# With $\vert\Psi^+\rangle$ the Bell state, can $\sqrt{\rho}\vert\Psi^+\rangle\langle\Psi^+\vert\sqrt{\rho}$ be simplified?

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 $$D\left(\Omega_{AB}(\rho)\|\rho_A\otimes\frac{I_B}{|B|}\right)$$

• 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
– glS
Dec 18, 2022 at 23:30
• You can use the identity (which you should try to prove) $$\mathrm{Tr}_B[(X_A \otimes \mathbb{I}) Y_{AB} (Z_A \otimes \mathbb{I})] = X_A \mathrm{Tr}_B[Y_{AB}] Z_A$$ Dec 19, 2022 at 7:47

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 $$N$$.