8.29 Dilation Theorem and Examples of Channels
Stinespring's Dilation Theorem
Every quantum channel is a unitary evolution when restrict to the subsystem(Trace the environment)
The environment can be mixed state, but it can be purified in large system
Given \(V:A\to AE\quad V^*V=I^A\quad V:=U(I^A\otimes |0\rang^E)\quad V|\psi\rang:=U(|\psi\rang^A|0\rang^E)\)
Note that we denote \(\mathcal{E}(\rho)=\text{Tr}_{E}[V\rho V^{*}]\) where \(V\rho V^{*}=U^{AE}(\rho^{A}\otimes|0\rangle\langle0|^{E})U^{*AE}\)
Theorem
Given \(\mathcal{E}\in \mathcal{L}(A\to B)\). \(\mathcal{E}\) is a CPTP (i.e. Quantum Channel) iff \(\exists E\) (Hilbert space) with \(|E|\leq |AB|\) and an isometry \(V:A\to BE\) s.t. \(\mathcal{E}(\rho)=\text{Tr}_{E}[V\rho V^{*}],\forall \rho\in\mathcal{L}(A)\)
Proof
\(\Leftarrow\)) Suppose \(\mathcal{E}(\rho)=\text{Tr}_{E}[V\rho V^{*}]\)
Trace Preserving: \(\text{Tr}[\mathcal{E}(\rho)]=\text{Tr}[V\rho V^*]=\text{Tr}[V^*V\rho]=\text{Tr}[\rho]\)
Complete Positive: \(\mathcal{E}(\rho)=\sum_{z=1}^{|E|}\underbrace{\lang^E z|V}_{M_z}\rho \underbrace{V^*|z\rang^E} _{M_z^*}\) where \(V^{*}: BE \rightarrow A\), \(M_{z}^{*}:=V^{*}\left(I^{B}\otimes|z\rangle^{E}\right)\), \(M_{z}^{*}: B \rightarrow A\) and \(M_{z}^{*}|\psi_{B}\rangle:=V^{*}|\psi_{B}\rangle|z\rangle\)
Then \(\mathcal{E}(\rho)=\sum_{z=1}^{|E|}M_{z}\rho M_{z}^{*}\), this is complete positive since this
\(\mathcal{E}(\rho) = \sum_{z=1}^{|E|} M_z \rho M_z^*\)is TP iff \(\sum M_{z}^{*} M_{z} = I^{A}\) and \(\sum_{z=1}^{|E|} M_z^* M_z = \sum_{z=1}^{|E|} V^* |z\rangle^E \langle z|^E V = V^* \left( \sum_z |z\rangle \langle z|^E \right) V = V^* V = I\)
\(J_{\mathcal{E}}^{AB}=\sum(I\otimes M_{z})\Omega^{A\tilde{A}}(I\otimes M_{z}^{*}) \ge0\) CP
\(\Rightarrow\)) Suppose \(\mathcal{E}\) is CPTP, then since this we have \(\mathcal{E}(\rho)=\sum_{z=1}^{m}M_{z}\rho M_{z}^{*}\) and \(\sum_{z=1}^{m}M_{z}^{*}M_{z}=I^{A}\) where \(E:=\mathbb{C}^m\)
We define \(V:A\to BE\), \(V:=\sum^{m}_{z=1}M_{z}\otimes |z\rang^{E}\) and \(V|\psi^A\rang =\sum_z\underbrace{M_z|\psi^A\rang}_{\in B} |z\rang^E\)
Then we need to check \(V\) is isometry: \(V^*V=\)
\(=\sum_{z,z'\in[m]}M_z^* M_{z'}\underbrace{\lang z|z'\rang}_{\delta_{zz'}}=\sum_{z\in [m]}M_z^*M_z=I^A\)
Then \(\text{Tr}_{E}[V\rho V^{*}]=\sum_{z=1}^{m}\underbrace{\langle^{E}z|V}_{M_{z}}\rho \underbrace{V^{*}|z\rangle^{E}}_{M_{z}^{*}}=\sum_{z=1}^{m}M_{z}\rho M_{z}^{*}\) since \(V^{*}|z\rangle^{E}=\sum_{z^{\prime}=1}^{m}M_{z^{\prime}}^{*}\otimes\left\langle z^{\prime}|z\right\rangle^{E}=M_{z}^{*}\)
Example
\(\mathcal{E}(\rho)=\frac{p}{2}\text{Tr}[\rho]I+(1-p)\rho=\sum_{z=0}^3M_z\rho M_z\) where \(M_{0}= \sqrt{1 - \frac{3p}{4}}I\) and \(M_{x}= \frac{\sqrt{p}}{2}\sigma_{x}\) for \(x = 1, 2, 3\)
Since this, we can define \(V=\sqrt{1-\frac{3p}{4}}I\otimes |0\rang^{E}+\frac{\sqrt{p}}{2}(\sigma_{1}\otimes \ket{1}+\sigma_{2}\otimes \ket{2}+\sigma_{3}\otimes \ket{3})\)
One way: calculate for example \(\bra{00}V\ket{0}=\sqrt{1-\frac{3p}{4}}\) to get the entry
Another: Direct expand
Column sums to one, thus this is column stochastic matrix
The Completely Dehasing Channel
\(\Delta\in CPTP(A\to A)\) and \(\Delta(\rho):=\sum_{x=1}^{|A|}\mathinner{\langle{x}|}\rho\mathinner{|{x}\rangle} \mathinner{|{x}\rangle}\mathinner{\langle{x}|}=\sum_{x=1}^{|A|}\underbrace{\ket{x}\bra{x}} _{M_x}\rho \underbrace{\ket{x}\bra{x}}_{M_x^*}\) where \(\rho= \begin{pmatrix} \rho_{11} & \rho_{12} & - & - \\ \rho_{21} & \rho_{22} & - & - \\ \vdots & \vdots & \ddots & \\ & & & \end{pmatrix}\) and \(\Delta(\rho)= \begin{pmatrix} \rho_{11} & & & \placeholder{} \\ & \rho_{22} & & \placeholder{} \\ & & \ddots & \placeholder{} \\ & & & \rho_{dd} \end{pmatrix}\)
And it's idempotent: \(\Delta\circ \Delta=\Delta\)
Classical Channel
\(\mathcal{E}\in CPTP(A\to A)\) is called a classical channel if \(\Delta\circ \mathcal{E}\circ \Delta=\mathcal{E}\)
We only consider the diagonal input since \(\Delta\) is completely Dehasing Channel.
\(\sigma:=\mathcal{E}(\rho)=\mathcal{E}\left(\sum^{d}_{x=1}t_{x}|x\rang\lang x|\right )=\sum^{d}_{x=1}t_{x}\mathcal{E}(\ket{x}\bra{x})\) where \(\rho = \begin{pmatrix} t_{1} & & 0 \\ & \ddots & \\ 0 & & t_{d} \end{pmatrix}\)
Then \(\sigma\) is also diagonal since \(\mathcal{E}(\rho)=\Delta(\mathcal{E}(\rho))\) by definition
Then \(\sum_{y=1}^{d} S_{y}|y\rangle\langle y| = \sum_{x=1}^{d} t_x \mathcal{E}(|x\rangle\langle x|)\), then after multiply \(\bra{y'},\ket{y'}\) on both sides, we get \(S_{y}=\sum_{x=1}^{d}t_{x}\underbrace{\langle y|\mathcal{E}(|x\rangle\langle x|)|y\rangle} _{p_{y|x}}\) where \(p_{y|x}:= \langle y|\mathcal{E}(|x\rangle\langle x|)|y\rangle\)
This satisfy \(\vec{S} = E\vec{t}\). Also \(\Delta(\sigma)=\sigma\)
POVM Channel
\(\mathcal{E} \in CPT P(A \to A)\) is a POVM channel if \(\mathcal{E} = \Delta \circ \mathcal{E}\) where POVM: \(\{\Lambda_{x}\}_{x \in [m]}, \sum_{x} \Lambda_{x} = I\)
\(\mathcal{E}(\rho)=\Delta\circ\mathcal{E}(\rho)=\sum_{x=1}^{m}\langle x|\mathcal{E} (\rho)|x\rangle|x\rangle\langle x|=\sum_{x=1}^{m}\text{Tr}\left[|x\rangle\langle x|\mathcal{E}(\rho)\right]|x\rangle\langle x|\)
\(=\sum_{x=1}^{m}\text{Tr}\left[ \mathcal{E}^{*}(|x\rangle \langle x|) \rho \right ] |x\rangle \langle x|\) since \(\text{Tr}\left[|x\rangle\langle x|\mathcal{E}(\rho)\right]=\langle|x\rangle\langle x|,\mathcal{E}(\rho)\rangle_{HS}=\langle\mathcal{E^{*}}(|x\rangle\langle x|),\rho \rangle_{HS}=\text{Tr}[\mathcal{E}^*(|x\rangle\langle x|)\rho]\) the last step since \(\mathcal{E}^{*}(|x\rang\lang x|)\geq 0\)(exercise in moodle), then it's output is positive, then hermitian, then the adjoint vanished.
Let \(\Lambda_{x}:=\mathcal{E}^{*}(\ket{x}\bra{x})\), then \(\mathcal{E}(\rho)=\sum_{x=1}^{m}\text{Tr}[\Lambda_{x}\rho]\ket{x}\bra{x}\)
And \(\sum_{x=1}^{m}\Lambda_{x}= \sum_{x=1}^{m}\mathcal{E}^{*}(\ket{x}\bra{x}) = \mathcal{E} ^{*}(\sum_{x}\ket{x}\bra{x}) = \mathcal{E}^{*}(I) = I\) since \(\mathcal{E}\) is trace preserving, then \(\mathcal{E}^*\) is unital, take identity to identity(exercise in moodle)