Proof that $\frac{1}{d} | \text{Tr}(\mathcal{P}_i^\dagger \mathcal{E}(\mathcal{P}_j))| \leq 1$ for superoperators

Question

I have two Pauli operators $\frac{1}{\sqrt{d}} \mathcal{P}_i$, $\frac{1}{\sqrt{d}} \mathcal{P}_j$, and an arbitrary quantum channel $\mathcal{E}$ (in the superoperator/Liouville representation) all acting on a $d=2^n$ dimensional Hilbert space. I want to show that the matrix element $\frac{1}{d} | \text{Tr}(\mathcal{P}_i^\dagger \mathcal{E}(\mathcal{P}_j))| \leq 1$.

Is this true? If so, how can one show this? Unfortunately, all inequalities for trace known to me don't work, because neither $\mathcal{P}_i$ or $\mathcal{E}(\mathcal{P}_j)$ are positive semidefinite.

Related question: Does any quantum channel satisfy ${\rm Tr}(\mathcal E^\dagger \mathcal E) \in[0, d^2]$? Essentially I am looking for a proof that $\frac{1}{d} | \text{Tr}(\mathcal{P}_k^\dagger \mathcal{E}(\mathcal{P}_{k'}))| = |\chi_\mathcal{E} (k, k')| \leq 1$.

Answer 1

Here's a one-line proof using Hölder's inequality: $$ |\mathrm{tr}\left( \mathcal{P}_i^\dagger \mathcal{E}(\mathcal{P}_i) \right)| \leq \| \mathcal{P}_i^\dagger \|_\infty \|\mathcal{E}\|_{1\rightarrow1} \| \mathcal{P}_j \|_1 = \| \mathcal{P}_j \|_1 = \mathrm{tr}(I) = d. $$ We only used that quantum channels have unit $1\rightarrow1$ norm, as well that unitaries have unit spectral norm and trace norm $d$. Note this also works for trace non-increasing CP superoperators (and, more generally, for any superoperator in the $1\rightarrow1$ norm ball, or in the dual $\infty\rightarrow\infty$ ball).

Note: $\|\cdot\|_p$ denotes the Schatten $p$-norms, and $\| \cdot \|_{p\rightarrow q}$ is the induced norm on superoperators, i.e. $$ \| \Phi \|_{p\rightarrow q} := \sup_{\|X\|_p\leq 1} \|\Phi(X)\|_q \,. $$

Answer 2

TL;DR: It's true. You have a good sense that positive semi-definiteness would help. We don't have it for the Paulis, but we have it for the eigenprojectors.

Let $P_+$ and $P_-$ denote the projectors onto the eigenspaces of $\mathcal{P}_i$ associated with the $+1$ and $-1$ eigenvalues, respectively, and similarly for $R_\pm$ and $\mathcal{P}_j$. Also, set $Q_\pm:=\mathcal{E}(R_\pm)$. Then \begin{align} \frac{1}{d}\mathrm{tr}(\mathcal{P}_i^\dagger\mathcal{E}(\mathcal{P}_j)&=\frac{1}{d}\mathrm{tr}((P_+-P_-)^\dagger\mathcal{E}(R_+-R_-))\tag1\\ &=\frac{1}{d}[\mathrm{tr}(P_+Q_+)-\mathrm{tr}(P_-Q_+)-\mathrm{tr}(P_+Q_-)+\mathrm{tr}(P_-Q_-)].\tag2 \end{align} Now, $P_\pm$ and $R_\pm$ are positive semi-definite. Moreover, $\mathcal{E}$ is completely positive, so $Q_\pm$ are positive semi-definite, too. Hilbert-Schmidt inner product of two positive semi-definite operators is non-negative, so $\mathrm{tr}(P_\pm Q_\pm)\geqslant 0$. Therefore, \begin{align} -\mathrm{tr}(P_+Q_+)-\mathrm{tr}(P_-Q_+)-\mathrm{tr}(P_+Q_-)-\mathrm{tr}(P_-Q_-)\tag3\\ \leqslant\mathrm{tr}(P_+Q_+)-\mathrm{tr}(P_-Q_+)-\mathrm{tr}(P_+Q_-)+\mathrm{tr}(P_-Q_-)\tag4\\ \leqslant\mathrm{tr}(P_+Q_+)+\mathrm{tr}(P_-Q_+)+\mathrm{tr}(P_+Q_-)+\mathrm{tr}(P_-Q_-).\tag5 \end{align} However, \begin{align} &\mathrm{tr}(P_+Q_+)+\mathrm{tr}(P_-Q_+)+\mathrm{tr}(P_+Q_-)+\mathrm{tr}(P_-Q_-)\tag6\\ &=\mathrm{tr}((P_++P_-)\mathcal{E}(R_++R_-))\tag7\\ &=\mathrm{tr}(\mathcal{E}(I))=2^n=d\tag8 \end{align} where we used the fact that $\mathcal{E}$ is trace preserving. Substituting into the inequalities above and reintroducing normalization constant $\frac{1}{d}$, we obtain \begin{equation} -1\leqslant \frac{1}{d}\mathrm{tr}(\mathcal{P}_i^\dagger\mathcal{E}(\mathcal{P}_j))\leqslant+1\tag9 \end{equation} as claimed.