2
$\begingroup$

New to quantum and ran into the block-encoding. Having a bit of trouble understanding $|0\rangle \otimes I$.

$|0\rangle$ is just a vector but $I$ is an $n$ by $n$ matrix? Not clear how vector can be tensorproducted with a matrix? I know I am missing something here.

Any help could be appreciated.

enter image description here

$\endgroup$
2
  • $\begingroup$ You should treat $|v\rangle$ as a $m\times 1$ matrix, where $m$ is the dimension of the Hilbert space in which the vector lives. $\endgroup$
    – Rammus
    Jun 30, 2021 at 7:05
  • $\begingroup$ Could you add the reference to the paper where you copied the definition above about block encodings? $\endgroup$
    – Jadzia
    Sep 2, 2022 at 14:18

1 Answer 1

2
$\begingroup$

This is probably best done with an example. Let's consider a $4\times 4$ matrix $U$ which acts on two qubits. The $|0\rangle\otimes I$ is equivalent to $$ \left(\begin{array}{c} 1 \\ 0 \end{array}\right)\otimes\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)=\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \end{array}\right) $$ (if you don't know where this comes from, go back to the definition of the tensor product). This is a $4\times 2$ matrix, meaning it's the right size for you to pre-multiply by $U$. Similarly, $$ \langle 0|\otimes I\equiv \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right) $$ so that, overall your matrix $A$ comes out as $2\times 2$ (it's the action of what happens to the second qubit if the first qubit starts and ends in state $|0\rangle$).

$\endgroup$
3
  • $\begingroup$ Thank you. A follow-up question if you don't mind. I assume other elements in the 4x4 matrix are zero and U has the same rank as A? If so, how U can be a unitary? The paper says A doesn't need to be Unitary. $\endgroup$ Jun 30, 2021 at 12:07
  • $\begingroup$ No, I think you have to fill in the rest of the matrix $U$ so that it becomes unitary. $\endgroup$
    – DaftWullie
    Jun 30, 2021 at 12:19
  • $\begingroup$ You may want to look at: quantumcomputing.stackexchange.com/questions/5167/… $\endgroup$
    – DaftWullie
    Jun 30, 2021 at 12:20

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.