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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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