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.

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  • $\begingroup$ 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
    Commented 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
    Commented Sep 2, 2022 at 14:18

1 Answer 1


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 $|0\rangle$).

  • $\begingroup$ 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. $\endgroup$ Commented 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
    Commented Jun 30, 2021 at 12:19
  • $\begingroup$ You may want to look at: quantumcomputing.stackexchange.com/questions/5167/… $\endgroup$
    – DaftWullie
    Commented Jun 30, 2021 at 12:20

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