Padding a quantum circuit to increase the amplitude by a constant

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

$$$$\langle 0^n|\mathsf{Q}|0^n\rangle = \frac{\#0_f - \#1_f}{\sqrt{2^n}},$$$$ for some Boolean function $$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 $$$$\frac{\#0_f - \#1_f~+k}{\sqrt{2^n}},$$$$ for some constant integer $$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.

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.
• How can you modify $C$ without adding extra input bits? Oct 4, 2022 at 19:54
• In general, you do need to add extra bits to modify the circuit $C$. Oct 5, 2022 at 5:18