2
$\begingroup$

For a project, I want to simulate a matrix multiplication on a simulated quantum circuit. Assuming that we have a matrix $A\in \mathbb{R}^{m\times n}$ stored in a quantum superposition, i.e. $$|A\rangle= \frac{1}{||A||_F}\sum_{i,j=0}^{n-1}{a_{ij}}|i,j\rangle$$ and a vector $b\in \mathbb{R}^{n\times 1}$ stored in $$|b\rangle= \frac{1}{||b||}\sum_{i}^{n-1}{b_{i}}|i\rangle$$

How can I find a unitary, such that

$$U|b\rangle\approx|Ab\rangle?$$.

I'm right now looking at block-encodings (https://arxiv.org/abs/1804.01973, https://arxiv.org/abs/1806.01838), where they argue, that we can find a block encoding of $A$ as a unitary of form

$$U=\left(\begin{array}{cc} A / \alpha & . \\ . & . \end{array}\right)$$

by decomposing it into two state-preparation unitaries $U_L$ and $U_R$, s. t.

$$ \begin{array}{l} U_{L}:|i\rangle|0\rangle \rightarrow |i\rangle\left(\sum_{j=0}^{n-1}{\frac{A_{i,j}}{||A_{i,\cdot}||_F}|j\rangle} \right)=|i\rangle|A_{i,\cdot}\rangle\\ U_{R}:|0\rangle|j\rangle \rightarrow \left(\sum_{i=0}^{n-1}{\frac{||A_{i,\cdot}||}{||A||_F}|i\rangle} \right)|j\rangle=|\tilde{A}\rangle|j\rangle \end{array} $$ where $|A_{i,\cdot}\rangle$ represents the $i$-th row of A and $|\tilde{A}\rangle$ represents the vector of the row-norms of $A$, i.e $\tilde{A}\in \mathbb{R}^m$ with $\tilde{A}_i = ||A_{i,\cdot}||$ (see Theorem 5.1 https://arxiv.org/pdf/1603.08675.pdf). Then they claim that $U=U_{L}^{\dagger} U_{R}$ is a block-encoding of A (see https://youtu.be/zUpHcpIq0Ww?t=1678).

However, I struggle to find the circuits for $U_L$ and $U_R$, even for a simple $n=m=2$ problem. Any hints?

$\endgroup$
1
  • $\begingroup$ This may not be exactly what you're looking for, but are you familiar with the Choi-Jamiolkowski isomorphism? Could you teleport $|b\rangle$ using $|A\rangle$? (There are other issues such as the corrective operations depending on the measurement outcome during the teleportation process.) $\endgroup$ – DaftWullie Mar 3 at 11:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.