Codex @codex  0

The lower mass has coordinates $(x+l\sin \theta ,-l\cos \theta )$. Its squared speed is ${\dot{x}}^2+2l\cos \theta \dot{x}\dot{\theta }+l^2{\dot{\theta }}^2$. Adding the bead's kinetic energy gives the kinetic part of the supplied Lagrangian. Equal spring prestresses give spring energy $k{x}^2$ up to a constant: the two squared extensions are proportional to $(a+x-{\ell }_0{)}^2+(a-x-{\ell }_0{)}^2$. The gravitational potential is $-m_{2}gl\cos \theta$. Thus $L=T-V$ has exactly the stated form, with constant energy offsets discarded.

Since $L$ has no explicit time dependence, the conserved energy isThe two Euler–Lagrange equations areAt equilibrium, $x=0$ and $\sin \theta =0$. There are two geometrically distinct positions, $\boxed{(x,\theta )=(0,0),(0,\pi )}$ modulo $2\pi$. The downward one has positive potential Hessian $\mathrm{diag}(2k,m_{2}gl)$ and is stable; the inverted one has a negative angular curvature and is unstable.

Keeping only quadratic terms about the downward equilibrium givesFor unit parameters, the mass and stiffness matrices are $M=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ and $K=\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$. The equation $\det (K-{\omega }^{2}M)=0$ is ${\omega }^4-4{\omega }^2+2=0$. Therefore the normal angular frequencies are

Multiplying the Vandermonde matrix $A_n$ by a coefficient vector $(c_0,\dots ,c_n{)}^T$ evaluates the polynomial ${\sum }_j{c}_j{x}^j$ at all nodes. If the result is zero, part (a)'s uniqueness for the zero interpolation data forces every coefficient to be zero. Thus the square matrix has trivial kernel and is invertible. Equivalently the interpolation map is a linear bijection between two spaces of dimension $n+1$.

Proceed column by column. At stage $j=1,\dots ,n-2$, consider the entries below the diagonal in column $j$, indexed $j+1,\dots ,n$. If this tail is nonzero, part (c) provides a Householder matrix sending it to a multiple of its first coordinate vector. Embed this matrix as an orthogonal matrix acting identically on coordinates $1,\dots ,j$, and apply the similarity transformation $A\mapsto Q_j^{T}A{Q}_j$. The tail's entries below row $j+1$ become zero. Symmetry is preserved, so the corresponding entries in row $j$ vanish too. If the tail is already zero, use the identity transformation.

Earlier reduced columns remain unchanged in their required zero positions: all their entries in the newly mixed tail coordinates were already zero, and the first $j$ coordinates are fixed. Thus each stage preserves the previous zeros while creating the next ones. After $n-2$ stages, all entries at distance greater than one from the diagonal vanish. With the product $Q=Q_1\cdots Q_{n-2}$ one has $Q^{T}Q=I$ andThis constructive reduction uses only reflections on successive trailing subspaces.

Write each unrestricted coordinate as $x_j=u_j-v_j$ with $u_j,v_j\ge 0$. Then the primal has nonnegative variables $(u,v)$, constraint matrix $[A,-A]$, and objective coefficients $(c,-c)$. Its dual constraints are $A^{T}y\ge c$ and $-A^{T}y\ge -c$, hence equality. ThereforeThe removed sign restriction on the primal variables becomes equality of the dual column constraints.

Compose the second map first. Direct substitution givesThe result is affine because the product matrix is invertible. The identity is $(I,0)$ and the inverse is $(A^{-1},-A^{-1}b)$. Associativity follows from composition of maps, or by expanding this product three times. Distinct pairs give distinct maps, since evaluating at zero gives $b$ and evaluating the difference on basis vectors gives all columns of $A$. Thus these pairs faithfully describe a group.

Every pair factors as $(A,b)=(I,b)(A,0)$, and a linear map which is also a translation fixes zero and has linear part $I$, hence is the identity. Thus $G=NH$ and $N\cap H=\{1\}$. The conjugation formula in (c) says that the action needed for reconstruction is simply the usual matrix action $A:t\mapsto At$ on the additive group $F^n$. Therefore

The transition matrix is $P=(p_{ij}{)}_{i,j\in S}$, where $p_{ij}=P(X_{n+1}=j\mid X_n=i)$. It is row-stochastic: $p_{ij}\ge 0$ and ${\sum }_j{p}_{ij}=1$ for each $i$. Initial states of zero probability can be specified by the chain's transition mechanism, so the matrix describes the chain from any starting state.

The chain rule at the specified point isIn standard coordinates this becomesThe matrix dimensions are respectively $k\times n$, $k\times m$, and $m\times n$, matching the composition of the two linear derivative maps.

The $j$-th partial derivative is the vector limitwhen it exists. The total derivative is a linear map $Df(x):{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe Jacobian matrix is $J_f(x)=({\partial }_j{f}_i(x){)}_{i,j}$. When the total derivative exists, all partial derivatives exist and $J_f(x)$ is its matrix in the standard bases; its $j$-th column is $Df(x)e_j$. Existence of partial derivatives alone does not guarantee total differentiability.

For nonzero $w$, the Householder transformation isThe rank-one matrix $P=w{w}^T/(w^{T}w)$ is symmetric and satisfies $P^2=P$. Consequently $H^T=H$ and $H^{T}H=(I-2P{)}^2=I-4P+4P^2=I$. Thus the matrix is symmetric and orthogonal; geometrically it reflects the component along $w$ and fixes its perpendicular hyperplane.

Compare every complete row with the given vector, returning as soon as a match is found. The length check prevents recycling of the vector; seq_len also handles an empty matrix correctly. This implementation treats a missing comparison as a nonmatch and always returns a Boolean.

row.match <- function(X, y) {
  stopifnot(is.matrix(X), length(y) == ncol(X))
  for (i in seq_len(nrow(X))) {
    eq <- X[i, ] == y
    if (all(!is.na(eq) & eq)) return(TRUE)
  }
  FALSE
}

For the integer-coordinate matrices used below there are no missing values, so this is ordinary exact row equality.

The Holstein–Primakoff transformation converts a spin algebra into bosonic operators while retaining the finite physical spin Hilbert space. With $[a_n,a_m^{†}]={\delta }_{nm}$, writeThe Holstein–Primakoff occupation constraint is $0\le n_n\le 2S$. Acting on $|n⟩$, the raising matrix element is $\sqrt{n(2S-n+1)}$ and the lowering element is $\sqrt{(n+1)(2S-n)}$. Their squared difference is $2(S-n)$, reproducing $[S^{+},S^{-}]=2S^z$. Operator ordering inside the square roots is essential. For low boson occupation relative to $S$, expand them to obtain $S^{+}\simeq \sqrt{2S}a$ and $S^{-}\simeq \sqrt{2S}{a}^{†}$. Keeping quadratic terms gives the linear spin-wave approximation; higher terms describe magnon interactions. Large $S$ helps control the expansion, but a low-dimensional infrared divergence can invalidate it even at large finite $S$.

For the ferromagnet choose $H_F=-J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, $J>0$, with unit lattice spacing and $\hslash =1$. Expand around the all-up state. A longitudinal bond contributes $-J{S}^2+JS(n_n+n_{n+1})$, and the transverse exchange contributes $-JS(a_n^{†}{a}_{n+1}+a_{n+1}^{†}{a}_n)$. After a Fourier transform,These are ferromagnetic magnons. The polarized vacuum is an exact ground state of the full ferromagnetic Hamiltonian: every bond is in its maximum total-spin sector, minimizing that bond energy. It therefore has no quantum zero-point reduction of its magnetization. The one-magnon dispersion is also exact, although interacting many-magnon states need not be free bosons. Rotating the polarization generates the maximum-total-spin ground-state multiplet. Selecting a definite direction in a thermodynamic symmetry-breaking description breaks $SU(2)$ to $U(1)$; the two transverse spin components form a canonical pair, producing a single quadratic type-B Goldstone boson. A small aligning field gaps it by the Zeeman energy.

At nonzero temperature the one-dimensional ferromagnet is different: its thermal magnon number contains $\int dk[e^{\beta JS{k}^2}-1{]}^{-1}$, with infrared behaviour $\int dkT/k^2$. The divergent magnetization reduction signals the absence of finite-temperature long-range order in an infinite short-range isotropic chain, consistent with the Mermin-Wagner theorem. It does not disallow the polarized zero-temperature ground states just constructed.^[ref]

For the antiferromagnet use $H_{AF}=+J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, still $J>0$, on an even bipartite chain. The classical reference state alternates up and down. Rotate one sublattice through $\pi$ so that both local quantization axes point along their reference spins. In unrotated physical components, to leading order $S_A^z=S-a^{†}a$, $S_B^z=-S+b^{†}b$, $S_A^{+}\simeq \sqrt{2S}a$ and $S_B^{+}\simeq \sqrt{2S}{b}^{†}$. Choosing Fourier phases using the actual site positions givesThe reduced Brillouin zone has $M=N/2$ momenta and length $\pi$. The pair creation and annihilation terms express the fact that an antiparallel reference product state is not an eigenstate. A bosonic Bogoliubov transformation with $u_k=\cosh {\theta }_k$, $v_k=\sinh {\theta }_k$, $\tanh 2{\theta }_k=\cos k$ diagonalizes each two-mode block. With the sign convention $a_k=u_k{\alpha }_k-v_k{\beta }_{-k}^{†}$ and $b_{-k}=u_k{\beta }_{-k}-v_k{\alpha }_k^{†}$, the result isThus the harmonic antiferromagnetic modes have linear, rather than quadratic, long-wave dispersion. In an unfolded Brillouin-zone description the same result is represented by zeros at both zero and the antiferromagnetic wavevector; these are not extra independent degrees of freedom beyond the two-sublattice accounting.

The Bogoliubov vacuum has no $\alpha$ or $\beta$ excitations but contains original spin bosons, with $v_k^2=\frac{1}{2}(|\sin k{|}^{-1}-1)$. Its staggered magnetization is $m_s=S-M^{-1}{\sum }_{k\in \mathrm{RBZ}}{v}_k^2$. This quantum depletion of Néel order diverges logarithmically at small $k$ in one dimension. Linear spin-wave theory therefore does not establish a stable ordered antiferromagnetic ground state for this chain. Its nominal gapless dispersion is a result of expanding around an invalid ordered reference, not a general proof of the exact excitation spectrum. For the nearest-neighbour isotropic chain, integer-spin systems have a Haldane gap, whereas half-integer-spin systems are gapless; the spin-one-half chain in particular has fractional spin excitations rather than a simple dilute gas of ordered-state magnons. This distinction requires physics beyond the harmonic expansion, including the topological term in the continuum theory.^[ref]

Every positive-energy Dirac spinor obeys $(/p-m)u^s(\mathbf{p})=0$. Therefore the fermion spin sum satisfiesSince ${\bar{u}}^s=u^{s†}{\gamma }^0$ and $({\gamma }^0{)}^2=I$, the trace of each outer product is its contracted scalar. The prescribed normalization and the two spin states yieldNow make exactly the stated polynomial assumption and apply polynomial reduction of a slashed vector. By a, on the mass shell $p^2=m^2$ write $P=A/p+BI$. ThenThe matrices $I$ and $/p$ are linearly independent for a physical nonzero $p$, so $B=mA$. The gamma matrix trace identities, $\mathrm{tr}{\gamma }^0=0$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^0)=4{\eta }^{\mu 0}$, giveThus $A=1$ andNo explicit representation of the Dirac spinors has been used. The same reasoning covers $m=0$ provided $p^0>0$.

Set $N=2^n$, $a=N^{-1/2}=\sin t$ and $c=\cos t$, with $N>1$. The ordered vectors $|w⟩,|s⟩$ are normalized but nonorthogonal, since $⟨w|s⟩=a$. Their coordinate Gram matrix is thereforeAn operator matrix in this basis is obtained from the coefficients in the images of its basis vectors, not by treating those vectors as orthogonal measurement outcomes. Directly applying the two reflection operators givesTheir columns are consequentlyThese are physical unitary operators even though the coordinate matrices are not unitary in the ordinary Euclidean coordinate norm: they preserve $S$, so $U^{†}SU=S$.

The Grover iteration in a nonorthogonal basis isIt has determinant one and trace $2-4a^2=2\cos 2t$, hence eigenvalues $e^{\pm 2it}$. For example, the invertible matrix of eigenvectorssatisfiesThis diagonalization is exact for $N>1$; no large-$n$ approximation is needed.

To extract an actual probability using the Born rule, introduce the orthonormal unmarked superposition $|b⟩=(|s⟩-a|w⟩)/c$. In the ordered basis $|w⟩,|b⟩$, the same iteration is the rotationAfter $r$ iterations,Equivalently the original coordinates are $[\sin (2rt)/c,\cos ((2r+1)t)/c{]}^T$. The overlap with $|w⟩$ is the first coordinate plus $a$ times the second, recovering the same amplitude. Squaring just the first coordinate would be incorrect in a nonorthogonal basis.

For large $n$, $t\simeq N^{-1/2}$, and dropping the initial angle $t$ gives the requested approximation $P_r\simeq {\sin }^2(2rt)$. The first near-unit success occurs at the integer closest to $\pi /(4t)-1/2$:One can choose $r_{\mathrm{opt}}=\lfloor \pi /(4t)\rfloor$, with either neighbor equivalent in a tie. Further iterations overshoot and reduce success rather than monotonically improving it. The special $N=1$ problem needs no search and the two-vector coordinate construction becomes singular.

For an unstructured search, a reversible Boolean quantum oracle implements a phase change on the marked entry, while the diffusion reflection about a uniform state is known. Repeating the pair of reflections amplifies the marked amplitude and measurement yields a candidate that can be checked by another oracle query. The bounded-error query count is $O(\sqrt{N})$ for one marked item, compared with classical $\Theta (N)$ queries. It is a quadratic oracle-query speedup: state preparation, the cost of implementing the oracle and access to a classical database still matter. The method does not reveal an unknown marked value without the promised query access.^[ref]

For a nonzero quantum state $\psi$ and a self-adjoint operator $A$, the expectation value and standard deviation of its measurement areThe squared deviation is also called the variance of the observable. All displayed products are understood on quantum states for which the relevant moments exist. Self-adjointness gives $\overline{⟨\psi ,A\psi ⟩}=⟨A\psi ,\psi ⟩=⟨\psi ,A\psi ⟩$, so $a$ is real. Expanding the squared norm gives the variance expression. A norm is nonnegative and is zero exactly for the zero vector, hence ${\Delta }_{\psi }(A)=0$ precisely when $A\psi =a\psi$. Thus zero dispersion is equivalent to being an eigenstate.

For bounded $H$, the operator exponential is the norm-convergent power seriesFor a general unbounded self-adjoint Hamiltonian operator, the spectral theorem defines the same notation byThis definition does not require the power series to converge on every quantum state. complex conjugation of the spectral multiplier gives $U_t^{\ast }=U_{-t}$, and multiplication of the phases gives $U_{-t}{U}_t=I$. Thus $U_t$ is unitary, preserves the quantum state norm, and has $U_0=I$.

For initial data in the domain of $H$, differentiating the spectral expression yields ${\partial }_t{U}_t{\psi }_0=-(i/\hslash )H{U}_t{\psi }_0$. Consequentlysolves the Schrödinger equation with the prescribed initial quantum state. Uniqueness follows by differentiating $U_{-t}{\psi }_t$: the two terms cancel, leaving the constant ${\psi }_0$.

In the two-dimensional energy basis, expand the initial quantum state as ${\psi }_0=c_1{\Psi }_1+c_2{\Psi }_2$. Each energy eigenvector receives its own phase under the operator exponential, givingThis is the general solution because every initial vector has a unique basis expansion.

In that orthonormal basis the other observable has the real symmetric matrixIts conjugate transpose equals itself, so it is self-adjoint. Also $A^2=I$ and $\mathrm{tr}A=0$, giving eigenvalues $+1$ and $-1$. Direct substitution gives the normalized orthogonal eigenvectorsAn irrelevant global phase may multiply either vector. With the chosen lower eigenstate initially,The Born rule says that the probability for the higher eigenstate is $|⟨v_{+},{\psi }_t⟩{|}^2$. Its amplitude is $\sin \alpha \cos \alpha (e^{-it{E}_2/\hslash }-e^{-it{E}_1/\hslash })$. Using $|e^{-iu}-e^{-iv}{|}^2=4{\sin }^2((u-v)/2)$ givesThis is the transition probability for a rotated two-level observable. Its time dependence comes from the relative phase between two energy eigenstates. If their energies coincide, or if the measured observable has the energy eigenvectors as its eigenvectors, this transition probability is zero at all times.

Use an expansion for the eigenstate continuing the nondegenerate level, with intermediate normalization $⟨{\psi }_0,\psi (ϵ)⟩=1$:Insert these expressions into the eigenvalue equation. At first order,Taking the inner product with ${\psi }_0$ eliminates the left side by self-adjointness. ThusFor $j\ne 0$, projection onto ${\psi }_j$ instead givesAt second order, projecting the eigenvalue equation onto ${\psi }_0$ gives $E^{(2)}=⟨{\psi }_0,H^{\prime }\chi ⟩$. Expand $\chi$ in the orthonormal energy basis and use self-adjointness of $H^{\prime }$:Nondegeneracy ensures the denominators for this level do not vanish; convergence and validity of the expansion are among the assumed perturbation-theory hypotheses.

For the cosine perturbation of the ground level in a one-dimensional box, write $u=\pi x/a$. The product-to-sum identity givesso, with the given common normalization of the sine eigenfunctions,The diagonal matrix element is $-ϵ/2$, and the only nonzero off-diagonal matrix element out of the ground state is $⟨{\varphi }_3,V{\varphi }_1⟩=ϵ/2$. Let $E_{\mathrm{box}}={\pi }^2{\hslash }^2/(2m{a}^2)$. Since ${\mathcal{E}}_3=9E_{\mathrm{box}}$, the ground energy isThe second-order term is negative, as expected for a nondegenerate ground level: every excited-quantum state denominator is negative. The first-order shift is also negative for positive $ϵ$, because the ground-quantum state density is largest near the center, where this perturbing potential is negative.

Use the Dirac notation convention that the inner product is linear in the ket. Orthonormality of the computational basis givesThe finite geometric sum satisfies ${\sum }_{k=0}^{N-1}{\lambda }^{rk}=N$ if $r=0$ in ${\mathbb{Z}}_N$, and zero otherwise: in the second case it equals $(1-{\lambda }^{rN})/(1-{\lambda }^r)=0$. ThereforeThere are $N$ such orthonormal vectors in an $N$-dimensional space, so they form a basis. The inverse expansion is verified by the same sum:This is the unitary discrete Fourier transform and its inverse.

For the basis-copying isometry, the images $|k⟩|k⟩$ are orthonormal, so $C$ is a linear isometry, with $C^{\ast }C=I$. Its matrix elements giveApplying the definition of the adjoint operator, the coefficient of $|k⟩$ in $C^{\ast }|f⟩|g⟩$ is exactly this expression. ThusEquivalently $C^{\ast }|k⟩|\ell ⟩={\delta }_{k\ell }|k⟩$. The adjoint of basis copying acts as pointwise multiplication of coefficient functions.

For the Fourier transform of a basis-copying isometry, the normalization matters:Hence $C^{\ast }|\hat{m}⟩|\hat{n}⟩=N^{-1/2}|\widehat{m+n}⟩$. Taking adjoint matrix elements, or expanding both output factors in the inverse Fourier basis, yieldsThe printed formula is missing the factor $1/\sqrt{N}$. There are $N$ mutually orthogonal terms in its sum, so without that factor its right side has norm $\sqrt{N}$, whereas its left side has norm one because $C$ is an isometry. The formula above is the normalized identity consistent with all the preceding definitions. For $N=1$ the distinction disappears, but the no-cloning assertion is substantive for $N\ge 2$.

The normalized output is not $|\hat{j}⟩|\hat{j}⟩$. Its coefficient matrix pairs every Fourier label $m$ with $j-m$; tracing out either factor gives $I/N$. Thus it is a maximally entangled state, while two clones of a pure Fourier quantum state would have pure reduced quantum states. A copier for the computational basis therefore does not copy the Fourier basis.

More generally, suppose a linear isometry with a fixed blank input could clone every normalized quantum state. Its restriction to $|\psi ⟩|\mathrm{blank}⟩$ would define a linear map $D$ with $D|k⟩=|k⟩|k⟩$ on every computational basis quantum state. Linearity forces $D=C$, which has just been shown to fail on Fourier quantum states. This proves the no-cloning theorem in the pure-quantum state isometric setting.

The inner-product proof makes the obstruction independent of the chosen basis. If an isometry cloned two quantum states with overlap $s=⟨\psi |\varphi ⟩$, preservation of inner products would give $s=s^2$, or, allowing irrelevant output phases, $|s|=|s{|}^2$. Distinct nonorthogonal rays have $0<|s|<1$, which is impossible. Even allowing an output environment does not help: unitary evolution would give $|s|=|s{|}^2|⟨e_{\psi }|e_{\varphi }⟩|\le |s{|}^2$, the same contradiction. Arbitrary unknown quantum states cannot be cloned, although any fixed orthonormal basis can be copied and a one-dimensional quantum state space is trivial.

A density operator is a positive self-adjoint trace-class operator with $\boxed{\mathrm{Tr}\rho =1}$. A finite expectation is $\boxed{⟨A{⟩}_{\rho }=\mathrm{Tr}(\rho A)}$. It represents a pure state exactly when $\rho =|v⟩⟨v|$ for a unit vector, equivalently ${\rho }^2=\rho$ or $\mathrm{Tr}{\rho }^2=1$.

For a finite-dimensional spectral decomposition $\rho ={\sum }_j{p}_j|v_j⟩⟨v_j|$, introduce an auxiliary space with orthonormal vectors $f_j$ and use the enlarged space $\mathcal{K}=\mathcal{H}\otimes \mathcal{E}$. DefineThen $\Vert {\Omega }_{\rho }{\Vert }^2=\sum p_j=1$ and $⟨{\Omega }_{\rho },\tilde{A}{\Omega }_{\rho }⟩=\sum p_j⟨v_j,A{v}_j⟩=\mathrm{Tr}(\rho A)$. This purification shows that subsystem mixedness can arise from entanglement with an unobserved environment even when the total state is pure. Subsystem measurements alone cannot distinguish these two descriptions.

A nonzero pure bipartite vector is entangled when it cannot be written as a tensor product of two vectors; equivalently its Schmidt rank exceeds one. For mixed bipartite states, entanglement means absence of a convex decomposition into product density operators.

For the nonorthogonal expansion in the question, the unnormalized reduced operator isIndeed $⟨\Psi ,(A\otimes I)\Psi ⟩={\sum }_{j,k}⟨{\xi }_k,A{\xi }_j⟩⟨{\eta }_k,{\eta }_j⟩=\mathrm{Tr}(A{S}_1)$. Positivity follows because $⟨v,S_{1}v⟩=\Vert {\sum }_j⟨v,{\xi }_j⟩{\eta }_j{\Vert }^2\ge 0$, and $\mathrm{Tr}{S}_1=\Vert \Psi {\Vert }^2=:Z$. Thus $\boxed{{\rho }_1=S_1/Z}$ is a density operator; if $\Psi$ is already normalized, $Z=1$. The Gram factors cannot be dropped merely because the vectors are called bases. For unnormalized $\Psi$, physical expectations likewise divide its raw quadratic form by $Z$.

For the Schmidt decomposition diagonalize $S_1$ in an orthonormal basis ${\xi }_j$ with its positive eigenvalues $c_j^2$. Expand $\Psi ={\sum }_j{\xi }_j\otimes w_j$ with $w_j=(⟨{\xi }_j|\otimes I)\Psi$. The reduced matrix gives $⟨w_k,w_j⟩=c_j^2{\delta }_{jk}$. Therefore ${\eta }_j=w_j/c_j$ are orthonormal andZero-eigenvalue components vanish because their squared norms are zero. This proves the decomposition for an arbitrary nonzero vector; the zero vector uses the empty sum. For a unit vector $\sum c_j^2=1$.

In the numerical example the coefficient matrix in the two product bases is $C=\left(\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\end{array}\right)$. Taking the first-subsystem trace yieldsThe vector $(1,-2,1{)}^{\mathsf{T}}$ is in its nullspace, so rank is at most two; its leading $2\times 2$ minor has determinant one, so $\boxed{\mathrm{rank}{S}_2=2}$. This is the rank of $C$ and the Schmidt rank, proving entanglement. The displayed source ket has norm squared $19$, and the displayed reduced matrix has trace $19$. The physical density operator is $\boxed{{\rho }_2=S_2/19}$; its nonzero eigenvalues are $(19\pm \sqrt{337})/38$, since $C{C}^{\mathsf{T}}=\left(\begin{array}{cc}14& 8\\ 8& 5\end{array}\right)$. The raw matrix is correct for the unnormalized quadratic form actually printed.

Put $d=2^N$ and $\lambda =e^{2\pi i/d}$, with nonnegative integer $N$ so the register exists. The geometric sum satisfiesConsequently the matrix $F_{+}$ with entries $d^{-1/2}{\lambda }^{jk}$ has $F_{+}^{\ast }{F}_{+}=I$. This proves unitarity andFor a binary output string $j={\sum }_{r=1}^N{j}_r{2}^{N-r}$, expansion of the printed tensor product gives coefficientThus the printed product is exactly $F_{+}|k⟩$. It is a correct product decomposition of a Fourier phase state, but the printed recovery identity has inconsistent signs:by the square of the quantum Fourier transform. To recover $f(j)$ instead, use $F_{-}=F_{+}^{\ast }$, replacing every product phase by its complex conjugate. Equivalently one may reverse the sign in the definition of $\hat{f}$. The discrepancy is visible for $d\ge 4$ by taking $f$ concentrated at $1$: the printed expression returns the basis vector at $d-1$. The recovery display also uses $k$ as the summation variable on its right while its summand uses $j$; that dummy variable should be $j$.

For factorization first check $\gcd (3,91)=1$. Successive residues areThe first return is $\boxed{r=6}$. Its half power is $3^3=27$, which is neither $1$ nor $-1$ modulo $91$. Therefore

In Shor's algorithm prepare a uniform exponent register of size $Q=2^n$, conveniently $91^2\le Q<2\cdot 91^2$, and reversibly compute $3^x\mathrm{mod}91$ into a second register. Measuring that second register, or tracing it out, leaves periodic exponent correlations of period $r$. Conditional on one measured residue, the first register is a superposition over an arithmetic progression of exponents separated by $r$. The quantum Fourier transform converts this periodic structure into peaks near $k/Q=s/r$. Continued fractions recover $s/r$; when $s$ is coprime to $r$, its denominator is $r$. Verify by modular exponentiation and repeat when the reduced denominator is insufficient. For this example $Q=16384$ is suitable, and useful peaks include those near $Q/6$ and $5Q/6$. Either transform sign gives the same period information, so the source's inverse-sign error does not invalidate the order-finding mechanism.

In the ordered basis $|+⟩,|-⟩$, the transverse magnetic field producesThe off-diagonal complex exponentials follow directly from $\cos \omega t\mp i\sin \omega t$. This matrix is Hermitian at every time.