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Let's say Bob prepares a bipartite quantum state $\rho_{AB}$ to be shared between him and Alice. Bob sends Alice's part to her lab. Alice measures her subsystem $A$ in the computational basis $\mathcal{Z}$. Now we want to know the uncertainty in Alice's system given Bob's quantum memory register($B$). This quantity is denoted as $H(Z|B)$.

I was reading this paper by Berta et al. where they mention in a footnote (page 2, footnote 4) that $H(R|B)$ is the conditional von Neumann entropy of the following state: $$ \left(\sum_j |\psi_j \rangle \langle\psi_j| \otimes \mathbb{1} \right) \rho_{AB} \left(\sum_j |\psi_j \rangle \langle\psi_j| \otimes \mathbb{1} \right), $$ where $|\psi_j \rangle$ is the eigenvector of the measurement $\mathcal{R}$. In our case, this would be $\mathcal{Z}$.

My confusion now is this. We know that the sum of eigenvectors of the computational basis $\mathcal{Z}$ makes up the identity matrix $\mathbb{1}$ again. In this case, isn't it becoming the following?

$$ \left(\mathbb{1} \otimes \mathbb{1} \right) \rho_{AB} \left( \mathbb{1} \otimes \mathbb{1} \right) = \rho_{AB}, $$ in which case the conditional von Neumann entropy is simply $H(A|B)$. In the end, are we getting $H(Z|B) = H(A|B)$? This does not seem correct though. Thanks in advance.

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    $\begingroup$ Was it still there by the time it was published? Nature link to abstract, and ArXiv link to newer version. $\endgroup$
    – Mark S
    Jun 20 at 15:46
  • $\begingroup$ I see, that must be it. Thanks Mark. $\endgroup$ Jun 20 at 15:52
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It is almost certainly meant to be the post measurement state $$ \sum_j \left(|\psi_j\rangle \langle \psi_j| \otimes \mathbb{1}\right) \rho_{AB} \left(|\psi_j\rangle \langle \psi_j| \otimes \mathbb{1}\right). $$ Alternatively you may see such a state written as $$ \sum_{j} |j \rangle \langle j | \otimes \rho_B(j) $$ where $\rho_B(j) = \mathrm{tr}_B[ |\psi_j\rangle \langle \psi_j | \rho_{AB}]$.

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