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


I asked a similar question here, but used a wrong definition of $ B_j $. Specifically, I divided by $ tr(H) $, but should have divided by $ tr(H^2) $ as in equation (4) of this paper.

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  • $\begingroup$ As @Adam Zalcman described below, this property relies on the positive definiteness of H. On the other hand the quantum MacWilliams identitiy between the A_j and B_j still holds, even if H is not positive semidefinite. $\endgroup$ Mar 15, 2023 at 8:33

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TL;DR: No. We can actually blow $A_j$ up to infinity while simultaneously sinking $B_j$ negative.

Sneaky plan

Coefficients $A_j$ cannot be negative, but if $H$ squares to identity and anticommutes with a Pauli $E$ then $\mathrm{tr}(EHE^\dagger H)=-2$ suggesting that we may be able to make $B_j$ negative (if the commuting terms don't contribute too much).

Helpful obstacle

Now, the problem is that any such $H$ is necessarily traceless and the normalization coefficient in $A_j$ blows up due to division by zero. But this is even better! Let's blow $A_j$ up! We'll then have $A_j\to +\infty$ and $B_j$ finite and maybe even negative.

Counterexample

We make the idea rigorous by constructing the following single-qubit countrexample. Set $H=\epsilon I+X$ for some $\epsilon>0$. Compute $$ \begin{align} H^2&=(1+\epsilon^2)I+2\epsilon X\tag1\\ EHE^\dagger H&=(\epsilon I-X)(\epsilon I+X)=(-1+\epsilon^2)I\tag2\\ \mathrm{tr}(H)&=2\epsilon\tag3\\ \mathrm{tr}(H^2)&=2+2\epsilon^2\tag4\\ \mathrm{tr}(EHE^\dagger H)&=-2+2\epsilon^2\tag5 \end{align} $$ where $E\in\{Y,Z\}$ and $$ \begin{align} A_1&=\frac{1}{4\epsilon^2}(4+0+0)=\frac{1}{\epsilon^2}\tag6\\ B_1&=\frac{1}{2+2\epsilon^2}(2+2\epsilon^2-2+2\epsilon^2-2+2\epsilon^2)=\frac{-2+6\epsilon^2}{2+2\epsilon^2}\tag7 \end{align} $$ so $$ \begin{align} \lim_{\epsilon\to 0}A_j&=+\infty\tag8\\ \lim_{\epsilon\to 0}B_j&=-1\tag9 \end{align} $$ as anticipated. This means that $H$ is a counterexample for sufficiently small $\epsilon$.

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  • $\begingroup$ oh nice this example is really cool and sneaky. You are awesome! $\endgroup$ Mar 14, 2023 at 0:08
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This is just fleshing out some of the themes from Adam Zalcman's answer, which I have already accepted.

The standard proof that $ B_j\geq A_j $ for a projection $ H $ uses the spectral theorem to write $$ H=\sum_{a} |a><a| $$ where $ \{ |a> \} $ is a basis for the codespace. From there one shows that $$ A_j=\frac{1}{K^2} \sum_{E \in \mathcal{E}_j } \Bigg| \sum_{a} <a|E|a> \Bigg|^2 $$ and $$ B_j=\frac{1}{K} \sum_{E \in \mathcal{E}_j } \sum_{a,b} \Bigg|<a|E|b> \Bigg|^2 $$ then shows that $ A_j \leq B_j $ using Cauchy Schwarz for the $ K^2 $ length vectors $ <a|E|b> $ and $ \frac{1}{K}\delta_{ab} $.

If $ H $ is not a projection but merely a generic Hermitian operator then we can again use spectral theorem to write $$ H=\sum_{a} a|a><a| $$ where $ \{ |a> \} $ is an orthonormal eigenbasis for $ H $. From there one shows that $$ A_j=\frac{1}{K^2} \sum_{E \in \mathcal{E}_j } |a|^2\Bigg| \sum_{a} <a|E|a> \Bigg|^2 $$ and $$ B_j=\frac{1}{K} \sum_{E \in \mathcal{E}_j } \sum_{a,b} ab\Bigg|<a|E|b> \Bigg|^2 $$ if $ H $ is positive semidefinite then we can still make the proof of $ A_j \leq B_j $ work using Cauchy Schwarz for the $ K^2 $ length vectors $ \sqrt{a}\sqrt{b}<a|E|b> $ and $ \frac{1}{K}\delta_{ab} $. This is a related, but slightly stronger, result than Theorem 3 of https://arxiv.org/abs/quant-ph/9611001

However if $ H $ has a mix of positive and negative eigenvalues then the proof totally breaks down and not only do we not have $ 0 \leq A_j \leq B_j $ we may even get $ B_j < 0 \leq A_j $ as in the example from Adam Zalcman above.

Indeed looking back at the example from Adam Zalcman we see that $ H $ has eigenvalues $ \epsilon \pm 1 $. So for $ 0< \epsilon <1 $ we indeed have negative eigenvalues. And for $ \epsilon=\frac{1}{2} $ it is already clear that $ B_1<0\leq A_1 $.

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