Moments of Pauli coefficients of Haar-random states

Question

I want to evaluate the quantity $\sum_{P\in \rm{P}^n}\text{Tr}^{\alpha}(\rho P)$, where $P$ is an element of n-qubit Pauli group $\rm{P}^n$ and $\rho$ is a density matrix of a Haar random state. It is well known that when $\alpha=2$, the quantity will be $2^n$, as long as $\rho$ is pure. But when $\alpha=4$, the quantity can not be easily calculated. From my own numerical experiments, the quantity is finite and seems really close to 4 when n gets large. I am wondering if there is a theorem about it or some possible ways to prove it, for $\alpha=4$ or maybe higher moments.

Answer 1

Let us compute the value for $\alpha=4$, averaged over Haar-random states. We have the following identity: $$ \sum_{P\in\mathcal{P}_n} \mathrm{tr}(\rho P)^4= \sum_{P\in\mathcal{P}_n} \mathrm{tr}(\rho^{\otimes 4} P^{\otimes 4}) \,. $$ Note that the fourth power cancels the possible phase in front of the Paulis, so we're actually summing over the $4^n$ Pauli operators. Again, due to the cancellation of phases, the set $$ V_{n,4} = \{ P^{\otimes 4} \; | \; P \in \{I,X,Y,Z\}^{\otimes n} \} $$ forms a stabilizer group of rank $2n$. The projector onto the corresponding code space is $$ P_{n,4} = \frac{1}{4^n} \sum_{P \in \{I,X,Y,Z\}^{\otimes n}} P^{\otimes 4}\,. $$ Moreover, recall that $(d=2^n)$ $$ \mathbb{E}_{\psi\sim\mathrm{Haar}} |\psi\rangle\langle\psi|^{\otimes 4} = \int_{U(d)} U^{\otimes 4} |0\rangle\langle 0| U^{\otimes 4 \dagger} \mathrm{d}U = \frac{P_{[4]}}{ D_{[4]}} \,, $$ where $P_{[4]}$ is the projector onto the totally symmetric subspace of $(\mathbb{C}^d)^{\otimes 4}$ and $D_{[4]} = \frac{d(d+1)(d+2)(d+3)}{24}$ is its dimension.

Hence, we can rewrite the expectation value over Haar-random states as $$ \frac{\mathrm{tr}(P_{[4]}P_{n,4})}{D_{[4]}} \,. $$ Note that this is not exactly your quantity as you do not average over Paulis. Hence, you have to multiply the result with either $4 d^2$ or $d^2$, depending what your index set is exactly (Pauli group or Pauli operators?).

By definition, $P_{[4]}$ and $P_{n,4}$ commute and thus their product is the projector onto the intersection of their ranges. The latter is shown to be an irrep for the fourth tensor power of the Clifford group in Zhu et al.. In the same paper, the authors also computed its dimension $D_{[4]}^+ = \frac{(d+1)(d+2)}{6}$ (see Tab. 1) which yields $$ \frac{\mathrm{tr}(P_{[4]}P_{n,4})}{D_{[4]}} = \frac{D_{[4]}^+}{D_{[4]}} = \frac{4}{d(d+3)} \,. $$

For more details, see the linked paper. In principle, a similar calculation should be possible if $\alpha$ is a multiple of 4.