Infidelity as distance measure

Question

Let $\mathcal{X} \in {\rm CP}(\mathcal{H}, \mathcal{K})$ and unital (completely positive and unital maps). Let $\mathcal{Y} \in {\rm CPT}(\mathcal{H}, \mathcal{K})$ (completely positive and trace preserving). We can derive $$ \langle {\rm id}, \mathcal{Y} \rangle \leq \Vert {\rm id} \Vert_F \Vert \mathcal{Y} \Vert_F \leq d^2,\tag1 $$ where the equality holds when ${\rm id}=\mathcal{Y}$ (This can be found in the proof of lemma 48 / arXiv). Now we aim to use the above inequality to evaluate $$ \langle \mathcal{X}, \mathcal{Y} \rangle = \langle {\rm id}, \mathcal{X}^\dagger\circ \mathcal{Y} \rangle \leq d^2,\tag2 $$ where the inequality is due to the fact that $\mathcal{X}$ is unital and CP, so $\mathcal{X}^\dagger$ is CPT, then $\mathcal{X}^\dagger \circ \mathcal{Y}$ is CPT. By Cauchy-Schwarz inequality, the equality holds when $\mathcal{X}^\dagger \circ \mathcal{Y}={\rm id}$. Also, if we directly use Cauchy-Schwarz inequality for $\langle \mathcal{X}, \mathcal{Y} \rangle$, the equality holds when $\mathcal{X}=\mathcal{Y}$. Then we can deduce that $\mathcal{X} = \mathcal{Y}$ is equavalent to $\mathcal{X}^\dagger \circ \mathcal{Y}={\rm id}$. However, it is not generally true. So, what are the problems of these derivations?

Answer 1

tl;dr. In Eq. (2) in the original question there is an intermediate step missing which is where equality actually holds when Cauchy-Schwarz is saturated: linear dependence of $\mathcal X,\mathcal Y$ is equivalent to $\langle \mathcal{X}, \mathcal{Y} \rangle=\|X\|_F\|Y\|_F$ and not to $\langle \mathcal{X}, \mathcal{Y} \rangle = \langle {\rm id}, \mathcal{X}^\dagger\circ \mathcal{Y} \rangle =d^2$. So as $\|X\|_F\|Y\|_F$ has of course no reason to be equal to $d^2$ so the equality in this case reads $$ \langle\mathcal X,\mathcal Y\rangle\color{red}{\bf=}\|X\|_F\|Y\|_F\leq d^2 $$

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The two equalities in question are not equivalent because they satisfy different bounds; this point should become quite obvious with the following simple example: consider $\mathcal X=\mathcal Y=\mathcal D$ to be the qubit dephasing map $$ \mathcal D\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}:=\begin{pmatrix}a_{11}&0\\0&a_{22}\end{pmatrix}. $$ This map is CP, trace preserving, and unital and has corresponding representation matrix (see also here and here) $$ \widehat{\mathcal D}=\begin{pmatrix}1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&1\end{pmatrix}\,. $$ Now let us look at the two different applications of Cauchy-Schwarz:

1. The quantity${}^1$ \begin{align*} \langle{\rm id},\mathcal X^\dagger\circ\mathcal Y\rangle=\langle{\bf1},\widehat{\mathcal X}^\dagger\widehat{\mathcal Y}\rangle=\langle{\bf1},\widehat{\mathcal D}^\dagger\widehat{\mathcal D}\rangle=\langle{\bf1},\widehat{\mathcal D}\rangle={\rm tr}(\widehat{\mathcal D})={\rm tr}\begin{pmatrix}1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&1\end{pmatrix}=2 \end{align*} is strictly smaller than its Cauchy-Schwarz upper bound $$ \|{\bf1}\|_F\|\|\widehat{\mathcal X}^\dagger\widehat{\mathcal Y}\|_F=\|{\bf1}\|_F\|\|\widehat{\mathcal D}\|_F=\sqrt4\cdot\sqrt2=2\sqrt2 $$ because ${\rm id}$ and $\mathcal X^\dagger\circ\mathcal Y=\mathcal D$ are not linearly dependent. As a result neither of these expressions is equal to $d^2=4$.
2. The quantity $\langle\mathcal D,\mathcal D\rangle=\langle\mathcal X,\mathcal Y\rangle=\langle{\rm id},\mathcal X^\dagger\circ\mathcal Y\rangle=2$ is equal to its upper bound \begin{align*} \|\widehat{\mathcal D}\|_F\|\|\widehat{\mathcal D}\|_F=\sqrt2\cdot\sqrt2=2 \end{align*} because, trivially, $\mathcal X$ and $\mathcal Y$ are linearly dependent. Of course while these two are equal this quantity has nothing to do with $d^2=4$ because there is no identity channel involved (and $\|{\rm id}\|_F=d$ is what gets the upper bound to $d^2$ in the first place).

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_{1: We will frequently use that switching to the representation matrix is a unitary operation, that is, $\langle\Phi,\Psi\rangle=\langle\widehat\Phi,\widehat\Psi\rangle_{\rm HS}$ for all linear maps $\Phi,\Psi$}