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.

  • $\begingroup$ how do you show that $\sum_P \operatorname{tr}(\rho P)^2$, averaged over Haar random (I assume pure?) states $\rho$, equals $2^n$? $\endgroup$
    – glS
    Commented Oct 18, 2023 at 11:06
  • $\begingroup$ @glS That works for any pure state. Basically you can prove that using n-qubit swap test of the n copies of $\rho$ itself. Or you can use the decomposition $\rho = \sum_P\frac{1}{2^n}tr(\rho P)P$. $\endgroup$
    – Feng Pan
    Commented Oct 18, 2023 at 12:53
  • 1
    $\begingroup$ Note the relation to the stabilizer Renyi entropy link.aps.org/doi/10.1103/PhysRevLett.128.050402 which may give additional meaning to that quantity. In particular, your $\alpha=4$ case corresponds to the $\alpha=2$ case discussed in that paper, for which some results are derived. Are you interested in the value averaged over Haar-random states? I think this can be done using that the projector $2^{-n} \sum_P P^{\otimes 4}$ plays a role in the rep theory of the Clifford group. I can try to elaborate on this. $\endgroup$ Commented Oct 19, 2023 at 6:48
  • $\begingroup$ @MarkusHeinrich Thanks for the reference! Yes, I am interested in the average case over Haar random states. $\endgroup$
    – Feng Pan
    Commented Oct 20, 2023 at 2:56

1 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.


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