# In the context of block-encoding, what does $|0\rangle\otimes I$ represent?

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.

• 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. Jun 30 '21 at 7:05

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 $$|0\rangle$$).
• No, I think you have to fill in the rest of the matrix $U$ so that it becomes unitary. Jun 30 '21 at 12:19