8.26 Choi Matrix

Corollary

A linear map \(\mathcal{E} \in \mathcal{L}(A \rightarrow B)\) is \(|A|\)-positive if and only if it is completely positive.

Proof

By definition if \(\mathcal{E}\) is a CP map then it is \(|A|\)-positive. Conversely, if \(\mathcal{E}\) is \(|A|\)-positive, then its Choi matrix \(\mathcal{E}^{\tilde{A} \rightarrow B}\left(\Omega^{A \tilde{A}}\right)\) is positive semi-definite. Hence, from the theorem above \(\mathcal{E}\) is a CP map.

Lemma

\(\mathcal{E}^{A\to B}\) is Trace-Preserving iff \(J_{\mathcal{E}}^{A}=I^{A}\) where \(J_{\mathcal{E}}^{A}:=\text{Tr}_{B}[J_{\mathcal{E}}^{AB}]\)

Proof

\(\Rightarrow\)) Suppose \(\mathcal{E}\) is trace preserving and set \(m:=|A|\). Then, \(J_{\mathcal{E}}^{A}=\operatorname{\mathrm{Tr}}_{B}\left[J_{\mathcal{E}}^{AB}\right ]=\text{Tr}_{B}[\mathcal{E}^{\tilde{A}\to B}(\Omega^{A\tilde{A}})]\)

\(=\text{Tr}_{B}\left[\sum_{x,y\in\left\lbrack m\right\rbrack}|x\rangle\langle y|^{A} \otimes\mathcal{E}^{\tilde{A}\to B}(|x\rangle\langle y|^{\tilde{A}})\right]=\sum_{x,y\in\left\lbrack m\right\rbrack}|x\rangle\langle y|^{A}\text{Tr}_{B}\left[\mathcal{E}^{\tilde{A}\to B}(|x\rangle\langle y|^{\tilde{A}})\right]\)

Since \(\mathcal{E}\) is trace preserving, then \(J_{\mathcal{E}}^{A}=\sum_{x,y\in[m]}|x\rangle\langle y|\operatorname{\mathrm{Tr}} [|x\rangle\langle y|)]=\sum_{x,y\in[m]}|x\rangle\langle y|\delta_{xy}=I^{A}.\)

\(\Leftarrow\)) Suppose \(J_{\mathcal{E}}^{A}=I^{A}\), then \(\forall\rho \in \mathcal{L}(A)\), NTP: \(\text{Tr}[\rho]=\operatorname{\mathrm{Tr}}[\mathcal{E}(\rho)]\), we start from \(\text{Tr}[\mathcal{E}(\rho)]\)

Since this and we already have a full trace, then \(\text{Tr}[\mathcal{E}(\rho)]=\text{Tr}[J^{AB}\left((\rho^{A})^{T}\otimes I^{B}\right )]\)

Claim: \(\text{Tr}[J^{AB}\left((\rho^{A})^{T}\otimes I^{B}\right)]=\text{Tr}[J_{\mathcal{E}}^{A}(\rho^{A})^{T}]\)

Since \(J_{\mathcal{E}}^{A}=I^{A}\), then \(\operatorname{\mathrm{Tr}}\left[J_{\mathcal{E}}^{A}\rho^{T}\right]=\operatorname{\mathrm{Tr}} \left[\rho^{T}\right]=\operatorname{\mathrm{Tr}}[\rho].\)

Proof of claim
Let \(J^{AB}=\sum_{j}\mu_{j}^{A}\otimes\zeta_{j}^{B}\), we get \(\text{Tr}[\mathcal{E}(\rho)]=\text{Tr}\left[\left(\sum_{j}\mu_{j}^{A}\otimes\zeta _{j}^{B}\right)\left((\rho^{A})^{T}\otimes I^{B}\right)\right]=\text{Tr}\left[\left (\sum_{j}\mu_{j}^{A}(\rho^{A})^{T}\otimes\zeta_{j}^{B}\right)\right]=\sum_{j}\text{Tr} [\zeta_{j}^{B}]\text{Tr}[\mu_{j}^{A}(\rho^{A})^{T}]\)
Since \(J_\mathcal{E}^A=\text{Tr}_B[J_\mathcal{E}^{AB}]=\sum_j\text{Tr}[\zeta_j^B]\mu_j^A\), then \(\sum_{j}\text{Tr}[\zeta_{j}^{B}]\text{Tr}[\mu_{j}^{A}(\rho^{A})^{T}]=\text{Tr}[\sum _{j}\text{Tr}[\zeta_{j}^{B}]\mu_{j}^{A}(\rho^{A})^{T}]=\text{Tr}[J_{\mathcal{E}}^{A} (\rho^{A})^{T}]\)

---

Conclusion

\(\mathcal{E}\in CPTP(A\to B)\iff J_{\mathcal{E}}^{AB}\geq 0 \text{ and }J_{\mathcal{E}} ^{A}=I^{A}\)

Examples

1. Choi matrix of the transpose map \(\Tau:A\to A,\Tau(\rho)=\rho^{T},J_{\Tau}^{AB}=\Tau^{\tilde{A}\to A}(\Omega^{A\tilde{A}} )=\sum_{x,y}|x\rang\lang y|\otimes (|x\rang\lang y|)^{T}\)
   Then \(J_{\mathrm{T}}^{AB}=\sum_{x,y}|x\rangle\langle y|\otimes|y\rangle\langle x|=F^{AB}\)
   Then \(F^{AB}|x'\rang|y'\rang=\sum(|x\rang\lang y|\otimes |y\rang\lang x|)(|x'\rang| y'\rang)=|y'\rang|x'\rang\) this is flip operator
   Exercise: \(F^{AB}|\psi\rang|\phi\rang=|\phi\rang|\psi\rang\) and \(F^{2}=I^{AB}\)(Write it in basis)

2. \(\sigma\in \mathfrak{D}(B)\) (Density)
   \(\mathcal{E}(\rho^{A})=\sigma^{B}\,\,\forall \rho\in \mathcal{L}(A)\) is not a quantum channel since this is not trace preserving since \(\sigma\) is density matrix but \(\rho\) not

   Let \(\mathcal{E}(\rho^A):=\text{Tr}[\rho^A]\sigma^B,\forall \rho\in\mathcal{L}(A)\)
   Clearly this is trace preserving because
   \(J_{\mathcal{E}}^{AB}=\mathcal{E}^{\tilde{A}\to B}(\Omega^{A\tilde{A}})=\sum_{x,y} |x\rang\lang y|\otimes \mathcal{E}(|x\rang\lang y|)=\sum_{x,y}|x\rang\lang y|\otimes \text{Tr}[|x\rang\lang y|]\sigma^{B}=\sum_{x,y}|x\rang\lang y|\otimes \delta_{xy} \sigma^{B}\)
   \(=\sum_{x}|x\rang\lang x|\otimes \sigma^{B}=I^{A}\otimes \sigma^{B}\), then \(J_{\mathcal{E}}^{A}=\text{Tr}_{B}[J_{\mathcal{E}}^{AB}]=\text{Tr}_{B}[I^{A}\otimes \sigma^{B}]=I^{A}\text{Tr}[\sigma^{B}]=I^{A}\) since \(\sigma\) is density matrix

Exercise

Let \(\rho\in\mathfrak{D}(AB)\), show that there exists a channel \(\mathcal{E}\in CPTP(A\to B)\) and a pure state \(\psi^{A\tilde{A}}\) s.t.

\[ \rho^{AB}=\mathcal{E}^{\tilde{A}\to B}(\psi^{A\tilde{A}}) \]

Proof

We need to check two conditions

• \(\rho^{AB}\geq 0\) by definition of a density matrix
• If \(\rho^A=I^A\), then \(\text{Tr}\rho^A=\text{Tr}I^A\Rightarrow 1=|A|\) which is not true, thus we need to fix this

Let \(J^{AB}:= (\underbrace{(\rho^{A})^{-1/2}\otimes I^{B}}_{M^*}) J^{AB}(\underbrace{(\rho^{A})^{-1/2}\otimes I^{B}}_{M})\)

• \(J^{AB} \ge 0\) by theorem
• \(J^{A}= \text{Tr}_B[J^{AB}] = (\rho^{A})^{-1/2}\text{Tr}_B[\rho^{AB}] (\rho^{A} )^{-1/2}= (\rho^{A})^{-1/2}\rho^{A} (\rho^{A})^{-1/2}= I^{A}\)

Then by theorem there exists quantum channel s.t. \(J^{AB}=J_{\mathcal{E}}^{AB}\)

\(J_{\mathcal{E}}^{AB}= \mathcal{E}^{\tilde{A} \to B}(\Omega^{A\tilde{A}}) =\left( \left (\rho^{A}\right )^{-1/2}\otimes I^{B}\right)\rho^{AB}\left(\left(\rho^{A}\right)^{-1/2}\otimes I^{B}\right)\)
Then \(\left(\sqrt{\rho^{A}}\otimes I^{B}\right) \mathcal{E}^{\tilde{A} \to B}(\Omega^{A\tilde{A}} ) \left(\sqrt{\rho^{A}}\otimes I^{B}\right) = \rho^{AB}\)

Then \(\mathcal{E}^{\tilde{A}\to B}\left(\left(\sqrt{\rho^{A}}\otimes I^{B}\right)\Omega^{A\tilde{A}} \left(\sqrt{\rho^{A}}\otimes I^{B}\right)\right)=\rho^{AB}\)
Let \(|\psi^{A\tilde{A}}\rangle := \sqrt{\rho^A} \otimes I^{\tilde{A}} |\Omega^{A\tilde{A}}\rangle\), then \(\mathcal{E}^{\tilde{A} \to B} (\psi^{A\tilde{A}}) = \rho^{AB}\)

---

Note that \(\rho^A=\psi^A\) where \(\rho^{A}:=\text{Tr}_{B}[\rho^{AB}],\psi^{A}:=\text{Tr}_{\tilde{A}}[\psi^{A\tilde{A}} ]\)

This is because a theorem: \(\text{Tr}_B\circ \mathcal{E}^{A\to B}=\text{Tr}_A\)

\(\rho^{A}:=\text{Tr}_{B}[\rho^{AB}]=\text{Tr}_B\circ \mathcal{E}^{\tilde{A}\to B}(\psi^{A\tilde{A}})=\text{Tr}_{\tilde{A}}[\psi^{A\tilde{A}}]=\psi^A\)

Proof of Theorem

\(\text{Tr}_{B}\circ\mathcal{E}^{\tilde{A}\to B}(\phi^{A\tilde{A}})=\text{Tr}_{B}\left [\sum_{x,y,x^{\prime},y^{\prime}\in[d]}|x\rangle\langle y|^{A}\otimes\mathcal{E}^{\tilde{A}\to B}(|x^{\prime}\rangle\langle y^{\prime}|^{\tilde{A}})\right]\) \(=\sum_{x,y,x^{\prime},y^{\prime}\in\left\lbrack d\right\rbrack}|x\rangle\langle y|^{A}\text{Tr}[\mathcal{E}^{\tilde{A}\to B}|x^{\prime}\rangle\langle y^{\prime}| ^{\tilde{A}}]\)

\(\text{Tr}_{\tilde{A}}[\phi^{A\tilde{A}}]=\sum_{{x,y,x^{\prime},y^{\prime}\in\left\lbrack d\right\rbrack}} |x\rangle\langle y|^{A}\text{Tr}[|x^{\prime}\rangle\langle y^{\prime}|^{\tilde{A}} ]\)

They are equal since \(\mathcal{E}\) is trace preserving

---

Exercise: Show that the multiply of two positive(density) matrix is no longer positive even hermitian

Counter example

\(\begin{pmatrix} 1 & i \\ 0 & 3 \end{pmatrix}\times \begin{pmatrix} 1 & i \\ 0 & 3 \end{pmatrix}^{*}= \begin{pmatrix} 2 & 3i \\ -3i & 9 \end{pmatrix}\) which is positive by theorem
Then \(\begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}\) is also positive, but \(\begin{pmatrix} 2 & 3i \\ -3i & 9 \end{pmatrix}\cdot \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}=\begin{pmatrix} 2&9i\\ -3i&27 \end{pmatrix}\) which is not hermitian