Choi-Kraus Theorem
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The proof of Choi-Kraus theorem and understanding
This post originates from the assignment of the course “Introduction to Quantum Information”, where a state damping form of the quantum system and reservoir is given.
\[\ket{0}\ket{e_0}\rightarrow\ket{0}\ket{e_0}\] \[\ket{1}\ket{e_o}\rightarrow\sqrt{q}\ket{0}\ket{e_1}+\sqrt{1-p}\ket{1}\ket{e_0}\]Using partial trace we can convert it into the density matrix form.
\[Tr_e(\ket{after}\bra{after})\] \[\begin{bmatrix} p_{00}+pp_{11}&\sqrt{1-p}p_{01}\\ \sqrt{1-p}p_{10}&(1-p)p_{11} \end{bmatrix}\]But from this form, how should we derive the Kraus operator is a question.
This leads to the Choi-Kraus Theorem
Now what is Choi-Kraus theorem? The definition goes here
To prove it we inroduce the Choi state which is
\[N_{A-B}\ket{\Gamma}\bra{\Gamma}\]where
\[\ket{\Gamma}=\sum_{i}^{d_A-1}\ket{i}_R\otimes\ket{i}_A\]The R is the auxiliary space.
And because the requirement of CPTP the target of mapping is positive semidefinite can be diagonalized(some doubt here)
it goes like
\[\sum_{i}^{d-1}\ket{\phi_l}\bra{\phi_l}\]take one $\ket{\phi}$ out and suppose it as the most fundamental form
\[\ket{\phi}=\sum_{i}^{d_A-1}\sum_{j}^{d_B-1}\alpha_{ij}\ket{i}_R\ket{j}_B\]check a possible V form as
\[V=\sum_{i}^{d_A-1}\sum_{j}^{d_B-1}\alpha_{ij}\ket{j}_B\bra{i}_A\]And we see
\[(I\otimes V)\ket{\Gamma}=\sum_{i}\sum_{j}\sum_k\alpha_{jk}\ket{i}_R\otimes\ket{k}_B\bra{j}_A\ket{i}_A=\ket{\phi}\]Then it obviously leads to
\[(I\otimes V)\ket{\Gamma}\bra{\Gamma}(I\otimes V)^{\dagger}=N_{A-B}(\ket{\Gamma}\bra{\Gamma})\]then do
\[\bra{i}_R\ket{\phi}\bra{\phi}\ket{j}_R=V\ket{i}\bra{j}V^{\dagger}=\bra{i}_R N_{A-B} (\ket{\Gamma}\bra{\Gamma})\ket{j}_R\]and because of linearity it applies to any density matrix
now back to the question we mentioned earlier that how can we find the right Kraus decomposition form. The answer lays with the the Choi state. in our situation, it only invovles two level system
so our Choi state is
\[(\ket{00}+\ket{11})(\bra{00}+\bra{11})\]after our mapping mentioned in the beginning
\[\ket{00}\bra{00}+\sqrt{1-p}\ket{11}\bra{00}+\sqrt{1-p}\ket{00}\bra{11}+p\ket{01}\bra{01}+(1-p)\ket{11}\bra{11}\]so we can suddenly see the matrix form of our Kraus operator
\[\begin{bmatrix} 0 & \sqrt{p}\\ 0 & 0\\ \end{bmatrix}\] \[\begin{bmatrix} 1 & 0\\ 0 & \sqrt{1-p}\\ \end{bmatrix}\]