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

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 $$\frac{1}{d} | \text{Tr}(\mathcal{P}_k^\dagger \mathcal{E}(\mathcal{P}_{k'}))| = |\chi_\mathcal{E} (k, k')| \leq 1$$.

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 $$$$-1\leqslant \frac{1}{d}\mathrm{tr}(\mathcal{P}_i^\dagger\mathcal{E}(\mathcal{P}_j))\leqslant+1\tag9$$$$ as claimed.

• Perfect! Thank you for the proof and the very cool technique :-) May 12, 2023 at 2:50

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. $$\| \Phi \|_{p\rightarrow q} := \sup_{\|X\|_p\leq 1} \|\Phi(X)\|_q \,.$$

• how's the "$1\to 1$ norm" defined here?
– glS
May 12, 2023 at 7:44
• @glS it is the operator norm induced by the Schatten 1-norm (trace norm) on both domain and image. You can also take the diamond norm if you like (which is the "stabilized" $1\rightarrow 1$ norm). May 12, 2023 at 8:28
• Just to make sure I understand, as this is very new to me. $|\text{Tr}(\mathcal{P}_i^\dagger \mathcal{E}(\mathcal{P}_j))| = \left\Vert \mathcal{P}_i^\dagger \mathcal{E}(\mathcal{P}_j))\right\Vert_1 \leq \left\Vert \mathcal{P}_i^\dagger \right\Vert_\infty \left\Vert\mathcal{E}(\mathcal{P}_j)) \right\Vert_1$ is Hölder's inequality on the Schatten norm. The induced norm (as defined in arxiv.org/pdf/quant-ph/0411077.pdf) is $\left\Vert \mathcal{E} \right\Vert_{q \rightarrow p} = \sup_{X \neq 0} \frac{\left\Vert \mathcal{E}(X) \right\Vert_p}{\left\Vert X \right\Vert_q}$. May 12, 2023 at 13:41
• Therefore $\left\Vert\mathcal{E}(\mathcal{P}_j)) \right\Vert_1 \leq \left\Vert \mathcal{E} \right\Vert_{1 \rightarrow 1} \left\Vert \mathcal{P}_j \right\Vert_1$, and we substitute this into the first inequality. Correct? May 12, 2023 at 13:42
• @JedBurkat correct, except that the first equality in your first comment is generally an inequality. The operator norm is in general defined for a linear operator between normed spaces, see also en.m.wikipedia.org/wiki/Operator_norm May 13, 2023 at 16:08