8.22 Quantum Channels and Choi Representation

Quantum Channels

Axiomatic Approach. Examples

1. Unitary Evolution: \(\mathcal{N}(\rho) := U\rho U^{*}\) where \(U\)-unitary matrix.
   NTP: \(\mathcal{N} \in CPTP(A \to B)\) where \(\mathcal{L}(A) \to \mathcal{L}(B)\)(Note, if unitary, \(A=B\), if isometry, \(A,B\))

   Check: \(\text{Tr}[\mathcal{N}(\rho)]=\text{Tr}[U\rho U^{*}]=\text{Tr}[U^{*}U\rho]=\text{Tr} [\rho]\)

   We need to show that \(\mathcal{N}\) is C.P. Let \(R\) be a Hilbert space.

   NTP: \((id^{R\to R}\otimes \mathcal{N}^{A \to A})(\rho^{RA}) \ge 0\iff\mathcal{N}^{A \to A}(\rho^{RA}) \ge 0\) (just abbreviation) if \(\rho^{RA} \ge 0\)

   \(\mathcal{N}(\rho) = \underbrace{(I^R \otimes U^A)}_{\Lambda} \rho^{RA} \underbrace{(I^R \otimes U^A)^*}_{\Lambda^*} = \Lambda \rho \Lambda^*\) which is positive by homework before

   Comment: \(\mathcal{E}(\rho) := \Lambda \rho \Lambda^* \leftarrow\) completely positive for all complex \(\Lambda\) when we change \(U^A\)

   Also, usually, we denote it as \(\mathcal{V}(\rho) = V \rho V^{*}\), \(\forall \rho \in \mathcal{L}(A)\) where \(\mathcal{V}\) is isometry

2. \(\mathcal{E}^{(k)}(\eta) := k \text{Tr}[\eta] I_{d}- \eta \quad \forall \eta \in C^{d \times d}\) \(k \in [\mathbb{d}]\) choose \(k < d\)

   \(id_{k} \otimes \mathcal{E}^{(k)}\) is a positive map but \(id_{k+1}\otimes \mathcal{E}^{(k)}\) is not positive where \(id_{k}: \mathbb{C}^{k \times k}\rightarrow \mathbb{C}^{k \times k}\)

The Dual/Adjoint of a linear map

\(\mathcal{E} \in \mathcal{L}(A \rightarrow B) := \{T: \mathcal{L}(A) \rightarrow \mathcal{L}(B) \}\) but \(\mathcal{L}(A,B) := \{T: A \rightarrow B\}\)
Then the dual map is \(\mathcal{E}^* \in \mathcal{L}(B \rightarrow A)\)
We have \(\langle \sigma^A, \mathcal{E}^*(\rho^B) \rangle_{HS} = \langle \mathcal{E}(\sigma^A), \rho^B \rangle_{HS} \quad \forall \rho \in \mathcal{L}(B), \sigma \in \mathcal{L}(A)\)
Also \(\mathcal{E}\) is self-Adjoint if \(\mathcal{E} = \mathcal{E}^*\)

Example

\(\mathcal{E}(\rho) := \Lambda \rho \Lambda^*\) for \(\Lambda\) is some fixed matrix where \(\Lambda \in \mathcal{L}(A, B)\), \(\forall \rho \in \text{Herm}(A) \quad \sigma \in \text{Herm}(B)\) recall this
\(\langle \sigma, \mathcal{E}(\rho) \rangle = \text{Tr}[\sigma \mathcal{E}(\rho)] = \text{Tr}[\sigma \Lambda \rho \Lambda^*] = \text{Tr}[\Lambda^* \sigma \Lambda \rho] = \text{Tr}[\mathcal{E}^*(\sigma) \rho]\)
\(\langle \mathcal{E}^*(\sigma), \rho \rangle_{HS} = \text{Tr}[\mathcal{E}^*(\sigma) \rho]\), then \(\text{Tr}[(\Lambda^{*}\sigma\Lambda-\mathcal{E}^{*}(\sigma))\rho]=0\)
Since \(\Lambda^{*}\sigma\Lambda-\mathcal{E}^{*}(\sigma)\) is Hermitian, then take \(\rho=\Lambda^{*}\sigma\Lambda-\mathcal{E}^{*}(\sigma)\)
Then \(\langle\Lambda^{*}\sigma\Lambda-\mathcal{E}^{*}(\sigma),\Lambda^{*}\sigma\Lambda -\mathcal{E}^{*}(\sigma)\rangle_{HS}=0=||\Lambda^{*}\sigma\Lambda-\mathcal{E}^{*}(\sigma) ||^2_{HS}\)
Then \(\Lambda^{*}\sigma\Lambda-\mathcal{E}^{*}(\sigma)=0\). Thus \(\mathcal{E}^{*}(\sigma) = \Lambda^{*} \sigma \Lambda\)

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We have \(\mathcal{E}(\rho) = \Lambda\rho\Lambda^*\), \(\mathcal{E}^*(\sigma) = \Lambda^*\sigma\Lambda\)
Trace Non-Increasing means \(\text{Tr}[\mathcal{E}(\rho)] \leq \text{Tr}[\rho], \forall \rho \geq 0\).

\(\text{Tr}[\mathcal{E}(\rho)] = \text{Tr}[\Lambda\rho\Lambda^{*}] = \text{Tr}[\Lambda ^{*}\Lambda\rho] \leq \text{Tr}[\rho]\). If \(\Lambda^*\Lambda \leq I\), then \(\mathcal{E}\) is T.N.I.

The Choi Representation

The Choi representation offers a simpler way to identify complete positivity
We will adopt shorthand notation \(\mathcal{E}^{B \rightarrow B'}(\rho^{AB})\) to denote \((\text{id}^{A} \otimes \mathcal{E}^{B \rightarrow B'}) (\rho^{AB})\)

Given \(\mathcal{E} \in \mathcal{L}(A \rightarrow B)\) where \(\mathcal{E}: \mathcal{L}(A) \rightarrow \mathcal{L}(B)\)

Let \(d:=|A|, d':=|B|\), then \(\Omega^{A\tilde{A}}= |\Omega^{A\tilde{A}}\rang\lang\Omega^{A\tilde{A}}|\) \(= \sum_{x=1}^{d}|xx\rang \sum_{y=1}^{d}\lang yy|\)
​\(=\sum_{x,y\in[d]}|xx\rangle\langle yy|=\sum_{x,y\in[d]}|x\rangle\langle y|^{A}\otimes |x\rangle\langle y|^{\tilde{A}}\)

Definition

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

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

\(=\sum_{x,y\in\left\lbrack d\right\rbrack}|x\rangle\langle y|^{A}\otimes\mathcal{E} ^{\tilde{A}\to B}(|x\rangle\langle y|^{\tilde{A}})\in\mathbb{C}^{dd^{\prime}\times dd^{\prime}}\)

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

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

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Proposition:

If given \(\mathcal{E}\), we have \(J\). If given \(J\), we have \(\mathcal{E}\)

It's enough to prove one side because both space has same dimension. Let \(J_{\mathcal{E}}^{AB}\) as in \((*)\). Then, \((**)\)

Proof
​\(\text{Tr}_{A} \left[ J_{\mathcal{E}}^{AB}((\rho^{A})^{T} \otimes I^{B}) \right] = \text{Tr}_{A} \left[ \left( \sum_{x,y \in [d]}|x \rangle \langle y | \otimes \mathcal{E}(|x \rangle \langle y |) \right) ((\rho^{A})^{T} \otimes I^{B}) \right]\)

\(= \sum_{x,y \in [d]}\text{Tr}_{A} \left[ (|x \rangle \langle y | \otimes \mathcal{E}(|x \rangle \langle y |)) ((\rho^{A})^{T} \otimes I^{B}) \right] = \sum_{x,y \in [d]} \text{Tr} \left[ |x \rangle \langle y | \rho^{T} \right] \mathcal{E}(|x \rangle \langle y |)\)

\(= \sum_{x,y \in [d]}\langle y | \rho^{T}| x \rangle \mathcal{E}(|x \rangle \langle y |) = \sum_{x,y \in [d]}\langle x | \rho | y \rangle \mathcal{E}(|x \rangle \langle y |) = \mathcal{E}\left( \underbrace{\sum_{x,y \in [d]} \langle x | \rho | y \rangle |x \rangle \langle y |}_{\rho}\right)\)

Theorem

Let \(\mathcal{E}\in \mathcal{L}(A\to B)\), then \(\mathcal{E}^{A\to B}\) is complete positive iff \(J_{\mathcal{E}}^{AB}\geq 0\)

Proof

\(\Rightarrow\)) Suppose \(\mathcal{E}^{A\to B}\) is completely positive, then by definition \(J_{\mathcal{E}}^{AB}:=\mathcal{E}^{\tilde{A}\to B}(\Omega^{A\tilde{A}})\geq 0\) \(\forall \Omega^{A\tilde{A}}\geq 0\) where \(\Omega^{A\tilde{A}}\) has one eigenvalue \(d\)​

Proof
Let \(|ij\rangle\) be basis of \(A \otimes B\), then \(\Omega^{\bar{A}\bar{A}}|ij\rangle=\sum_{{x,y\in\left\lbrack d\right\rbrack}}|x\rangle \langle y|^{A}\otimes|x\rangle\langle y|^{B}|ij\rangle=\sum_{x,y\in\left\lbrack d\right\rbrack} |x\rangle\delta_{yi}\otimes|x\rangle\delta_{yj}\)
When \(i=j=y\), \(\Omega^{A\tilde{A}}|ij\rangle=\Omega^{A\tilde{A}}|ii\rangle=\sum_{x\in\left\lbrack d\right\rbrack}|x\rangle\langle x|:=|\phi\rangle\)
When \(i \ne j\), \(\Omega^{A\tilde{A}}|ij\rangle = 0\)

Then \(d\) vectors satisfy \(\Omega^{A\tilde{A}}|ii\rangle = |\phi\rangle\), \(\forall i\). Clearly \(\Omega^{A\tilde{A}}|\phi\rangle = d|\phi\rangle\). \(d-1\) vectors are sent to \(0\).
And \(d^2-d\) vectors are eigenvectors with eigenvalue \(0\)

Then \(1\) eigenvalue \(d\), \(d^2+d-1=d^2-1\) eigenvalues \(0\).

\(\Leftarrow\)) Suppose \(J_{\mathcal{E}}^{AB}\geq0\). NTP: \(\mathcal{E}\) is C.P.

Let \(R\) be a Hilbert space. NTP \(\mathcal{E}^{A\to B}(\rho^{RA})\geq 0,\forall \rho\geq 0\)

It's sufficient to show that \(\mathcal{E}^{A\to B}(\psi^{RA})\geq 0\) since \(\mathcal{E}^{A\to B}(\rho^{RA})\geq0\iff \mathcal{E}^{A\to B}(\sum p_{x}\psi_{x} ^{RA})\geq0 \iff \sum p_{x}\mathcal{E}^{A\to B}(\psi_{x}^{RA})\geq0\) because every mixed state can be written as the convex combination of pure states

We know \(|\psi^{RA}\rang = M\otimes I^{A}|\Omega^{\tilde{A}A}\rang\)

Then NTP \(\mathcal{E}^{A\to B}((M\otimes I^{A})\Omega^{\tilde{A}A}(M\otimes I^{A})^{*})\geq 0,\forall M:\tilde{A}\to R\)

Since \(\mathcal{E}\) only acts on \(\Omega^{\tilde{A}A}\), then NTP \((M\otimes I^{B})\mathcal{E}^{A\to B}(\Omega^{\tilde{A}A})(M\otimes I^{B})^{*}\geq 0\)

And \((M\otimes I^{B})\mathcal{E}^{A\to B}(\Omega^{\tilde{A}A})(M\otimes I^{B})^{*}=NJ _{\mathcal{E}}^{\tilde{A}B}N^{*}\geq0\) since \(J_{\mathcal{E}}^{\tilde{A}B}\geq 0\)

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