2
$\begingroup$

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 $\cos^2 \theta$?

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $\mu = e^{-\ln(4)} = 1/4$.

$\endgroup$
6
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.