Does $\mathbb P^1 \times \mathbb P^1$ admit an Ulrich bundle?

Question

In an answer to a MathOverflow question on the following link Vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$, it is mentioned that $\mathbb P^1 \times \mathbb P^1$ has an Ulrich sheaf. However, I am not able to see a proof that it does. Can someone please suggest how to prove this or cite a reference?

EDIT: Let me add the definition of an Ulrich sheaf: Let $X \hookrightarrow \mathbb P^n$ be a scheme of dimension $d$. $X$ is said to admit an Ulrich sheaf $\mathcal F$ if $\mathcal F$ is a coheren sheaf on $X$ such that $\pi_*\mathcal F \simeq \mathcal O_{\mathbb P^d}^r$, for some $r$, and for some $\pi: X \to \mathbb P^n \to \mathbb P^d$, which is the inclusion followed by a general linear projection.

Answer 1

View $\mathbb{P}^1\times \mathbb{P}^1$ as a quadric $Q$ in $\mathbb{P}^3$; let $\pi :Q\rightarrow \mathbb{P}^2$ be a general projection. Let $D$ be a line contained in $Q$. Then $\pi _*\mathcal{O}_Q(D)\cong \mathcal{O}_{\mathbb{P}^2}^2$, so $\mathcal{O}_Q(D)$ is a Ulrich bundle.