# Properties of composition of isometry and a perturbed adjoint

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

• 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. Nov 11, 2020 at 13:23
• @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... Nov 11, 2020 at 14:44
• Btw, once you have the equality, the inequality follows from the neat little identity that $\langle \Phi_{AR} | I_A \otimes M_R | \Phi_{AR} \rangle = \mathrm{Tr}[M_R]$. Nov 12, 2020 at 12:20
• @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? Nov 12, 2020 at 13:03
• yep, woops, sorry. Nov 12, 2020 at 13:11

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.

• ~ Nice bound! :) Nov 13, 2020 at 13:50
• @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? Nov 21, 2020 at 16:53

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.