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How do I prove that for a general tri-partite state $\rho_{ABE}$, the following holds: $$ H(\rho_{AB}) = H(\rho_{E}), H(\rho_{AE}) = H(\rho_{B}), $$ where, $H$ is the Von Neumann entropy. Would Schmidt decomposition help? But I can only do it in a bi-partite scenario. Thanks!

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This is not true for general tripartite states. Take the trivial example where $ABE$ share a maximally mixed state and each parties subsystem is of dimension $d$. The reduced states of a two-party subsystem are maximally mixed of dimension $d^2$ and of a single party system are maximally mixed of dimension $d$. As the dimensions are different and they are maximally mixed they cannot have the same entropy.

However, the result does hold if $\rho_{ABE} = |\psi\rangle \langle \psi|$ is a pure state. Moreover, we can use the Schmidt decomposition by identifying a two-party subsystem with just a single party. For example lets call $H_{A} \otimes H_B$ just $H_{D}$. Then we can view $|\psi\rangle$ as a state in $H_D \otimes H_E$ and use the Schmidt decomposition. That is we know there exists orthonormal bases $\{|i\rangle_D\}$ and $\{|i\rangle_E\}$ of $D$ and $E$ respectively such that $$ |\psi \rangle_{DE} = \sum_i \sqrt{\lambda_i} |ii\rangle. $$ As a consequence the reduced states have the same spectrum, $ \rho_D = \sum_i \lambda_i |i \rangle \langle i |$ and $\rho_E = \sum_i \lambda_i |i \rangle \langle i |$ and hence $H(D) = H(E)$. If you want to make this proof more formal you can do the identifying $H_A \otimes H_B$ with a single Hilbert space $H_D$ step using an isometry map $V : H_{A} \otimes H_{B} \rightarrow H_D$ and then note that the von Neumann entropy is invariant under isometries.

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  • $\begingroup$ Thanks! My initial understanding was flawed. Thanks again for the clarification. $\endgroup$ Nov 16 '20 at 18:36

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