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Suppose $\vert\Phi\rangle_{AR} = \frac{1}{\sqrt{|D|}}\sum_{i\in D} \vert ii\rangle_{AR}$ is the maximally entangled state. Let $V_{A\rightarrow BE}$ and $\tilde{V}_{A\rightarrow BE}$ be two isometries from $H_A$ to $H_B\otimes H_E$ such that

$$\langle\Phi_{AR}\vert I_R\otimes \tilde{V}^\dagger V\vert\Phi_{AR}\rangle\geq 1- \varepsilon$$

My questions are about the map $\tilde{V}^\dagger V$.

  1. Is $\tilde{V}^\dagger V\leq I_A$ in a positive semidefinite sense? Following Rammus' comment, perhaps this doesn't make sense. Instead, how can one show that $\|I - \tilde{V}^\dagger V\|_1 = \text{tr}(I - \tilde{V}^\dagger V)$?

  2. How can one show that $\|I_A - \tilde{V}^\dagger V \|_1\leq \varepsilon|D|$?

Both these claims are made in Lemma 14 of this paper

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    $\begingroup$ To comment on 1. most times when we say something is PSD we make an assumption that it is also Hermitian. However, here $\tilde{V}^* V$ is not necessarily Hermitian. $\endgroup$
    – Rammus
    Nov 11 '20 at 13:23
  • $\begingroup$ @Rammus I see! The specific statement made in the paper is $\|I - \tilde{V}^\dagger V\|_1 = \text{Tr}(I - \tilde{V}^\dagger V) \leq \varepsilon|D|$. My two questions are basically about the equality and the inequality. I assumed that the trace was equal to the 1-norm only when the argument is PSD. Perhaps that's incorrect... $\endgroup$ Nov 11 '20 at 14:44
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    $\begingroup$ Btw, once you have the equality, the inequality follows from the neat little identity that $\langle \Phi_{AR} | I_A \otimes M_R | \Phi_{AR} \rangle = \mathrm{Tr}[M_R]$. $\endgroup$
    – Rammus
    Nov 12 '20 at 12:20
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    $\begingroup$ @Rammus That's a nice identity, thank you for pointing it out. I assume there is a factor of the dimension missing in your comment above? $\endgroup$ Nov 12 '20 at 13:03
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    $\begingroup$ yep, woops, sorry. $\endgroup$
    – Rammus
    Nov 12 '20 at 13:11
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After @Rammus answer explaining that the given inequality does not hold in general, i'll try to prove a weaker statement.

Define $ \Delta = V - \tilde{V} $. The assumption is equivalent to $$ \text{Tr}[I - \tilde{V}^{\dagger} V] \leq \epsilon \cdot |D| \implies \text{Tr}[V^{\dagger}V - \tilde{V}^{\dagger} V] \leq \epsilon \cdot |D| \implies \text{Tr}[\Delta^{\dagger} V] \leq \epsilon \cdot |D| $$

Since our assumption still holds if we exchange $V, \tilde{V}$ we get $ \hspace{0.3em} \text{Tr}[(- \Delta)^{\dagger} \tilde{V}] \leq \epsilon \cdot |D| $. The two inequalities combined give $ \text{Tr}[\Delta^{\dagger}\Delta] \leq 2 \epsilon \cdot |D| $.

So \begin{align*} & ||I - \tilde{V}^{\dagger} V ||_1 = ||\Delta^{\dagger} V||_1 = ||\Delta^{\dagger}||_1 = \\ & \sum_k s_k \leq \sqrt{|D|} \cdot \sqrt{\sum_k s_k^2} = \sqrt{|D|} \cdot \sqrt{\text{Tr}[\Delta^{\dagger}\Delta]} \leq \sqrt{2\epsilon} \cdot |D| \end{align*}

where $ s_k $ the singular values of $ \Delta $ and we used Cauchy-Schwarz inequality and the fact that trace norm is isometrically invariant.

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  • $\begingroup$ ~ Nice bound! :) $\endgroup$
    – Rammus
    Nov 13 '20 at 13:50
  • $\begingroup$ @tsgeorgios could you add a reference for the claim that the trace norm is isometrically invariant as used here? In Mark Wilde's book, arxiv.org/pdf/1106.1445.pdf in Property 9.1.4, it is shown that $\|A V^\dagger\|_1 = \|A\|_1$ where $V$ is an isometry. You are using the adjoint here i.e. $\|AV\|_1 = \|A\|_1$ so why does that hold? $\endgroup$ Nov 21 '20 at 16:53
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I'll use $W$ instead of $\tilde V$. I believe there must be more constraints in the paper as to relation between $W$ and $V$ because as it stands the inequality is not true. For example take $V$ to be the identity matrix and let $W$ also be the identity matrix except we make the penultimate and final elements on the diagonal of $W$ be $e^{i \alpha}$ and $e^{-i \alpha}$ for some $\alpha \in [0,2\pi)$. Then $$ \mathrm{Tr}[ W^* V] = (|D|-2) + 2 \cos(\alpha). $$ So, using the identity $\langle \Phi_{AR} | I_A \otimes M_R | \Phi_{AR} \rangle = \frac{1}{|D|} \mathrm{Tr}[M_R]$, the constraint in the question is equivalent to $$ (|D|-2) + 2 \cos(\alpha) \geq |D|(1-\epsilon). $$ Rearranging we find this is equivalent to $$ \epsilon \geq \frac{2 - 2 \cos(\alpha)}{|D|} = \frac{4 \sin(\alpha/2)^2}{|D|}. $$

Now let $K = \begin{pmatrix} 1-e^{-i \alpha} & 0 \\ 0 & 1-e^{i \alpha} \end{pmatrix}$, then $$ \| I - W^* V \|_1 = \| K \|_1 = 4 \sin(\alpha / 2). $$

The claim is that if $|D| \epsilon \geq 4 \sin(\alpha/2)^2$ then we also have $|D|\epsilon \geq 4 \sin(\alpha/2)$ but notice that if $\alpha \in (0, \pi/2)$ then $\sin(\alpha/2)^2 < \sin(\alpha/2)$. Thus we can choose values of $\epsilon$ such that the constraint holds but the conclusion does not.

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