Phase Kickback

Phase kickback is the coherent conversion of a target eigenvalue into a relative phase on a selector register. Nothing travels backward through the circuit: one joint unitary acts on both registers, the target returns on a common ray, and the selector branches retain the eigenvalue phases. Those phases become useful only when a later interference experiment makes their differences observable.

We will derive the eigenstate condition, convert Boolean and modular-addition oracles into phases, and read those phases through interference. The controlled operation is assumed to be available; constructing that operation and proving an algorithm’s advantage are separate questions.

Helpful background. Controlled operations explains coherent selector branches and their relative phases.

Eigenvalues become selector phases

Let a selector register have orthonormal basis $\{|x\rangle:x\in X\}$, and let $V_x$ be a unitary on a target register. The block-controlled operator is

$$W = \sum_{x\in X} |x\rangle\langle x|\otimes V_x.$$

For a selector state and normalized target

$$|\psi\rangle = \sum_x\alpha_x|x\rangle, \qquad \langle\eta|\eta\rangle=1,$$

suppose every populated branch sees the same target as an eigenstate:

$$V_x|\eta\rangle = e^{i\theta_x}|\eta\rangle \quad\text{whenever}\quad \alpha_x\ne0.$$

Linearity then gives

$$\begin{aligned} W|\psi\rangle|\eta\rangle &= \sum_x\alpha_x|x\rangle V_x|\eta\rangle\\ &= \left( \sum_x\alpha_xe^{i\theta_x}|x\rangle \right)|\eta\rangle. \end{aligned}$$

Thus the target returns exactly, while the selector undergoes the diagonal unitary

$$D_\theta = \sum_xe^{i\theta_x}|x\rangle\langle x|.$$

The important step is the common target factor: after the operation, no target state distinguishes one selector branch from another. The selector can therefore retain a coherent phase pattern.

Eigenstates, Selectors, and Relative Phase

The common-eigenstate condition is sufficient for exact target return. A slightly weaker condition is sufficient only for factorization. If there is one normalized state $|\xi\rangle$ such that

$$V_x|\eta\rangle = e^{i\theta_x}|\xi\rangle$$

on every populated branch, then

$$W|\psi\rangle|\eta\rangle = D_\theta|\psi\rangle|\xi\rangle.$$

The registers factor, but the target has changed from $|\eta\rangle$ to $|\xi\rangle$. Calling this target return would be inaccurate unless the two states lie on the same ray. If an identity branch is populated, that branch outputs $|\eta\rangle$, so factorization already fixes the common output ray to the input target ray.

For a generic target, tracing it out gives the selector state

$$\rho_D' = \sum_{x,x'} \alpha_x\alpha_{x'}^* \langle\eta|V_{x'}^\dagger V_x|\eta\rangle |x\rangle\langle x'|.$$

The matrix of target overlaps is a Gram matrix. Under exact kickback its entries are

$$\langle\eta|V_{x'}^\dagger V_x|\eta\rangle = e^{i(\theta_x-\theta_{x'})}.$$

A magnitude below one suppresses selector coherence and signals correlation with the target. A unit-modulus entry with the wrong argument is instead a coherent phase error. Computational-basis probabilities and target-return probability alone can miss both distinctions.

For one control and one target unitary,

$$C_1(U) \left(a|0\rangle+b|1\rangle\right)|u\rangle = \left(a|0\rangle+be^{i\theta}|1\rangle\right)|u\rangle, \qquad U|u\rangle=e^{i\theta}|u\rangle.$$

Only phase differences among populated branches are observable. A common shift $\theta_x\mapsto\theta_x+\gamma$ changes the whole joint state by one global phase, but rephasing only one controlled branch changes $D_\theta$. In particular,

$$C_1(e^{i\gamma}U) = \left(P(\gamma)\otimes I\right)C_1(U), \qquad P(\gamma) = |0\rangle\langle0|+e^{i\gamma}|1\rangle\langle1|.$$

The phase of an isolated state vector remains ray redundancy, as explained by Rays and Global Phase. The controlled construction promotes a target-operator phase into a selector-relative phase; it does not make an independently observable global phase. Nor does the existence of the block matrix imply that an unknown channel automatically supplies coherent controlled access.

Boolean XOR Oracles and the Minus Ancilla

A clean Boolean XOR oracle has the exact interface

$$O_f|x\rangle|y\rangle = |x\rangle|y\mathbin{\oplus}f(x)\rangle, \qquad f(x)\in\{0,1\}.$$

The target operation on branch $x$ is $X^{f(x)}$. Define the two $X$ eigenstates

$$|\chi_s\rangle = \frac{|0\rangle+(-1)^s|1\rangle}{\sqrt2}, \qquad s\in\{0,1\},$$

so that

$$X|\chi_s\rangle = (-1)^s|\chi_s\rangle.$$

The oracle therefore acts as

$$O_f|x\rangle|\chi_s\rangle = (-1)^{s f(x)}|x\rangle|\chi_s\rangle.$$

For $s=1$, the target is

$$|\chi_1\rangle = |-\rangle = \frac{|0\rangle-|1\rangle}{\sqrt2},$$

and the data branch receives $(-1)^{f(x)}$. For $s=0$, the target is $|+\rangle$ and every branch eigenvalue is $+1$, so no phase is marked. In both cases the target returns exactly; the difference is entirely phase-sensitive.

For a data superposition $\sum_x\alpha_x|x\rangle$, one clean query gives

$$O_f \left( \sum_x\alpha_x|x\rangle \right)|-\rangle = \left( \sum_x\alpha_x(-1)^{f(x)}|x\rangle \right)|-\rangle.$$

This is not a readable list of all function values. The phase pattern must be followed by an interference operation chosen for the task. It also assumes that the declared oracle is clean: if an implementation leaves $x$-dependent workspace, the data register need not carry a pure phase pattern by itself.

Modular Addition and Character States

Boolean kickback is the two-level member of a more general character-state construction. Let an $m$-qubit target encode $M=2^m$ basis states in increasing integer order, and define modular translation by

$$T_a|y\rangle = |y+a\bmod M\rangle.$$

Using the positive-sign convention of the quantum Fourier transform, define the negative-exponent character

$$|\chi_k^-\rangle = \frac1{\sqrt M} \sum_{y=0}^{M-1} e^{-2\pi i k y/M}|y\rangle = F_M^\dagger|k\rangle.$$

Changing variables to $z=y+a\bmod M$ gives

$$\begin{aligned} T_a|\chi_k^-\rangle &= \frac1{\sqrt M} \sum_y e^{-2\pi i k y/M}|y+a\rangle\\ &= e^{+2\pi i k a/M} \frac1{\sqrt M} \sum_z e^{-2\pi i k z/M}|z\rangle\\ &= e^{+2\pi i k a/M}|\chi_k^-\rangle. \end{aligned}$$

Thus a coherent addition oracle

$$A_f = \sum_x|x\rangle\langle x|\otimes T_{f(x)}$$

obeys

$$A_f|x\rangle|\chi_k^-\rangle = e^{+2\pi i k f(x)/M}|x\rangle|\chi_k^-\rangle.$$

Both signs matter. Replacing $T_a$ by $T_{-a}$ conjugates the eigenvalue, and using $F_M|k\rangle$ instead of $F_M^\dagger|k\rangle$ also reverses the exponent. Reducing $ka$ modulo $M$ is safe only after the translation and Fourier conventions have been fixed.

For a known classical integer $k$, this character is a product state. In tensor order $a_{m-1}\otimes\cdots\otimes a_0$,

$$|\chi_k^-\rangle =\bigotimes_{j=m-1}^{0} \frac{|0\rangle+e^{-2\pi i k2^j/M}|1\rangle}{\sqrt2}.$$

Thus one can prepare it using a Hadamard and a known one-qubit phase on each target wire; a full inverse QFT applied to $|k\rangle$ is valid but unnecessary. This shortcut assumes classical knowledge of $k$. Applying $F_M^\dagger$ to an arbitrary quantum superposition of labels is a different task. In a finite gate alphabet, synthesizing the preparation phases to the required accuracy still has a cost. Ideal kickback leaves the prepared character available for reuse; imperfect preparation or shifts can accumulate errors.

Interference and Phase Readout

For one returned control phase, write

$$|\psi_\theta\rangle = \frac{|0\rangle+e^{i\theta}|1\rangle}{\sqrt2}.$$

Computational-basis measurement gives one half for each outcome, independently of $\theta$. It cannot distinguish correct kickback from a coherent phase bias. With

$$|+_Y\rangle = \frac{|0\rangle+i|1\rangle}{\sqrt2},$$

the complementary expectations are

$$\langle X\rangle=\cos\theta, \qquad \langle Y\rangle=\sin\theta,$$

and therefore

$$p_{X+} = \frac{1+\cos\theta}{2}, \qquad p_{Y+} = \frac{1+\sin\theta}{2}.$$

An $X$-basis readout applies $H$ and then measures $Z$. With the $|+_Y\rangle$ convention above, a $Y$-basis readout applies $S^\dagger$, then $H$, then measures $Z$. These are incompatible settings and require separate ensembles. Measurement in circuits explains these basis rotations and the sampling uncertainty of their estimates.

A phase measurement can combine complementary selector observables with a target or joint-state check. Correct $X/Y$ means can coexist with untested leakage outside a promised subspace, while target-return probability alone can miss a coherent phase offset. For finite shots, report the number of trials in each basis, the interval or uncertainty rule, and a prespecified acceptance tolerance.

Compute–Phase–Uncompute

A useful predicate is often implemented by a reversible evaluator rather than delivered as a clean phase oracle. Suppose

$$C_f |x\rangle|0^w\rangle|0\rangle = |x\rangle|g_x\rangle|f(x)\rangle,$$

where $|g_x\rangle$ is workspace. Applying $Z$ to the output bit gives

$$|x\rangle|g_x\rangle|f(x)\rangle \longmapsto (-1)^{f(x)} |x\rangle|g_x\rangle|f(x)\rangle.$$

Then the inverse evaluator restores the auxiliaries:

$$C_f^\dagger (I\otimes I\otimes Z) C_f |x\rangle|0^w\rangle|0\rangle = (-1)^{f(x)} |x\rangle|0^w\rangle|0\rangle.$$

The data phase survives because the inverse removes correlations, not the scalar multiplying each branch. On a superposition, omitting the inverse leaves data coherences weighted by the full auxiliary overlap $\langle g_{x'}|g_x\rangle\langle f(x')|f(x)\rangle$. Orthogonal garbage can erase the very interference the phase was meant to produce.

Reversible computation explains how to construct and undo the evaluator. This sequence uses one forward evaluator, one phase gate and one inverse evaluator. If an oracle directly supplies the complete clean phase operation, that operation may instead count as a single query. Query counts describe access to a supplied operation; they do not specify its internal gate count or hardware time.

Imperfect phase transfer

For a desired branch phase $\theta_x$, define

$$\epsilon_x = \left\| V_x|\eta\rangle - e^{i\theta_x}|\eta\rangle \right\|.$$

Selector orthogonality gives the exact state-error identity

$$\left\| |\Psi_{\mathrm{actual}}\rangle - |\Psi_{\mathrm{ideal}}\rangle \right\|^2 = \sum_x|\alpha_x|^2\epsilon_x^2.$$

The corresponding ideal-state fidelity is

$$\left| \sum_x|\alpha_x|^2 e^{-i\theta_x} \langle\eta|V_x|\eta\rangle \right|^2.$$

These are metrics for the chosen input. A uniform bound on every populated branch gives a corresponding uniform bound over selector states with the same prepared target, but does not characterize the operation on arbitrary target inputs. The phase of the ideal vector is fixed in the norm comparison; the squared overlap is invariant under a common global phase.

Worked example: a Boolean phase pattern

Let $s=101_2$ and $f_s(x)=x_2\oplus x_0$, with data order $q_2q_1q_0$ followed by the target. Starting from $|000\rangle|0\rangle$, prepare $|+\rangle^{\otimes3}|-\rangle$. One clean XOR query gives

$$O_{f_s}|+\rangle^{\otimes3}|-\rangle =\frac1{\sqrt8}\sum_{x=0}^{7}(-1)^{s\cdot x}|x\rangle|-\rangle.$$

In increasing binary order the signs are

$$(+1,-1,+1,-1,-1,+1,-1,+1).$$

The target returns, but measuring the data immediately would still give a uniform distribution. Apply $H^{\otimes3}$ to the data first. Its amplitude at $z$ is

$$\frac18\sum_x(-1)^{(s\oplus z)\cdot x} =\prod_{j=0}^{2}\frac{1+(-1)^{s_j\oplus z_j}}2 =\delta_{s,z}.$$

The interference therefore returns $101$ with certainty. If the target were mistakenly prepared in $|+\rangle$, the query would make no phase pattern and the same decoder would return $000$.

There are four qubits, seven Hadamards, one initial $X$ on the target, one clean XOR query and three data measurements. For this explicitly known function the query itself can be implemented by two CNOTs onto the target. For an unknown function supplied as an oracle, its internal implementation is not specified by the one-query description. The Bernstein–Vazirani algorithm explains the unknown-function problem and its classical comparison.

Worked example: a modular shift

Take $M=8$, $k=3$ and a positive translation $T_5|y\rangle=|y+5\bmod8\rangle$. Prepare the target

$$|\chi_3^-\rangle=\frac1{\sqrt8}\sum_{y=0}^{7}e^{-2\pi i3y/8}|y\rangle.$$

The shift eigenvalue is $e^{2\pi i(3)(5)/8}=e^{-i\pi/4}$. With a coherent identity reference branch,

$$C_1(T_5)|+\rangle|\chi_3^-\rangle =\frac{|0\rangle+e^{-i\pi/4}|1\rangle}{\sqrt2} \otimes|\chi_3^-\rangle.$$

Consequently, separate $X$ and $Y$ measurements of the control give

$$p_{X+}=\frac{2+\sqrt2}{4}\simeq0.853553, \qquad p_{Y+}=\frac{2-\sqrt2}{4}\simeq0.146447.$$

Reversing the Fourier sign leaves the $X$ probability unchanged but changes $p_{Y+}$ to $0.853553$. This is why checking a single real quadrature does not determine the phase sign.

The target can be prepared directly in $a_2a_1a_0$ order as

$$|\chi_3^-\rangle =\frac{|0\rangle-|1\rangle}{\sqrt2} \otimes\frac{|0\rangle+i|1\rangle}{\sqrt2} \otimes\frac{|0\rangle+e^{-3\pi i/4}|1\rangle}{\sqrt2}.$$

Without the coherent control, $T_5|\chi_3^-\rangle$ differs from its input only by a global phase. There would then be no control observable to measure. The construction uses one supplied controlled-shift query, while preparation, repeated measurements and any circuit implementation of the shift are additional costs.

Common pitfalls

Target return alone does not test the phase. Both Boolean ancillas $|+\rangle$ and $|-\rangle$ return, but only the latter produces the sign pattern. Use interference and check the phase convention.

A generic target can retain information about the selector. Distinct branch states can reduce selector coherence. For a pure product input this means output entanglement; with a mixed target, classical correlations can also reduce visibility.

Controlled access is an assumption. A known, phase-fixed circuit can be controlled with additional gates. An uncontrolled unknown channel does not determine its phase relative to an identity branch; quantum oracles discusses the access distinction.

One interference step is not a complete algorithm. Phase estimation adds controlled powers and decoding; Grover search adds repeated reflections. Their guarantees depend on those additional operations and on the problem being solved.

Exercises

1. Derive the Exact Return Criterion

For $W=\sum_j|j\rangle\langle j|\otimes V_j$ and $|\psi\rangle=\sum_j\alpha_j|j\rangle$, prove exact returning kickback when $V_j|\eta\rangle=e^{i\theta_j}|\eta\rangle$ on the populated support. Distinguish this from the weaker common-output condition $V_j|\eta\rangle=e^{i\theta_j}|\xi\rangle$. Why does an occupied identity branch make the two conditions coincide up to a common phase?

Solution

Direct substitution gives

$$\begin{aligned} W|\psi\rangle|\eta\rangle &= \sum_j\alpha_j|j\rangle V_j|\eta\rangle\\ &= \sum_j\alpha_je^{i\theta_j}|j\rangle|\eta\rangle\\ &= D_\theta|\psi\rangle|\eta\rangle. \end{aligned}$$

The target is exactly the input ray. Under the weaker condition, the same calculation ends in $D_\theta|\psi\rangle|\xi\rangle$: the registers factor, but the target ray may have changed.

If an occupied branch has $V_{j_0}=I$, its output is $|\eta\rangle$. Common-output factorization also says that output equals $e^{i\theta_{j_0}}|\xi\rangle$, so $|\xi\rangle=e^{-i\theta_{j_0}}|\eta\rangle$. The common target is therefore the input ray, and its common phase can be absorbed into all selector phases.

2. Promote a Target Phase to a Branch Phase

Prove $C_1(e^{i\gamma}U)=(P(\gamma)\otimes I)C_1(U)$. If $U|u\rangle=e^{i\theta}|u\rangle$ and the control begins in $|+\rangle$, find $p_{X+}$. Explain why $U$ and $e^{i\gamma}U$ define the same isolated channel but different controlled operations.

Solution

Using $P_b=|b\rangle\langle b|$,

$$\begin{aligned} C_1(e^{i\gamma}U) &= P_0\otimes I+e^{i\gamma}P_1\otimes U\\ &= \left[ (P_0+e^{i\gamma}P_1)\otimes I \right] \left(P_0\otimes I+P_1\otimes U\right)\\ &= (P(\gamma)\otimes I)C_1(U). \end{aligned}$$

On $|+\rangle|u\rangle$, the active branch receives phase $e^{i(\theta+\gamma)}$, so

$$p_{X+} = \frac{1+\cos(\theta+\gamma)}{2}.$$

Conjugation by $U$ and by $e^{i\gamma}U$ gives the same isolated channel because the scalar cancels against its conjugate. After control, the scalar multiplies only one branch and becomes observable relative phase.

3. Choose the Boolean Ancilla

Apply a clean XOR oracle to $|x\rangle|+\rangle$ and to $|x\rangle|-\rangle$. Give a phase-sensitive test that rejects the wrong ancilla for a branch with $f(x)=1$.

Solution

Since $X|+\rangle=|+\rangle$,

$$O_f|x\rangle|+\rangle = |x\rangle|+\rangle$$

for both values of $f(x)$. Since $X|-\rangle=-|-\rangle$,

$$O_f|x\rangle|-\rangle = (-1)^{f(x)}|x\rangle|-\rangle.$$

If both function values occur, place the data in an equal superposition of one $f=0$ branch and one $f=1$ branch. If $f$ is constant, this interference test instead needs an additional coherent control that compares the query with an identity branch: the sign on a fixed target eigenstate is global without that reference. This does not prevent reading a constant Boolean value by using the XOR oracle on a computational-basis target. With the coherent reference, the minus ancilla produces relative phase $\pi$ when $f=1$, so an $X$-basis measurement swaps the constructive and destructive outcomes. The plus ancilla produces relative phase zero. Measuring only the target in the computational basis would not distinguish the intended phase pattern.

4. Clean a Dirty Predicate

Suppose

$$C_f|x\rangle|0^w\rangle|0\rangle = |x\rangle|g_x\rangle|f(x)\rangle.$$

Show that applying $Z$ to the output and then $C_f^\dagger$ produces a clean sign-phase oracle. Count the abstract operations and explain the interference failure if the inverse is omitted.

Solution

The output bit is a $Z$ eigenstate, so

$$(I\otimes I\otimes Z) |x\rangle|g_x\rangle|f(x)\rangle = (-1)^{f(x)} |x\rangle|g_x\rangle|f(x)\rangle.$$

Applying $C_f^\dagger$ restores both auxiliary registers:

$$C_f^\dagger(I\otimes I\otimes Z)C_f |x\rangle|0^w\rangle|0\rangle = (-1)^{f(x)}|x\rangle|0^w\rangle|0\rangle.$$

The sequence contains one forward evaluator, one $Z$, and one inverse evaluator. Without the inverse, tracing out the workspace multiplies a data coherence between $x$ and $x'$ by $\langle g_{x'}|g_x\rangle$ together with any output-bit overlap. Orthogonal garbage can remove that coherence completely.

5. Reverse a Character Convention

For $M=8$, $k=1$, and $a=3$, derive the eigenvalues of $T_3$ on $F_8^\dagger|1\rangle$ and on $F_8|1\rangle$. For a controlled shift, calculate $p_{X+}$ in each convention and show how the $Y$ expectation changes.

Solution

The negative-exponent character is $F_8^\dagger|1\rangle$, so the positive translation has eigenvalue

$$\lambda_- = e^{+2\pi i(1)(3)/8} = e^{+3\pi i/4}.$$

The positive-exponent character $F_8|1\rangle$ has the conjugate eigenvalue

$$\lambda_+ = e^{-2\pi i(1)(3)/8} = e^{-3\pi i/4}.$$

In both cases $\cos(\pm3\pi/4)=-1/\sqrt2$, hence

$$p_{X+} = \frac{1+\cos(\pm3\pi/4)}2 = \frac{2-\sqrt2}{4}.$$

The quadrature is sign-sensitive:

$$\langle Y\rangle_- = \sin(3\pi/4) = \frac1{\sqrt2}, \qquad \langle Y\rangle_+ = \sin(-3\pi/4) = -\frac1{\sqrt2}.$$

Thus $X$ alone cannot distinguish the two Fourier conventions, while a separate $Y$-basis ensemble can.

6. Diagnose a Non-Eigenstate Target

Apply controlled-$Z$ to

$$|+\rangle_C \left( \cos\alpha|0\rangle + \sin\alpha|1\rangle \right)_T.$$

Derive the selector overlap and reduced-control purity. Evaluate the purity at $\alpha=\pi/8$ and interpret the result.

Solution

Let

$$|\eta\rangle = \cos\alpha|0\rangle + \sin\alpha|1\rangle.$$

The two control branches carry $|\eta\rangle$ and $Z|\eta\rangle$, whose overlap is

$$z = \langle\eta|Z|\eta\rangle = \cos^2\alpha-\sin^2\alpha = \cos2\alpha.$$

Tracing out the target gives

$$\rho_C = \frac12 \begin{pmatrix} 1 & z\\ z & 1 \end{pmatrix}, \qquad \operatorname{Tr}\rho_C^2 = \frac{1+z^2}{2} = \frac{1+\cos^2(2\alpha)}2.$$

At $\alpha=\pi/8$, $z=1/\sqrt2$ and the purity is $3/4$. Because it is below one, the output is entangled: the target has not returned to a common ray, and no single pure control phase describes the result.

7. An unchanged mixed target

Prepare $|+\rangle\langle+|\otimes I/2$ and apply controlled-$Z$. Find both reduced output states. Is the output entangled? Does the unchanged target imply perfect phase transfer?

Solution

Resolving the target mixture into $Z$ eigenstates gives

$$\rho_{CT}'=\frac12|+\rangle\langle+|\otimes|0\rangle\langle0| +\frac12|-\rangle\langle-|\otimes|1\rangle\langle1|.$$

This is explicitly a mixture of product states, so it is separable. Both reduced states are $I/2$: the target marginal is unchanged, but the control has lost its initial coherence. The two target components contribute opposite control phases. Thus marginal target return is weaker than returning a pure common target while preserving the desired selector phase.

8. Prepare and read a four-qubit character

Use a $|+\rangle$ control, $M=16$, $k=5$, $a=7$ and the negative-exponent character. Derive the phase from one controlled-$T_7$ query and calculate the separate $X$ and $Y$ probabilities. Give a direct product-state preparation of the known character and compare it with preparing $|5\rangle$ followed by an exact inverse QFT.

Solution

The phase reduces to

$$\theta=\frac{2\pi(5)(7)}{16} =\frac{35\pi}{8}=\frac{3\pi}{8}\pmod{2\pi}.$$

The final state is $(|0\rangle+e^{3\pi i/8}|1\rangle)|\chi_5^-\rangle/\sqrt2$, and hence

$$p_{X+}=\frac{2+\sqrt{2-\sqrt2}}4\simeq0.691342, \qquad p_{Y+}=\frac{2+\sqrt{2+\sqrt2}}4\simeq0.961940.$$

For direct preparation, apply $H$ to each of $a_3,a_2,a_1,a_0$, followed respectively by $P(\pi)$, $P(-\pi/2)$, $P(3\pi/4)$ and $P(-5\pi/8)$. This uses four Hadamards and four phase gates in a continuous one-qubit alphabet, with no entangling gates. Phase synthesis into a finite alphabet is a separate step.

Alternatively, prepare $|5\rangle=|0101\rangle$ with two $X$ gates and apply the inverse QFT: four Hadamards, six controlled inverse phases and two SWAPs in the construction of the previous page. This is correct but uses unnecessary two-qubit gates for a known basis label. Both preparations require the same five-qubit control-plus-target system. Neither count includes implementing the controlled shift or taking repeated measurements.

References

• E. Bernstein and U. Vazirani, “Quantum Complexity Theory”, SIAM Journal on Computing 26(5), 1411–1473 (1997).
• R. Cleve, A. Ekert, C. Macchiavello, and M. Mosca, “Quantum Algorithms Revisited”, Proceedings of the Royal Society A 454, 339–354 (1998).
• D. Deutsch, “Quantum Theory, the Church–Turing Principle and the Universal Quantum Computer”, Proceedings of the Royal Society A 400, 97–117 (1985).
• D. Deutsch and R. Jozsa, “Rapid Solution of Problems by Quantum Computation”, Proceedings of the Royal Society A 439, 553–558 (1992).
• T. G. Draper, “Addition on a Quantum Computer”, arXiv:quant-ph/0008033 (2000).
• L. K. Grover, “A Fast Quantum Mechanical Algorithm for Database Search”, in Proceedings of the 28th Annual ACM Symposium on Theory of Computing, 212–219 (1996).
• A. Yu. Kitaev, “Quantum Measurements and the Abelian Stabilizer Problem”, arXiv:quant-ph/9511026 (1995).
• N. D. Mermin, Quantum Computer Science: An Introduction, Cambridge University Press (2007).
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary edition, Cambridge University Press (2010).
• A. Ruiz-Perez and J. C. Garcia-Escartin, “Quantum Arithmetic with the Quantum Fourier Transform”, Quantum Information Processing 16, 152 (2017).
• P. W. Shor, “Polynomial-Time Algorithms for Prime Factorization and Discrete Logarithms on a Quantum Computer”, SIAM Journal on Computing 26(5), 1484–1509 (1997).
• V. Vedral, A. Barenco, and A. Ekert, “Quantum Networks for Elementary Arithmetic Operations”, Physical Review A 54, 147–153 (1996).