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# Tag Info

## New answers tagged density-matrix

2

Another way to see this is remembering that the density matrix of a bipartite pure state is essentially its covariance matrix, in the sense that if a state $|\psi\rangle$ has components $\psi_{ij}$ and $\rho=|\psi\rangle\!\langle\psi|$, then $\rho=\psi\psi^\dagger$ (where we are thinking of $\psi$ as a matrix, so that $(\psi\psi^\dagger)_{ij}=\sum_\ell \psi_{... 5$\newcommand{\ket}[1]{\vert#1\rangle}$First, write$\ket\psi$and$\ket{\tilde\psi}in their Schmidt decomposition: \begin{aligned} \ket\psi &= \sum \lambda_i \ket{a_i}\ket{b_i}\ , \\ \ket{\tilde\psi} & = \sum \tilde\lambda_i \ket{\tilde a_i}\ket{\tilde b_i}\ . \end{aligned} Let us assume for simplicty that the\lambda_i$are non-degenerate. ... 2 It is due to Schmidt decomposition. For some$|\psi \rangle_{AB} \in H_A \otimes H_B$, there exists a decomposition in terms of the orthonormal basis (Schmidt bases) of system A and B.$\lambda_i$are the Schmidt coefficients calculated from$Tr_B(|\psi\rangle \langle\psi|_{AB})$whose eigenvalues are$\lambda^2_i\$. Given below is the Schmidt decomposition, {...

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