Quantum abstract detecting systems

Abstract

In this paper, we study Quantum Abstract Detecting Systems (QADS), that generalize some key characteristics of the operators used in Grover’s algorithm, a wide variety of quantum walks and the quantum abstract search algorithm. A QADS is an algorithm that constructs a quantum state and a quantum operator that help testing whether a circuit-implemented boolean function f is identically zero. We also identify some relatively weak properties of QADS that lead to the construction of algorithms for the detection problem (i.e. determining whether there is a marked element in a given set). Our results provide not only a common framework to all the aforementioned search methods, and their transformation into algorithms for the detection problem, but also allow the development of new similar methods. As an example, we construct a modification of Grover’s algorithm (from the tensor product of controlled QADS) that shows improved detection probability.

Appendices

Detailed examples of QADS

In this appendix, we provide detailed examples of algorithms and procedures that can be seen as QADS.

Example 1

(Grover search [9]) Given an arbitrary boolean function $f:\{0,1\}^k\rightarrow \{0,1\}$, Grover’s algorithm requires a state space ${\mathcal {H}}=({\mathbb {C}}^2) ^{\otimes k}$ to look for marked elements (i.e., those in $W=\{x\in \{0,1\}^k\ |\ f(x)=1\}$), and the initial state $|{\psi _0}\rangle$, which is the superposition of all the elements of the computational basis $\frac{1}{\sqrt{2^k}}\sum _{x=0}^{2^k-1}|x\rangle$. The search iterates two operators that can be effectively constructed, namely:

• Oracle: $O(|x\rangle )=(-1)^{f(x)}|x\rangle$, i.e., $O=I-2\sum _{x\in W}|x\rangle \langle x|$.

• Diffusion operator: $G=2|{\psi _0}\rangle \langle {\psi _0}|-I$

The algorithm which constructs $U:=GO$ from f is a QADS because $U|{\psi _0}\rangle =|{\psi _0}\rangle$ if and only if

$$\begin{aligned} \frac{1}{\sqrt{2^k}}\left( \sum _{x\not \in W}|x\rangle -\sum _{x\in W}|x\rangle \right) =O|{\psi _0}\rangle =G^{-1}|{\psi _0}\rangle =G|{\psi _0}\rangle =|{\psi _0}\rangle \end{aligned}$$

which is equivalent to $W=\emptyset$.

Example 2

(Szegedy’s quantum walk [15]) Let $(V\subseteq \{0,1\}^k,E)$ be a connected, non-directed, non-bipartite graph. Szegedy’s quantum search requires a state space ${\mathcal {H}}\subseteq ({\mathbb {C}}^2)^{\otimes 2k}$ with basis $\{|x\rangle |y\rangle \ |\ x,y\in V\}$ representing potential edges of the graph. If for any $x\in V$, the set of adjacent vertices $y\in Y$ is denoted by $A_x=\{y\in Y\ |\ \{x,y\}\in E\}$, then the (doubly stochastic) matrix associated with the graph is $P\in {\mathcal {M}}_{|V|\times |V|}({\mathbb {R}})$ given by

$$\begin{aligned} P_{xy}=\left\{ \begin{array}{ll}\frac{1}{|A_x|}&{}\hbox {if }y\in A_x\\ 0&{}\hbox {otherwise}\end{array}\right. \end{aligned}$$

Searching for vertices marked according to a boolean function $f:\{0,1\}^k\rightarrow \{0,1\}$ (i.e., those in $W=\{x\in V\ |\ f(x)=1\}$), requires the matrix associated with the leaking graph (i.e, the directed graph obtained from V converting all outer arcs from a marked vertex in a loop):

$$\begin{aligned} P^W_{xy}=\left\{ \begin{array}{ll} P_{xy}&{}\hbox {if }x\not \in W\\ \delta _{x,y}&{}\hbox {otherwise}\end{array}\right. \end{aligned}$$

The initial state $|{\psi _0}\rangle$ is the weighted superposition of the potential graph edges

$$\begin{aligned} \frac{1}{\sqrt{|V|}}\sum _{x\in V}\frac{1}{\sqrt{|A_x|}}\sum _{y\in A_x}|x\rangle |y\rangle \end{aligned}$$

and the quantum walk iterates the following two reflection operators:

• $R_A^W=2\sum _{x\in V}|\Phi ^W_{x}\rangle \langle \Phi ^W_{x}|-I$, where $|\Phi ^W_{x}\rangle =|x\rangle \otimes \left( \sum _{y\in V}\sqrt{P ^W_{xy}}|y\rangle \right)$

• $R_B^W=2\sum _{y\in V}|\Psi ^W_{y}\rangle \langle \Psi ^W_{y}|-I$, where $|\Psi ^W_{y}\rangle =\left( \sum _{x\in V}\sqrt{P^W_{xy}}|x\rangle \right) \otimes |y\rangle$

If $|\Phi _{x}\rangle =|x\rangle \otimes \left( \sum _{y\in V}\sqrt{P _{xy}}|y\rangle \right) ,|\Psi _{y}\rangle =\left( \sum _{x\in V}\sqrt{P_{xy}}|x\rangle \right) \otimes |y\rangle$, then observe that $|{\psi _0}\rangle =\frac{1}{\sqrt{|V|}}\sum _{x\in V}|\Phi _x\rangle =\frac{1}{\sqrt{|V|}}\sum _{y\in V}|\Psi _y\rangle$, and that $|\Phi ^W_{x}\rangle =\left\{ \begin{array}{ll}|x\rangle |x\rangle &{}\hbox {if }x\in W\\ |\Phi _{x}\rangle &{}\hbox {if }x\not \in W\end{array}\right.$. Moreover, since the graph has no loops, $\{|\Phi _x\rangle \}_{x\in V}\cup \{|x\rangle |x\rangle \}_{x\in W}$ is an orthonormal family, and so

$$\begin{aligned} R_A^W(|{\psi _0}\rangle )=\frac{1}{\sqrt{|V|}}\left( \sum _{x\in V\setminus W}|\Phi _x\rangle -\sum _{x\in W}|\Phi _x\rangle \right) =\frac{1}{\sqrt{|V|}}\sum _{x\in V}(-1)^{f(x)}|\Phi _x\rangle \end{aligned}$$

(an analogous property holds for $R_B^W$). Both $R_A^W$ and $R_B^W$ can be algorithmically constructed from f, and so the algorithm which computes $U:=R_B^WR_A^W$ is a QADS (with respect to the graph (V, E)), since

$$\begin{aligned}&U|{\psi _0}\rangle =|{\psi _0}\rangle \Longleftrightarrow R_B^{W}|{\psi _0}\rangle =R_A^W|{\psi _0}\rangle \\&\quad \Longleftrightarrow \frac{1}{\sqrt{|V|}}\sum _{y\in V}(-1)^{f(y)}|\Psi _y\rangle =\frac{1}{\sqrt{|V|}}\sum _{x\in V}(-1)^{f(x)}|\Phi _x\rangle \\&\quad \Longleftrightarrow \frac{1}{\sqrt{|V|}}\sum _{x,y\in V}(-1)^{f(y)}\sqrt{P_{xy}}|x\rangle |y\rangle =\frac{1}{\sqrt{|V|}}\sum _{x,y\in V}(-1)^{f(x)}\sqrt{P_{xy}}|x\rangle |y\rangle \end{aligned}$$

which is true when $f\not =0$.

Example 3

(Santos’ quantum walk [13]) Since this procedure is Sgezedy’s quantum walk with queries, the set up for both methods is the same. The difference consists in that in Santos’ quantum walk the leaking graph is no longer used, and Grover’s oracle is used instead. Namely, the following operators are iterated:

• Oracle: $O\otimes I$

• Reflection $R_A=2\sum _{x\in V}|\Phi _{x}\rangle \langle \Phi _{x}|-I$

• Reflection $R_B=2\sum _{y\in V}|\Psi _{y}\rangle \langle \Psi _{y}|-I$

Any algorithm computing $U:=R_BR_AO$ is a QADS because $R_A|{\psi _0}\rangle =|{\psi _0}\rangle =R_B|{\psi _0}\rangle$ and so

$$\begin{aligned}&U|{\psi _0}\rangle =|{\psi _0}\rangle \Longleftrightarrow (O\otimes I)|{\psi _0}\rangle =|{\psi _0}\rangle \\&\quad \Longleftrightarrow \frac{1}{\sqrt{|V|}}\left( \sum _{x\in V\setminus W}|x\rangle -\sum _{x\in W}|x\rangle \right) \otimes \left( \sum _{y\in V}\sqrt{P _{xy}}|y\rangle \right) \\&\quad =\frac{1}{\sqrt{|V|}}\sum _{x\in V}|x\rangle \otimes \left( \sum _{y\in V}\sqrt{P _{xy}}|y\rangle \right) \end{aligned}$$

which is equivalent to $W=\emptyset$.

Example 4

(Quantum abstract search [2]) Let ${\mathcal {H}}=({\mathbb {C}}^2)^{\otimes k}$ be a state space containing two states $|{\psi _0}\rangle$ (with real amplitudes) and $|\psi _{good}\rangle$. In the quantum abstract search the later is the aim state to be found from the former. Two unitary transformations are used:

• $U_1=I-2|\psi _{good}\rangle \langle \psi _{good}|$

• $U_2$: described by a real unitary matrix such that $|{\psi _0}\rangle$ is the only eigenvector (up to phase change) with eigenvalue 1.

Because of the characteristics of such operators, non-orthogonality of the initial and the “good” states is implicitly assumed. For instance, $|{\psi _0}\rangle$ is usually taken as the uniform superposition of all basic states, and $|\psi _{good}\rangle$ is in the computational basis. In our detection setting, we allow $|\psi _{good}\rangle$ to be the zero state, corresponding to the non-existence of marked elements. Also, from the point of view of a detecting problem, the “good” states $\sqrt{\frac{5}{6}}|0\rangle +\sqrt{\frac{1}{6}}|1\rangle$ and ${\frac{1}{\sqrt{2}}}|0\rangle +{\frac{1}{\sqrt{2}}}|1\rangle$ behave similarly, as the key feature is that both are nonzero and they have the same supporting computational basis elements. Therefore, we shall assume that $|\psi _{good}\rangle$ is a uniform superposition of computational states.

With this setting in mind, let $f:\{0,1\}^k\rightarrow \{0,1\}$ be an arbitrary boolean function. Any algorithm taking f as an input that outputs a constructible initial state $|{\psi _0}\rangle$ and the constructible operator $U:=U_2U_1$ where $|\psi _{good}\rangle =\left\{ \begin{array}{cc}\frac{1}{\sqrt{|W|}}\sum _{x\in W}|x\rangle &{} \hbox {if }W\not =\emptyset \\ 0&{}\hbox {otherwise}\end{array}\right.$, is a QADS since

$$\begin{aligned} W=\emptyset \Longrightarrow |{\psi _0}\rangle \in |\psi _{good}\rangle ^\perp \Longrightarrow U_1|{\psi _0}\rangle =|{\psi _0}\rangle \Longrightarrow U|{\psi _0}\rangle =|{\psi _0}\rangle \end{aligned}$$

Example 5

(Wong’s quantum walk [16]) This type of search is applied to 2D grids of $2^k$ vertices $\{|0\rangle ,\ldots ,|2^k-1\rangle \}$ when they are walked by a “lackadaisical” quantum walk (i.e., where each vertex has a self-loop of weight $0<l$). Detection of marked vertices by a boolean function $f:\{0,1\}^k\rightarrow \{0,1\}$ requires the construction of the initial state $|{\psi _0}\rangle =\frac{1}{\sqrt{2^k}}\sum _{x=0}^{k-1}|x\rangle |s_c\rangle \in {\mathcal {H}}=({\mathbb {C}}^2)^{\otimes k}\otimes {\mathbb {C}}^5$ of this weighted quantum walk, where

$$\begin{aligned} |s_c\rangle =\frac{1}{\sqrt{4+l}}(|\rightarrow \rangle +|\leftarrow \rangle +|\uparrow \rangle +|\downarrow \rangle +\sqrt{l}|\circlearrowleft \rangle ) \end{aligned}$$

The grid is walked by iteration of the following two operators:

• Grover diffusion coin for a weighted graph: $I\otimes C$, where $C=2|s_c\rangle \langle s_c|-I$.

• Flip-flop shift: S such that $S(|x\rangle |\rightarrow \rangle )=|x+e_1\rangle |\leftarrow \rangle ,S(|x\rangle |\leftarrow \rangle )=|x-e_1\rangle |\rightarrow \rangle ,S(|x\rangle |\uparrow \rangle )=|x+e_2\rangle |\downarrow \rangle ,S(|x\rangle |\downarrow \rangle )=|x-e_2\rangle |\uparrow \rangle$, and $S(|x\rangle |\circlearrowleft \rangle )=|x\rangle |\circlearrowleft \rangle ,$ where $e_1,e_2$ denote the basic vector directions of the 2D grid.

Notice that $|{\psi _0}\rangle$ is an $1-$eigenvector of both operators, i.e., $(I\otimes (2|s_c\rangle \langle s_c|-I))|{\psi _0}\rangle =|{\psi _0}\rangle =S|{\psi _0}\rangle$. On the other hand, the detection for marked elements in $W=\{x\in \{0,1\}^k\ |\ f(x)=1\}$ is carried out by one of the following two oracles:

• Grover’s oracle: $O\otimes I$

• SKW oracle: $O_{SKW}=I-2\sum _{x\in W}|x,s_c\rangle \langle x,s_c|$

Observe that $(O\otimes I)|{\psi _0}\rangle =|{\psi _0}\rangle \Longleftrightarrow W=\emptyset$ (just like in Santos’s quantum walk), and notice that

$$\begin{aligned} |{\psi _0}\rangle= & {} (I-2\sum _{x\in W}|x,s_c\rangle \langle x,s_c|)|{\psi _0}\rangle \\= & {} \frac{1}{\sqrt{2^k}}\left( \sum _{x\not \in W}|x\rangle |s_c\rangle -\sum _{x\in W}|x\rangle |s_c\rangle \right) \Longleftrightarrow W=\emptyset \end{aligned}$$

Therefore, any algorithm computing $|{\psi _0}\rangle$ and $U:=(O\otimes I)S(I\otimes C)$ or $U:=(O\otimes I)SO_{SKW}$ is a QADS.

As noticed in [16], the laickadaisical quantum walk based on the Grover oracle does not fit into the general machinery of the quantum abstract search. However, we have seen that it fits into our definition. Also, the laickadaisical quantum walk for 2D grids can be naturally considered for regular graphs of torus type. This generalization can be also described in terms of our QADS.

Example 6

(Deutsch-Jozsa’s algorithm [7]) Given a boolean function $f:\{0,1\}^k\rightarrow \{0,1\}$ which is promised to be either constant or balanced (i.e., $f(x)=1$ for exactly half of all possible x), Deutsch–Jozsa’s algorithm prepares an initial state $|{\psi _0}\rangle =|0\rangle ^{\otimes k}|1\rangle$ in the Hilbert space ${\mathcal {H}}=({\mathbb {C}}^2)^{\otimes (k+1)}$, and uses the following two constructible operators:

• Hadamard transform: $H^{\otimes (k+1)}$, where H is the Hadamard gate.

• Unitary version of f: $U_f(|x\rangle |y\rangle )=|x\rangle |y\oplus f(x)\rangle$.

The algorithm which constructs $U:=H^{\otimes (k+1)}U_fH^{\otimes (k+1)}$ from f is a QADS since

$$\begin{aligned} |{\psi _0}\rangle =H^{\otimes (k+1)}U_fH^{\otimes (k+1)}|{\psi _0}\rangle \end{aligned}$$

if and only if

$$\begin{aligned} \sum _{x=0}^{2^k-1}\frac{|x\rangle |-\rangle }{\sqrt{2^k}}=U_f\sum _{x=0}^{2^k-1}\frac{|x\rangle |-\rangle }{\sqrt{2^k}}=\sum _{x=0}^{2^k-1}\frac{(-1)^{f(x)}|x\rangle |-\rangle }{\sqrt{2^k}} \end{aligned}$$

where $|-\rangle =\frac{|0\rangle -|1\rangle }{\sqrt{2}}$. This is equivalent to $\forall x\in \{0,1\}^k:\ f(x)=0$, i.e., $W=\emptyset$ .

Example 7

(Oracle) In examples 1, 3, 4 or 5, any algorithm that simply generates the oracle O generating (alt. $O\otimes I$, $U_1$, $O\otimes I$ or $O_{SKW}$), together with the initial state $|{\psi _0}\rangle$, is also a QADS.

Formal description of procedures in the algorithmic closure of QADS

In this appendix, we provide a formal description of the procedures contained in Table 1, all of them in the algorithmic closure of QADS.

Proposition 1

Consider a QADS that generates a pair $(|{\psi _0}\rangle )\ ,\ U)\in {\mathcal {H}}\times {\mathcal {U}}({\mathcal {H}})$ for any given boolean input f from a set of inputs ${\mathcal {M}}$, where ${\mathcal {U}}({\mathcal {H}})$ is the group of unitary operators on the Hilbert space ${\mathcal {H}}$.

• 1. Algorithms generating the following pairs of initial state/unitary transformation, are also QADS.

  • (a) Extension: $(|{\psi _0}\rangle |0\rangle ^{\otimes l}\ ,\ U\otimes I)\in {\mathcal {H}}'\times {\mathcal {U}}({\mathcal {H}}')$, where ${\mathcal {H}}'={\mathcal {H}}\otimes ({\mathbb {C}}^2)^l$.

  • (b) Inversion: $(|{\psi _0}\rangle \ ,\ U^{\dag })\in {\mathcal {H}}\times {\mathcal {U}}({\mathcal {H}})$.

  • (c) Powers: $(|{\psi _0}\rangle \ ,\ U^{n_f})\in {\mathcal {H}}\times {\mathcal {U}}({\mathcal {H}})$, for all $n_f\in {\mathbb {N}}$.

  • (d) Roots: $(|{\psi _0}\rangle \ ,\ U^{1/n_f})\in {\mathcal {H}}\times {\mathcal {U}}({\mathcal {H}})$, for all $n_f\in {\mathbb {N}}$.

  • (e) Conjugation: $(T|{\psi _0}\rangle \ ,\ TUT^\dag )\in {\mathcal {H}}\times {\mathcal {U}}({\mathcal {H}})$, for all $T\in {\mathcal {U}}({\mathcal {H}})$. Moreover, conjugation induces an equivalence relation on the set of possible outputs of a QADS for a given input $f\in {\mathcal {M}}$.

  • (f) Controlled detecting operator: $(|+\rangle |{\psi _0}\rangle \ ,\ U_c)\in {\mathbb {C}}^2\otimes {\mathcal {H}}\times {\mathcal {U}}({\mathbb {C}}^2\otimes {\mathcal {H}})$, where $|+\rangle =\frac{|0\rangle +|1\rangle }{\sqrt{2}}$ and $U_c|i\rangle |x\rangle =|i\rangle U^i|x\rangle$.

• 2. If a second QADS generates pairs $(|{\psi _0}'\rangle \ ,\ U')\in {\mathcal {H}}'\times {\mathcal {U}}({\mathcal {H}}')$ for boolean functions from the same set of inputs ${\mathcal {M}}$, then:

  • (a) A QADS tensor product of QADS can be realized: $(|{\psi _0}\rangle |{\psi _0}'\rangle \ ,\ U\otimes U')\in {\mathcal {H}}\otimes {\mathcal {H}}'\times {\mathcal {U}}({\mathcal {H}}\otimes {\mathcal {H}}')$.

  • (b) If ${\mathcal {H}}'={\mathcal {H}}$ and $|{\psi _0}'\rangle =|{\psi _0}\rangle$, then a product of detecting operators can be considered as a QADS: $(|{\psi _0}\rangle \ ,\ U'U)\in {\mathcal {H}}\times {\mathcal {U}}({\mathcal {H}})$.

  • (c) The pair of QADS can be doubly controlled according to the following scheme: $(|+\rangle |{\psi _0}\rangle |{\psi _0}'\rangle \ ,\ (U\otimes U')_{dc})\in {\mathbb {C}}^2\otimes {\mathcal {H}}\otimes {\mathcal {H}}'\times {\mathcal {U}}({\mathbb {C}}^2\otimes {\mathcal {H}}\otimes {\mathcal {H}}')$, where $|+\rangle =\frac{|0\rangle +|1\rangle }{\sqrt{2}}$ and $(U\otimes U')_{dc}|i\rangle |x\rangle |x'\rangle =|i\rangle U^i|x\rangle U'^{1-i}|x'\rangle$. It is also a QADS.

Proof

• 1. Assume $W=\{x\in \{0,1\}^k\ |\ f(x)=1\}=\emptyset$, so that $U|{\psi _0}\rangle =|{\psi _0}\rangle$. Then:

  • (a) $(U\otimes I)(|{\psi _0}\rangle |0^{\otimes l}\rangle )=|{\psi _0}\rangle |0^{\otimes l}\rangle$.

  • (b) $|{\psi _0}\rangle =U^\dag |{\psi _0}\rangle$.

  • (c) $U^{n_f}|{\psi _0}\rangle =U^{n_f-1}|{\psi _0}\rangle =\cdots =|{\psi _0}\rangle$.

  • (d) If $U=\sum _{j=1}^te^{\lambda _j i}|x_j\rangle \langle x_j|$, with $\lambda _1=0$ and $|x_1\rangle =|{\psi _0}\rangle$, then $U^{1/n_f}=\sum _{j=1}^te^{\lambda _j i/n_f}|x_j\rangle \langle x_j|$ and so $U^{1/n_f}|{\psi _0}\rangle =|{\psi _0}\rangle$.

  • (e) $(TUT^\dag )(T|{\psi _0}\rangle )=T|{\psi _0}\rangle$.

  • (f) $U_c(|+\rangle |{\psi _0}\rangle )=\frac{|0\rangle |{\psi _0}\rangle +|1\rangle U|{\psi _0}\rangle }{\sqrt{2}}=|+\rangle |{\psi _0}\rangle$.

• 2.

  • (a) We have $(U\otimes U')(|{\psi _0}\rangle |{\psi _0}'\rangle )=|{\psi _0}\rangle |{\psi _0}'\rangle$ whenever $U|{\psi _0}\rangle =|{\psi _0}\rangle$ and $U'|{\psi _0}'\rangle =|{\psi _0}'\rangle$, which is always the case when $W=\emptyset$.

  • (b) If $f=0$, then $U'U|{\psi _0}\rangle =U'|{\psi _0}\rangle =|{\psi _0}\rangle$.

  • (c) Finally, $(U\otimes U')_{dc}(|+\rangle |{\psi _0}\rangle |{\psi _0}'\rangle )=\frac{|0\rangle |{\psi _0}\rangle U'|{\psi _0}'\rangle +|1\rangle U|{\psi _0}\rangle |{\psi _0}'\rangle }{\sqrt{2}}=|+\rangle |{\psi _0}\rangle |{\psi _0}'\rangle$ if $U|{\psi _0}\rangle =|{\psi _0}\rangle$ and $U'|{\psi _0}'\rangle =|{\psi _0}'\rangle$, which is true when $f=0$. $\square$

Corollary 1

If $|{\psi _0}\rangle$ is the initial space of a QADS, then the set G of detecting operators (for any possible QADSwith the same initial state) is a group under the product, which is a subgroup of the stabiliser of $|{\psi _0}\rangle$, under the action of the unitary group.

Proof

Since $|{\psi _0}\rangle$ is the initial state of a QADS, the set G is nonempty. Moreover, since “Inversion”, and “Product” are in the algorithmic closure of QADS, G is a subgroup of the unitary group, and any of its elements stabilise $|{\psi _0}\rangle$. $\square$

On the efficient constructibility of QADS

In this appendix, we give examples of efficient and non-efficient constructible QADS.

Example 8

Let us provide an example of a non-efficiently constructible QADS (unless $P=NP$). Consider the algorithm that, on any input $f:\{0,1\}^k\rightarrow \{0,1\}$, outputs the zero state of ${\mathcal {H}}={\mathbb {C}}^{2}$, together with the unitary operator realized by a gateless circuit, when $W=\emptyset$, and by a circuit consisting on a single X gate, when $W\not =\emptyset$. The output construction of the QADS encapsulates the detection problem of an element $x\in \{0,1\}^k$ such that $f(x)=1$, and so it could be used in a polynomial-time reduction of SAT.

Example 9

For the examples of the previous section we have the following results on efficient constructibility:

• 1. The QADS for Grover search is efficiently constructible, as its output can be computed in O(n) time [12, Section 6.1.2]. Observe that the oracle can be straightforwardly implemented from an actual implementation of the input function f, which can be assumed to be given in a reversible form [12, Section 3.2.5].

• 2. The initial state and the unitary operator of Deutsch-Jozsa’s algorithm require also O(n) to be computed, and so the corresponding QADS is also efficiently constructible.

• 3. Szegedy’s quantum walk initial state and operator for Ambaini’s Element Distinctness problem [1] can be realised in $O(n\log n)$ depth, according to [8, 11], and so the QADS of Example 2 are efficient constructible.

• 4. The QADS of the quantum abstract search is efficiently constructible whenever the initial and good states, together with the operator $U_2$ can be computed in $O(\hbox {poly}(n))$.

• 5. The oracles of the previous examples are, in particular, also efficiently constructible QADS.

Example 10

Consider a QADS that is efficiently constructible. Then, the following QADS from Proposition 1 are also constructible:

• 1. Extension, provided that the number t is $O(\hbox {poly}(n))$.

• 2. Inversion (just by inversion of the quantum circuit describing U).

• 3. Powers, as long as $n_f$ is $O(\hbox {poly}(f))$.

• 4. Conjugation, provided that the circuit for the unitary operator T has size $O(\hbox {poly}(n))$.

• 5. Controlled detecting operator, because of [12, Section 4.3].

The tensor product of two QADS is also efficiently constructible assuming both QADS are. The same property holds for the product and the doubly controlled QADS.

On the detection rate and efficient detection of QADS

In this appendix, we explore in detail aspects on the detection rate of QADS. In particular, we provide $\delta -$detecting functions for some QADS appearing in the main text, we prove the existence of $\delta -$detecting functions for some procedures in the algorithmic closure of QADS, and finally we show a relation between the $\delta -$detecting function and the quantum hitting time for quantum walks.

Example 11

(A QADS with no $\delta$-detecting function) Consider the QADS associated to Grover’s oracle, and let ${\mathcal {M}}=\{f:\{0,1\}^k\rightarrow \{0,1\}\ |\ f(x_0)=1\hbox { for exactly {none} or one value }x_0\}$. It is straigtforward to see that $O^t|{\psi _0}\rangle =\left\{ \begin{array}{ll}|{\psi _0}\rangle &{}\hbox { if }t\hbox { even}\\ \frac{1}{\sqrt{2^k}}\left( \sum _{x\not =x_0}|x\rangle -|x_0\rangle \right) &{}\hbox { if }t\hbox { odd}\end{array}\right.$, and so $\langle {\psi _0}|O^t|{\psi _0}\rangle =1-\frac{t\hbox { mod }2}{2^{k-1}}\ge 1-\frac{1}{2^{k-1}}$. Hence, for all $t\in {\mathbb {N}}$, $\frac{\sum _{t=0}^{T}|\langle {\psi _0}|O^t|{\psi _0}\rangle |^2}{T+1}\ge \left( 1-\frac{1}{2^{k-1}}\right) ^2\ge 1-\frac{1}{2^{k-2}}$. Therefore, for any $\delta >0$, no function $T:{\mathbb {N}}\rightarrow {\mathbb {N}}$ can be $\delta$-detecting for such a QADS.

Example 12

(A QADS with constant $\delta$-detecting function) The QADS in Example 8, has a constant $\delta -$detecting function (namely, $T(k)=1$ is $\frac{1}{2}-$detecting). Observe that, however, such a QADS encapsulates the detecting problem in the construction of the operator U, and so it can not be efficiently constructible (unless $P=NP$). This shows the importance for a QADSto be simultaneously efficiently constructible, and to have a $\delta -$detecting fucntion.

Example 13

For the examples mentioned in the paper, we have the following results on $\delta -detection$:

• 1. Let us consider the QADS of Grover search for

  $$\begin{aligned} {\mathcal {M}}=\{f:\{0,1\}^k\rightarrow \{0,1\}\ |\ f(x_0)=1\hbox { for exactly none or one value }x_0\ ,\ k\ge 2\} \end{aligned}$$

  For all $k\ge 2$, let $\theta _k=2\arccos \left( \sqrt{\frac{2^k-1}{2^k}}\right)$, and let $R_k$ be the closest integer to $\frac{\arccos \left( \sqrt{\frac{1}{2^k}}\right) }{\theta _k}$ so that $\cos \left( {\theta _k}+R_k\theta _k\right) \le \frac{\sqrt{2}}{2}$ [12]. Define, for all $k\in {\mathbb {N}}$, T(k) the smallest natural number greater than $\frac{8}{\sin \left( {\theta _k}\right) }-1$ such that $T(k)\equiv 2R_k\ (\hbox {mod } 2\pi )$. The function T is $\frac{\sqrt{2}-1}{4\sqrt{2}}-$detecting for the QADS since

  $$\begin{aligned}&\frac{\sum _{t=0}^{T(k)}|\langle {\psi _0}|U^t|{\psi _0}\rangle |^2}{T(k)+1}\\&\quad =\frac{\sum _{t=0}^{T(k)}\left( \sqrt{\frac{2^k-1}{2^k}}\cos \left( \frac{2t+1}{2}\theta _k\right) +\sqrt{\frac{1}{2^k}}\sin \left( \frac{2t+1}{2}\theta _k\right) \right) ^2}{T(k)+1}\\&\quad =\frac{\sum _{t=0}^{T(k)}\left( {\frac{2^k-1}{2^k}}\cos ^2\left( \frac{2t+1}{2} \theta _k\right) +{\frac{1}{2^k}}\sin ^2\left( \frac{2t+1}{2}\theta _k\right) +2\frac{\sqrt{2^k-1}}{2^k}\cos \left( \frac{2t+1}{2}\theta _k\right) \sin \left( \frac{2t+1}{2}\theta _k\right) \right) }{T(k)+1}\\&\quad \le \frac{\sum _{t=0}^{T(k)}\left( {\frac{2^k-1}{2^k}}\cos ^2 \left( \frac{2t+1}{2}\theta _k\right) +{\frac{1}{2^k}}\left( 1-\cos ^2 \left( \frac{2t+1}{2}\theta _k\right) \right) +\frac{1}{\sqrt{2^{k-2}}} \cos \left( \frac{2t+1}{2}\theta _k\right) \sin \left( \frac{2t+1}{2} \theta _k\right) \right) }{T(k)+1}\\&\quad ={\frac{1}{2^k}}+\frac{\sum _{t=0}^{T(k)}\left( {\frac{2^k-2}{2^k}} \left( \frac{1+\cos \left( ({2t+1})\theta _k\right) }{2}\right) +\frac{1}{\sqrt{2^{k-2}}} \frac{\sin \left( (2t+1)\theta _k\right) }{2}\right) }{T(k)+1} \end{aligned}$$

  (because $\cos ^2(\frac{\alpha }{2})={\frac{1+\cos (\alpha )}{2}},\sin ^2(\frac{\alpha }{2})={\frac{1-\cos (\alpha )}{2}}$)

  $$\begin{aligned}= & {} {\frac{1}{2^k}}+{\frac{2^{k-1}-1}{2^k}}+\frac{\sum _{t=0}^{T(k)} \left( {\frac{2^k-2}{2^{k+1}}}\cos \left( ({2t+1})\theta _k\right) +\frac{1}{\sqrt{2^{k}}}\sin \left( (2t+1)\theta _k\right) \right) }{T(k)+1}\\\le & {} \frac{1}{4}+\frac{1}{2}+\frac{\sin \left( (T(k)+1)\theta _k\right) \left( {\frac{2^k-2}{2^{k+1}}}\cos \left( (T(k)+1)\theta _k\right) +\frac{1}{\sqrt{2^k}} \sin \left( (T(k)+1)\theta _k\right) \right) }{\sin \left( \theta _k\right) (T(k)+1)}\\\le & {} \frac{3}{4}+\frac{1\cdot \left( 1\cdot \frac{\sqrt{2}}{2}+\sqrt{\frac{1}{2}} \cdot 1\right) }{\sin \left( \theta _k\right) (T(k)+1)}\le \frac{3}{4}+\frac{1}{4\sqrt{2}} =1-\frac{\sqrt{2}-1}{4\sqrt{2}} \end{aligned}$$

  Moreover, $T(k)\in O(\sqrt{2^k})$ [3, Proof of Theorem 3].

• 2. From Proposition 1, we can construct a QADS from the QADS of Grover search by simply taking as output the same initial state and an $\sqrt{2^k}-$th root of the operator U. For any input function $f:\{0,1\}^k\rightarrow \{0,1\}$ this new QADS requires $\sqrt{2^k}$ iterations of the detecting operator to replicate a single iteration in Grover’s QADS. Therefore, the same bounding technique of the previous example shows that a $\frac{\sqrt{2}-1}{4\sqrt{2}}-$detecting function $T\in O(2^k)$ for the QADS can be considered.

• 3. Consider the QADS of Deutsch–Jozsa’s algorithm, and let $T(k)=1$. We shall show that T is $\frac{1}{2}-$detecting for such a QADS. Namely, if f is balanced, then

  $$\begin{aligned} \frac{\sum _{t=0}^{T(k)}|\langle {\psi _0}|U^t|{\psi _0}\rangle |^2}{T(k)+1}=\frac{|\langle {\psi _0}|{\psi _0}\rangle |^2+|\langle {\psi _0}|H^{\otimes (k+1)}U_fH^{\otimes (k+1)}|{\psi _0}\rangle |^2}{2} =\frac{1}{2} \end{aligned}$$

  because $H^{\otimes (k+1)}|{\psi _0}\rangle =\sum _{x=0}^{2^k-1}\frac{|x\rangle |-\rangle }{\sqrt{2^k}}$ and $U_fH^{\otimes (k+1)}|{\psi _0}\rangle =\frac{\left( \sum _{f(x)=0}|x\rangle -\sum _{f(x)=1}|x\rangle \right) |-\rangle }{\sqrt{2^k}}$.

• 4. The controlled QADS of Szegedy’s quantum walk for

  $$\begin{aligned} {\mathcal {M}}=\{f:\{0,1\}^k\rightarrow \{0,1\}\ |\ f(x_0)=1\hbox { for less than half values}\} \end{aligned}$$

  has, because of Proposition 3, a $\frac{1}{4}-$detecting time $T\in O\left( \frac{1}{\sqrt{\theta (P)}}\right)$, where $\theta (P)$ is the eigenvalue gap of the matrix P associated with the graph, by application of the quantum hitting time [15].

• 5. For some specific graphs, the non-controlled QADS of Szegedy’s quantum walk can also be shown to have a $\delta -$detecting time. For instance, let us consider

  $$\begin{aligned} \scriptstyle {\mathcal {M}}=\{f:\{0,1\}^k\rightarrow \{0,1\}\ |\ f(x_0)=1\hbox { for exactly {none} or one value }x_0\ ,\ k\ge 3\} \end{aligned}$$

  for the complete graph of $2^k$ vertices. Explicit expressions for $U^t|{\psi _0}\rangle$ and the quantum hitting time Q(T) of this graph, can be found in [14]. We can use them to get

  $$\begin{aligned} \langle {\psi _0}|U^t|{\psi _0}\rangle= & {} \frac{2(2^k-1)^2T_{2t}\left( \frac{2^k-2}{2^k-1} \right) +2^k-2}{2^k(2^{k+1}-3)}\\ Q(T)= & {} \frac{2(2^k-1)^2\left( 2T+1-U_{2t}\left( \frac{2^k-2}{2^k-1} \right) \right) }{2^k(2^{k+1}-3)(T+1)}\ge \frac{7}{10}\\&\cdot \left( 2-\frac{1+U_{2t} \left( \frac{2^k-2}{2^k-1}\right) }{T+1}\right) \end{aligned}$$

  where $T_{2t}$ and $U_{2t}$ are the Chebyshev polynomial of the first and second kinds. Taking $T(k)=\left\lceil \frac{\frac{1}{\sin \arccos \left( \frac{2^k-2}{2^k-1}\right) }-1}{2}\right\rceil$ we have that the real scalar product $\langle {\psi _0}|U^t|{\psi _0}\rangle$ is nonnegative for all $0\le t\le T(k)$, and $Q(T(k))\ge \frac{1}{5}$. Therefore, by Proposition 3, T(k) is $\frac{1}{10}-$detecting for the QADS.

• 6. For the same set ${\mathcal {M}}$, the controlled QADS of Santos’ quantum walk with respect to the complete graph with $2^k$ vertices, has also a $\delta -$detecting time. According to [13, Section 3.1], the Hilbert space ${\mathcal {H}}$ can be decomposed in an $U-$invariant $3-$dimensional subspace and its orthogonal complement (over which U is the identity). Therefore, $U=-|v_{-1}\rangle \langle v_{-1}|+e^{i\theta }|v_+\rangle \langle v_+|+e^{-i\theta }|v_-\rangle \langle v_-|$ where $|v_+\rangle$ and $|v_-\rangle$ are complex conjugates, and $\cos \theta =\frac{1+\cos ^2\phi }{2}$ with $\cos \phi =\frac{2^k-3}{2^k-1}$. Since $|{\psi _0}\rangle =\lambda _{-1}|v_{-1}\rangle +\lambda _+|v_+\rangle +\overline{\lambda _-}|v_-\rangle$, with $\lambda _+=\frac{-i}{\sqrt{2}},\lambda _-=\overline{\lambda _+}$, we have

  $$\begin{aligned}&\frac{\sum _{t=0}^{T(k)}\Vert U^t|{\psi _0}\rangle -|{\psi _0}\rangle \Vert ^2}{T(k)+1}\\&\quad =\frac{\sum _{t=0}^{T(k)}\left( |\lambda _{-1}|^2|(-1)^t-1|^2 +|\lambda _{+}|^2|e^{it\theta }-1|^2+|\lambda _{-}|^2|e^{-it\theta }-1|^2\right) }{T(k)+1}\\&\quad \ge \frac{\sum _{t=0}^{T(k)}2\left( \frac{1}{2}|e^{it\theta }-1|^2\right) }{T(k)+1}=\frac{\sum _{t=0}^{T(k)}2(1-\cos {(t\theta )})}{T(k)+1}\\&\quad =2\left( 1-\frac{\sin \left( \frac{(T(k)+1)\theta }{2}\right) \cos \left( \frac{T(k)\theta }{2}\right) }{\sin \left( \frac{\theta }{2}\right) (T(k)+1)}\right) \ge 2\left( 1-\frac{ \cos \left( \frac{T(k)\theta }{2}\right) }{\sin \left( \frac{\theta }{2}\right) T(k)}\right) \end{aligned}$$

  Take now $T(k)=\lceil \frac{\pi }{2\theta }\rceil$, so that $\cos \left( \frac{T(k)\theta }{2}\right) \le \frac{\sqrt{2}}{2}$ and $\sin \left( \frac{\theta }{2}\right) \cdot T(k)\ge \frac{\sqrt{2}}{2}\varepsilon$ for some $1<\varepsilon <\frac{6}{5}$. Hence, $\frac{\sum _{t=0}^{T(k)}\Vert U^t|{\psi _0}\rangle -|{\psi _0}\rangle \Vert ^2}{T(k)+1}\ge 4\cdot \frac{1}{48}$, and so $T\in O(\sqrt{2^k})$ is $\frac{1}{24}-$detecting for the controlled QADS, because of Proposition 3.

Proposition 2

(Detecting times for procedures in the algorithmic closure) Consider a $\delta -$detecting time $T\ge 1$ for a given QADS. Then,

• 1. T is $\delta -$detecting time for the extension, inversion and conjugation QADS.

• 2. $S(k)=T(k)\cdot n_f$ is $\frac{\delta }{2L}-$detecting time for the root QADS, provided that $n_f$ depends only on the input size k of f, and that $n_f\le L$ for all $f\in {\mathcal {M}}$. As a consequence, if $T'$ is $\delta '-$detecting for the power QADS with $L'\ge n_f$ depending only on the input size k of f, then $\frac{T'}{n_f}$ is $\frac{\delta '}{2L'}-$detecting time for the original QADS.

• 3. T is $\left( \frac{1+\frac{\delta }{2}-\sqrt{1-\delta }}{2}\right) -$detecting time for the QADS based on the controlled detecting operator.

• 4. If T is also $\delta '-$detecting for a second QADS, then: T is $\max \{\delta ,\delta '\}-$detecting for the tensor product of both QADS, and $\left( \frac{1+\frac{\delta +\delta '}{2}-\sqrt{1-\delta }\sqrt{1-\delta '}}{2}\right) -$detecting for the doubly controlled QADS.

Proof

• 1. This easily follows from the following equalities:

  $$\begin{aligned} \langle {\psi _0}|U^t|{\psi _0}\rangle= & {} \langle |{\psi _0}\rangle |0\rangle ^{\otimes l}|(U\otimes I)^t||{\psi _0}\rangle |0\rangle ^{\otimes l}\rangle \\= & {} \langle {\psi _0}|(U^{\dag })^t|{\psi _0}\rangle =\langle T|{\psi _0}\rangle |(TUT^\dag ) ^t|T|{\psi _0}\rangle \rangle . \end{aligned}$$

• 2. For all $\frac{T(k)+1}{S(k)+1}\ge \frac{1}{2n_f}\ge \frac{1}{2L}$. Now, if $A=\{0, n_f,2\cdot n_f\ldots ,T(k)\cdot n_f\}$, then

  $$\begin{aligned}&\frac{\sum _{t=0}^{S(k)}|\langle {\psi _0}|U^{\frac{t}{n_f}}|{\psi _0}\rangle |^2}{S(k)+1}\\&\quad =\frac{\sum _{t\in A}|\langle {\psi _0}|U^{\frac{t}{n_f}}|{\psi _0}\rangle |^2+\sum _{t\in \{0,\ldots , S(k)\}\setminus A}|\langle {\psi _0}|U^{\frac{t}{n_f}}|{\psi _0}\rangle |^2}{S(k)+1}\\&\quad \le \frac{\sum _{t=0}^{T(k)}|\langle {\psi _0}|U^{{t}}|{\psi _0}\rangle |^2}{T(k)+1}\cdot \frac{T(k)+1}{S(k)+1}+\frac{(S(k)-T(k))}{S(k)+1}\\&\quad \le (1-\delta )\cdot \frac{T(k)+1}{S(k)+1}+\frac{(S(k)-T(k))}{S(k)+1}\le 1-\frac{\delta }{2L} \end{aligned}$$

• 3. Let $|{\psi _0}_c\rangle =|+\rangle |{\psi _0}\rangle$. For all $t=0,\ldots , T$, we have $U_c^t|{\psi _0}_c\rangle =\frac{|0\rangle |{\psi _0}\rangle +|1\rangle U^t|{\psi _0}\rangle }{\sqrt{2}}$, and so $\langle {\psi _0}_c|U_c^t|{\psi _0}_c\rangle =\frac{1+\langle {\psi _0}|U^t|{\psi _0}\rangle }{2}$, and $|\langle {\psi _0}_c|U_c^t|{\psi _0}_c\rangle |\le \frac{1+|\langle {\psi _0}|U^t|{\psi _0}\rangle |}{2}$. Therefore:

  $$\begin{aligned} \frac{\sum _{t=0}^{T(k)}|\langle {\psi _0}_c|U_c^t|{\psi _0}_c\rangle |^2}{T(k)+1}= & {} \frac{\left( \sqrt{\sum _{t=0}^{T(k)}\left( \frac{1+|\langle {\psi _0}|U^t |{\psi _0}\rangle |}{2}\right) ^2}\right) ^2}{T(k)+1}\\\le & {} \left( \frac{\sqrt{\sum _{t=0}^{T(k)}1^2}+\sqrt{\sum _{t=0}^{T(k)}| \langle {\psi _0}|U^t|{\psi _0}\rangle |^2}}{2\sqrt{T(k)+1}}\right) ^2\\\le & {} \left( \frac{1+\sqrt{1-\delta }}{2}\right) ^2=1-\left( \frac{1+\frac{\delta }{2}-\sqrt{1-\delta }}{2}\right) \end{aligned}$$

  and so T is $\left( \frac{1+\frac{\delta }{2}-\sqrt{1-\delta }}{2}\right) -$detecting time for the QADS based on the controlled detecting operator.

• 4. For all $t=0,\ldots , T$, we have

  $$\begin{aligned} \langle |{\psi _0}\rangle |{\psi _0}'\rangle |(U\otimes U')^t||{\psi _0}\rangle |{\psi _0}'\rangle \rangle =\langle {\psi _0}|U^t|{\psi _0}\rangle \langle {\psi _0}'|U'^t|{\psi _0}'\rangle \end{aligned}$$

  Therefore:

  $$\begin{aligned}&\frac{\sum _{t=0}^{T(k)}|\langle |{\psi _0}\rangle |{\psi _0}'\rangle |(U\otimes U')^t||{\psi _0}\rangle |{\psi _0}'\rangle \rangle |^2}{T(k)+1}\\&\quad \le \min \left\{ \frac{\sum _{t=0}^{T(k)}|\langle {\psi _0}|U^t|{\psi _0}\rangle |^2}{T(k)+1},\frac{\sum _{t=0}^{T(k)}|\langle {\psi _0}'|U'^t|{\psi _0}'\rangle |^2}{T(k)+1}\right\} \\&\quad \le \min \left\{ 1-\delta ,1-\delta '\right\} =1-\max \{\delta ,\delta '\} \end{aligned}$$

  because $|\langle {\psi _0}|U^t|{\psi _0}\rangle |,|\langle {\psi _0}'|U'^t|{\psi _0}'\rangle |\le 1$.

  Finally, for the doubly controlled QADS, let $|{\psi _0}_{dc}\rangle =|+\rangle |{\psi _0}\rangle |{\psi _0}'\rangle$. For all $t=0,\ldots , T$, we have

  $$\begin{aligned} (U\otimes U')^t_{dc}(|+\rangle |{\psi _0}\rangle |{\psi _0}'\rangle )= \frac{|0\rangle |{\psi _0}\rangle U'^t|{\psi _0}'\rangle +|1\rangle U^t|{\psi _0}\rangle |{\psi _0}'\rangle }{\sqrt{2}} \end{aligned}$$

  and so $\langle {\psi _0}_{dc}|(U\otimes U')^t_{dc}|{\psi _0}_{dc}\rangle =\frac{\langle {\psi _0}'|U'^t|{\psi _0}'\rangle +\langle {\psi _0}|U^t|{\psi _0}\rangle }{2}$. Therefore:

  $$\begin{aligned}&\frac{\sum _{t=0}^{T(k)}|\langle {\psi _0}_{dc}|(U\otimes U')^t_{dc}|{\psi _0}_{dc}\rangle |^2}{T(k)+1}\\&\quad =\frac{\left( \sqrt{\sum _{t=0}^{T(k)}\left( \frac{|\langle {\psi _0}'|U'^t |{\psi _0}'\rangle |+|\langle {\psi _0}|U^t|{\psi _0}\rangle |}{2}\right) ^2}\right) ^2}{T(k)+1}\\&\quad \le \left( \frac{\sqrt{\sum _{t=0}^{T(k)}|\langle {\psi _0}'|U'^t|{\psi _0}'\rangle |^2} +\sqrt{\sum _{t=0}^{T(k)}|\langle {\psi _0}|U^t|{\psi _0}\rangle |^2}}{2\sqrt{T(k)+1}}\right) ^2\\&\quad \le \left( \frac{\sqrt{1-\delta '}+\sqrt{1-\delta }}{2}\right) ^2 =1-\left( \frac{1+\frac{\delta +\delta '}{2}-\sqrt{1-\delta }\sqrt{1-\delta '}}{2}\right) \end{aligned}$$

$\square$

Motivated by the quantum hitting time of quantum walks, an alternative definition of $\delta -$detecting time for the detecting operator U of a QADS was introduced in [6, Section 4], namely a $R\in {\mathbb {N}}$ such that

$$\begin{aligned} \frac{1}{R+1}\sum _{t=0}^R\Vert U^t|{\psi _0}\rangle -|{\psi _0}\rangle \Vert ^2\ge 4\delta \end{aligned}$$

(1)

Next, we shall show a relation between both definitions.

Proposition 3

• 1. Let Q be a QADS, and let $T:{\mathbb {N}}\rightarrow {\mathbb {N}}$ be such that for all nonzero $f\in {\mathcal {M}}$ of input size k, the detecting operator U provided by Q on input f satisfies Eq. (1) for $R=T(k)$. If $\langle {\psi _0}|U^t|{\psi _0}\rangle$ is real and nonnegative for all $t\le T(k)$, then T is $2\delta$-detecting for the QADS. In particular, this is true when Q outputs controlled operators such that $\langle {\psi _0}|U^t|{\psi _0}\rangle$ is real for all $t\in {\mathbb {N}}$.

• 2. Reciprocally, if $T:{\mathbb {N}}\rightarrow {\mathbb {N}}$ is $\delta '-$detecting for a QADS Q, then for all nonzero $f\in {\mathcal {M}}$ of input size k, the detecting operator U provided by Q on input f satisfies Eq. (1) for $R=T(k)$ and $\delta =\frac{1-\sqrt{1-\delta '}}{2}$.

Proof

• 1. Because $\Vert U^t|{\psi _0}\rangle -|{\psi _0}\rangle \Vert ^2=2(1-\mathfrak {R}\langle {\psi _0}|U^t|{\psi _0}\rangle )$ we have

  $$\begin{aligned} 4\delta\le & {} \frac{\sum _{t=0}^{T(k)}\Vert U^t|{\psi _0}\rangle -|\psi (0\rangle \Vert ^2}{T(k)+1}\\= & {} 2\left( 1-\frac{\sum _{t=0}^{T(k)}\mathfrak {R}\langle {\psi _0}|U^t| {\psi _0}\rangle }{T(k)+1}\right) =2\left( 1-\frac{\sum _{t=0}^{T(k)} \langle {\psi _0}|U^t|{\psi _0}\rangle }{T(k)+1}\right) \end{aligned}$$

  since $\langle {\psi _0}|U^t|{\psi _0}\rangle$ is a real number. Now, because such a number is nonnegative, we have

  $$\begin{aligned} \frac{\sum _{t=0}^{T(k)} |\langle {\psi _0}|U^t|{\psi _0}\rangle |^2}{T(k)+1}\le \frac{\sum _{t=0}^{T(k)} \langle {\psi _0}|U^t|{\psi _0}\rangle }{T(k)+1}\le 1-2\delta \end{aligned}$$

  In particular, if Q outputs controlled operators such that $\langle {\psi _0}|U^t|{\psi _0}\rangle$ is real, for all $t\in {\mathbb {N}}$, because of the proof of Proposition 2 we know that $\langle {\psi _0}_c|U_c^t|{\psi _0}_c\rangle =\frac{1+\langle {\psi _0}|U^t|{\psi _0}\rangle }{2}$ is always real and nonnegative.

• 2. Cauchy–Schwarz inequality gives us

  $$\begin{aligned} \sum _{t=0}^{T(k)}\frac{1}{T(k)+1} |\langle {\psi _0}|U^t|{\psi _0}\rangle |\le \frac{1}{\sqrt{T(k)+1}}\sqrt{\sum _{t=0}^{T(k)} |\langle {\psi _0}|U^t|{\psi _0}\rangle |^2} \end{aligned}$$

  Since $\mathfrak {R}\langle {\psi _0}|U^t|{\psi _0}\rangle \le |\langle {\psi _0}|U^t|{\psi _0}\rangle |$ we have

  $$\begin{aligned}&\frac{\sum _{t=0}^{T(k)}\Vert U^t|{\psi _0}\rangle -|\psi (0\rangle \Vert ^2}{T(k)+1}\\&\quad =2\left( 1-\frac{\sum _{t=0}^{T(k)}\mathfrak {R}\langle {\psi _0}|U^t|{\psi _0}\rangle }{T(k)+1}\right) \ge 2\left( 1-\frac{\sum _{t=0}^{T(k)} |\langle {\psi _0}|U^t|{\psi _0}\rangle |}{T(k)+1}\right) \\&\quad \ge 2\left( 1-\sqrt{\frac{\sum _{t=0}^{T(k)} |\langle {\psi _0}|U^t|{\psi _0}\rangle |^2}{T(k)+1}}\right) \ge 4\left( \frac{1-\sqrt{1-\delta '}}{2}\right) \end{aligned}$$

$\square$

The condition real and nonnegative in the previous proposition is necessary. Just observe that a QADS that provides, for nonzero inputs, a detecting operator U satisfying $U|{\psi _0}\rangle =\lambda |{\psi _0}\rangle$ with $\lambda \ne 1$, satisfies Eq. (1) with $R=1$ for $\delta =\frac{|\lambda -1|^2}{8}$. However, $T=1$ is not $\delta '-$detecting time for any $\delta '>0$.

On the detection with a QADS

In this final appendix, we prove the main theorem of the paper, and we comment on the definition of QADS introduced in [6].

Proof of the main theorem

If $f=0$, then $U|{\psi _0}\rangle =|{\psi _0}\rangle$, and so for all $t\in \{0,\ldots , T\}$ the measurement of $U^t|{\psi _0}\rangle$ is $|{\psi _0}\rangle$. Hence, the algorithm’s output NO is always correct. When $f\not =0$, NO is an output on integer t with a probability $|\langle {\psi _0}|U^t|{\psi _0}\rangle |^2$. So, the error probability of the algorithm is the average $\frac{\sum _{t=0}^T|\langle {\psi _0}|U^t|{\psi _0}\rangle |^2}{T+1}$. When the QADS in efficiently constructible and has $\delta -$detecting time, the number of iterations of the algorithm to be taken is T(k). This yields the promised one-sided error quantum algorithm requiring $O(\hbox {poly}(n))$ precomputation time. $\square$

Remark 1

The relation between the detecting scheme for QADS and the one given in [15] for quantum walks, is implicitly contained in Proposition 3 (Appendix D). The error probability in the later case is bounded by $\delta$ of Eq. (1) [6], while in the former case it is bounded by $2\delta$. An explanation for this gap is that the inner product $\langle {\psi _0}|U^t|{\psi _0}\rangle$ tests the state $U_c^t|{\psi _0}\rangle$ for both $|0\rangle$ and $|1\rangle$ in the auxiliary register, while the norm $\Vert U^t|{\psi _0}\rangle -|{\psi _0}\rangle \Vert$ only accounts for the control qubit $|1\rangle$. In order to boost up the success probability for quantum walks, a second measurement is carried out in the final state. If the measured vertex is marked, then the error probability is improved. This was the rationale for the map m in the (preliminary) definition of a QADS in [6], since a quantum-walk-like detecting scheme was assumed. Such map was thought to provide a boosting up of the detection probability, but this can be achieved by the detecting scheme given in Sect. 5 when the QADS provides a controlled detecting operator (Sect. 4). The broader approach adopte in this paper includes the boosting up probability directly in the detection scheme.