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For the phase gate $R_a=\mathrm{diag}(1,e^{2\pi ia/2^n})$, apply powers $R_a^{2^{n-1}},\dots ,R_a$ to a uniform quantum superposition on $n$ qubits. Binary positional weights multiply the phases into $e^{2\pi iax/2^n}$ on basis vector $|x⟩$. The inverse quantum Fourier transform returns $|a⟩$ exactly, by orthogonality of roots of unity. Preparing the powers by repeated applications uses $2^n-1$ copies of $R_a$; this does not claim a number of gate calls polynomial in $n$. If $a$ is odd, its powers generate the entire set of dyadic phase gates because $a$ has a modular inverse modulo $2^n$.

The Hadamard gate is the unitary operatorwhere $I$ is the identity operator. It maps the computational basis to the Hadamard basis:

These outputs have equal computational-basis measurement probabilities, but their relative quantum phase differs. Applying another Hadamard gate sends them to different definite outputs. Calling this gate a classical coin toss would discard precisely the information that makes this possible. ^[ref].

Starting from $|0⟩$, compare two quantum circuits, reading matrix products from right to left:Here $Z$ is the Pauli Z gate. After the first Hadamard gate, the probability amplitudes are $(1/\sqrt{2},1/\sqrt{2})$. Inserting $Z$ changes them to $(1/\sqrt{2},-1/\sqrt{2})$ without changing their squared magnitudes. At the final gate, the contributions to output zero cancel and those to output one add. Without $Z$, cancellation occurs at output one instead. This is controlled interference, with the quantum phase determining which answer emerges.

The same calculation works for any phase gate $P_{\varphi }=\mathrm{diag}(1,e^{i\varphi })$:For example, $\varphi =\pi /3$ gives $P(1)=1/4$, while $\varphi =\pi /2$ gives $P(1)=1/2$. The output-one amplitude at $\varphi =\pi /2$ is $(1-i)/2$: its two complex contributions add before the Born rule is applied. A phase gate alone leaves the populations unchanged; recombination converts its relative quantum phase into observable probabilities.

If quantum decoherence completely removes the computational basis coherence between the gates, the intermediate density matrix becomes $I/2$. Both the phase gate and final Hadamard gate leave $I/2$ unchanged, so $P(1)=1/2$ for every $\varphi$. The disappearing fringe explains why preserving quantum coherence matters: it enables the gate sequence to carry phase information into the final measurement.

For the printed equatorial-basis convention,It is the Hadamard gate after a phase gate. The identities $J(\alpha )Z=XJ(\alpha )$ and $J(\alpha )X=e^{i\alpha }ZJ(-\alpha )$ follow by multiplying their matrices. For $\alpha =0,\pi /2$ these are Clifford gates. In particular $J(-\pi /2)=XJ(\pi /2)$, permitting a sign change of the measurement angle to be absorbed into a Pauli frame.

The J gate is $J(\alpha )=HP(\alpha )$, where $P(\alpha )=\mathrm{diag}(1,e^{i\alpha })$: the phase gate acts first and the Hadamard gate acts second. Unitary invariance of the operator norm reduces its angle error to the single nonzero diagonal difference of the phase gate. Its norm is $|e^{i\delta }-1|=2|\sin (\delta /2)|\le |\delta |$. The quantum circuit gate-error telescoping bound then makes an angle tolerance of $ϵ/k$ sufficient for $k$ such imperfect gates and exact other gates.

Use CNOT gates to encode one logical qubit and $r-1$ zero ancillas into the repetition subspace spanned by $|0^r⟩,|1^r⟩$. Apply the phase gate in parallel to all $r$ physical qubits, then undo the encoding. The logical phase is multiplied by $r$ in one oracle-time layer. This copies computational-basis labels coherently rather than cloning an arbitrary quantum state. Blocks of sizes $1,2,4,\dots ,2^{n-1}$ prepare a dyadic Fourier phase state using $2^n-1$ parallel oracle applications.

Take the Hadamard gate to be $H=2^{-1/2}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$ and the phase gate to be $P(\varphi )=\mathrm{diag}(1,e^{i\varphi })$. This fixes the sign convention for the interference phase.

For the coherent quantum circuit, the successive quantum states areThe Born rule therefore gives the concise answer

With the stated quantum decoherence after the phase gate, the joint pure state before the last Hadamard gate isAfter that Hadamard gate, its environment coefficient at output zero is $(|e_0⟩+e^{i\varphi }|e_1⟩)/2$. Squaring its norm sums over every unobserved environment outcome, soThe Cauchy-Schwarz inequality gives $0\le v\le 1$. In this single-qubit interference with conditional environment states, $v$ is the interference visibility: orthogonal environment states erase the fringe, while a unit-modulus overlap shifts its phase without reducing visibility. The plus sign in $\varphi +\alpha$ follows from the specified overlap $⟨e_0|e_1⟩$, rather than its complex conjugate.

If the same quantum decoherence occurs before the phase gate, the joint pure state is first $(|0⟩|e_0⟩+|1⟩|e_1⟩)/\sqrt{2}$. The phase gate then produces exactly the same state used above. Equivalently, the controlled environment interaction commutes with the diagonal phase gate. The probability is unchanged.

For the Deutsch algorithm, use the reversible Boolean quantum oracleThe algebraic oracle map printed in the source has its arguments misplaced: taken literally, it sends two distinct answer inputs to the same output for constant $f$, so it cannot be a unitary operator. The corrected map above agrees with the diagram's input-control, answer-target arrangement. Put the answer qubit in $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Since the Pauli X gate satisfies $X|-⟩=-|-⟩$, quantum phase kickback givesWrite $b=f(0)\oplus f(1)$. Up to the irrelevant global phase $(-1{)}^{f(0)}$, the input qubit acquires relative phase $(-1{)}^b$, so its ideal final computational-basis measurement returns $b$: zero for a constant function and one for a balanced function.

Let the specified quantum decoherence occur in the interference portion between the two Hadamard gates. It commutes with the oracle's effective diagonal phase on the input, so the same analysis applies whether it occurs just before or just after the query. The two likelihoods areThus the diagram's usual zero-for-constant decision rule hasThis is the reliability for either promised class separately, with no prior needed. For equally likely classes, if $v\cos \alpha$ is known to be negative, reversing the interpretation gives $(1+|v\cos \alpha |)/2$. A known environment phase can instead be compensated by the phase gate $P(-\alpha )$ before the final Hadamard gate, giving success probability $(1+v)/2$. In particular, $v=0$ gives no information and $v=1$ permits certainty after phase compensation. These refinements distinguish a shifted fringe from lost quantum coherence; with the unmodified diagram a phase shift alone can reverse the apparent answer. Quantum decoherence after the final Hadamard gate leaves its computational populations unchanged, so that different placement would not spoil the measured answer.

Write both the input and output integer labels in most-significant-bit-first order, $x={\sum }_{j=1}^n{x}_j{2}^{n-j}$ and $y={\sum }_{j=1}^n{y}_j{2}^{n-j}$. Factoring the Fourier phase over the bits of $y$ gives the product decomposition of a Fourier phase state:Here $0.x_l\cdots x_n$ denotes the binary fraction ${\sum }_{r=l}^n{x}_r/2^{r-l+1}$; the integer part of $x/2^j$ does not affect the phase.

Build a dyadic quantum Fourier transform circuit as follows. Process wires $j=1,\dots ,n$ in that order. First apply a Hadamard gate to wire $j$. Then, for each $l=j+1,\dots ,n$, apply the controlled phase gate $R_{l-j+1}$ with control wire $l$ and target wire $j$. On a computational-basis state, every unprocessed control is still the bit $x_l$, so wire $j$ becomesOnce a wire has been processed it is not touched by later stages. Reversing the order of the output wires gives exactly the Fourier product above. Agreement on every computational-basis state proves agreement as linear operators on all superpositions, even though the intermediate operation on a general input can create entanglement.

There are $n$ Hadamard gates and $n(n-1)/2$ controlled phase gates. Output reversal may be treated as wire relabelling. If it must instead be implemented physically while using only the requested gate family, use $\lfloor n/2\rfloor$ swap operators. Each swap is three controlled-NOT gates, andThus the swaps use only Hadamard gates and Controlled-Z gates, which are controlled $R_1$ gates. An exact physical network haselementary gates from the stated family. This implements the positive-exponent Fourier transform with the correct output order in quadratic size. Omitting the output reversal without relabelling the wires would produce the reversed-order transform instead.

For the perturbation bound, the operator norm in the source is equivalently $\Vert A\Vert ={\sup }_{\Vert \psi \Vert =1}\Vert A\psi \Vert$. It satisfies the triangle inequality, is unchanged by left or right multiplication by a unitary operator, and a unitary operator has norm one. The exact telescoping identity isIt follows by replacing one factor at a time; alternatively all intermediate products cancel in the sum. Taking the operator norm and using unitary invariance gives the quantum circuit gate-error telescoping boundThe original PDF assumes strict $\Vert U_j-V_j\Vert <ϵ$, so its strict conclusion is justified. If the hypothesis were weakened to a non-strict inequality, the conclusion would likewise be $\le mϵ$. The TeX aid misreads this hypothesis; the PDF resolves it.

Apply the non-strict version to each approximate unitary gate in the Fourier network. Extending a gate by the identity on other wires leaves its operator norm error unchanged. Including the physical swap decompositions, all gates are of the required type and can obey the stated accuracy. HenceWith exact output relabelling rather than physical swaps the numerator is just $n(n+1)/2$. The proof retains gate phases and makes no assumption of cancellation between coherent errors.

For every normalized input the output-state norm error is at most this same bound. If $M$ is any positive operator satisfying $0\le M\le I$, the change in its measurement in quantum mechanics outcome probability is at most $2\delta$, where $\delta =\Vert U_n-{\mathrm{QFT}}_{2^n}\Vert$: expand the difference of the two expectations into two terms, each bounded by $\delta$ using the Cauchy-Schwarz inequality. Thus inverse-polynomial gate accuracy gives an inverse-polynomial total error, rather than requiring exponential precision in the number of qubits. For the phase gates this corresponds to specifying angles with $O(\log n)$ bits of precision, since $|e^{i\theta }-e^{i{\theta }^{\prime }}|\le |\theta -{\theta }^{\prime }|$. This is a robustness statement for unitary gate approximations; it does not by itself correct quantum decoherence, correlated stochastic faults, or limitations of hardware connectivity.

The PDF circuit has its upper wire controlling both CNOT gates. Its successive states areThe first arrow is the upper Hadamard gate; the middle phase arises because both ones receive a phase. The second CNOT disentangles and resets the lower wire. Hence the output isThis implements doubled logical phase using two parallel phase gates, rather than two serial applications on one qubit.

Choose the phase gateIt is unitary. Acting on both qubits leaves $|00⟩$ unchanged and multiplies $|11⟩$ by $i^2=-1$, so $S\otimes S$ sends ${\Psi }_{+}$ to ${\Psi }_{-}$ with exactly the requested phase.

Using the Hadamard gate $H$, chooseIt is unitary and symmetric, andFor any matrix $U$, applying $U\otimes U$ to $(|00⟩+|11⟩)/\sqrt{2}$ gives the coefficient matrix $U{U}^T/\sqrt{2}$. Here $U_2{U}_2^T=U_2^2=X$, henceBoth displayed transformations are exact, not merely equal up to global phase. These common local unitaries permute the triplet Bell projectors while preserving the singlet projector by the preceding determinant identity.

Using the printed measurement-vector convention, the unitary operator isThus it is a phase gate followed by a Hadamard gate, with angle sign opposite to the parameter in the catalogued J gate. Its rows are the conjugate transposes of the orthonormal measurement vectors, so it is unitary. Right multiplication by the Pauli Z gate changes the sign of the second column, which is exactly the effect of left multiplication by the Pauli X gate, exchanging the two rows. Explicitly,Similarly, right multiplication by the Pauli X gate exchanges columns, givingTherefore the exact identities, retaining the global phase in the second one, areThese supply Pauli-frame propagation along a measurement wire.

Put $N=2^n$ and label the binary digits of $x$ by $x_{n-1},\dots ,x_0$, with $x={\sum }_{j=0}^{n-1}{2}^j{x}_j$. The Walsh-Hadamard transform prepares the uniform quantum superposition $N^{-1/2}{\sum }_x|x⟩$. Applying the phase gate $R_a^{2^j}$ to the qubit for digit $x_j$ multiplies its basis amplitude by $\exp (2\pi ia{2}^j{x}_j/N)$. Taking their product givesWith the most significant qubit written first, this provesEach power is implemented by repeated uses of the supplied gate. The total is ${\sum }_{j=0}^{n-1}{2}^j=N-1$ uses. This realizes dyadic phase-gate identification without needing an unknown controlled gate or asserting a gate-call cost polynomial in $n$.

Write the normalized qubit as $|\varphi ⟩=c_0|0⟩+c_1|1⟩$, with $|c_0{|}^2+|c_1{|}^2=1$. Only the $|1⟩$ amplitude changes between the two phase gates, soSince $|e^{i\delta }-1|=2|\sin (\delta /2)|\le |\delta |$,The phase-gate discretization error is uniform over all input states; its exact operator norm is $2|\sin (\delta /2)|$. To choose an approximating gate, round $N\theta /(2\pi )$ to the nearest integer and reduce it modulo $N$. Taking the shortest circular angle difference gives $|\delta |\le \pi /N$, and hence a uniform state-vector error at most $\pi /2^n$. The circular choice handles the identified endpoints $0$ and $2\pi$ correctly.