9.4 LOCC and Mojorization

Entanglement Theory of Pure States

\(E(\psi^{AB}) \geq E(\phi^{AB})\) if \(\psi^{AB} \xrightarrow{\text{LOCC}} \phi^{AB}\) by definition of LOCC

Let consider Reversible LOCC: \(U^A \otimes V^B |\psi^{AB}\rangle := |\phi^{AB}\rangle \implies E(\psi^{AB}) \geq E(\phi^{AB})\)

Also \(U^*\otimes V^*\) is LOCC, then \(U^* \otimes V^* |\phi^{AB}\rangle = |\psi^{AB}\rangle \implies E(\phi^{AB}) \geq E(\psi^{AB})\)
Thus \(E(\psi^{AB}) = E(\phi^{AB})\)

All the states in \(\left\{U\otimes V|\psi\rangle\right\}_{\text{U,V-unitaries.}}\) have the same entanglement as \(\psi^{\text{AB}}\)
By Schmidt Decomposition: \(|\psi^{\text{AB}}\rangle=\sum_{x=1}^{d}\sqrt{P_{x}}|Ux\rangle^{\text{A}}|Vx\rangle ^{\text{B}}=U\otimes V|\tilde{\psi}^{\text{AB}}\rangle\)
Let \(|\tilde{\psi}^{\text{AB}}\rangle:=\sum_{x=1}^{d}\sqrt{P_{x}}|x\rangle^{\text{A}} |x\rangle^{\text{B}}\)
Thus there is a probability function related to entanglement: \(E(\psi^{\text{AB}}) = f(\vec{P}) \quad \vec{P} = \{P_{x}\}_{x\in[d]}\)

\(f(1,0,...,0)=0\) since we will get \(|x\rang |x\rang\) which is not entangled

---

​

Proposition

We are trying to prove we can get \(\tilde{\phi}^{AB}\) by LOCC from \(\tilde{\psi}^{AB}\), then since \(U\) and \(V\) is reversible, we can get \(\phi^{AB}\) by LOCC from \(\psi^{AB}\)

Let \(|\tilde{\psi}^{AB}\rang=\sum_x\sqrt{p_x}|xx\rang =\sqrt{\rho}\otimes I^B|\Omega^{AB}\rang\) and \(|\tilde{\phi}^{AB}\rang=\sum_y\sqrt{q_y}|yy\rang=\sqrt{\sigma}\otimes I|\Omega^{AB}\rang\) where \(\rho= \begin{pmatrix} p_{1} & & & \\ & p_{2} & & \\ & & \ddots & \\ & & & p_{d} \end{pmatrix}>0\) and \(\sigma= \begin{pmatrix} q_{1} & & & \\ & q_{2} & & \\ & & \ddots & \\ & & & q_{d} \end{pmatrix}>0\)

After outcome \(x\)-occurred, the post-measurement state is \(\frac{1}{\sqrt{r_x}}M_x\otimes U_x|\tilde{\psi}\rang^{AB}=|\tilde{\phi}\rang^{AB},\forall x\) where \(r_x:=\lang \tilde{\psi}^{AB}|M_x^*M_x\otimes I|\tilde \psi^{AB}\rang\)

\(\iff\frac{1}{\sqrt{r_{x}}}(M_{x}\otimes U_{x})(\sqrt{\rho}\otimes I^{B})|\Omega^{AB} \rangle=(\sqrt{\sigma}\otimes I^{B})|\Omega^{AB}\rangle\)

\(\iff\frac{1}{\sqrt{r_{x}}}(M_{x}\sqrt{\rho}\otimes U_{x})|\Omega^{AB}\rangle=(\sqrt{\rho} \otimes I)|\Omega^{AB}\rangle\)

\(\iff\left(\frac{1}{\sqrt{r_{x}}}M_{x}\sqrt{\rho}U_{x}^{T}\otimes I\right)|\Omega^{AB} \rangle=(\sqrt{\sigma}\otimes I)|\Omega^{AB}\rangle\)

Thus we need to ensure \(\frac{1}{\sqrt{r_{x}}}M_{x}\sqrt{\rho}U_{x}^{T}=\sqrt{\sigma}\iff M_{x}\sqrt{\rho} =\sqrt{r_{x}}\sqrt{\sigma}V_{x}\text{ where }V_{x}:=\bar{U}_{x}\)
\(\iff M_{x}=\sqrt{r_{x}}\sigma^{1/2}V_{x}\rho^{-1/2}\)

Once \(M_x\) satisfy this, we can get \(\tilde{\phi}^{AB}\) by LOCC from \(\tilde{\psi}^{AB}\). But we need to ensure \(M_x\) is a measurement

Check: \(I=\sum_{x}M_{x}^{*}M_{x}=\sum^{m}_{x=1}r_{x}\rho^{-\frac{1}{2}}V^{*}_{x}\sigma^{\frac{1}{2}} \sigma^{\frac{1}{2}}V_{x}\rho^{-\frac{1}{2}}=\sum_{x=1}^{m}r_{x}\rho^{-\frac{1}{2}} V_{x}^{*}\sigma V_{x}\rho^{-\frac{1}{2}}\)

Then we mutiply \(\rho^{\frac{1}{2}}(...)\rho^{\frac{1}{2}}\) on both sides, we get \(\rho=\sum^m_{x=1}r_xV_x^*\sigma V_x\)

Thus we can \(\tilde{\phi}^{AB}\) by LOCC from \(\tilde{\psi}^{AB}\) iff we can find unitary \(V_x\) and probability vectors such that \(\rho=\sum^m_{x=1}r_xV_x^*\sigma V_x\)

Let \(K_x=\sqrt{r_x}V_x^*\), then \(\rho=\mathcal{E}(\sigma)=\sum_{x}K_{x}\sigma K_{x}^{*}\) is a quantum channel

Definition

\(\vec{q}\succ\vec{p}\;\char"27FA \;q_{1}^{\downarrow}\geq p_{1}^{\downarrow},\,\, q_{1}^{\downarrow}+q_{2}^{\downarrow}\geq p_{1}^{\downarrow}+p_{2}^{\downarrow},\, \,...\,\,\sum_{x=1}^{k}q_{x}^{\downarrow}\geq\sum_{x=1}^{k}p_{x}^{\downarrow}\) where \(\vec p=\begin{pmatrix} p_1\\p_2\\\vdots\\p_d \end{pmatrix},\vec p^\downarrow=\begin{pmatrix} p_1^\downarrow\\p_2^\downarrow\\\vdots\\p_d^\downarrow \end{pmatrix}\) and \(p_1^\downarrow\geq ...\geq p_d^\downarrow,\,\,p_x^\downarrow=p_{\pi(x)}\)

We can see that \(\begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}\succ \begin{pmatrix} p_{1} \\ p_{2} \\ \vdots \\ p_{d} \end{pmatrix}\succ \begin{pmatrix} \frac{1}{d} \\ \frac{1}{d} \\ \vdots \\ \frac{1}{d} \end{pmatrix}\), this is a partial order since \(\begin{pmatrix} \frac{2}{5} \\ \frac{2}{5} \\ \frac{1}{10} \\ \frac{1}{10} \end{pmatrix}\begin{matrix} \nprec\\\nsucc \end{matrix} \begin{pmatrix} \frac{1}{2} \\ \frac{1}{4} \\ \frac{1}{4} \\ 0 \end{pmatrix}\)

We know \(p_1\geq \frac{1}{d}\), then \(p_1+p_2\geq \frac{2}{d}\iff \frac{1}{2}(p_1+p_2)\geq \frac{1}{d}\) and since \(\min(p_1+p_2)=\frac{2}{d}\), then it's true

Games of Chance(Majorization)

If we need to guess one number and \(p_{1}^{\downarrow} > q_{1}^{\downarrow}\) we choose the black dice.

If we need to guess two number and \(p_1^\downarrow + p_2^\downarrow \ge q_1^\downarrow + q_2^\downarrow\) we choose black dice.

If we need to guess all numbers and \(\vec{p}\succ\vec{q}\) we choose black dice.

Pre-Lemma

\(\vec{p}\succ\vec q\iff \exists D\in \R^{d\times d}_{+}\,\,\text{ s.t }\,\,\vec{q}= D\vec {p}\) where \(D - \text{doubly stochastic matrix}\)

---

Example of doubly stochastic matrix
\(D = \begin{pmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{4} & \frac{1}{2} \end{pmatrix}\) \(D = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}\) \(D= \begin{pmatrix} p & 1-p \\ 1-p & p \end{pmatrix}\)

They can be written as convex combination

• \(D= \begin{pmatrix} p & 1-p \\ 1- p& p \end{pmatrix}=p \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}+(1-p) \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\)
• \(D = \begin{pmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{4} & \frac{1}{2} \end{pmatrix}=\frac{1}{2}\underbrace{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}}_{P_1}+\frac{1}{4}\underbrace{ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}}_{P_2}+\frac{1}{4}\underbrace{ \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}}_{P_3}\)

Thus \(D_3=\sum^3_{x=1}p_xP_x\)

Theorem(Birkhoff)

\(D\) is doubly stochastic iff it can be written as convex combination of permutation matrix
​\(D=\sum_{x=1}^{m}p_{x}\prod_{x}\) where \(\prod_{x}\) is permutation matrix and \(p_x\) is probability

‍