9.9

Mixed-state Entanglement

\(\text{SEP}(AB):=\{\rho\in \mathfrak{D}(AB):\rho^{AB}=\sum^{m}_{x=1}p_{x}\omega_{x} ^{A}\otimes \tau_{x}^{B},\sum p_{x}=1,p_{x}\geq 0,\omega_{x}\in\mathfrak{D}(A),\tau _{x}\in\mathfrak{D}(B)\}\)

Clearly, \(\text{SEP}(AB)\subset \mathfrak{D}(AB)\)

If \(\rho\in \mathfrak{D}(AB)\), it's hard to say if it is entanglement when the size is big

Detection of Entanglement

Entanglement Witness

Definition

An operator \(\Gamma\in \text{Herm}(AB)\) is called Entanglement Witness if:

1. \(\text{Tr}[\Gamma^{AB}\sigma^{AB}]\geq 0,\,\,\forall\sigma\in \text{SEP}(AB)\)

2. \(\exists \rho^{AB}\in \mathfrak{D}(AB)\text{ s.t. }\text{Tr}[\Gamma^{AB}\rho^{AB} ]<0\)

What do these two conditions tell us?

1. Observe that if \(\text{Tr}[\Gamma^{AB}(\omega \otimes \tau)]\geq 0,\forall \omega,\tau\), then \(\text{Tr}[\Gamma^{AB}\sigma^{AB}]\geq 0,\forall \sigma\in \text{SEP}(AB)\) since definition of \(\text{SEP}(AB)\)

   And we know any density matrix can be written as convex combination of pure state, then it's enough to assume that \(\text{Tr}[\Gamma^{AB}(\psi^{A}\otimes \phi^{B})]=\lang \psi^A\otimes \phi^B|\Gamma^{AB}|\psi^A\otimes \phi^B\rang\geq 0,\forall \psi,\phi\in \text{Pure}\)

2. \(\Gamma\) is not positive, otherwise \(\text{Tr}[\Gamma^{AB}\rho^{AB}]\geq 0\) since \(\text{Tr}[\Gamma\rho] = \text{Tr}[\Gamma \sum p_{x} \psi_{x}]\)​\(= \sum p_{x} \text{Tr}[|\psi_{x}\rangle\langle\psi_{x}|\Gamma]\) \(= \sum p_{x} \langle\psi_{x}|\Gamma|\psi_{x}\rangle\)

If separable state is pure, then it looks like \(|\psi^{AB}\rang=|\chi^A\rang \otimes |\phi^B\rang\) since the rank 1

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\(u^{AB}=\frac{1}{|AB|}I^{AB}=\frac{1}{|A|}I^A\otimes \frac{1}{|B|}I^B=u^A\otimes u^B\)

\(\mathcal{B}^{\varepsilon}(u^{AB}):=\{\rho\in \mathfrak{D}(AB):\frac{1}{2}||u^{AB} -\rho^{AB}||\leq \varepsilon\}\)

Lemma

\(\text{SEP}(AB)^*=\{\Gamma^{AB}\in \text{Herm}(AB):\text{Tr}[\Gamma\sigma]\geq 0,\forall \sigma\in \text{SEP}(AB)\}\), \(\text{SEP}(AB)^{**}=\text{SEP}(AB)\)

\(\forall\) entangled \(\rho^{AB}\), \(\exists\) entanglement witness \(\Gamma\) s.t. \(\text{Tr}[\rho\Gamma]<0\)

Theorem

There exists \(\varepsilon>0\) s.t. \(\mathcal{B}^{\varepsilon}(u^{AB})\subset \text{SEP}(AB)\)

Proof

By contradiction, suppose \(\forall \varepsilon>0\), \(\mathcal{B}^{\varepsilon}(u^{AB})\not \subset \text{SEP}(AB)\), that is \(\exists\) sequence of emtamgled states \(\{\tau^{AB}\}_{n\in \N}\) s.t. \(\lim_{n\to\infty}\frac{1}{2}||\tau^{AB}-u^{AB}||_1=0\)

By Lemma, \(\exists\) entanglement witness s.t. \(\text{Tr}[\Gamma_n\tau^{AB}_n]<0\) and assume \(\text{Tr}[\Gamma_n^2]=1\) otherwise we can absorb it

Since set \(\{\Gamma\in \text{Herm}(AB):\text{Tr}[\Gamma^2]=1\}\) is compact, then \(\{\Gamma_n\}_{n\in \N}\) has converging subsequence s.t. \(\lim_{n\to\infty}\Gamma_n=\Gamma\)

Then \(0\geq \lim_{n\to\infty}\text{Tr}[\Gamma_n\tau_n]=\text{Tr}[\Gamma u^{AB}]\)

But \(\Gamma\) is entanglement witness since \(\Gamma_n\) are entanglement witness and if \(\text{Tr}[\Gamma_{n}^{AB}\sigma^{AB}]\geq 0\), then \(\text{Tr}[\Gamma^{AB}\sigma^{AB}]\geq 0\)

Thus \(0\geq \lim_{n\to\infty}\text{Tr}[\Gamma_{n}\tau_{n}]=\text{Tr}[\Gamma u^{AB}]\geq 0\), then \(\text{Tr}[\Gamma]=0\)

\(\text{Tr}[\Gamma]=0=\text{Tr}[\Gamma^{AB}]=\text{Tr}[\Gamma^{AB}(I^{A}\otimes I^{B} )]=\sum_{xy}\text{Tr}[\Gamma^{AB}(|x\rangle\langle x|\otimes|y\rangle\langle y|)]\)

Then \(\text{Tr}[\Gamma^{AB}|x\rangle\langle x|\otimes|y\rangle\langle y|]=0\), then \(\langle xy| \Gamma^{AB} |xy \rangle = 0 \implies \Gamma = 0\)

But \(\text{Tr}(\Gamma^{2})=1\), contradiction

Theorem

\(\Gamma\in \text{Herm}(AB)\) is Entanglement Witness iff \(\Gamma\) is the Choi matrix of a positive map that is not completely positive

Proof

\(\Rightarrow\)) Suppose \(\Gamma=\mathcal{E}^{\tilde{A}\to B}(\Omega)^{A\tilde{A}}\) where \(\mathcal{E}^{\tilde{A}\to B}\) is positive but not completely positive. Therefore \(\Gamma^{AB}\ngeq 0\)

\(\text{Tr}[\Gamma^{AB}(\rho^{A}\otimes\sigma^{B})]=\text{Tr}[J_{\mathcal{E}}^{AB} (\rho^{A}\otimes\sigma^{B})]=\text{Tr}_{B}[\sigma^{B}\text{Tr}_{A}[J_{\mathcal{E}} ^{AB}(\rho^{A}\otimes I^{B})]]\)
​\(=\text{Tr}_{B}[\sigma^{B}\text{Tr}_{A}[J_{\mathcal{E}}^{AB}((\rho^{A})^{T})^{T}\otimes I^{B}]]]=\text{Tr}[\sigma\mathcal{E}(\rho^{T})]\ge0\) since we can let \(\gamma := \mathcal{E}(\rho^{T}) \ge 0\) and \(\text{Tr}[\sigma \gamma] \ge 0\) since \(\sigma,\gamma\geq 0\)

\(\Leftarrow\)) Conversely, suppose \(\Gamma^{AB}\) is entanglement witness.
Let \(\mathcal{E} \in \mathcal{L}(A) \to \mathcal{L}(B)\) s.t. \(\Gamma^{AB} = J_\mathcal{E}^{AB}\) \(\Rightarrow \mathcal{E} \notin CP(A \to B)\) because \(\Gamma^{AB}\ngeq 0\)
Since \(\Gamma^{AB}\) is Entanglement Witness \(\text{Tr}[\Gamma^{AB}(\rho^{A}\otimes \sigma^{B})]=\text{Tr}[\sigma^{B} \mathcal{E}(\rho^{A})^{T}] \ge 0\), \(\forall \sigma, \forall \rho\)

Then we can take \(\sigma\) as eigenvectors of \(\mathcal{E}(\rho)^T\), then we get \(\mathcal{E}(\rho^A)^T\geq 0\), then \(\mathcal{E}(\rho^A)\geq 0\)

Theorem

Let \(\Gamma^{AB}\) be an entanglement witness. Suppose \(|AB| \le 6\). Then, there exists \(\eta_1^{AB}, \eta_2^{AB} \ge 0\) s.t.

\[ \Gamma^{AB}= \eta_{1}^{AB}+ (\eta_{2}^{AB})^{T_B} \]

where \(T_B\) is transpose only on \(B\)

Proof: Suppose \(|A|=2, |B|=3\). Every positive map \(\mathcal{E}^{A \to B}\) has the form \(\mathcal{E}^{A \to B}= \mathcal{N}_{1}^{A \to B}+ T^{B \to B}\circ \mathcal{N}_{2}^{A \to B}\) where \(T^{B\to B}\) is transpose on \(B\) and \(\mathcal{N}_{1}, \mathcal{N}_{2} \in CP\).

\(\Gamma^{AB}=\mathcal{E}^{\tilde{A}\to B}(\Omega^{A\tilde{A}})=\underbrace{J_{\mathcal{N}_1}^{AB}}_{\eta_1} +\underbrace{(J_{\mathcal{N}_2}^{AB})^{T_B}}_{\eta_2}\)

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If \(\rho^{AB}\) is separable then \(\rho^{AB} = \sum P_x \omega_x^A \otimes \tau_x^B\) and \((\rho^{AB})^{T_B} = \sum_{x \in \{m\}} P_x \omega_x^A \otimes (\tau_x^B)^T \ge 0\)

Theorem

Let \(\rho \in D(AB)\) with \(|AB| \leq 6\). Then, \(\rho^{AB}\) is separable if and only if \((\rho^{AB})^{T_B} \geq 0\) (Peres-Horodecki Criterion)

Proof

Suppose \((\rho^{AB})^{T_B} \geq 0\). By contradiction, suppose \(\rho^{AB}\) is entangled \(\Rightarrow \exists E.W. \Gamma^{AB}\) s.t. \(\text{Tr}[\Gamma^{AB}\rho^{AB}] < 0\).
From the previous theorem, \(\Gamma^{AB} = \eta_1^{AB} + (\eta_2^{AB})^{T_B}\)

So that \(0>\text{Tr}[\eta^{AB}\rho^{AB}]=\text{Tr}[\eta_{1}^{AB}+(\eta_{2}^{AB})^{T_{B}}) \rho^{AB}]\) \(=\underbrace{\text{Tr}[\eta_{1}^{AB}\rho^{AB}]}_{\geq0}+\underbrace{\text{Tr}[(\eta_{2}^{AB})^{T_{B}}\rho^{AB}]} _{=\text{Tr}[\eta_{2}^{AB}(\rho^{AB})^{T_{B}}]]\ge0}\) since this partial transpose is a self-adjoint

Comment: \(((\rho^{AB})^{T_B}) ^T= (\rho^{AB})^{T_A}\)

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