|10\rangle=\tfrac{1}{\sqrt{2}}(|\Psi_{+}\rangle -|\Psi_{-}\rangle),$
  Then $|\Psi_-^{AB}\rangle|\Psi_-^{CD}\rangle=\tfrac{1}{2}\Big(-|\Phi_+^{AD}\rangle|\Phi_+^{BC}\rangle+|\Phi_-^{AD}\rangle|\Phi_-^{BC}\rangle+|\Psi_+^{AD}\rangle|\Psi_+^{BC}\rangle-|\Psi_-^{AD}\rangle|\Psi_-^{BC}\rangle\Big).$
  
  So $|\Phi_+\rangle^{BC} \rightarrow |\Phi_+\rangle^{AD}$, $|\Phi_-\rangle^{BC} \rightarrow |\Phi_-\rangle^{AD}$, $|\Psi_+\rangle^{BC} \rightarrow |\Psi_+\rangle^{AD}$, $|\Psi_-\rangle^{BC} \rightarrow |\Psi_-\rangle^{AD}$ each with probability $\tfrac{1}{4}$. 2. Show that the singlet state in $AD$ can be generated by quantum teleportation between system $BC$ and system $D$.
  
  Idea: The singlet state is $|\Psi_{-}\rang^{AD}$, then we need to measure on $BC$ and send classical message to $D$, then perform unitary. It's better to remember the order
  
  Proof
  
  Using Pauli matrices $\sigma_{1}= \begin{pmatrix} 0 & 1 \\[2 pt] 1 & 0 \end{pmatrix},\sigma_{2}: \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix},\sigma_{3}: \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
  - First apply a Bell measurement on $BC$. Then send a classical outcome $x$ to $D$:
  If $BC$ gets $|\Psi_{-}\rangle$, send $0$
  
  If $BC$ gets $|\Phi_{-}\rangle$, send $1$
  
  If $BC$ gets $|\Phi_{+}\rangle$, send $2$
  
  If $BC$ gets $|\Psi_{+}\rangle$, send $3$ - Corrections at $D$ (Pauli-type operators $\sigma_0,\sigma_1,\sigma_2,\sigma_3$ on $D$):
  
  If $D$ gets $0$, apply $\sigma_0$: $(I^{A}\!\otimes\!\sigma_{0}^{D})\,|\Psi_{-}\rangle^{AD}=|\Psi_{-}\rangle^{AD}$
  
  If $D$ gets $1$, apply $\sigma_1$: $(I^{A}\!\otimes\!\sigma_{1}^{D})\,|\Phi_{-}\rangle^{AD}=|\Psi_{-}\rangle^{AD}$
  
  If $D$ gets $2$, apply $\sigma_{2}$: $(I^{A}\!\otimes\!\sigma_{2}^{D})\,|\Phi_{+}\rangle^{AD}=i|\Psi_{-}\rangle^{AD}$
  
  If $D$ gets $3$, apply $\sigma_{3}$: $(I^{A}\!\otimes\!\sigma_{3}^{D})\,|\Psi_{+}\rangle^{AD}=-|\Psi_{-}\rangle^{AD}$
Suppose Alice and Bob share a composite quantum system in the state $\rho \in \mathcal{D}(A B)$. Alice performs a measurement on her system described by a POVM $\left\{\Lambda_{x}\right\}_{x=1}^{m}$, and record the outcome $x$ in a classical system $X$ (with $|X|=m)$. Show that the post-measurement state can be expressed as a classical-quantum state of the form $\sigma^{X B}=\sum_{x=1}^{m} p_{x}|x\rangle\langle x|^{X} \otimes \sigma_{x}^{B} .$
Express the probabilities $p_{x}$, and density matrices $\sigma_{x}^{B}$, in terms of $\Lambda_{x}$ and $\rho^{A B}$.

Idea: Directly express $\sigma^{XB}$ where in $B$ there is the postmeasurement state: $\mathrm{Tr}_{A}\left\lbrack \left(\sqrt{\Lambda_{x}}\otimes I_{B}\right)\rho^{AB}\left(\sqrt{\Lambda_{x}}\otimes I_{B}\right)^{*}\right\rbrack$

Proof

If Alice measures $A$ with POVM $\{\Lambda_x\}_{x=1}^m$ and stores the outcome in a classical register $X$, the joint post-measurement state is $\sigma^{XB}=\sum_{x=1}^{m}|x\rangle\!\langle x|^{X}\otimes\mathrm{Tr}_{A}\left\lbrack \left(\sqrt{\Lambda_{x}}\otimes I_{B}\right)\rho^{AB}\left(\sqrt{\Lambda_{x}}\otimes I_{B}\right)^{*}\right\rbrack$
Since $\rho^{AB}=\sum_{y}\eta_{y}^{A}\otimes\zeta_{y}^{B}$, then $\mathrm{Tr}_{A}\left\lbrack \left(\sqrt{\Lambda_{x}}\otimes I_{B}\right)\left(\sum _{y}\eta_{y}^{A}\otimes\zeta_{y}^{B}\right)\left(\sqrt{\Lambda_{x}}\otimes I_{B}\right )^{*}\right\rbrack =\mathrm{Tr}_{A}\left\lbrack\sum_{y}\sqrt{\Lambda_{x}}\eta_{y} ^{A}\sqrt{\Lambda_{x}}\otimes\zeta_{y}^{B}\right\rbrack =\sum_{y}\text{Tr}[\sqrt{\Lambda_{x}} \eta_{y}^{A}\sqrt{\Lambda_{x}}]\cdot \zeta_{y}^{B}=\sum_{y}\text{Tr}[\Lambda_{x}\eta _{y}^{A}]\cdot \zeta_{y}^{B}$Then $=\sum_{x=1}^{m}|x\rangle\!\langle x|^{X}\otimes\operatorname{\mathrm{Tr}} _{A}\!\left[(\Lambda_{x}\otimes I_{B})\,\rho^{AB}\right]=\sum_{x=1}^{m}p_{x}\,|x\rangle \!\langle x|^{X}\otimes\sigma_{x}^{B}$
Where $p_{x}=\operatorname{\mathrm{Tr}}\!\left[(\Lambda_{x}\otimes I_{B})\,\rho^{AB} \right]=\operatorname{\mathrm{Tr}}\!\left[\Lambda_{x}\,\rho^{A}\right],\sigma_{x} ^{B}=\frac{1}{p_{x}}\,\operatorname{\mathrm{Tr}}_{A}\!\left[(\Lambda_{x}\otimes I _{B})\,\rho^{AB}\right]$and $\rho^{A}=\operatorname{Tr}_B \rho^{AB}$.
Set $d := |A|$ and let $\{\Lambda_x\}_{x \in [d^2]}$ be an informationally complete POVM in Herm($A$). Denote also let its dual basis by $\{\Gamma_y\}_{y \in [d^2]}$.
1. Show that $\mathrm{Tr}[\Gamma_y] = 1$ for all $y \in [d^2]$.
  
  We know that $\mathrm{Tr}[\Lambda_{x}\Gamma_{y}]=\delta_{xy}$, then $\sum_{x}\mathrm{Tr}[\Lambda_{x}\Gamma_{y}]=\sum_{x}\delta_{xy}$, then $\mathrm{Tr}\!\left[\left(\sum_{x}\Lambda_{x}\right)\Gamma_{y}\right]=1$
  
  Thus $\mathrm{Tr}[\Gamma_{y}]=1$ since $\sum_{x}\Lambda_{x}=I$ 2. Show that at least one of the matrices in $\{\Gamma_y\}_{y \in [d^2]}$ is not positive semidefinite.
  
  Idea: $\text{Tr}[\Lambda_{x}\cdot\Gamma_{x}]\leq \text{Tr}[\Lambda_{x}]$, then to get contradiction
  
  Suppose $\forall y, \Gamma_y \succeq 0$. We know $\mathrm{Tr}[\Lambda_{x} \Gamma_{x}] =1$ and $\text{Tr}[\Lambda_{x}\cdot\Gamma_{x}]=\sum_{y}p_{y}\text{Tr}[\Lambda_{x}\psi_{y} ]\leq\sum_{y}\text{Tr}[\Lambda_{x}\cdot\psi_{y}]=\text{Tr}[\Lambda_{x}]$
  
  Thus $1\leq \text{Tr}[\Lambda_x]$, then $\sum_{x=1}^{d^2}\text{Tr}[\Lambda_{x}] \geq d^{2}$. But $\sum_{x=1}^{d^2}\text{Tr}[\Lambda_{x}] = \text{Tr}[\sum_{x=1}^{d^2}\Lambda_{x}] = d$
  
  Then $d \geq d^{2}$ is a contradiction. 3. Show that if each $\Lambda_x$ is rank 1, $a := \mathrm{Tr}[\Lambda_x] = \mathrm{Tr}[\Lambda_y]$, and $b := \mathrm{Tr}[\Lambda_x \Lambda_y] = \mathrm{Tr}[\Lambda_x \Lambda_z]$ for any 3 distinct numbers $x, y, z \in [d^2]$, then necessarily $a = \frac{1}{d} \quad \text{and} \quad b = \frac{1}{d^2(d+1)}.$
  
  We know $\mathrm{Tr}\!\left[\sum_{x=1}^{d^2}\Lambda_{x}\right] = d$. And $\mathrm{Tr}\!\left[\sum_{x=1}^{d^2}\Lambda_{x}\right]=\sum_{x=1}^{d^2}\mathrm{Tr} [\Lambda_{x}]=ad^{2}\Rightarrow ad^{2}=d\Rightarrow a=\frac{1}{d}$
  
  We know $\text{Tr}[\Lambda_x]=a$, then $\mathrm{Tr}[\Lambda_{x}\sum_{y=1}^{d^2}\Lambda_{y}]=a$
  
  $\;\Rightarrow\;\sum_{x=1}^{d^2}\operatorname{\mathrm{Tr}}[\Lambda_{x}\Lambda_{y}]=a$ $\Rightarrow b(d^{2}-1) + \operatorname{Tr}[\Lambda_{x}^{2}] = \frac{1}{d}$
  
  Since $\Lambda_x$ is rank 1, then $\text{Tr}[\Lambda_{x}^2]=a^{2}$ $\Rightarrow b(d^2-1) + a^2 = \frac{1}{d}$ $\Rightarrow b = \frac{\frac{1}{d} - a^2}{d^2 - 1}= \frac{\frac{1}{d} - \frac{1}{d^2}}{d^2 - 1}= \frac{1}{d^2(d+1)}$ 4. Show that if we remove from part (3) the condition $\mathrm{Tr}[\Lambda_x] = \mathrm{Tr}[\Lambda_y]$ for $x \neq y$ then $\mathrm{Tr}[\Lambda_x]$ can take only two values that depend only on $d$ and $b$; what they are?
  
  $a^{2}-a+bd^{2}-b=0\Rightarrow a=\frac{1\pm\sqrt{1-4bd^{2}+4b}}{2}$
Let $|\Phi^{AB}\rangle := \frac{1}{\sqrt{m}}\sum_{z\in[m]} |zz\rangle$ be a maximally entangled state shared between Alice and Bob. Show that by sending her quantum system A to Bob, Alice can transmit to Bob $2\log_2(m)$ classical bits.

Idea: Describe super dense coding

Proof

Alice encodes his message by unitary matrix. From above we know $U_{xy}=T^{x}S^{y}$ where $|\psi_{xy}^{AB}\rangle=(T^{x}S^{y}\otimes I)|\Phi^{AB}\rangle$ is the state that Alice send to Bob after encoding.

This is orthogonal since $\langle \Phi|(S^{y'})^{*}(T^{x'})^{*}T^{x}S^{y}\otimes I|\Phi \rangle= \frac{1}{m} \text{Tr}[T^{x-x'}S^{y-y'}] = \delta_{xx'}\delta_{yy'}$ which is orthogonal since $T,S$ is unitary then $S^y(S^{y'})^* = S^y S^{-y'} = S^{y-y'}$

Then Bob has $m^2$ possibilities: $\left\{|\psi_{xy}^{AB}\rangle=(T^{x}S^{y}\otimes I)|\Phi^{AB}\rangle\right\}_{x,y\in[m]\times [m]}$

Then Bob perform the measurement $\{|\underbrace{\psi_{xy}\rangle\langle\psi_{xy}}_{\pi_{xy}}|\}_{x,y\in\left\lbrack m\right\rbrack\times [m]}$ to get the $x,y$

Thus Alice can transmit to Bob $\log_{2}m^{2}=2\log_{2}(m)\text{ bits}$
Let $\psi^{A \tilde{A}} := |\psi\rangle\langle\psi|$ with $|\psi^{A \tilde{A}}\rangle = \sum_{x=1}^{|A|} \sqrt{p_x} |x\rangle|x\rangle \in A \otimes \tilde{A}$. Find the eigenvalues and eigenvectors of $\text{id}^A \otimes \mathcal{T}(\psi^{A \tilde{A}})$, where $\mathcal{T}$ is the transpose map. For which values of $p_x$, the matrix $\text{id}^A \otimes \mathcal{T}(\psi^{A \tilde{A}})$ is positive semidefinite?

Idea: Consider basis $B=\{\ket{11},\ket{22},...,\ket{dd},\ket{12},\ket{21},...,\}$ and group $|ij\rang,\ket{ji}$ to calculate eigenvalues

Proof

Let $d=\lvert A\rvert$ and $M:= (\mathrm{id}^A\otimes \mathcal T)(\psi^{A\tilde A})= \sum_{i,j=1}^{d}\sqrt{p_i p_j}\,\lvert i j\rangle\langle j i\rvert .$

In the ordered basis $\mathcal B=\big\{\lvert 11\rangle,\dots,\lvert dd\rangle,\; \lvert 12\rangle,\lvert 21\rangle,\; \lvert 13\rangle,\lvert 31\rangle,\;\dots\big\},$ the matrix of $M$ is block diagonal with
- $1\times 1$ blocks: $[p_i]$ on each $\lvert ii\rangle$,
- $2\times 2$ blocks on $\mathrm{span}\{\lvert ij\rangle,\lvert ji\rangle\}$ for $i<j$: $B_{ij}=\begin{pmatrix}0 & \sqrt{p_i p_j}\\\sqrt{p_i p_j} & 0\end{pmatrix}.$
Characteristic polynomial: $\chi_M(\lambda)=\det(\lambda I - M)= \prod_{i=1}^{d}(\lambda - p_i)\;\prod_{1\le i<j\le d}\det\!\begin{pmatrix}\lambda & -\sqrt{p_i p_j}\\-\sqrt{p_i p_j} & \lambda\end{pmatrix}= \prod_{i=1}^{d}(\lambda - p_i)\;\prod_{i<j}(\lambda^2 - p_i p_j).$

Eigenvalues (roots of $\chi_M$): $p_i$ for each $i$ or $\pm \sqrt{p_i p_j}$ for each $i<j$.
- $i=j$: $(M-p_i I)\lvert ii\rangle=0\quad\Rightarrow\quad\ker(M-p_i I)=\mathrm{span}\{\lvert ii\rangle\}.$
- $i\neq j$, write $v=\alpha\lvert ij\rangle+\beta\lvert ji\rangle$. Then $(M-\lambda I)v=\begin{pmatrix}-\lambda & \sqrt{p_i p_j}\\\sqrt{p_i p_j} & -\lambda\end{pmatrix}\begin{pmatrix}\alpha\\ \beta\end{pmatrix}=0.$
  Hence: For $\lambda=+\sqrt{p_i p_j}$: $\alpha=\beta$, $\ker(M-\lambda I)=\mathrm{span}\left\{\tfrac{1}{\sqrt2}(\lvert ij\rangle+\lvert ji\rangle)\right\}.$
For $\lambda=-\sqrt{p_i p_j}$: $\alpha=-\beta$, $\ker(M-\lambda I)=\mathrm{span}\left\{\tfrac{1}{\sqrt2}(\lvert ij\rangle-\lvert ji\rangle)\right\}.$

Hence $M\succeq 0$ iff $p_i p_j=0$ for all $i\ne j$, i.e., exactly one $p_r=1$ and the rest $0$.
Let $\mathcal{E}: \mathcal{L}(A) \rightarrow \mathcal{L}(B)$ be a linear map.
1. Show that $\mathcal{E}$ is trace preserving iff its dual $\mathcal{E}^{*}$ is unital; i.e. $\mathcal{E}^{*}(I^{B}) = I^{A}$.
  
  Idea: Express trace preserving in inner product. When we have $\lang \mathcal{E}^*(I)-I|\rho\rang =0$, let $\rho=\mathcal{E}^*(I)-I$ to get $\mathcal{E}^*(I)=I$
  
  Proof
  
  $\Rightarrow$) $\operatorname{Tr}\big[\mathcal{E}(\rho)\big]=\operatorname{Tr}[\rho]\ \Rightarrow\ \langle \mathcal{E}(\rho), I\rangle=\langle \rho, I\rangle\ \ (\forall \rho)$
  $\langle \mathcal{E}(\rho), I\rangle=\langle \rho, \mathcal{E}^{*}(I)\rangle=\langle \rho, I\rangle$
  $\forall \rho:\ \langle \rho,\ \mathcal{E}^{*}(I)-I\rangle=0,\ \ \text{take }\rho =\mathcal{E}^{*}(I)-I$ $\Rightarrow\ \mathcal{E}^{*}(I)=I.$
  
  $\Leftarrow$) $\operatorname{\mathrm{Tr}}[\mathcal{E}(\rho)]=\langle I\mid\mathcal{E}(\rho)\rangle =\langle\mathcal{E}^{*}(I)|\rho\rangle=\langle I|\rho\rangle=\operatorname{\mathrm{Tr}} [\rho]$. 2. Show that $\mathcal{E}$ is trace non-increasing $(i.e. \text{Tr}[\mathcal{E}(\eta)] \leq \text{Tr}[\eta] \text{ for all } \eta \in \text{Pos}(A))$ iff its dual $\mathcal{E}^{*}$ is sub-unital; i.e. $\mathcal{E}^{*}(I^{B}) \le I^{A}$.
  
  Idea: Similar to 1 but we also need to use $\rho=\sum\lambda_{i}|\psi_{i}\rang \lang \psi_{i}|$
  
  Proof
  
  $\Rightarrow$) We know $\operatorname{\mathrm{Tr}}[\mathcal{E}(\rho)] \le \operatorname{\mathrm{Tr}}[\rho]=1$.
  
  Then $\operatorname{\mathrm{Tr}}\!\left[(\mathcal{E}^{*}(I^{B})^{*}\rho\right]\le\text{Tr} [\rho]\implies\operatorname{\mathrm{Tr}}\!\left[((\mathcal{E}^{*}(I^{B}))^{*}-I^{A} )\,\rho\right]\le0$.
  
  Then $\langle \mathcal{E}^{*}(I^{B})-I^{A}\mid \rho\rangle \le 0,\ \forall \rho\ge 0$. Then $\rho=\sum\lambda_{i}|\psi_{i}\rang \lang \psi_{i}|$ with $\lambda_{i}\geq 0$
  
  Then $\sum\lambda_{i}\lang \psi_{i}|I-\mathcal{E}^{*} (I)|\psi_{i}\rang\geq 0$, then $\mathcal{E}^{*}(I^{B}) \le I^{A}$.
  
  $\Leftarrow$) $\operatorname{\mathrm{Tr}}[\mathcal{E}(\rho)]-\text{Tr}[\rho]=\langle\mathcal{E} ^{*}(I^{B})-I^{A}\mid\rho\rangle$ where $\rho\geq 0$ but $\mathcal{E}^{*}(I^{B})-I^{A}\leq 0$
  
  Then $\sum\lambda_{i}\lang \psi_{i}|\mathcal{E}^{*} (I)-I|\psi_{i}\rang\leq 0$, then $\langle\mathcal{E}^{*}(I^{B})-I^{A}\ket {\rho}\leq 0$, then $\operatorname{\mathrm{Tr}}[\mathcal{E}(\rho)]\leq\text{Tr}[\rho]$ 3. Show that $\mathcal{E}$ is positive if and only if $\mathcal{E}^{*}$ is positive.
  
  Idea: Definition and relate it to trace
  
  Proof
  
  $\Rightarrow$) We know $\mathcal{E}^{A\to B}(\rho^{A}) \ge 0,\ \forall \rho^{A}$. Then Use definition to prove.
  Then, $\langle\psi|\mathcal{E}^{*}(\rho^{B})|\psi\rangle=\text{Tr}[\psi\cdot\mathcal{E} ^{*}(\rho^{B})]=\text{Tr}[\mathcal{E}(\psi^{*})^*\cdot\rho^{B}]=\text{Tr}[\mathcal{E}(\psi^{*})\cdot \rho^B ]$ since $\mathcal{E}(\psi^*) \ge 0$.
  Then $\text{Tr}[\mathcal{E}(\psi^{*})\cdot\rho^{B}]=\sum_x\text{Tr}[p_x\psi_x\cdot\rho^{B}]=\sum_xp_x\lang\psi_x|\rho^{B}|\psi_x\rang\geq 0$ since spectral decomposition and $\rho \geq 0$
  
  $\Leftarrow$) Apply (⇒) on $\mathcal{E}^{*}$. 4. Show that $\mathcal{E}$ is completely positive if and only if $\mathcal{E}^{*}$ is completely positive.
  
  Proof
  
  We know $\mathcal{E}$ is positive iff $\mathcal{E}^{*}$ is positive.
  
  Then $\operatorname{id}^{R\to R}\otimes \mathcal{E}^{A\to B} = \mathcal{F}^{RA\to RB}$. Apply (3), we got $\mathcal{F}$ is positive iff $\mathcal{F}^{*}$ is positive.
  
  Then $\operatorname{id}^{R\to R}\otimes \mathcal{E}^{A\to B}$ is positive iff $(\operatorname{\mathrm{id}}^{R\to R}\otimes\mathcal{E}^{A\to B})^{*}=\operatorname{\mathrm{id}} ^{R\to R}\otimes\mathcal{E}^{*\,B\to A}$ is positive.
A generalized dephasing channel $\mathcal{N} \in \text{CPTP}(A \rightarrow A)$ is a channel that transmit some preferred basis $\{|x\rangle\}_{x \in [m]}$ of $A$ without error. That is, $\mathcal{N}(|x\rangle\langle x|) = |x\rangle\langle x| .$
1. Show that the completely dephasing channel $\Delta$ is a generalized dephasing channel.
  
  Proof
  
  We know $\Delta(\ket x\bra x)=\sum_{x'=1}^{|A|}\mathinner{\langle{x'}|}(\ket x\bra x)\mathinner{|{x'}\rangle}\mathinner {|{x'}\rangle}\mathinner{\langle{x'}|}=\sum_{x'=1}^{|A|}\mathinner{\langle{x'}|}\delta_{x,x'}\mathinner{|{x'}\rangle}\mathinner {|{x'}\rangle}\mathinner{\langle{x'}|}=\ket x\bra x$ 2. Show that $\mathcal{N}$ above has a Stinespring isometry $V_{\mathcal{N}}|x\rangle^{A} = |x\rangle^{A}|\phi_{x}^{E}\rangle \quad \forall x \in [m] ,$ where $\{|\phi_x^E\rangle\}_{x \in [m]}$ are some normalized vectors in $E$ (note that if $\{|\phi_x^E\rangle\}$ is an orthonormal set then $\mathcal{N} = \Delta$).
  
  Idea: By theorem: $\exists$ an isometry $V:A\to AE$ s.t. $\mathcal{N}(\rho)=\text{Tr}_{E}[V\rho V^{*}],\forall \rho\in\mathcal{L}(A)$
  Then combining with definition of $\mathcal{N}$, we get $\text{Tr}_{E}[V|x\rang\lang x |V^{*}]=|x\rang\lang x|$
  Then define $V|x\rangle^{A}= |\psi_{x}\rang$. NTP: $|\psi_{x}\rang=|x\rangle^{A}|\phi_{x}^{E}\rangle$
  Using schmidt decomposition $|\psi_{x}^{AE}\rang=\sum\sqrt{p_{y}}|y^{A}\rangle\left|\phi_{y}^{E}\rangle\right.$ to show that
  
  Proof
  
  Since $\mathcal{N}$ is a channel, then by theorem $\exists$ an isometry $V:A\to AE$ s.t. $\mathcal{N}(\rho)=\text{Tr}_{E}[V\rho V^{*}],\forall \rho\in\mathcal{L}(A)$
  
  By definition $\mathcal{N}(|x\rang\lang x|)=|x\rang\lang x|$, then $\mathcal{N}(|x\rang\lang x|)=\text{Tr}_{E}[V|x\rang\lang x |V^{*}]=|x\rang\lang x|$
  
  We define $|\psi_{x}\rangle^{AE}:=V|x\rangle$, then $\text{Tr}_{E}[\psi_{x}^{AE}]=|x\rangle\langle x|$
  
  We have $|\psi_{x}^{AE}\rang=\sum\sqrt{p_{y}}|y^{A}\rangle\left|\phi_{y}^{E}\rangle\right.$ by spectral decomposition
  
  Then $\text{Tr}_{E}[\psi_{x}^{AE}]=\sum p_{y}\ket y\bra y\cdot\text{Tr}_{E}[\phi_{y}^{E} ]=\sum p_{y}\ket y\bra y=\ket x\bra x$
  $\implies$When $y \ne x$, $p_{y} = 0$. When $y=x$, $p_{y} = 1$.
  
  Then $\psi_{x}^{AE}=\mathinner{|{x^{A}}\rangle}\mathinner{|{\phi_{x}^{E}}\rangle}\implies V\mathinner{|{x}\rangle}=\mathinner{|{x}\rangle}\mathinner{|{\phi_{x}^{E}}\rangle}$
  
  If $\{|\phi_x^E\rangle\}$ is an orthonormal set, then $\mathcal{N}(|x\rang\lang y|)=\text{Tr}_E[V|x\rang\lang y|V^*]=\text{Tr}_E[|x\rang|\phi_x\rang\cdot \lang y|\lang \phi_y|]$
  $=|x\rangle\langle y|\text{Tr}_{E}[|\phi_{x}\rangle\langle\phi_{y}|]=|x\rangle\langle y|\delta_{x,y}=0$ 3. Show that $\Delta \circ \mathcal{N}= \mathcal{N}\circ \Delta = \Delta .$
  
  Proof
  
  $\Delta(\rho)=\sum_{x=1}^{|A|}\mathinner{|{x}\rangle}\rho_{xx}\mathinner{\langle{x}|}$
  $\Delta\circ\mathcal{N}(\rho)=\Delta\left(\sum_{x,y=1}^{|A|}\rho_{xy}\text{Tr}_{E} [|\phi_{x}\rangle\langle\phi_{y}|]|x\rangle\langle y|\right)=\sum_{x,y=1}^{|A|}\rho _{xy}\text{Tr}_{E}[|\phi_{x}\rangle\langle\phi_{y}|]\cdot\Delta\left(|x\rangle\langle y|\right)$
  $=\sum_{x,y=1}^{|A|}\rho_{xy}\text{Tr}_{E}[|\phi_{x}\rangle\langle\phi_{y}|]\cdot |x\rangle\langle x|=\sum_{x=1}^{|A|}\mathinner{|{x}\rangle}\rho_{xx}\mathinner{\langle{x}|}$
  
  $\mathcal{N}(\Delta(\rho))=\mathcal{N}(\sum_{x=1}^{|A|}\mathinner{|{x}\rangle}\rho _{xx}\mathinner{\langle{x}|})=\sum_{x=1}^{|A|}\mathinner{|{x}\rangle}\rho_{xx}\mathinner {\langle{x}|}$
Show that a quantum channel $\mathcal{E} \in \text{CPTP}(A \rightarrow B)$ is a measurement-prepare channel if and only if it is entanglement breaking.

Idea: $\Rightarrow$) NTP: $\mathcal{E}^{A\to B}(\psi^{RA})$ is separable and use $|\psi^{RA}\rang =(M\otimes I^{A})|\Omega^{\tilde{A}A}\rang$ and change $\mathcal{E}(\Omega)$ to $J_{\mathcal{E}}^{AB}$ and use $J_{\mathcal{E}}^{AB}=\sum_{z=1}^{m}\Lambda_{z}^{T}\otimes\sigma_{z}^{B}$ of measurement-prepare channel, then change it to separable state
$\Leftarrow$) Use $J_{\mathcal{E}}^{AB}=\mathcal{E}^{\tilde{A}\rightarrow B}\left(\Omega^{A\tilde{A}} \right)$ to prove it, then check if the first one is POVM

Proof

$\Rightarrow$) Since $\rho$ is convex combination of pure states, then it's enough to prove it for all pure state

Then NTP: $\mathcal{E}^{A\to B}(\psi^{RA})$ is separable $\forall$ pure states $\psi^{RA}$

We know $|\psi^{RA}\rang =(M\otimes I^{A})|\Omega^{\tilde{A}A}\rang$, then $\mathcal{E}^{A\to B}(\psi^{RA})=(M\otimes I^{A})\mathcal{E}\left(\Omega^{\tilde{A}A} \right)(M\otimes I^{A})^{*}$

$=\left(M\otimes I^{A})J_{\mathcal{E}}^{AB}\right.(M\otimes I^{A})^{*}$, then since this, $=\left(M\otimes I^{A}\right)\left(\sum_{z=1}^{m}\Lambda_{z}^{T}\otimes\sigma_{z} ^{B}\right)\left(M\otimes I^{A}\right)^{*}$

$=\sum_{z=1}^{m}q_{z}\frac{M\Lambda_{z}M^{*}}{q_{z}}\otimes\sigma_{z}$ where $q_{z}:=\operatorname{Tr}\left[M \Lambda_{z} M^{*}\right]=\operatorname{Tr}\left[M^{*} M \Lambda_{z}\right]$

Thus this is separable

$\Leftarrow$) Let $J_{\mathcal{E}}^{AB}=\mathcal{E}^{\tilde{A}\rightarrow B}\left(\Omega^{A\tilde{A}} \right)=\sum_{z}q_{z}\omega_{z}^{A}\otimes\tau_{z}^{B}$, let $\eta_{z}^{A}:=q_{z}\omega_{z}^{A}$, then $J_{\mathcal{E}}^{AB}=\sum_{z}\eta_{z}^{A}\otimes\tau_{z}^{B}$.
Then we just check that $\eta_{z}^{A}$ is a POVM: $\text{Tr}_{B}\left\lbrack J_{\mathcal{E}}^{AB}\right\rbrack=\sum_{z}\eta_{z}^{A} \cdot\text{Tr}[\tau_{z}^{B}]=\sum_{z}\eta_{z}^{A}$
Since $\mathcal{E}$ is quantum channel, then $J_\mathcal{E}^A=I^A$. Thus $\sum_z\eta_z^A=I_A$
Consider a POVM channel $\mathcal{E} \in \mathrm{CPTP}(A \rightarrow B)$ s.t. $\mathcal{E}(\rho) := \sum_{y=1}^{|B|} \operatorname{Tr}\left[\Lambda_{y} \rho^{A}\right]|y\rangle\langle y|^{B}$ where $\left\{\Lambda_{y}\right\}_{y=1}^{|B|}$ is a POVM in $\operatorname{Pos}(A)$.
1. Find an operator sum representation (i.e. a Kraus decomposition) of $\mathcal{E}$.
  
  Idea: $\Lambda\to \sqrt{\Lambda}\sqrt{\Lambda}$ and expand trace
  
  Proof
  
  Let $m:= |B|$, then $\mathcal{E}(\rho) = \sum_{y=1}^{m}\text{Tr}[\Lambda_{y}\rho] |y\rang\lang y|$ $=\sum_{y=1}^{m}\text{Tr}[\sqrt{\Lambda_{y}}\rho\sqrt{\Lambda_{y}}]|y\rang\lang y|$
  $= \sum_{y=1}^{m}\sum_{x=1}^{d}\lang x|\sqrt{\Lambda_{y}}\rho\sqrt{\Lambda_{y}}|x\rang |y\rang\lang y|$ $=\sum_{x,y}\underbrace{|y\rangle\langle x|\sqrt{\Lambda_{y}}}_{M_{xy}}\rho\underbrace{\sqrt{\Lambda_{y}}|x\rangle\langle y|}_{M_{xy}^*}$
  $\left\{M_{xy}\right\}_{y\in\{m\},\,x\in\{d\}}$ check that $\sum_{x,y}M_{xy}^{*}M_{xy}=\sum_{x,y}\sqrt{\Lambda_{y}}|x\rangle\langle y|y\rangle \langle x|\sqrt{\Lambda_{y}}=I$ 2. Find a Stinespring representation of $\mathcal{E}$.
  
  Idea: $\mathcal{E}(\rho) = \sum M_{z}\rho M_{z}^{*}= \text{Tr}_{E}[V \rho V^{*}]$ and just define $V$
  
  $\mathcal{E}(\rho) = \sum M_{z} \rho M_{z}^{*}$ $= \text{Tr}_{E}[V \rho V^{*}]$ where $V = \sum_{z} M_{z} \otimes |z\rangle$
  
  Then define $V = \sum_{x,y} |y\rangle \langle x| \sqrt{\Lambda_y} \otimes |xy\rangle^E$ and $|E|=md$. And $V^*V=I$
Consider $n$ bipartite states $\left\{\left|\psi_{z}^{A B}\right\rangle\right\}_{z \in[n]}$ in $\mathbb{C}^{d} \otimes \mathbb{C}^{d}$. Find an optimal state $\left|\phi^{A B}\right\rangle \in \mathbb{C}^{d} \otimes \mathbb{C}^{d}$ with the following two properties:
1. The state $\left|\phi^{A B}\right\rangle$ can be converted by LOCC to $\left|\psi_{z}^{A B}\right\rangle$, for all $z \in[n]$.
  
  We know given $|\psi_{z}^{AB}\rangle=\sum_{x=1}^{d}\sqrt{p_{x}^{z}}|\phi_x\rang\otimes |\psi_x\rangle,\,\,|\phi^{AB}\rangle =\sum_{x=1}^{d}\sqrt{q_{x}}|\phi_x'\rang\otimes |\psi_x'\rangle$, then $\phi^{AB}\xrightarrow{LOCC}\psi_{z}^{AB}\,\text{ if and only if }\,\,\vec{p}^{z} \succ\vec{q}$
  
  Then $\vec{p}^{z}\succ\vec{q}$ 2. If another state, $\left|\chi^{A B}\right\rangle$, can be converted by LOCC to $\left|\psi_{z}^{A B}\right\rangle$, for all $z \in[n]$, then $\left|\chi^{A B}\right\rangle$ can also be converted to $\left|\phi^{A B}\right\rangle$. That is, $\left|\phi^{A B}\right\rangle$ is optimal.
  
  Idea: use $\min S_k^z$ and let $q_i=s_i-s_{i-1}$
  
  Given $\left|\chi^{A B}\right\rangle=\sum^{d}_{x=1}\sqrt{r_{x}}|\phi_x''\rang\otimes |\psi_x''\rangle$, then same reason if $\vec{p}^{z}\succ\vec{r}$, then $\vec q\succ\vec r$
  
  Consider $\vec{p_z}= (p_{1}^{z}, p_{2}^{z}, \dots, p_{d}^{z})$, $S_{1}^{z}= p_{1}^{z}, \dots, S_{d}^{z}= \sum_{x=1}^{d}p_{x}^{z}$
  
  $\vec{p_{z}}\succ\vec{r}\Rightarrow p_{1}^{z}>r_{1}(S_{1}^{z}\geq r_{1}),\,S_{2}^{z} \ge r_{1}+r_{2},\dots,S_{d}^{z}\ge r_{1}+\dots+r_{d}$
  
  Let $\min_{z \in [n]}S_{k}^{z}= s_{k}$.
  Since $\vec{p}^{z}\succ\vec{r}$, then $s_{1}\ge r_{1},s_{2}\ge r_{1}+r_{2},\dots,s_{d}\ge r_{1}+\dots+r_{d}$
  
  And we want $\vec q\succ\vec r$
  
  Since $\vec{p}^{z}\succ\vec{q}$, then $s_{1} \ge q_{1}, s_{2} \ge q_{1} + q_{2}, \dots, s_{d} \ge q_{1} + \dots + q_{d}$
  
  Then we should let $q_{1}=s_{1},q_{1}+q_{2}=s_{2},...,q_{1}+...+q_{d}=s_{d}$
  
  Then $q_{i}=s_{i}-s_{i-1},\forall i\in [d],s_{0}:=0$, this can imply $\vec q\succ\vec r$
  
  Thus $|\phi^{AB}\rangle=\sum_{x=1}^{d}\sqrt{s_{i}-s_{i-1}}|xx\rangle$ where $s_{i}=\min_{z\in[n]}S_{i}^{z}$ and $S_i^z=\sum^i_{x=1}p_x^z$
Consider the entangled state shared between Alice and Bob, $|\Psi^{AB}\rangle = \sqrt{p_{1}}|1\rangle^{A}|1\rangle^{B}+ \dots + \sqrt{p_{n}} |n\rangle^{A}|n\rangle^{B}$ where $p_x > 0$ for all $x \in [n]$.
Prove the following theorem: Quantum teleportation of a $d$-dimensional qudit is possible if and only if none of the Schmidt coefficients are greater than $1/d$. This also implies that the Schmidt rank $n$ is greater or equal to $d$.

Proof

$|\Psi^{AB}\rangle=\sum_{i=1}^{n}\sqrt{p_{i}}|i\rangle^{A}|i\rangle^{B}$ and consider $|\Phi_{d}^{AB}\rangle=\sum_{i=1}^{d}\frac{1}{\sqrt{d}}|i\rangle^{A}|i\rangle^{B} =\frac{1}{\sqrt{d}}\sum_{i=1}^{d}|i\rangle^{A}|i\rangle^{B}$.

$\Leftarrow$) Suppose $p_{x}\le \tfrac{1}{d},\ \forall x\in[m]$. Then $p \prec \big(\tfrac{1}{d},\ldots,\tfrac{1}{d},0,\ldots,0\big)$

By Nielsen's Theorem $\Psi^{AB}\xrightarrow{\mathrm{LOCC}}\Phi^{AB}_{d}$. Then we can use Quantum Teleportation with $\Phi^{AB}_{d}$ to teleport a $d$-dimensional system.

$\Rightarrow$) Suppose $\Psi^{AB}$ can be used to teleport a qudit. We need to show $p_{x}\leq \frac{1}{d}$

It's possible that $|0\rang\lang 0|\xrightarrow{LOCC}\Phi_{d}^{A''A'}$ since we can throw $|0\rang\lang 0|$ and prepare $\Phi_{d}^{A''A'}$ since we just do local things

We prepare an entangled qudit state on $A'',A'$ and send one of it to $B$ by quantum teleportation.

$\Psi^{AB}\xrightarrow{LOCC}\Phi_{d}^{A''B}$

$LOCC\downarrow$ $\uparrow LOCC$

$\Phi_{d}^{A''A'}\otimes \Psi^{AB}\xrightarrow{LOCC}\Phi_{d}^{A''B}\otimes \omega^{AA'}$

Then by Nielsen's Theorem $p \prec \big(\tfrac{1}{d},\ldots,\tfrac{1}{d},0,\ldots,0\big)$

Then $p_{1}\le \frac{1}{d}$ and we know $p_{1}\geq p_{2}\geq ...\geq p_{n}$. Then $\frac{1}{d}\geq p_{1}\geq p_{2}\geq ...\geq p_{n}$.
Thus $p_{x}\leq \frac{1}{d}$.

Also by $p \prec \big(\tfrac{1}{d},\ldots,\tfrac{1}{d},0,\ldots,0\big)$, we get $p_{1}\le \frac{1}{d}$, $p_{1}+p_{2}\leq \frac{2}{d}$, . . . , $\sum_{i=1}^{k}p_{i}\le \frac{k}{d},\forall k \le n$$\implies 1 \le \frac{n}{d}$ when $k=n$. Then $d \le n$

Exercise in Lectures

Homework 1: Let A be a normed space, and let $\psi, \phi \in A$ with $\|\phi\|=1$. Show that
$\left\|\frac{1}{\|\psi\|}\psi-\phi\right\| \leq 2\|\psi-\phi\|$

Here

Idea: From left to right, we want get $\|\psi-\phi\|$, thus we construct such in left: $\left\|\frac{1}{\|\psi\|}\psi-\psi+\psi-\phi\right\|$
Homework 2: Show that $P\geq 0$ iff $P=M^*M$ for some complex matrix $M$

Here

From left to right: $P=U^*\sqrt D \sqrt DU$

From right to left: Definition of positive
Homework 4: Show that $H_\pm=\frac{|H|\pm H}{2}$ where $|H|:=\sqrt{H^*H}=\sqrt{H^2}$

Here

Hint $|H|=U|D|U^{*}$
Homework 6: If $P$ is Hermitian and $\text{Tr}P^{3}=\text{Tr}P^{2}=\text{Tr}P=1$, then $P$ is pure state

Here

Idea: Consider $\text{Tr}P^{2}-\text{Tr}P^{3}=0$, then use trace to get the scope of $\lambda_x^2\text{ and }1-\lambda_x$
Homework 1: $\hat{n}=(\sin\alpha\cos\beta, \sin\alpha\sin\beta, \cos\alpha)^{T}$. Prove:
1. $S_{\hat{n}}^2=\frac{1}{4}I$
2. The eigenvalues of $S_{\hat{n}}$ are $\pm\frac{1}{2}$
3. Show that if $S_{\hat{n}}|\psi\rangle=\frac{1}{2}|\psi\rangle$, then $|\uparrow_{\hat{n}}\rangle=|\psi\rangle$ $= \cos(\frac{\alpha}{2})|0\rangle+e^{i\beta}\sin(\frac{\alpha}{2})|1\rangle$
Here
Homework 2: $\forall$ unit vectors $\hat{n}$, $|\Psi^{AB}\rangle = \frac{1}{\sqrt{2}}(|\uparrow_{\hat{n}}\rangle^{A} |\downarrow _{\hat{n}}\rangle - |\downarrow_{\hat{n}}\rangle^{A} |\uparrow_{\hat{n}}\rangle)$

Here
1. Given $\{M_x\}_{x\in[m]}, \{N_y\}_{y\in[n]}$ - Generalized measurements. Prove that $\{M_x N_y\}_{x,y}$ is also a generalized measurement.
2. Given $M_{0}=a|+\rangle\langle0|,M_{1}=b|0\rangle\langle+|$. Find conditions on $a,b \in \mathbb{C}$ s.t. $\exists M_2$ with $\{M_0, M_1, M_2\}$ are Generalized Measurements.
Here
H.W. Show that for every Unitary matrix $\mathcal{E}(U\rho U^{*}) = U\mathcal{E}(\rho)U^{*}$ where $\mathcal{E}$ is depolarizing channel

Here
H.W. $(U \otimes \bar{U}) J_{\varepsilon}^{AB}(U \otimes \bar{U})^{*} = J_{\varepsilon} ^{AB}$ $\forall$Unitary matrices $U$

Here
H.W. Let $\mathcal{E} \in \text{CPTP}(A \rightarrow B)$ be a cq-channel as above, and set $m := |A|$ and $n := |B|$.
1. Show that $\{M_{xy}\}$ with $M_{xy} : A \rightarrow B$ and $M_{xy}:=\sqrt{\sigma_{x}}|y^{B}\rangle\langle x^{A}|,x\in[m]\,,\,y\in[n]$ forms an operator sum representation of $\mathcal{E}$.
  
  Here 2. Show that the map $V: A \rightarrow B \otimes \tilde{A}\tilde{B}$ given by $V=\sum_{x\in[m]}\left(\sqrt{\sigma_{x}^{B}}\otimes I^{\tilde{B}}\right)|\Omega^{B\tilde{B}} \rangle\otimes|x^{\tilde{A}}\rangle\langle x^{A}|$ is a Stinespring isometry (the environment system is $\tilde{A}\tilde{B}$) satisfying $\text{Tr}_{\tilde{A}\tilde{B}}[V\rho V^{*}]=\sum_{x\in[m]}p_{x}\sigma_{x}=\mathcal{E} (\rho)\text{ and }V^{*}V=I^{A}$
  
  Stinespring isometry(an isometry $V:A\to BE$ s.t. $\mathcal{E}(\rho)=\text{Tr}_{E}[V\rho V^{*}],\forall \rho\in\mathcal{L}(A)$)
  
  Here

$\lVert \psi \rVert_{p} = \big(\, |a_{1}|^{p} + \cdots + |a_{n}|^{p} \,\big)^{1/p}$ and $\lVert M \rVert_{p} = \big(\, \mathrm{Tr}\,|M|^{p} \,\big)^{1/p}$

$\sigma_{0} = I$, $\sigma_{1} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, $\sigma_{2} = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$, $\sigma_{3} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

To rotate a qudit, formula: let $\hat{n}=(\sin\alpha\cos\beta, \sin\alpha\sin\beta, \cos\alpha)^{T}, S_{\hat{n}}: = \frac{1}{2}\hat{n}\cdot\vec{\sigma}, T^{(\hat{n})}_{\theta} = e^{-i\theta S_{\hat{n}}}$
Example of Qubit
Let $|0\rangle = |\uparrow_z\rangle$, $|1\rangle = |\downarrow_z\rangle$
Then by formula, $|\uparrow_{x}\rangle = T^{\hat{y}}_{\pi/2}|0\rangle$, just like $\hat{x}=R^{(\hat{y})}_{(\pi/2)}\hat{z}$
$=T_{\pi/2}^{\hat{y}}|\uparrow_{z}\rangle=e^{-i\frac{\pi}{2}\hat{y}\cdot\vec{\sigma}} |0\rangle=e^{-i\frac{\pi}{4}\sigma_2}|0\rangle$ since $\hat{y}=(0,1,0)$ and $\vec{\sigma}=(\sigma_1, \sigma_2, \sigma_3)$
$= (\cos(\frac{\pi}{4})I - i\sin(\frac{\pi}{4})\sigma_2)|0\rangle = \left[\frac{\sqrt{2}}{2}I - i\frac{\sqrt{2}}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\right]|0\rangle$
$= \frac{\sqrt{2}}{2}\left(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\right)\begin{pmatrix} 1 \\ 0 \end{pmatrix}$
$= \frac{\sqrt{2}}{2}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \frac{\sqrt{2}}{2} \times \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$

$|\Psi_-\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$ is called singlet state

Measurements in open systems: Given $\{M_x\}_{x\in[m]}$ - Generalized measurement and $\rho$ - initial state
Probability for outcome $x$: $p_{x}=\mathrm{Tr}[M_{x}\rho M_{x}^{*}]$
Post-measurement state after outcome $x$ occurred is $\sigma_{x} = \frac{M_{x} \rho M_{x}^{*}}{p_{x}}$

$\mathcal{E}\in CPTP(A\to B)\iff J_{\mathcal{E}}^{AB}\geq 0 \text{ and }J_{\mathcal{E}} ^{A}=I^{A}$ where $J_{\mathcal{E}}^{A}:=\text{Tr}_{B}[J_{\mathcal{E}}^{AB}]$
completely positive and trace preserving

The Choi matrix of a linear map $\mathcal{E} \in \mathcal{L}(A \to B)$ is

\[ J_{\mathcal{E}}^{AB}:=\mathcal{E}^{\tilde{A}\to B}(\Omega^{A\tilde{A}})\quad(*) \]

Given Choi matrix $J_{\mathcal{E}}^{AB}\in C^{d d' \times d d'}$, we can define $\mathcal{E}$

\[ \forall\rho\in C^{d\times d}\cong L(A),\quad\mathcal{E}^{A\to B}(\rho^{A}):=\text{Tr} _{A}[J_{\mathcal{E}}^{AB}(\rho^{A})^{T}\otimes I^{B}]\quad (**) \]

A linear map $\mathcal{E}:\mathcal{L}(A)\to \mathcal{L}(B)$ is CPTP (i.e. a quantum channel) iff there exists a generalized measurement $\{M_x\}_{x\in [m]}$(satisfying $\sum^m_{x=1}M_xM^*_x=I$ automatically) such that $\forall \rho\in \mathcal{L}(A),\mathcal{E}(\rho)=\sum^{m}_{x=1}M_{x} \rho M_{x}^{*}$ (operator sum representation) where $M_x:A\to B$

Given $\mathcal{E}\in \mathcal{L}(A\to B)$. $\mathcal{E}$ is a CPTP (i.e. Quantum Channel) iff $\exists E$ (Hilbert space) with $|E|\leq |AB|$ and an Stinespring's isometry $V:A\to BE$ s.t. $\mathcal{E}(\rho)=\text{Tr}_{E}[V\rho V^{*}],\forall \rho\in\mathcal{L}(A)$

$V:=\sum^{m}_{z=1}M_{z}\otimes |z\rang^{E}$

The Completely Dehasing Channel: $\Delta(\rho):=\sum_{x=1}^{|A|}\mathinner{\langle{x}|}\rho\mathinner{|{x}\rangle} \mathinner{|{x}\rangle}\mathinner{\langle{x}|}$

Preparation Channel: $\mathcal{E}(\rho)=\mathcal{E}\circ \Delta(\rho)=\sum^{d}_{x=1}\lang x|\rho|x\rang \mathcal{E}(|x\rang\lang x|)=\sum^{d}_{x=1}\lang x|\rho|x\rang\sigma_{x}$ where $\sigma_x:=\mathcal{E}(|x\rangle\langle x|)$
$J_{\mathcal{E}}^{A\tilde{A}}=\mathcal{E}^{A\to \tilde{A}}=\mathcal{E}\circ\Delta (\Omega^{A\tilde{A}})=\sum_{x,y}|x\rang\lang y|\otimes \mathcal{E}\circ\Delta(|x\rang \lang y|)$ $ =\sum_{x}|x\rangle\langle x|\otimes\sigma_{x}$

Measurement-Prepare Channel: $\mathcal{E}:=\mathcal{P}^{X\rightarrow B}\circ M^{A\rightarrow X}$ where $M^{A\to X}(\rho)=\sum_{x=1}^{m}\operatorname{\mathrm{Tr}}[\underbrace{\Lambda_{x}} _{POVM}\rho]\underbrace{|x\rangle\langle x|^{X}}_{\text{Classical Outcome}}$

Then $\mathcal{E}(\rho) = \mathcal{P}\cdot M(\rho)$ $=\sum_{x=1}^{m}\operatorname{\mathrm{Tr}}[\Lambda_{x}\rho]\mathcal{P}(|x\rangle\langle x|)$
Let $\mathcal{E}(\rho)=\sum_{z=1}^{m}\operatorname{\mathrm{Tr}}\left[\Lambda_{z}\rho\right ]\sigma_{z}^{B}$, then Choi Matrix: $J_{\mathcal{E}}^{AB}=\sum_{z=1}^{m}\Lambda_{z}^{T}\otimes\sigma_{z}^{B}$

classical channel

$\mathcal{E}\in CPTP(A\to A)$ is called a classical channel if $\Delta\circ \mathcal{E}\circ \Delta=\mathcal{E}$

We only consider the diagonal input since $\Delta$ is completely Dehasing Channel.

$\sigma:=\mathcal{E}(\rho)=\mathcal{E}\left(\sum^{d}_{x=1}t_{x}|x\rang\lang x|\right )=\sum^{d}_{x=1}t_{x}\mathcal{E}(\ket{x}\bra{x})$ where $\rho = \begin{pmatrix} t_{1} & & 0 \\ & \ddots & \\ 0 & & t_{d} \end{pmatrix}$

Then $\sigma$ is also diagonal since $\mathcal{E}(\rho)=\Delta(\mathcal{E}(\rho))$ by definition

Then $\sum_{y=1}^{d} S_{y}|y\rangle\langle y| = \sum_{x=1}^{d} t_x \mathcal{E}(|x\rangle\langle x|)$, then after multiply $\bra{y'},\ket{y'}$ on both sides, we get $S_{y}=\sum_{x=1}^{d}t_{x}\underbrace{\langle y|\mathcal{E}(|x\rangle\langle x|)|y\rangle} _{p_{y|x}}$ where $p_{y|x}:= \langle y|\mathcal{E}(|x\rangle\langle x|)|y\rangle$

This satisfy $\vec{S} = E\vec{t}$. Also $\Delta(\sigma)=\sigma$