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

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.

• Please can you make the question self-contained. What is $\varepsilon$, $U$ and $d$? I also see no quantum channel in the question at the moment. Commented Jan 10, 2023 at 10:08
• What kind of trace is this, if those are channels? Commented Jan 10, 2023 at 10:58
• Thank you. I have updated the information. Commented Jan 10, 2023 at 11:15
• What is the "superoperator trace"? Commented Jan 10, 2023 at 20:56
• @NorbertSchuch I made some edits such that the question is more in line with the paper. I think this is self-explanatory, no? Traces are defined for linear maps and superoperators are linear maps. Commented Jan 11, 2023 at 9:54

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 $$\mathcal E$$.