How to compute k-moment of Haar averaging with n qubits

Question

Let us consider the following Haar averaging over $k$ copies of Pauli strings of $n$ qubits:

$\mathbb{E}_U \left[ U^{\otimes k}\sigma_{q_1} \otimes … \otimes \sigma_{q_k} (U^{\dagger})^{\otimes k}\right]$

where the averaging is done with respect to the Haar measure $U(2^n)$, and $\sigma_{q_j}$ is a string of Paulis on $n$ qubits,$\sigma_{q_j} \in \{I,X,Y,Z\}^{\otimes n}$. Therefore, in every copy there are $4^n$ possible strings, and there are $4^{nk}$ operators of the form $\sigma_{q_1} \otimes … \otimes \sigma_{q_k} $. My question is, for how many of these operators will the averaging above yield exactly $0$? Is there a simple lower bound for this number?

Thank you very much for your help.

Answer 1

In general, $$\mathbb{E}_U \left[ U^{\otimes k}M (U^{\dagger})^{\otimes k}\right]$$ equals to the projection of $M$ onto the subspace of permutation matrices $\Pi_\pi$, $\pi \in S_k$ (permutations of $k$ elements), defined by $$ \Pi_\pi |\psi_1\rangle \otimes \dots \otimes |\psi_k\rangle = |\psi_{\pi^{-1}(1)}\rangle \otimes \dots \otimes |\psi_{\pi^{-1}(k)}\rangle. $$ It's not that easy to compute this projection since $\Pi_\pi$ are not orthogonal to each other in the space of matrices. But the projection will be exactly $0$ iff projections of $M$ on each $\Pi_\pi$ are $0$. This is equivalent to $$ {\rm Tr}(M\Pi_\pi^\dagger) = 0, $$ for each $\pi \in S_k$.

There are ways to compute and estimate the dimension of the space spanned by all $\Pi_\pi$ (you can start here Schur–Weyl duality, also check this answer). Let's denote this dimension by $r(k,2^n)$. It follows that the number of $M=\sigma_{q_1} \otimes … \otimes \sigma_{q_k}$ which average to $0$ can't be bigger than $4^{nk} - r(k,2^n)$, since they are orthogonal to each other and to all $\Pi_\pi$. I'm not sure right now how to compute this exactly, but I hope this helps.