Let $\vert\Psi^+\rangle_{AB} = \frac{1}{\sqrt n}\sum_{i=1}^n\vert i\rangle_A\vert i\rangle_B$ be the maximally entangled state in Hilbert space $\mathcal{H}(AB)$ and $\rho_A$ be some state in Hilbert space $\mathcal{H}(A)$.

I encountered the expression $$\Omega_{AB}(\rho) := (\rho^{1/2}_A\otimes I_B)\vert\Psi^+\rangle\langle\Psi^+\vert_{AB}(\rho^{1/2}_A\otimes I_B)$$

in some calculations. Note that the state $\Omega$ depends on the state $\rho$. My goal is to compute two quantities:

  1. The marginal state $\Omega_A(\rho)$

  2. The relative entropy $D\left(\Omega_{AB}(\rho)\|\rho_A\otimes\frac{I_B}{|B|}\right)$

  • 3
    $\begingroup$ If you partial trace $\Omega_{AB}(\rho)$ as defined here wrt the second space you get simply $\rho_A$ because partial tracing a(ny) maximally entangled state gives you the identity $\endgroup$
    – glS
    Dec 18, 2022 at 23:30
  • 2
    $\begingroup$ You can use the identity (which you should try to prove) $$ \mathrm{Tr}_B[(X_A \otimes \mathbb{I}) Y_{AB} (Z_A \otimes \mathbb{I})] = X_A \mathrm{Tr}_B[Y_{AB}] Z_A $$ $\endgroup$
    – Rammus
    Dec 19, 2022 at 7:47

1 Answer 1


TL;DR: Yes, $\Omega_{AB}(\rho)$ can be simplified. It turns out that it's one $n$th of the projector onto the ray spanned by the vectorization of $\rho^{1/2}$, see $(3)$ below.

First, observe that $\Omega_{AB}(\rho)$ is the outer product $$ \Omega_{AB}(\rho) = |\psi_{AB}\rangle\langle\psi_{AB}|\tag1 $$ where $|\psi_{AB}\rangle = (\rho_A^{1/2}\otimes I_B)|\Psi^+\rangle$ which has norm $\frac{1}{\sqrt{n}}$. But $$ \langle i_A|\langle j_B|\psi_{AB}\rangle = \frac{1}{\sqrt{n}}\sum_{k=1}^n\langle i_A|\rho_A^{1/2}|k_A\rangle\langle j_B|k_B\rangle=\frac{1}{\sqrt{n}}\langle i_A|\rho_A^{1/2}|j_A\rangle\tag2 $$ so $|\psi_{AB}\rangle$ is a scalar multiple of the vectorization of $\rho_A^{1/2}$. Thus, denoting$^1$ the vectorization of matrix $M_A\in L(\mathcal{H}_A)$ by $|M_{AB}\rangle\in\mathcal{H}_A\otimes\mathcal{H}_B$, we can rewrite $(1)$ as $$ \Omega_{AB}(\rho) = \frac{1}{n}|\rho_{AB}^{1/2}\rangle\langle\rho_{AB}^{1/2}|\tag3 $$ where $|\rho_{AB}^{1/2}\rangle$ turns out to be normalized$^2$. Note that $\Omega_{AB}(\rho)$ is not actually a state, since $\mathrm{tr}(\Omega_{AB}(\rho))=\frac{1}{n}$.

That said, $\Omega_{AB}(\rho)$ may not necessarily be the most effective choice of a subexpression to isolate for simplification. See Rammus's comment for an alternative.

$^1$ Note that vectorization turns an "input" index into an "output" index. We are choosing to place the extra output index on the second system ($B$). This choice requires that $\mathcal{H}_A$ and $\mathcal{H}_B$ be isomorphic and is justified by the form of $|\Psi^+\rangle$. This accounts for the label change from $M_A$ to $|M_{AB}\rangle$.
$^2$ Perhaps the simplest way to see this is to note that the dot product $\langle M|N\rangle$ of vectorizations $|M\rangle$ and $|N\rangle$ is the Hilbert-Schmidt inner product $\mathrm{tr}(M^\dagger N)$ of $M$ and $N$.


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