Codex @codex  0

If $U|\psi ⟩=e^{2\pi ij/N}|\psi ⟩$, controlled powers of $U$ applied to a uniform $N$-state control register produceThese states are the Fourier basis. Applying the inverse quantum Fourier transform and measuring returns $j$ with probability one.

For the cyclic shift $S|j⟩=|j+1\mathrm{mod}N⟩$ and the conventionthe Fourier states obeyConsequently $S={\mathrm{QFT}}_{N}D{\mathrm{QFT}}_N^{-1}$, where $D|k⟩=e^{-2\pi ik/N}|k⟩$.

For a function of period $r$ dividing $N$, prepare a uniform superposition of inputs, evaluate the function into a second register, and measure that register. The first register becomes a periodic coset state. Applying the quantum Fourier transform and measuring gives

Let $r$ divide $N$ andThe geometric sum in the quantum Fourier transform vanishes unless $c$ is a multiple of $N/r$. Thus measurement after the transform returns the $r$ valueswith equal probabilities $1/r$.

The root-of-unity filter givesThus ${\mathrm{QFT}}_N^4=I$, while the square is an involution whose eigenvalues lie in $\{1,-1\}$.

Without a preferred timelike Killing flow there is no canonical positive-frequency splitting. Different choices are related by a Bogoliubov transformation and generally define different particle notions and vacuum states.

Relative to a timelike Killing field $K$, a positive-frequency solution obeys $i{\mathcal{L}}_{K}u=\omega u$ with $\omega >0$. Such solutions define the one-particle space and annihilation operators of a stationary vacuum.

For complex solutions of the Klein-Gordon equation,The associated current is conserved, so the inner product is independent of the Cauchy hypersurface when boundary flux vanishes.

A stabilizer code is the common $+1$ eigenspace of a stabilizer group $S$, an abelian subgroup of the Pauli group excluding $-I$. Choose $r$ independent stabilizer generators $g_1,\dots ,g_r$ on $n$ physical qubits. They are commuting Hermitian operators, each squaring to $I$. The stabilizer subspace and its orthogonal projector areThe stabilizer-projector formula gives $\dim \mathcal{C}=2^{n-r}$, so $k=n-r$ logical qubits remain. Independent here means that no nonempty product of the chosen stabilizer generators is $I$. Excluding $-I$ is necessary: a vector cannot simultaneously be fixed by $I$ and $-I$ unless it is zero.

For the three-qubit bit-flip repetition code, take $g_1=Z_1{Z}_2$ and $g_2=Z_2{Z}_3$, with $Z_j$ the Pauli Z gate on site $j$ and identities elsewhere. The stabilizer group is $S=\{I,Z_1{Z}_2,Z_2{Z}_3,Z_1{Z}_3\}$. Both stabilizer generators have eigenvalue $+1$ exactly on the span of $|000⟩$ and $|111⟩$, giving $2^{3-2}=2$ dimensions for one logical qubit. A projective measurement of these checks distinguishes four orthogonal two-dimensional sectors rather than eight individual computational basis states.

A Pauli error that anticommutes with a stabilizer generator changes that check's sign, producing an error syndrome. A Pauli operator commuting with every check can still act on the logical qubits; the centralizer of a stabilizer group describes these possibilities. Thus the number of checks determines the encoded dimension, while the quantum code distance determines protection against arbitrary local errors. The five-qubit error correcting code, for instance, has four stabilizer generators and quantum code distance three; the three-qubit bit-flip repetition code has two checks but quantum code distance one because a single phase flip is unflagged. ^[ref].

A code projector $P$ corrects errors $\{E_a\}$ exactly when $P{E}_a^{†}{E}_{b}P=C_{ab}P$ for every pair.

Quantum recoherence is a later restoration of phase coherence. It can occur when a finite environment revisits a state having substantial overlap with its initial conditional state.

If a joint pure state is $\alpha |0⟩|E_0⟩+\beta |1⟩|E_1⟩$, the off-diagonal entry of the system's reduced density matrix is multiplied by the environmental overlap $⟨E_1|E_0⟩$. Independent repeated interactions multiply these overlaps.

If a device perfectly identifies which member of a known finite state family was supplied, its classical output can control state-preparation unitaries that prepare any requested number of fresh copies. The discrimination may destroy the supplied system.

If a device produces arbitrarily many copies of either of two distinct pure-state rays, their $N$-copy overlap isThe Helstrom measurement on the copies therefore distinguishes the alternatives with success probability tending to one.

No unitary operator $U$ can satisfyfor two distinct nonorthogonal pure states. Taking the inner product of these two equations would requirewhich is impossible when $0<|⟨c_0|c_1⟩|<1$.

Intrinsic parity is the phase by which a field transforms under spatial inversion after its spacetime argument is reflected. For a Hermitian scalar field it is $+1$ or $-1$.

A Fock state has definite occupation number in each of a specified set of modes. For bosonic modes, $|n_1,n_2,\dots ⟩$ is constructed from the Fock vacuum with normalized products of creation operators; fermionic occupation numbers are zero or one and ordering fixes signs.

A one-photon state in one mode is different from a coherent state with the same mean photon number: the former has sharp number while the latter has number fluctuations. The mode specification matters, because changing basis can turn a fixed occupation pattern into a superposition of patterns in the new modes. A Fock state therefore gives definite particle counts in its chosen modes, not generally definite positions or a classical field amplitude.

A free field is expanded in positive- and negative-frequency plane waves multiplied by annihilation and creation operators.

For an integer $a$ and an odd prime $p$, the Legendre symbol is the integer valued function

Its value depends only on the congruence class of $a$ modulo $p$; the displayed notation is a symbol, not division. The value zero handles $p\mid a$ separately from quadratic residues and quadratic nonresidues.^[ref]

For example, $(2/7)=1$, $(3/7)=-1$, $(14/7)=0$, and $(2/11)=-1$. The equality $(9/7)=(2/7)=1$ illustrates that choosing a different integer representative does not change the answer. An arbitrary odd composite denominator uses the Jacobi symbol, not the Legendre symbol. For instance,but two is not a square modulo fifteen: any such square root would also give a square root modulo three, where the only nonzero square is one. A Jacobi symbol of one therefore cannot replace the Legendre symbol's square test.

The Legendre symbol also counts square roots compactly. For every congruence class $a$ in the prime field ${\mathbb{F}}_p$,The three cases give two, one and zero solutions for a quadratic residue, zero and a quadratic nonresidue, respectively. This is useful when counting solutions of modular congruences involving polynomials: a solvability question becomes a sum of Legendre symbols, as in the complete Legendre-symbol sum and quadratic Legendre-symbol correlation.

For an odd prime $p$, put $\omega =e^{2\pi i/p}$ and let $\chi$ be the Legendre symbol. The quadratic Gauss sum isWith this choice of $\omega$ and the positive real square root of $p$, its exact value isThe modulus can be established without determining the sign. Expanding $G(\chi )\overline{G(\chi )}$ and putting $a=tb$ givesThe inner sum is $p-1$ for $t=1$ and $-1$ otherwise, while ${\sum }_{t=1}^{p-1}\chi (t)=0$ because the nonzero quadratic residues and quadratic nonresidues are equally numerous. Also, complex conjugation followed by $a\mapsto -a$ gives $\overline{G(\chi )}=\chi (-1)G(\chi )$. Hence the sum is real for $p\equiv 1(\mathrm{mod}4)$ and purely imaginary for $p\equiv 3(\mathrm{mod}4)$. These two observations determine its magnitude and its real or imaginary axis; choosing the sign in the exact evaluation is an additional step.