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9.5

Nielsen Majorization Theorem

Given \(|\psi^{AB}\rang =\sum^d_{x=1}\sqrt{p_x}|xx\rang,\,\,|\phi^{AB}\rang =\sum^d_{x=1}\sqrt{q_x}|xx\rang\), then

\[ \psi^{AB}\xrightarrow{LOCC}\phi^{AB}\,\text{ if and only if }\,\,\vec q\succ\vec p \]

Proof

Let \(\rho= \begin{pmatrix} p_{1} & & & \\ & p_{2} & & \\ & & \ddots & \\ & & & p_{d} \end{pmatrix}>0\) and \(\sigma= \begin{pmatrix} q_{1} & & & \\ & q_{2} & & \\ & & \ddots & \\ & & & q_{d} \end{pmatrix}>0\)

So far we showed that \(\psi^{AB}\xrightarrow{LOCC}\phi^{AB}\) iff \(\exists \vec r\in\text{Prob}(m)\) and unitary matrices \(\{U_x\}\) such \(\rho=\sum_{x\in[m]}r_xU_x\sigma U_x^*\)

To finish the proof, we show that \(\rho=\sum_{x\in[m]}r_{x}U_{x}\sigma U_{x}^{*}\iff\vec q\succ \vec p\).

\(\Rightarrow\)) We first prove a lemma

Lemma

If \(\rho=\mathcal{E}(\sigma)\) and \(\mathcal{E}(I)=I\), then \(\sigma\succ\rho\)

Proof

The eigenvalues of \(\rho,\sigma\) are probability vectors since \(\rho,\sigma\) are density matrix(positive and trace 1)

Let \(\sum_xp_x\psi_x\) and \(\sigma=\sum q_x\phi_x\) are spectral decomposition

Then \(p_x=\bra{\psi_x}\rho\ket{\psi_x}=\bra{\psi_x}\mathcal{E}(\sigma)\ket{\psi_x}=\sum^d_{y=1}q_y\underbrace{\bra{\psi_x}\mathcal{E}(\phi_y)\ket{\psi_x}}_{a_{xy}}\)

Since \(\mathcal{E}(\phi_y)\) is positive, then \(a_{xy}\geq 0\), \(\sum^{d}_{x=1}a_{xy}=\text{Tr}[\mathcal{E}(\phi_y)]=1\) and \(\sum^d_{y=1}a_{xy}=1\) since \(\mathcal{E}(\sum_y\phi_y)=\mathcal{E}(I)=I\)

Thus \(A=[a_{xy}]_{x,y\in [d]}\) is doubly stochastic matrix, then \(p_{x}=(A\vec{q})_{x}\implies\vec{p}=A\vec{q}\)

Since this, then we get \(\vec{q}\succ\vec{p}\)

Then continuing proving: We apply this Lemma
Since \(\rho=\sum_{x\in[m]}r_{x}U_{x}\sigma U_{x}^{*}\) is a quantum channel, then \(\rho=\mathcal{E}(\sigma)\)
Also, \(\mathcal{E}(I)=\sum_{x\in[m]}r_{x}U_{x}IU_{x}^{*}=\sum_{x\in[m]}r_{x}I=I\) since \(r_x\) is probability.
Thus \(\vec{q}\succ\vec{p}\)

\(\Leftarrow\)) Suppose \(\vec{q}\succ\vec{p}\implies\vec p=D\vec q=\sum^{m}_{x=1}r_{x}\Pi_{x}\vec q\) since this and this respectively

Observing that \(\rho=\sum^m_{x=1} r_x\Pi_x\sigma \Pi_x^*\), then since \(\Pi_x\) is permutation, then unitary

Example

\(\prod_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\) and \(0 \le p \le 1\) such that \(\prod_x \begin{pmatrix} p \\ 1-p \end{pmatrix} = \begin{pmatrix} 1-p \\ p \end{pmatrix}\)

Then \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} p & 0 \\ 0 & 1-p \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1-p & 0 \\ 0 & p \end{pmatrix}\)

Consequence

If \(\vec p= \begin{pmatrix} \frac{1}{d} \\ \vdots \\ \frac{1}{d} \end{pmatrix}\), then \(\psi^{AB}=\frac{1}{\sqrt d}\sum^d_{x=1}|xx\rang=|\Phi^{AB}\rang\). Since \(\vec{q}\succ \begin{pmatrix} \frac{1}{d} \\ \vdots \\ \frac{1}{d} \end{pmatrix}\), then \(\Phi^{AB}\xrightarrow{LOCC}\phi^{AB}\)

Quantification of Entanglement

A function \(E: \text{Pure}(AB) \longrightarrow \mathbb{R}\) is a measure of entanglement if:

  1. \(E(\psi^{AB}) \geq E(\phi^{AB})\) whenever \(\psi^{AB}\xrightarrow{LOCC}\phi^{AB}\)

  2. \(E(|0\rangle\langle 0| \otimes |0\rangle\langle 0|) = 0\)

Properties of Entanglement measures

  1. \(E(\psi^{AB}) \ge 0\), \(\forall\) \(\psi^{AB}\in \text{Pure}(AB)\)

  2. Invariance Under Unitaries: \(E((U \otimes V)\psi^{AB}(U \otimes V)^*) = E(\psi^{AB})\)

  3. \(E(\psi^{AB}) = f(\vec{p})\) where \(\psi^{AB}= \sum_{x} \sqrt{p_{x}}|u_{x}\rangle|v_{x}\rangle\)

Thus \(f(\vec{p}) \geq f(\vec{q})\) whenever \(\vec{q} \succ \vec{p}\) (Schur-Concave)

Examples of measures of entanglement

  1. Entropy of Entanglement

    \(E(\psi^{AB}):=H(\vec p):=-\sum^{d}_{x=1}p_{x}\log_2p_{x}\) where \(\vec p\) is eigenvalues of reduced density matrix of \(\psi\)

    Example: \(E(\Phi^{AB})=\log_2(d)\) where \(|\Phi^{AB}\rang=\frac{1}{\sqrt d}\sum^d_{x=1}|x\rang^A|x\rang^B\) and \(E(\Phi_+^{AB})=\log_2(2)=1\text{ebit}\)

  2. \(\alpha\)-Entropy of entanglement

    \(E_{\alpha}(\psi^{AB})=H_{\alpha}(\vec p)\) where \(\alpha\)-Renyi-Entropy \(H_\alpha(\vec p)=\frac{1}{1-\alpha}\log_2\sum_{x=1}^{d}p_x^\alpha\) and \(\alpha\in [0,+\infty]\)
    Check that when \(\alpha=1\), by LoHpital, we get \(H_1(\vec p)=H(\vec p)\) and \(H_\alpha(\Phi^{AB})=\log_2(d)\)

    Check: \(H_\alpha(\vec p\otimes \vec q)=H_\alpha(\vec p)+H_\alpha(\vec q)\)

    This is the only additive one

  3. Concurrence Monotones

    \(f_2(\vec p)=\sum_{x<y}p_xp_y,\,\,f_3(\vec p)=\sum_{x<y<z}p_xp_yp_z\)

  4. Ky-Fam Norm

    \(E_{(k)}(\vec p):=1-\sum^{k}_{x=1}p_{x}^{\downarrow},\forall k\in[d]\)

    Check: \(E_{(k)}(\psi^{AB})\geq E_{(k)}(\phi^{AB}),\forall k\iff \psi^{AB}\xrightarrow{LOCC}\phi^{AB}\)

Catalysis of Entanglement

Example

\(|\psi^{AB}\rang=\sqrt{\frac{2}{5}}|00\rang+\sqrt{\frac{2}{5}}|11\rang+\sqrt{\frac{1}{10}}|22\rang+\sqrt{\frac{1}{10}}|33\rang\), \(|\phi^{AB}\rang=\sqrt{\frac{1}{2}}|00\rang+\sqrt{\frac{1}{4}}|11\rang+\sqrt{\frac{1}{4}}|22\rang\)

\(\psi^{AB}\cancel{\xrightarrow{LOCC}}\phi^{AB}\) and \(\phi^{AB}\cancel{\xrightarrow{LOCC}}\psi^{AB}\) because \(\vec q\nsucc\vec p\) and \(\vec p\nsucc \vec q\) where \(\vec p=\begin{pmatrix} \frac{2}{5}\\ \frac{2}{5}\\ \frac{1}{10}\\ \frac{1}{10} \end{pmatrix},\vec q=\begin{pmatrix} \frac{1}{2}\\ \frac{1}{4}\\ \frac{1}{4}\\ 0 \end{pmatrix}\)

Let \(\vec{r}= \begin{pmatrix} \frac{3}{5} \\ \frac{2}{5} \end{pmatrix}\), then claim: \(\vec{q}\otimes\vec{r}>\vec{p}\otimes\vec{r}\), let \(|\chi^{A'B'}\rangle = \sqrt{\frac{3}{5}}|00\rangle + \sqrt{\frac{2}{5}}|11\rangle\)

image​

Definition

\(\vec{q}>_{T} \vec{p}\) if \(\exists \vec{r}\) s.t. \(\vec{q}\otimes \vec{r}> \vec{p}\otimes \vec{r}\)

If \(\vec{q}\otimes \vec{r}> \vec{p}\otimes \vec{r}\) then \(H_{\alpha}(\vec{p}\otimes \vec{r}) \ge H_{\alpha}(\vec{q}\otimes \vec{r})\)
\(\iff H_{\alpha}(\vec{p}) + H_{\alpha}(\vec{r}) \ge H_{\alpha}(\vec{q}) + H_{\alpha} (\vec{r})\)​\(\iff H_{\alpha}(\vec{p}) \ge H_{\alpha}(\vec{q}) \quad \forall \alpha\)

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