I'm reading the lecture notes from Ronald de Wolf and got confused at the part where it introduces Block encoding, specifically by this notation at page 76:

More generally we can define an $a$-qubit block-encoding of $A$, which is an $(a + n)$-qubit unitary $U$ with the property that $(\langle 0^a|\otimes I)U(|0^a\rangle \otimes I)=A$.

How do I interpret $(\langle 0^a|\otimes I)U(|0^a\rangle \otimes I)$? Like having a tensor product between a ket vector and a $2^n\times 2^n$ matrix?


1 Answer 1


I will try to answer this question for myself. We know that an $a$-qubit quantum state $|\psi\rangle$ lives in a $2^a$-dimensional Hilbert space $\mathcal{H}$. The operator $\mathbb{1}$ acting on $|\psi\rangle$ is a map $\mathcal{H}\rightarrow\mathcal{H}$, but also a vector in $\mathcal{B}(\mathcal{H})$, the vector space of bounded linear operators over $\mathcal{H}$. We can define elements in the tensor product space $\mathcal{H}\otimes \mathcal{B}(\mathcal{H})$. If $\mathcal{H}$ and $\mathcal{B}(\mathcal{H})$ are both finite dimensional, then $\dim(\mathcal{H}\otimes \mathcal{B}(\mathcal{H})) = \dim(\mathcal{H})\times\dim(\mathcal{B}(\mathcal{H}))$. In our case, $\dim(\mathcal{H}) = 2^a$ and $\dim(\mathcal{B}(\mathcal{H}))=2^a\times2^a$, so $\dim(\mathcal{H}\otimes \mathcal{B}(\mathcal{H}))=2^a\times2^a\times2^a$.

Generally, if $A\in \mathbb{C}^{m\times n}$ and $B\in \mathbb{C}^{k\times l}$ are two matrices, we can define $A\otimes B$ as $$ A\otimes B =\begin{bmatrix}a_{11} B & \cdots & a_{1n}B \\ \vdots & & \vdots \\ a_{m1}B & \cdots & a_{mn}B\end{bmatrix}. $$ In the case of block-encoding (let's assume $a=1$ and $A\in\mathbb{C}^{2\times 2}$for simplicity), $$ \begin{align} \langle 0 | &= \begin{bmatrix}1 & 0\end{bmatrix}, \\ |0\rangle &= \begin{bmatrix}1 \\ 0\end{bmatrix} \\ \mathbb{1} &= \begin{bmatrix}1 & 0\\0 & 1 \end{bmatrix} \end{align} $$ hence $$ \begin{align} \langle 0 | \otimes \mathbb{1} &= \begin{bmatrix}1&0&0&0\\0&1&0&0\end{bmatrix} = \begin{bmatrix}\mathbb{1} & 0\end{bmatrix},\\ |0\rangle \otimes \mathbb{1} &= \begin{bmatrix} 1 & 0 \\ 0&1\\0&0\\0&0\end{bmatrix} = \begin{bmatrix}\mathbb{1} \\ 0\end{bmatrix}. \end{align} $$ Therefore, $$ (\langle 0 | \otimes \mathbb{1}) U (| 0 \rangle \otimes \mathbb{1}) = \begin{bmatrix}\mathbb{1} & 0\end{bmatrix} \begin{bmatrix}A&B\\C&D\end{bmatrix}\begin{bmatrix}\mathbb{1} \\ 0\end{bmatrix} = \begin{bmatrix}\mathbb{1} & 0\end{bmatrix} \begin{bmatrix}A \\ C\end{bmatrix} = A. $$ The case when $A\in \mathbb{C}^{2^n\times 2^n}$ and $a>1$ is similar to the above.


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