What is the isometry connecting the state of a system with its environment for weak dephasing? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T23:48:16Z http://mathoverflow.net/feeds/question/43098 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43098/what-is-the-isometry-connecting-the-state-of-a-system-with-its-environment-for-we What is the isometry connecting the state of a system with its environment for weak dephasing? Jess Riedel 2010-10-21T21:27:04Z 2010-10-22T02:21:55Z <p>This is a math question with a physics motivation. I hope bra-ket notation is acceptable.</p> <h2>Physics Motivation</h2> <p>Suppose we have a $n$-state system $S$ with environment $E$ that is weakly dephased with no feedback. The pure global state is</p> <p>$| \psi_0 \rangle = \left[\sum_i \sqrt{p_i} |i\rangle \right] |e_\emptyset \rangle \to | \psi \rangle = \sum_i \sqrt{p_i} |i\rangle |e_i\rangle$</p> <p>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</p> <p>$\rho_S = \mathrm{Tr}_E |\psi \rangle \langle \psi | = \sum_i \sum_{j} \sqrt{p_i p_{j}} \gamma_{i,j} |i\rangle \langle j|$</p> <p>$\rho_E = \mathrm{Tr}_S |\psi \rangle \langle \psi | = \sum_x p_x | e_x \rangle \langle e_x |$</p> <p>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 between $\rho_S$ and $\rho_E$?</p> <h2>Math Question</h2> <p>What is the matrix $B$ which connects the orthonormal basis $\{| \epsilon_i\rangle \}$ to the normalized but not orthogonal spanning set $\{ | e_i\rangle \}$, </p> <p>$| e_i \rangle = \sum_j B_{i,j} | \epsilon_j \rangle$,</p> <p>such that</p> <p>$\sum_i p_i | e_i \rangle \langle e_i | = \sum_i \sum_{j} \sqrt{p_i p_{j}} \langle e_j | e_i \rangle |\epsilon_i\rangle \langle \epsilon_j|$,</p> <p>where $p_i$ is a probability distribution?</p>