8.28 Operator Sum Representation

Operator Sum Representation(Class operator)

Consider a Generalized Measurement \(\{M_x\}_{x\in[m]}\) where \(\sum^m_{x=1}M_x^*M_x=I\)

Block the outcome, just record probabilities

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Then we forget \(x\), the outcome is an ensemble:

\(\sum_{x=1}^{m}p_{x}\sigma_{x}^{B}=\sum_{x}p_{x}\frac{1}{p_{x}}M_{x}\rho^{A}M_{x} ^{*}=\sum_{x}M_{x}\rho^{A}M_{x}^{*}\), then \(\rho^{A}\to \sum^{m}_{x=1}M_x\rho^AM_x^*\)

Above representation of a quantum channel is called the operator sum representation

Every quantum channel CPTP map can be considered as a measurement without looking at the outcome

Theorem

A linear map \(\mathcal{E}:\mathcal{L}(A)\to \mathcal{L}(B)\) is CPTP (i.e. a quantum channel) iff there exists a generalized measurement \(\{M_x\}_{x\in [m]}\)(satisfying \(\sum^m_{x=1}M_xM^*_x=I\) automatically) such that \(\forall \rho\in \mathcal{L}(A),\mathcal{E}(\rho)=\sum^{m}_{x=1}M_{x} \rho M_{x}^{*}\) (operator sum representation) where \(M_x:A\to B\)

Proof

\(\Leftarrow\)) Suppose \(\mathcal{E}(\rho)=\sum^m_{x=1}M_x\rho M_x^*\) where \(\{M_x\}_{x\in [m]}\) is generalized measurement

Trace preserving: \(\text{Tr}[\mathcal{E(\rho)}]=\text{Tr}\left[\sum_{x}M_{x}M_{x}^{*}\rho\right]=\text{Tr} [I\rho]=\text{Tr}[\rho]\)

\(\mathcal{E}\) is positive \(\iff\)Choi matrix \(J^{AB}_\mathcal{E}=\sum_{x=1}^mJ_{\mathcal{E}_x}^{AB}\) since \(\mathcal{E}=\sum^m_{x=1}\mathcal{E}_x\) and \(\mathcal{E}_x(\rho):=M_x\rho M_x^*\)

It's enough to show that \(J_{\mathcal{E}_x^{AB}}\geq 0\): \(J_{\mathcal{E}_{x}}^{AB}=\mathcal{E_{x}}^{\tilde{A}\to B}(\Omega^{A\tilde{A}})=M _{x}|\Omega^{A\tilde{A}}\rangle\langle\Omega^{A\tilde{A}}|M_{x}^{*}\geq0\)​

\(\Rightarrow\)) Suppose \(\mathcal{E}\) is a quantum channel, then \(J_\mathcal{E}^{AB}\geq 0,J_\mathcal{E}^A=I^A\)

We know \(J_{\mathcal{E}}^{AB}= \sum_{x=1}^{m}|\psi_{x}^{AB}\rangle\langle\psi_{x}^{AB}|\) by spectral decomposition(hermitian, diagonalizable) and absorb real eigenvalues where \(|\psi_x^{AB}\rang\) are unnormalized eigenvectors

We know \(|\psi_{x}^{AB}\rang=I^{A}\otimes M_{x}|\Omega^{A\tilde{A}}\rang\), we need to argue this \(M_x\) is generalized measurements

Then \(J_{\mathcal{E}}^{AB}=\sum_{x=1}^{m}\left(I^{A}\otimes M_{x}\right)\Omega^{A\tilde{A}} \left(I^{A}\otimes M_{x}\right)^{*}\), then \(I^{A}=J_{\mathcal{E}}^{A}=\text{Tr}_{B}\left[\sum_{x=1}^{m}\left(I^{A}\otimes M_{x} \right)\Omega^{A\tilde{A}}\left(I^{A}\otimes M_{x}\right)^{*}\right]\)

\(=\text{Tr}_{B}\left[\sum_{x=1}^{m}(M_{x}^{T}\otimes I)\Omega^{\tilde{B}B}(M_{x}^{T} \otimes I)^{*}\right]=\sum_{x=1}^{m}M_{x}^{T}\,\text{Tr}_{B}[\Omega^{\tilde{B}B}] (M_{x}^{T})^{*}\)
​\(=\sum_{x=1}^{m}M_{x}^{T}I^{\tilde{B}}(M_{x}^{T})^{*}=\sum_{x=1}^{m}M_{x}^{T}(M_{x} ^{T})^{*}\)

Thus we have \(I=\sum M^T_x\overline{M_x}\Rightarrow\bar I=\sum \overline{M}^T_x M_x\Rightarrow I=\sum M_x^*M_x\), then it's generalized measurement.

Since this, then \(\mathcal{E}\left(\rho\right)=\text{Tr}_{A}[J_{\mathcal{E}}^{AB}\left(\rho^{T}\otimes I^{B}\right)]=\sum_{x=1}^{m}\text{Tr}_{A}[\psi_{x}^{AB}\left(\rho^{T}\otimes I ^{B}\right)]\)

\(=\sum_{x=1}^{m}\text{Tr}_{A}\left[(I^{A}\otimes M_{x})\Omega^{A\tilde{A}}(I\otimes M_{x})^{*}(\rho^{T}\otimes I^{B})\right]=\sum_{x=1}^{m}\text{Tr}_{A}\left[(I^{A}\otimes M_{x})\Omega^{A\tilde{A}}(\rho^{T}\otimes M_{x}^{*})\right]\)

\(=\sum_{x=1}^{m}\text{Tr}_{A}\left[(I^{A}\otimes M_{x})\Omega^{A\tilde{A}}(I^{A}\otimes \rho M_{x}^{*})\right]\)​\(=\sum_{x=1}^{m}M_{x}\text{Tr}_{A}\left[\Omega^{A\tilde{A}}\right]\rho M_{x}^{*}= \sum_{x=1}^{m}M_{x}\rho M_{x}^{*}\)

Theorem

Let \(m,n\in\N,\{M_{x}\}_{x\in[m]},\{N_{y}\}_{y\in [n]}\) generalized measurements and \(m\leq n\). The following statement are equivalent:

1. \(\sum^{m}_{x=1}M_{x}\rho M_{x}^{*}=\sum^{n}_{y=1}N_{y}\rho N_{y}^{*}\quad \forall \rho\in \mathcal{L}(A)\)

2. \(\exists V:\mathbb{C}^m\to \mathbb{C}^n\) isometry \(V^*V=I_m\) s.t. \(N_{y}=\sum_{x=1}^{m}v_{yx}M_{x}\)

Proof

\((2\Rightarrow 1)\)

\(\sum_{y=1}^{n}N_{y}\rho N_{y}^{*}=\sum_{y=1}^{n}\left(\sum_{x=1}^{m}v_{yx}M_{x}\right )\rho\left(\sum_{x=1}^{m}v_{yx^{\prime}}M_{x^{\prime}}\right)^{*}=\sum_{x,x^{\prime}\in[m]} \left(\sum_{y=1}^{n}v_{yx}v_{yx^{\prime}}^{*}\right)M_{x}\rho M_{x^{\prime}}^{*}\)

\(=\sum_{x,x^{\prime}\in[m]}\left(\sum_{y=1}^{n}v_{x^{\prime}y}^{*}v_{yx}\right)M_{x} \rho M_{x^{\prime}}^{*}=\sum_{x,x^{\prime}\in[m]}\left(V^*V\right)_{xx'}M_{x}\rho M_{x^{\prime}}^{*}\)

\(=\sum_{x,x^{\prime}\in[m]}\left(I\right)_{xx^{\prime}}M_{x}\rho M_{x^{\prime}}^{*} =\sum_{x,x^{\prime}\in[m]}\delta_{xx^{\prime}}M_{x}\rho M_{x^{\prime}}^{*}=\sum_{x\in[m]} M_{x}\rho M_{x}^{*}\)

\((1\Rightarrow 2)\)

Since this, there exists a quantum channel s.t. \(\mathcal{E}(\rho)=\sum_{x=1}^{m}M_{x}\rho M_{x}^{*}\)

\(J_{\mathcal{E}}^{AB}=\mathcal{E}^{\tilde{A}\to B}(\Omega^{A\tilde{A}})=\sum^{m}_{x=1} (I\otimes M_{x})\Omega^{A\tilde{A}}(I\otimes M_{x}^{*})\)

We define \(|\psi_x\rang^{AB}:=(I\otimes M_x)|\Omega^{A\tilde{A}}\rang\), then \(J_{\mathcal{E}}^{AB}=\sum^{m}_{x=1}|\psi_{x}\rang\lang \psi_{x}|\)

Also define \(|\phi_{x}\rang^{AB}:=(I\otimes N_{y})|\Omega^{A\tilde{A}}\rang\), then \(J_{\mathcal{E}}^{AB}=\sum_{x=1}^{m}|\phi_{x}\rangle\langle\phi_{x}|\)

Then \(J_{\mathcal{E}}^{AB}=\sum_{x=1}^{m}|\phi_{x}\rangle\langle\phi_{x}|=\sum_{x=1}^{m} |\phi_{x}\rangle\langle\phi_{x}|\). Then since this

\(\exists V:\mathbb{C}^m\to\mathbb{C}^n\) s.t. \(|\phi_{y}^{AB}\rang =\sum^{m}_{x=1}v_{yx}|\psi_{x}^{AB}\rang\), then \(|\phi_{y}^{AB}\rangle=I\otimes\left(\sum_{x=1}^{m}v_{yx}M_{x}\right)|\Omega^{A\tilde{A}} \rang\)

Also we know \(|\phi_{x}\rangle^{AB}:=(I\otimes N_{y})|\Omega^{A\tilde{A}}\rangle\), then \((I\otimes N_{y})|\Omega^{A\tilde{A}}\rangle=I\otimes\left(\sum_{x=1}^{m}v_{yx}M_{x} \right)|\Omega^{A\tilde{A}}\rangle\)​

By theorem: \(I\otimes E|\Omega\rang =I\otimes F|\Omega\rang\iff E=F\), we get \(N_{y}=\sum_{x=1}^{m}v_{yx}M_{x}\)

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Proof of that theorem \(I\otimes E|\Omega\rang =I\otimes F|\Omega\rang\iff E=F\)

\(I\otimes E|\Omega\rangle=I\otimes F|\Omega\rangle\iff I\otimes E\sum_{x=1} ^{m}|x\rangle|x\rangle=I\otimes F\sum_{x=1}^{m}|x\rangle|x\rangle\)

\(\iff \sum^m_{x=1}|x\rang E|x\rang=\sum^m_{x=1}|x\rang F|x\rang\iff E|x\rang =F|x\rang,\forall x\iff E=F\)

The Quantum Bit Flip (Example 1)

\(\Phi^{+} = \frac{|00\rangle + |11\rangle}{\sqrt{2}}\), \(\Phi^- = \frac{|00\rangle - |11\rangle}{\sqrt{2}}\), \(\Psi^{+}=\frac{|01\rangle+|10\rangle}{\sqrt{2}}\), \(\Psi^{-}=\frac{|01\rangle-|10\rangle}{\sqrt{2}}\)

Because there exists noise in communication channel, then \(0\) may become \(1\)

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\(\mathcal{E}(\rho):=p\rho+(1-p)XPX=M_0\rho M_0^*+M_1\rho M_1^*\) where \(X=\begin{pmatrix} 0&1\\1&0 \end{pmatrix}\), \(M_0=\sqrt{p}I\) and \(M_1=\sqrt{1-p}X\)

We can check that \(\sum_{x=1}^2M_x^*M_x=I\), thus \(M\) is a generalized measurement

Then \(J_{\mathcal{E}}^{AB}=(I\otimes M_{0})\Omega^{A\tilde{A}}(I\otimes M_{0})^{*}+(I\otimes M_{1})\Omega^{A\tilde{A}}(I\otimes M_{1})^{*}\)

\(I\otimes M_{0}|\Omega^{A\tilde{A}}\rangle=\sqrt{P}|\Omega^{AB}\rangle=\sqrt{2}\sqrt{P} |\Phi_{+}^{AB}\rangle\)

\(I \otimes M_{1}|\Omega^{A\tilde{A}}\rangle = \sqrt{1-P}(I \otimes X)|\Omega^{A\tilde{A}} \rangle = \sqrt{2}\sqrt{1-P}|\Psi_+^{AB}\rang\) since \((I\otimes X)|\Omega^{A\tilde{A}}\rangle=(I\otimes X)(|00\rangle+|11\rangle)\) \(= |0\rang X|0\rang + |1\rang X|1\rang\) \(=|0\rangle|1\rangle+|1\rangle|0\rangle\) \(=\sqrt{2}|\Psi_{+}\rangle\)

Thus \(J_{\mathcal{E}}^{AB}= 2p\Phi_{+}^{AB}+ 2(1-p) \Psi_{+}^{AB}\)

The Depolarizing Channel

\(\mathcal{E}(\rho)=\frac{1}{2}p\text{Tr}[\rho]I^{A}+(1-p)\rho\) where \(p\in[0,1]\), \(A = \mathbb{C}^2\), \(\rho\in \mathfrak{D}(A)\) and \(I^A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

\(M_{0}= \sqrt{1 - \frac{3p}{4}}I\) and \(M_{x}= \frac{\sqrt{p}}{2}\sigma_{x}\) for \(x = 1, 2, 3\) where \(\sigma_1, \sigma_2, \sigma_3\)- Pauli matrices\(\sum_{x=0}^{3}M_{x}^{*}M_{x}=\sum_{x=0}^{3}M_{x}^{2}=(1-\frac{3p}{4})I+\frac{p}{4} (\sigma_{1}^{2}+\sigma_{2}^{2}+\sigma_{3}^{2})=(1-\frac{3p}{4})I+\frac{3p}{4}I=I\)

\(\sum_{x=0}^{3}M_{x}\rho M_{x}^{*}=(1-\frac{3p}{4})\rho+\frac{p}{4}(\sigma_{1}\rho \sigma_{1}+\sigma_{2}\rho\sigma_{2}+\sigma_{3}\rho\sigma_{3})\) 也可以设出\(\rho\)然后死算
Then we claim: \(\Lambda=\sum_{x=0}^{3}\sigma_{x}\rho\sigma_{x}=2I_{2}\operatorname{\mathrm{tr}}( \rho)\)
Then \(\sum_{x=0}^{3}M_{x}\rho M_{x}^{*}=(1-\frac{3p}{4})\rho+\frac{p}{4}\left(2I-\rho\right )=\frac{p}{2}I+\left(1-p\right)\rho=\mathcal{E}(\rho)\) since \(\text{Tr}[\rho]=1\)

\(J_{\mathcal{E}}^{AB}=(I\otimes M_{0})\Omega^{A\tilde{A}}(I\otimes M_{0})^{*}+\sum _{x=1}^{3}(I\otimes M_{x})\Omega^{A\tilde{A}}(I\otimes M_{x})^{*}\)

\(I\otimes M_{0}|\Omega^{A\tilde{A}}\rangle=I\otimes M_{0}\cdot\sum_{x=0}^{1}|xx\rangle =\sum_{x=0}^{1}|x\rangle M_{0}|x\rangle=\sum_{x=0}^{1}|x\rangle\sqrt{1-\frac{3p}{4}} I|x\rangle=\sqrt{1-\frac{3p}{4}}|\Omega^{A\tilde{A}}\rangle=\sqrt{2-\frac{3p}{2}} |\Phi_{+}\rangle\)

\(I\otimes M_{x}|\Omega^{A\tilde{A}}\rangle=I\otimes M_{x}\cdot\sum_{x^{\prime}=0} ^{1}|x^{\prime}x^{\prime}\rangle=\sum_{x^{\prime}=0}^{1}|x^{\prime}\rangle M_{x}| x^{\prime}\rangle=\sum_{x^{\prime}=0}^{1}|x^{\prime}\rangle\frac{\sqrt{p}}{2}\sigma _{x}|x^{\prime}\rangle=\frac{\sqrt{p}}{2}\sum_{x^{\prime}=0}^{1}|x^{\prime}\rangle \sigma_{x}|x^{\prime}\rangle\)

\(I\otimes M_{1}|\Omega^{A\tilde{A}}\rangle=\frac{\sqrt{p}}{2}\left(|01\rangle+|10 \rangle\right)=\sqrt{\frac{p}{2}}|\Psi_{+}\rangle\), \(I\otimes M_{2}|\Omega^{A\tilde{A}}\rangle=i\sqrt{\frac{p}{2}}|\Psi_{-}\rang\), \(I\otimes M_{3}|\Omega^{A\tilde{A}}\rangle=\sqrt{\frac{p}{2}}|\Phi_-\rang\)

Then \(J_{\mathcal{E}}^{AB}=\left(2-\frac{3p}{2}\right)\Phi_{+}+\frac{p}{2}\left(\Psi_{+} +\Psi_{-}+\Phi_{-}\right)\)

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Proof of \(\Lambda=\sum_{x=0}^{3}\sigma_{x}\rho\sigma_{x}=2I_{2}\operatorname{\mathrm{tr}}( \rho)\)

Let \(\sigma_{x'}\Lambda\sigma_{x'}=\Lambda\), because \(\sigma_{x^{\prime}}\Lambda\sigma_{x^{\prime}}=\sum_{x=0}^{3}\sigma_{x^{\prime}}\sigma _{x}\rho\sigma_{x}\sigma_{x^{\prime}}=\sum_{x''=0}^{3}\sigma_{x''}\rho\sigma_{x''}=\Lambda\)​

Then \(\sigma_{x^{\prime}}\Lambda\sigma_{x^{\prime}}\cdot\sigma_{x^{\prime}}=\Lambda\cdot \sigma_{x^{\prime}}\Rightarrow\sigma_{x^{\prime}}\Lambda=\Lambda\sigma_{x^{\prime}}\)

Then \(\left(\sum t_{x^{\prime}}\sigma_{x^{\prime}}\right)\Lambda=\sum t_{x^{\prime}}\Lambda \sigma_{x^{\prime}}\Rightarrow\left(\sum t_{x^{\prime}}\sigma_{x^{\prime}}\right) \Lambda=\Lambda\sum t_{x^{\prime}}\sigma_{x^{\prime}}\Rightarrow H\Lambda=\Lambda H,\forall H\)

Assume \(\Lambda = a_{0} I + a_{1} X + a_{2} Y + a_{3} Z\)
\([\Lambda,X]=0 \;\Rightarrow\; a_2 [Y,X] + a_3 [Z,X] = 0 \;\Rightarrow\; 2i(a_2 Z - a_3 Y)=0 \;\Rightarrow\; a_2=a_3=0,\)
\([\Lambda,Y]=0 \;\Rightarrow\; a_1 [X,Y] = 2i a_1 Z = 0 \;\Rightarrow\; a_1=0.\)
Hence \(\Lambda = a_0 I\). Writing \(c:=a_0\), we get \(\Lambda = c\,I_{2}\)
\(\operatorname{\mathrm{tr}}(\Lambda)=\sum_{a=0}^{3}\operatorname{\mathrm{tr}}(\sigma _{a}\rho\sigma_{a})=\sum_{a=0}^{3}\operatorname{\mathrm{tr}}(\rho)=4\operatorname{\mathrm{tr}} (\rho)\quad\Rightarrow\quad\operatorname{\mathrm{tr}}(cI_{2})=2c=4\operatorname{\mathrm{tr}} (\rho)\Rightarrow c=2\operatorname{\mathrm{tr}}(\rho)\)
So \(\Lambda=2\operatorname{\mathrm{tr}}(\rho)I_{2}\)​

H.W. Show that for every Unitary matrix \(\mathcal{E}(U\rho U^{*}) = U\mathcal{E}(\rho)U^{*}\) where \(\mathcal{E}\) is depolarizing channel

Proof

\(\mathcal{E}(U\rho U^{*})=\frac{1}{2}U\rho U^{*}\text{Tr}[U\rho U^{*}]I^{A}+(1-p) \left(U\rho U^{*}\right)=\frac{1}{2}U\rho U^{*}\text{Tr}[\rho]I^{A}+(1-p)\left(U\rho U^{*}\right)=U\left(\frac{1}{2}\rho\text{Tr}[\rho]I^{A}+(1-p)\rho\right)U^{*}=U\mathcal{E} (\rho)U^{*}\)

H.W. \((U \otimes \bar{U}) J_{\varepsilon}^{AB}(U \otimes \bar{U})^{*} = J_{\varepsilon} ^{AB}\) \(\forall\)Unitary matrices \(U\)

Proof

Since \(J_{\mathcal{E}}^{AB}=\left(2-\frac{3p}{2}\right)\Phi_{+}+\frac{p}{2}\left(\Psi_{+} +\Psi_{-}+\Phi_{-}\right)\), then \((U\otimes\bar{U})J_{\varepsilon}^{AB}(U\otimes\bar{U})^{*}=(U\otimes\bar{U})J_{\varepsilon} ^{AB}(U^{*}\otimes U^{t})\)

\(=(U\otimes\bar{U})\left(\left(2-\frac{3p}{2}\right)\Phi_{+}+\frac{p}{2}\left(\Psi _{+}+\Psi_{-}+\Phi_{-}\right)\right)(U^{*}\otimes U^{t})\)

To see it equals \(J_\mathcal{E}^{AB}\), we just to see \((U\otimes\bar{U})\Phi_{+}(U^{*}\otimes U^{t})=\Phi_{+}\)​

\((U\otimes\bar{U})\Phi_{+}(U^{*}\otimes U^{t})=(I\otimes U^{t}\bar{U})\frac{|00\rangle+|11\rangle}{\sqrt{2}} (I\otimes\bar{U}U^{t})=(I\otimes I)\frac{|00\rangle+|11\rangle}{\sqrt{2}}(I\otimes I)=\Phi_{+}\)

The same as \(\Psi_{+},\Psi_{-},\Phi_{-}\), thus \((U \otimes \bar{U}) J_{\varepsilon}^{AB}(U \otimes \bar{U})^{*} = J_{\varepsilon} ^{AB}\)

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