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What is the isometry connecting the state of a system with it's its environment for weak dephasing? |
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This is a math question with a physics motivation. I hope bra-ket notation is acceptable. Physics MotivationSuppose we have a $n$-state system $S$ with environment $E$ that is weakly dephased with no feedback. The pure global state is $| \psi_0 \rangle = \left[\sum_i \sqrt{p_i} |i\rangle \right] |e_\emptyset \rangle \to | \psi \rangle = \sum_i \sqrt{p_i} |i\rangle |e_i\rangle$ with $|i\rangle \in S$, $|e_i\rangle \in E$, $\sum_i p_i = 1$, and $i=1,2,\cdots,n$. If we define $\gamma_{i,j} = \langle e_{j} | e_{i} \rangle$, "Weakly" dephased means $\gamma_{i,j} \neq 0,1$ for $i \neq j$. The reduced states are $\rho_S = \mathrm{Tr}_E |\psi \rangle \langle \psi | = \sum_i \sum_{j} \sqrt{p_i p_{j}} \gamma_{i,j} |i\rangle \langle j| $ $\rho_E = \mathrm{Tr}_S |\psi \rangle \langle \psi | = \sum_x p_x | e_x \rangle \langle e_x | $ For large $n$ and arbitrary $\gamma_{i,j}$, it is "hard" to disagonalize these matrices. In particular, you can't write down a closed formula for the eigenvalues in terms of $\gamma_{i,j}$. But, since the global state is pure, the spectrums of $\rho_S$ and $\rho_E$ are the same even though they appear in different forms above. So my question is this: is there a nice way to write down the isometry which between $\rho_S$ and $\rho_E$? Math QuestionWhat is the matrix $B$ which connects the orthonormal basis $\{| \epsilon_i\rangle \}$ to the normalized but not orthogonal spanning set $\{ | e_i\rangle \}$, $| e_i \rangle = \sum_j B_{i,j} | \epsilon_i epsilon_j \rangle$, such that $\sum_i p_i | e_i \rangle \langle e_i | = \sum_i \sum_{j} \sqrt{p_i p_{j}} \langle e_j | e_i \rangle |\epsilon_i\rangle \langle \epsilon_j| $, where $p_i$ is a probability distribution? |
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What is the isometry connecting the state of a system with it's environment for weak dephasing?This is a math question with a physics motivation. I hope bra-ket notation is acceptable. Physics MotivationSuppose we have a $n$-state system $S$ with environment $E$ that is weakly dephased with no feedback. The pure global state is $| \psi_0 \rangle = \left[\sum_i \sqrt{p_i} |i\rangle \right] |e_\emptyset \rangle \to | \psi \rangle = \sum_i \sqrt{p_i} |i\rangle |e_i\rangle$ with $|i\rangle \in S$, $|e_i\rangle \in E$, $\sum_i p_i = 1$, and $i=1,2,\cdots,n$. If we define $\gamma_{i,j} = \langle e_{j} | e_{i} \rangle$, "Weakly" dephased means $\gamma_{i,j} \neq 0,1$ for $i \neq j$. The reduced states are $\rho_S = \mathrm{Tr}_E |\psi \rangle \langle \psi | = \sum_i \sum_{j} \sqrt{p_i p_{j}} \gamma_{i,j} |i\rangle \langle j| $ $\rho_E = \mathrm{Tr}_S |\psi \rangle \langle \psi | = \sum_x p_x | e_x \rangle \langle e_x | $ For large $n$ and arbitrary $\gamma_{i,j}$, it is "hard" to disagonalize these matrices. In particular, you can't write down a closed formula for the eigenvalues in terms of $\gamma_{i,j}$. But, since the global state is pure, the spectrums of $\rho_S$ and $\rho_E$ are the same even though they appear in different forms above. So my question is this: is there a nice way to write down the isometry which between $\rho_S$ and $\rho_E$? Math QuestionWhat is the matrix $B$ which connects the orthonormal basis $\{| \epsilon_i\rangle \}$ to the normalized but not orthogonal spanning set $\{ | e_i\rangle \}$, $| e_i \rangle = \sum_j B_{i,j} | \epsilon_i \rangle$, such that $\sum_i p_i | e_i \rangle \langle e_i | = \sum_i \sum_{j} \sqrt{p_i p_{j}} \langle e_j | e_i \rangle |\epsilon_i\rangle \langle \epsilon_j| $, where $p_i$ is a probability distribution?
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