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I am trying to find the transition matrix for a quantum walk on a cycle. The vertices are labelled $\lbrace 0,1,2,\ldots,n-1\rbrace$, where vertex $i$ is a neighbour of vertex $i \pm 1$. Lets say we start at vertex 0. Then we are interested in $\langle i \vert U(t) \vert 0 \rangle$.

I have calculated the following: The normalized adjacency matrix for A is a circulant matrix, hence (normalized) eigenvectors and eigenvalues are $$v_j= \dfrac{1}{\sqrt{n}}\left( \omega_j^0, \omega_j^1,\ldots,\omega_j^{n-1}\right) \text{ and }\\ \lambda_j= \cos(2 \pi j / n), \hspace{2cm} \text{ respecively.}$$.

Thus, $U(t)= \exp(iAt)= \sum\limits_{j=1}^n\exp(i t \lambda_j ) v_j v_j^\dagger$.

This gives,

$$ \langle i \vert U(t) \vert 0 \rangle = \langle i \vert \sum\limits_{j=1}^n\exp(i t \lambda_j ) v_j v_j^\dagger \vert 0 \rangle $$ $$ = \sum\limits_{j=1}^n \exp(i t \lambda_j ) \langle i \vert v_j v_j^\dagger \vert 0 \rangle $$ $$ = \sqrt{\dfrac{1}{n}}\sum\limits_{j=1}^n \exp(i t \lambda_j ) \langle i \vert v_j \rangle $$ $$ ={\dfrac{1}{n}}\sum\limits_{j=1}^n \exp(i t \lambda_j )\ \ {\omega_j}^{i}$$.

Is this calculation correct? Is there a way to further simplify this?

Is there another way to get compact expression (in terms of t) for the matrix $U(t)$?

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  • $\begingroup$ To clarify, you don't have a Hadamard coin here, and you are just evolving unitarily according to the Hamiltonian by $e^{-iAt}$, right? Here you are asking for transition probability (transition amplitude) of evolving from node $0$ to node $i$ at time $t$, right? Classically walking on a cycle you get to the uniform distribution in roughly $\mathcal{O}(\sqrt n)$ time $\endgroup$ Commented Sep 19, 2022 at 16:20
  • $\begingroup$ @Yes. That is correct. We have the standard definition of continuous time quantum walk. $\endgroup$
    – Root
    Commented Sep 19, 2022 at 16:59

1 Answer 1

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(Not much of an answer yet, just some tentative ideas about how to simulate a continuous-time random walk on a cycle for now. I describe how to take a small quantum walk for a short amount of time, but I'm otherwise getting stuck.)

Recall that the cycle graph on $n$ vertices is isomorphic to the Cayley graph of the additive group $\mathbb Z/n\mathbb Z$ with generators $\pm 1$, which is isomorphic to the rotations of the $n$-gon with clockwise or right $R$ and counterclockwise or left $L$ rotations.

We'll start off with $n=4$ for simplicity and convenience. We have $n=4$ qubits labeled $\{|0\rangle,|1\rangle,|2\rangle,|3\rangle\}$, and we can use the following SWAP circuits that perform the rotations:

SWAP circuits

There's nothing quantum about this circuit yet, but we can use phase estimation to take the roots of each of $L$ and $R$, as we know that the eigenvalues of $R$ and $L$ are $\{\pm 1, \pm i\}$ because $L^4=R^4=I$. These eigenvalues correspond to the $\lambda_j$ in the question.

Letting $|\psi\rangle$ be the wavefunction for registers $|0\rangle|1\rangle|2\rangle|3\rangle$, below are the $4$-th root of $R$ and $L$:

Fourth Roots of Rotations

It's the fourth root in particular because of the $T=\sqrt[4]Z$ and $S=\sqrt[2]Z$ gates in the middle of the circuit - if we want a smaller movement corresponding to a larger root, we'd use other roots of $Z$. Let's call this power the Trotter factor. As this factor increases to infinity, I think we're approaching continuous-time evolution.

We can go back and forth between the $r$th root of a clockwise right rotation and a counterclockwise left rotation. An iteration would be:

$$\sqrt[4]{R}\sqrt[4]{L}\sqrt[4]{R}\sqrt[4]{L}\sqrt[4]{R}\sqrt[4]{L}\sqrt[4]{R}\sqrt[4]{L}\cdots$$

I'd like to get this to be a continuous-time quantum walk, corresponding to a little walk to the right (clockwise), then a walk to the left (counterclockwise), then... etc. But, this ends up being the identity after only two repetitions, I think. Either I'm doing something wrong (highly likely), or such a walk on the graph cycles quickly back and forth between the identity, perhaps?

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  • $\begingroup$ Thanks for your answer but here I am more interested in the transition probabilities rather than the actual circuit or implementation. What I am looking for is the probability that quantum walk is at vertex j at time t. I have included a calculation in the question, I want to know a) if it is correct b) Is there a way to further simplify this> $\endgroup$
    – Root
    Commented Sep 22, 2022 at 6:27
  • $\begingroup$ @Root have you thought about explicitly working out $U(t)$ for small $n$, such as $n=4$? $\endgroup$ Commented Sep 22, 2022 at 12:33

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