Weight enumerators for Hermitian operator (wrong $ B_j $ definition)

Question

Let $ H $ be Hermitian operator on an $ n $ qubit Hilbert space $ \mathbb{C}^{2^n} $. Define the weight enumerator coefficients $$ A_j=\frac{1}{({\rm tr}(H))^2} \sum_{E \in \mathcal{E}_j} |{\rm tr}(EH)|^2 $$ where $ \mathcal{E}_j $ is the set of all Paulis of weight $ j $. And define the dual weight enumerator coefficients $$ B_j=\frac{1}{{\rm tr}(H)} \sum_{ E \in \mathcal{E}_j} {\rm tr}(EH E^{\dagger} H). $$ Is it the case that $$ B_j \geq A_j $$ for all $ j $?

Note that if $ H $ is a projector, i.e. $ H^\dagger=H $ and $ H^2=H $ then this is a standard result from the theory of weight enumerators. I am wondering if it is still true when $ H $ is Hermitian but not necessarily a projection.

Answer 1

TL;DR: No. Suppose that $H=\alpha H'$ for some scale factor $\alpha\in\mathbb{R}$. The key observation is that $A_j$ is independent of $\alpha$, but $B_j$ is linear in $\alpha$. We can use this to construct counterexamples for any $j$.

For instance, we can proceed as follows for $j=0$ . Let $H=\alpha |\psi\rangle\langle\psi|$ be a scalar multiple of a rank one projector with $\alpha\in\mathbb{R}\setminus\{0\}$. Then $$ \begin{align} A_0&=\frac{1}{(\mathrm{tr}(H))^2}|\mathrm{tr}(H)|^2=1\tag1\\ B_0&=\frac{1}{\mathrm{tr}(H)}\mathrm{tr}(H^2)=\alpha\tag2 \end{align} $$ and we obtain a counterexample for any $\alpha < 1$.