Does any quantum channel satisfy ${\rm Tr}(\mathcal E^\dagger \mathcal E) \in[0, d^2]$?

Question

I am reading the paper "Direct Fidelity Estimation from Few Pauli Measurements".

According to the paper, the entanglement fidelity between the a unitary channel $\mathcal U$ and a quantum channel $\mathcal E$, is defined as \begin{align} F_e = {\rm Tr}[\mathcal U^\dagger \mathcal E]/d^2, \tag{1} \end{align} where $d$ is dimension of the underlying Hilbert space and ${\rm Tr}(\cdot)$ is the superoperator trace. My aim is to verify that $F_e \in [-1,1]$. We know that ${\rm Tr}(\mathcal U^\dagger \mathcal U)=d^2$. After Eq. (41), it is directly provided that for any quantum channel, ${\rm Tr}(\mathcal E^\dagger \mathcal E) \in[0, d^2]$ and ${\rm Tr}(\mathcal U^\dagger \mathcal E) \in[-d^2, d^2]$. How to prove this?

According to Eq. (41), we have \begin{align} {\rm Tr}(\mathcal U^\dagger \mathcal E) = \sum_{k=1,k=1}^{d^2, d^2}\chi_{U}(k,k') \chi_{\mathcal E}(k,k'), \tag{2} \end{align} where $\chi_{\mathcal E}(k,k')=\frac{1}{d}{\rm Tr}(W_k^\dagger \mathcal E(W_{k'}))$ and $W_k$ are Pauli strings. It can be proved that $\chi_{U}(k,k')$ and $\chi_{\mathcal E}(k,k')$ take values in [-1,1] due to Pauli strings. However, following this way, I obtain ${\rm Tr}(\mathcal U^\dagger \mathcal E)\in [-d^4,d^4]$, which is inconsistent with paper.

Answer 1

Note that the bound $\mathrm{Tr}(\mathcal E^\dagger \mathcal E) \geq 0$ is trivial since $\mathrm{Tr}(\mathcal E^\dagger \mathcal E) = \| \mathcal E \|_2^2$ is the square of the Schatten 2-norm of $\mathcal E$.

The following is slightly simpler if we use the cyclicity of the trace $\mathrm{Tr}(\mathcal E^\dagger \mathcal E) = \mathrm{Tr}(\mathcal E \mathcal E^\dagger)$ and consider the superoperator $\mathcal E \mathcal E^\dagger$ instead.

To prove the bound $\mathrm{Tr}(\mathcal E \mathcal E^\dagger) \leq d^2$, note that the trace is the sum of diagonal entries, say in the Pauli basis $W_k$. Thus, we can w.l.o.g. assume that $\mathcal E \mathcal E^\dagger$ is diagonal in the Pauli basis. Note that any quantum channel $\mathcal A$ which is diagonal in the Pauli basis is automatically unital (since $\mathcal A^\dagger = \mathcal A$ and $\mathcal A$ is trace-preserving by definition). It is well-known that the spectral norm of unital and trace-preserving quantum channels is $\|\mathcal A\|_\infty = 1$ (see e.g. the book by Watrous, Thm. 4.27).

Now, $\mathcal E \mathcal E^\dagger$ is completely positive but not necessarily trace-preserving (or unital, which is the same as we assumed that $\mathcal E \mathcal E^\dagger$ is diagonal). This is determined by the first diagonal entry $$ \frac1d(\mathbb 1|\mathcal E \mathcal E^\dagger| \mathbb 1) = \frac1d \|\mathcal E^\dagger(\mathbb 1)\|_2^2. $$ Here, $\| \cdot \|_2$ is the Schatten 2-norm (or Hilbert-Schmidt / Frobenius norm if you prefer). Now recall that $\mathcal E$ is trace-preserving and thus $\mathcal E^\dagger$ is unital. Hence, we have $$ \frac1d(\mathbb 1|\mathcal E \mathcal E^\dagger| \mathbb 1) = \frac1d \|\mathcal E^\dagger(\mathbb 1)\|_2^2 = \frac1d \|\mathbb 1\|_2^2 = \frac1d \mathrm{tr}(\mathbb 1) = 1. $$ Hence, we have shown that $\mathcal E \mathcal E^\dagger$ is a diagonal quantum channel and thus $\| \mathcal E \mathcal E^\dagger \|_\infty = 1$. From this, it immediately follows that $$ \| \mathcal E \|_2^2 = \mathrm{Tr}(\mathcal E \mathcal E^\dagger) \leq d^2 \| \mathcal E \mathcal E^\dagger \|_\infty = d^2. $$

Finally, we have $\| \mathcal U \|_2^2 = \mathrm{Tr}(\mathcal U^\dagger \mathcal U) = \mathrm{Tr}(\mathrm{Id}) = d^2$ for any unitary channel $\mathcal U$. Hence, by the Cauchy-Schwarz inequality, we have $$ |\mathrm{Tr}(\mathcal U^\dagger \mathcal E)| \leq \| \mathcal U \|_2 \|\mathcal E\|_2 \leq d^2, $$ for any unitary channel $\mathcal U$ and arbitrary channel $\mathcal E$.