# Computing $H(Z|B)$ in a bipartite density matrix $\rho_{AB}$

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.

• Was it still there by the time it was published? Nature link to abstract, and ArXiv link to newer version. Jun 20, 2021 at 15:46
• I see, that must be it. Thanks Mark. Jun 20, 2021 at 15:52

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 $$\rho_B(j) = \mathrm{tr}_B[ |\psi_j\rangle \langle \psi_j | \rho_{AB}]$$.