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Let us be given the description of a quantum circuit $\mathsf{Q}$, acting on $n$ qubits, such that

\begin{equation} \langle 0^n|\mathsf{Q}|0^n\rangle = \frac{\#0_f - \#1_f}{\sqrt{2^n}}, \end{equation} for some Boolean function $f : \{0, 1\}^n \rightarrow \{0, 1\}$, where $\#0_f$ and $\#1_f$ are the number of inputs for which $f$ evaluates to $0$ and $1$ respectively.

Let's say one could find a description of $f$ when given a description for $\mathsf{Q}$. Now, is there a way to efficiently construct the circuit $\mathsf{Q_k}$ such that the amplitudes of $\mathsf{Q_k}$ look like \begin{equation} \frac{\#0_f - \#1_f~+k}{\sqrt{2^n}}, \end{equation} for some constant integer $k$?

In this paper (page $10$), it is mentioned that this is what is termed a "padding argument," and the paper seems to contend that this can be done by using "$k$ additional inputs." But it wasn't immediately clear to me how to do this.

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In the paper that above question references, it is assumed that we are given a classical circuit $C$ which computes the function $f: \{0,1\}^n \rightarrow \{0,1\}$ and an another classical algorithm which takes $C$ as the input and outputs a diagonal quantum circuit $Q$ which uses $\mathrm{T} = $ poly$(n,|C|)$ qubits such that

\begin{align} \langle 0^n|Q|0^n\rangle = \frac{\#0_f - \#1_f}{2^T} \end{align}

The author of the paper claims that you can modify the circuit $C$, not $Q$, by adding $k$ additional input (by the increasing the domain size by $k$, not the number of input bits) to make a new circuit $C[k]$ such that $C(x) = 1$ for $x$ in the newly added domain. Now you can pass this new circuit to the classical algorithm which converts into a quantum circuit $Q_k$ as the question asks.

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  • $\begingroup$ How can you modify $C$ without adding extra input bits? $\endgroup$
    – Tom Clancy
    Oct 4, 2022 at 19:54
  • $\begingroup$ In general, you do need to add extra bits to modify the circuit $C$. $\endgroup$
    – Chaithanya
    Oct 5, 2022 at 5:18
  • $\begingroup$ Could you provide a sketch of how in your answer? $\endgroup$
    – Tom Clancy
    Oct 6, 2022 at 6:02

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