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Consider the XQUATH conjecture for random quantum circuits, as mentioned here.

(XQUATH, or Linear Cross-Entropy Quantum Threshold Assumption). There is no polynomial-time classical algorithm that takes as input a quantum circuit $C \leftarrow D$ and produces an estimate $p$ of $p_0$ = Pr[C outputs $|0^{n}\rangle$] such that \begin{equation} \mathbb{E}[(p_0 − p)^{2}] = \mathbb{E}[(p_0 − 2^{−n})^{2}] − Ω(2^{−3n}) \end{equation} where the expectations are taken over circuits $C$ as well as the algorithm’s internal randomness.

The authors remark that

The simplest way to attempt to refute XQUATH might be to try $k$ random Feynman paths of the circuit, all of which terminate at $|0^{n}\rangle$, and take the empirical mean over their contributions to the amplitude.

However, this approach will only yield an improvement in mean squared error over the trivial algorithm that decays exponentially with the number of gates in the circuit, rather than the number of qubits.

What is a proof of the second line (for random quantum circuits)? I tried using a Chernoff bound --- it does not work.

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The paper does not specify the exact algorithm or class of distributions $\mathcal{D}$ for which such algorithm fails to refute XQUATH, and some classes of distributions $\mathcal{D}$ do not satisfy XQUATH, so some additional assumptions about $\mathcal{D}$ are needed. In this answer we will try to follow the idea of the paper and try to show that it fails for a wide range of distributions $\mathcal{D}$. This answer is currently incomplete though (see the bottom).

We assume that circuits $C \in \mathcal{D}$ are expressed as $C = \pi_m U_m \pi_{m-1} U_{m-1} \cdots \pi_1 U_1 \pi_0$ where

  1. $\pi_l$ are "classical" in a sence that their matrix elements satisfy $\left|(\pi_l)_{ij}\right| \in \{0, 1\}$.
  2. Operators $U_{l}$ are random unitaries operating on a small number of qubits.
  3. For each structure of $C$, which includes $m$, $\{\pi_l\}_{l=0..m}$, and qubits on which $U_l$ operate, unitaries $U_l$ are independent and each of them is picked from a distribution $\mathcal{D}_l$.
  4. $\mathcal{D}_l$ is closed under multiplication by diagonal matrices with $\pm 1$ or $\pm i$ on the diagonal.
  5. For each fixed structure of $C$ the expectation of $p_0$ over the choice of $\{U_l\}$ is $2^{-n}$.
  6. Some additional assumptions may be needed to show $\beta_2 = O(1/N^\epsilon)$ for $\epsilon > 0$ below.

The assumption 5 is to exclude circuits where some bits of the output can be recovered using a classical simulation ignoring quantum effects (i.e. ignoring $U_l$-s). E.g. consider a circuit where qubit 1 is only used by the following operations: (1) NOT gates; (2) control gates where qubit 1 is used as a control for some classical operation on other qubits; (3) a single CCNOT operation on qubit 1 controlled by conjunction of at least 2 other qubits. In this case we can predict the state of qubit 1 at the end of the circuit with probability of at least 3/4 assuming control qubits for the operation (3) are as likely to be in $\left|11\right>$ state as in any other of 4 states, and, thus, refute XQUATH for that particular $\mathcal{D}$.

We want to maximize $I = 2^{2n}\mathbb{E}\left((p_0 - 2^{-n})^2 - (p_0 - p)^2\right)$. In particular, if we manage to find an algorithm providing $p$ in polynomial time and satisfying $I = \Omega(2^{-n})$ then we will refute XQUATH. Let's assume for now that the structure of $C$ is fixed and the expectation in the formula for $I$ is with respect to the choice of $U_l$. Let $p_0 = 2^{-n} (1 + \tilde p_0)$ and let's search for $p$ in the form of $p = 2^{-n} (1 + \alpha_1 p_1)$ where $\alpha_1$ depends only on the structure of $C$ and not on $U_l$ or the random choices within our algorithm. Then $$I = 2 \alpha_1 \mathbb{E}(\tilde p_0 p_1) - \alpha_1^2 \mathbb{E}(p_1^2).$$ The optimal choice of $\alpha_1$ is $\alpha_1 = \mathbb{E}(\tilde p_0 p_1) / \mathbb{E}(p_1^2)$ which gives $$I = \left.\left(\mathbb{E}(\tilde p_0 p_1)\right)^2\middle/\mathbb{E}(p_1^2)\right..$$

In order to come up with the estimator $p_1$, following the idea from the paper, we consider the Feynman paths $z = (z_0, \dots, z_{2m+1})$. We can write $p_0 = \left|\sum_z \left<0\right|C_z\left|0\right>\right|^2$ where $$C_z = \left|z_{2m+1}\right>\left<z_{2m+1}\right|\pi_m \left|z_{2m}\right> \left<z_{2m}\right| U_{m} \left|z_{2m-1}\right>... \left<z_0\right|.$$ Some of these paths trivially result in $\left<0\right|C_z\left|0\right> = 0$, so we can restrict our attention to nontrivial ones, i.e. where $z_{2m+1} = 0$, $z_0 = 0$, and $\left|\left<z_{2l+1}\right|\pi_l \left|z_{2l}\right>\right| = 1$. Assume there are $N$ nontrivial Feynman paths and pick random $k$ of them to compute $p_1$. In order to simplify the notation, we assume that the picking procedure works in the following way: first, we pick a random permutation $\sigma$, apply it to the list $\{2^{n/2}\left<0\right|C_z\left|0\right>\}_{z}$ of coefficients resulting from nontrivial Feynman paths. Then denote the resulting list of coefficients as $a_1,\dots,a_N$, and take $a_1,\dots,a_k$ as our sample. Given that $$\tilde p_0 = \left|\sum_{i=1}^N a_i\right|^2 - 1 = \sum_{i=1..N} \left(\left|a_i\right|^2 - \frac1N\right) + \sum_{i,j=1..N;i\neq j} a_i \bar a_j,$$ a reasonable choice of $p_1$ is $$p_1 = \alpha_2 p_2 + \alpha_3 p_3\textrm{, where}\; p_2 = \sum_{i=1..k} \left(\left|a_i\right|^2 - \frac1N\right)\!,\; p_3 = \sum_{i,j=1..k;i\neq j} a_i \bar a_j.$$ Let $$ \mathbb{E}\left(\left|a_1\right|^2 - \frac1N\right)^2 = \beta_1/N^2,\qquad \mathbb{E}\left(\left(\left|a_1\right|^2 - \frac1N\right)\left(\left|a_2\right|^2 - \frac1N\right)\right) = \beta_2/N^2.$$ Then if we can show that $\beta_2 = O(1/N)$, then by computing $\mathbb{E}(\tilde p_0 p_s)$ and $\mathbb{E}(p_s^2)$ for $s=2,3$ we can show that $I = O(k^2/N^2)$ and, hence, show that Feynman path averaging algorithm above does not refute XQUATH (since $\log_2 N = \sum_{l=1}^{m} \log_2 \dim U_l$). It is not clear, though, which additional assumptions are needed to show $\beta_2 = O(1/N)$ or, at least, $\beta_2 = O(1/N^\epsilon)$ for some $\epsilon > 0$.

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  • $\begingroup$ Why did you define $p_0$ as $p_0 = 2^{-n}(1 + \tilde p_0)$ and how did you guess the optimal choice of $I$? $\endgroup$
    – BlackHat18
    Jun 29 at 4:53
  • $\begingroup$ Also, what is $p_1$ in the definition of $p$? $\endgroup$
    – BlackHat18
    Jun 29 at 5:09
  • $\begingroup$ @BlackHat18 1. $\tilde p_0$ is a rescaled version of $p_0$ so that $\tilde p_0$ has mean of 0 and possibly variance which has a finite limit as $n\to\infty$. 2. $I$ is quadratic function of $\alpha_1$. Quadratic function with negative coefficient near $\alpha_1^2$ has exactly one maximum - where it's derivative with respect to $\alpha_1$ is 0. That would be the equation for $\alpha_1$ you need to solve to maximize $I$. 3. $p_1$ (as well as $p$) is something we need to find. We find it below (up to 2 coefficients): $p_1=\alpha_2 p_2 + \alpha_3 p_3$. $\endgroup$
    – fiktor
    Jun 29 at 6:00
  • $\begingroup$ Here's what I don't understand. From the Feynman method, $p_0 = |\sum_{i=1}^{N} a_i |^{2}$. Then shouldn't $\tilde p_0 = 2^{n} |\sum_{i=1}^{N} a_i |^{2} - 1$? $\endgroup$
    – BlackHat18
    Jun 29 at 7:13
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    $\begingroup$ This $2^n$ is already included in my definition of $a_i$: it is defined as $a_i = 2^{n/2} \left<0\right|C_z\left|0\right>$ for the corresponding $z$. $\endgroup$
    – fiktor
    Jun 29 at 7:23

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