In this first paper on the Rodeo algorithm, there is an argument on the second page about the suppression of "spectral weights" that I don't really understand.

In short, the algorithm is designed to find energy eigenvalues and prepare energy eigenstates. There are $N$ ancilla qubits, starting in the $| 1 \rangle$ state, which become entangled with the system of interest through stochastic controlled time evolution operators. As shown on page 2, for a system initially in the eigenstate with energy $E_{obj}$, the final probability of measuring all ancillas in the $| 1 \rangle$ state is $$\prod_{n=1}^{N} \cos^2 \left[ \left( E_{obj} - E \right) \frac{t_n}{2} \right],$$

Where $E$ is some chosen "target" energy and the $t_n$ are random times (a normal distribution is used/assumed).

The argument below this equation, labelled $\left( 3 \right)$ on page 2, is the part I'm struggling to understand:

"If we now take random values of $t_n$, we have an energy filter for $E_{obj} = E$. The geometric mean of $\cos^2 \theta$ when sampled uniformly over all $\theta$ is equal to $\frac{1}{4}$. Therefore the spectral weight for any $E_{obj} \neq E$ is suppressed by a factor of $\frac{1}{4^N}$ for large $N$."

Does "spectral weight" mean the probability of measuring the state with a particular energy? If so, how is it possible to recognise that the suppression factor is related to the geometric mean of $\cos^2 \theta$?


1 Answer 1


I believe the following is the key sentence for understanding their reasoning:

we can describe the action of the rodeo algorithm for each individual eigenvector of $H_{obj}$ with energy $E_{obj}.$

So, it seems it is best to look at individual eigenvectors.

Let $|\psi\rangle$ be an eigenvector of $H_{obj}$ with eigenvalue $E_{obj}$. For simplicity, let's consider the best case when $E_{obj}=E$. Then, the probability of measuring $|1\ldots 1\rangle |\psi \rangle$ is: $$\tag{1} \prod_{n=1}^{N} \cos^2 \left[ \left( E_{obj} - E \right) \frac{t_n}{2} \right] = \prod_{n=1}^{N} \cos^2 \left( 0 \right) = 1.$$

So, the eigenvector whose eigenvalue is very close to $E$ will have a high probability of being observed. If our initial guess is a superposition of eigenvectors of $H_{obj}$ then the vector whose eigenvalue is closest to $E$ gets a probability boost.

The authors claim that for every other eigenvector that satisfy $E_{obj} \neq E$, we have:

$$\tag{2} \prod_{n=1}^{N} \cos^2 \left[ \left( E_{obj} - E \right) \frac{t_n}{2} \right] = \prod_{n=1}^{N} \cos^2 \left[ c \frac{t_n}{2} \right] = \mu^{N}.$$

In this equation, $c = E_{obj} - E \neq 0$ and $\mu$ denotes the geometric mean defined as $\mu = \sqrt[N]{\cos^2(\theta_1)\cdots \cos^2(\theta_N)}$ where $\theta$ is some random number. Since $c\frac{1}{2}$ is a constant and $t_n$ is random, we can put $\theta_n = c\frac{t_n}{2}$. Then it follows that

$$\tag{3} \mu = \sqrt[N]{\cos^2(\theta_1)\cdots \cos^2(\theta_N)} \approx \frac{1}{4}.$$

Now, if we look at (2), we see that $\mu^{N} = \frac{1}{4^N}$. This means that any eigenvector whose eigenvalue is not $E$ has an exponentially decreasing probability of being observed.

Sketch proof for the geometric mean:
We can verify that $\mu$ is indeed 1/4. The geometric mean of a positive continuous random variable $X$ is: $$\tag{4} \mu = e^{\mathbb{E}[\ln X]}.$$

Let's assume $X$ is a uniformly distributed random variable, i.e. $X \sim U(0, \pi/2)$. Then $\cos^2(X)$ is a function of a random variable $X$, so is $\ln (\cos^2(X))$. Then $$\mathbb{E}\left[\ln (\cos^2(X))\right] = \frac{2}{\pi}\int_{0}^{\pi/2} \ln(\cos^2(x))dx=-\ln 4. $$ From the definition in (4) we get $\mu = e^{-\ln(4)} = 1/4$.

  • $\begingroup$ I can see that the argument is consistent, but it seems I'm struggling to match it to my intuition, which is that I would expect the suppression factor to be $1/2^N$, which comes from taking the product of arithmetic means of the squared cosines. It must be that there is something wrong with my intuition, but I can't see what. $\endgroup$ Commented Aug 3, 2022 at 22:17
  • $\begingroup$ Overall, I think the exact value of the suppression factor doesn't matter as long as $\mu$ is strictly less than 1. From (4) it became clear to me that 1/4 is only valid in the asymptotic sense (for N very large). On top of that, we take expectation over all possible values of a random variable $X$ to get 1/4. So this result is very theoretical. $\endgroup$
    – MonteNero
    Commented Aug 3, 2022 at 22:29
  • $\begingroup$ Sure, the argument only holds for large $N$, but so it seems does the reasoning for $1/2^N$, so I find it confusing that this discrepancy exists... $\endgroup$ Commented Aug 3, 2022 at 23:00
  • $\begingroup$ I don't understand what discrepancy you are talking about. Could you elaborate more, perhaps with some formal mathematical argument? I don't understand where you got 1/2. Equation (4) and what follows after shows that the root of product of cos squared is 1/4 $\endgroup$
    – MonteNero
    Commented Aug 4, 2022 at 4:56
  • $\begingroup$ It's possible that I'm misunderstanding what "suppression factor" means, but I'll try to explain what my intuition is telling me: away from $E_{obj} = E$, it seems to me that the mean value of the probability of measuring all zeroes is $1/2^N$, as the probability of measuring any single ancilla as zero is the average value of $\cos^2 \left( \theta_n \right)$, which is $1/2$ for random values of $\theta_n$. $\endgroup$ Commented Aug 4, 2022 at 5:53

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