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The spectral norm is the operator norm induced by the Euclidean norm:The Pauli operators $X_1$ and $Z_2$ act on different qubits, and each is a unitary operator. Their product is therefore unitary and preserves the norm of every vector. Hence

Let ${\omega }_d=e^{2\pi i/d}$. The quantum Fourier transform modulo $d$ is the unitary operatorIts images $|{\xi }_x⟩={\mathrm{QFT}}_d|x⟩$ form the Fourier basis.

Let $X|x⟩=|x+1⟩$ on the input register and let $N|y⟩=|-y⟩$ on the target register. Both are fixed unitary operators independent of $f$. Conjugating the query by the input shift givesbecause the intermediate query is made at $x+1$. Conjugating by target negation givessince $y\mapsto -y\mapsto -y+f(x)\mapsto y-f(x)$. Each construction uses $U_f$ exactly once.

Bloch theorem states that the energy eigenstates of a Hamiltonian invariant under translations by a Bravais lattice can be chosen as Bloch statesIndeed, the unitary operators $T_r$ commute with one another because $T_r{T}_s=T_{r+s}=T_s{T}_r$, and they commute with $H$ by hypothesis. The simultaneous diagonalization theorem therefore lets us diagonalize all translations within each energy eigenspace. Their eigenvalues form a unitary character $\chi$ of the additive lattice:Writing $\chi (a_j)=e^{-ik\cdot a_j}$ on a primitive basis gives $\chi (r)=e^{-ik\cdot r}$. In position space, with $(T_r\psi )(x)=\psi (x-r)$, this implies ${\psi }_k(x+r)=e^{ik\cdot r}{\psi }_k(x)$. Hence $u_k(x)=e^{-ik\cdot x}{\psi }_k(x)$ is lattice-periodic. Adding a reciprocal lattice vector to $k$ leaves the character unchanged, so the crystal momentum lies in a Brillouin zone.

Forthe equations $a_i\cdot b_j=2\pi {\delta }_{ij}$ give the reciprocal basisThe reciprocal lattice is triangular, so the first Brillouin zone is its Wigner-Seitz cell, a regular hexagon. Its six corners areReciprocal-lattice translations identify these corners in two classes of three. Representatives areFor example, $(b_1-b_2)/3=K^{\prime }-b_2$, while the other equivalences follow by symmetry and reciprocal translations. This gives the requested sketch: a regular hexagon with the vertical edge from $K^{\prime }$ to $K$ at $k_x=2\pi /(\sqrt{3}a)$ and alternating $K,K^{\prime }$ corner classes.

For the tight-binding model, introduce the normalized Bloch sumEach hop by $\pm a_j$ multiplies this state by $e^{\mp ik\cdot a_j}$. Thus the two-direction nearest-neighbour tight-binding dispersion isAlong the boundary edge from $K^{\prime }$ to $K$, $k_x=2\pi /(\sqrt{3}a)$ and $|k_y|\le 2\pi /(3a)$, soFor $\lambda >0$, it rises from $E_0+2\lambda$ at either corner to $E_0+4\lambda$ at the midpoint; for $\lambda <0$ the ordering reverses. Globally the two cosines in the first expression can simultaneously equal $1$ or $-1$, and therefore

Each orbital band contains two one-electron states per lattice site because an electron has two spin states. By band filling, a valency of one leaves this band half-filled and the material conducts, whereas a valency of two fills it. A filled band can be insulating only if it is separated from every empty band by a positive band gap.

The first band's maximum is $E_0+4|\lambda |$. If the second band's minimum is $E_0+\Delta$, a gap exists precisely when $\Delta >4|\lambda |$. Consequently, among the nontrivial fillings described here,For valency one the first band is partially filled. If $\Delta \le 4|\lambda |$, the bands overlap or touch, so even at valency two the overlapping energy bands prevent a band insulator.

For $x\in X$,The map $S$ defined by $(Sx{)}_n=x_{n-1}$ satisfies $ST=TS=I$, so $T$ is a surjective isometry and hence a unitary operator.

If $Tx=\lambda x$, thenFor $\lambda \ne 0$, this gives $x_n={\lambda }^n{x}_0$ for every integer $n$. When $|\lambda |>1$ the positive tail is not square summable, when $|\lambda |<1$ the negative tail is not square summable, and when $|\lambda |=1$ neither tail is square summable unless $x_0=0$. The case $\lambda =0$ also forces every coordinate to vanish. Thus

The reverse triangle inequality gives, for every $x$,Hence $T-\lambda I$ is bounded below whenever $|\lambda |\ne 1$, so such $\lambda$ do not belong to the approximate point spectrum.

Now let $|\lambda |=1$ and define the unit vectorThe equation $x_{n+1}^{(N)}=\lambda x_n^{(N)}$ fails only at the two endpoints, soTherefore the bilateral shift operator has

In the Parity basis of a one-dimensional S-matrix, parity acts as $\mathcal{P}=\mathrm{diag}(1,-1)$. Parity invariance requiresand thereforeIndependently, unitarity givesCombining parity and unitarity leaves two phases,

For the symmetric double-delta potential, consider the odd channel on $x\ge 0$. Choose the asymptotic conventionand write $psi(x)=C\sin (kx)$ for $0<x<a$; the sine enforces odd parity at the origin. Continuity and the delta potential derivative jump at $x=a$ giveEliminating $C$ yields the Odd-parity S-matrix of a symmetric double-delta potentialUsing $2i\sin (ka)=e^{ika}-e^{-ika}$ rewrites this as the requested expressionFor real $k$, the two factors in the ratio are complex conjugates, directly confirming $|S_{--}|=1$.

A bound state occurs at a pole $k=i\kappa$, $kappa>0$. Setting the denominator to zero givesorFor $U_0<0$, write $g=-U_0>0$. The right-hand side $g(1-e^{-2\kappa a})$ is increasing and concave, starts at zero with slope $2ga$, and approaches $g$, whereas the left-hand side is the straight line $2\kappa$. A positive intersection therefore exists exactly whenThis is the odd bound state of a symmetric double-delta potential; equality is its zero-energy threshold.

For $U_0>0$, analytically continue the same pole equation into the lower half of the complex $k$-plane. PutIt becomesWhen $G\gg 1$, the lowest odd mode lies near the hard-wall value $z=\pi$. Writing $z=\pi +\eta$ and expanding givessoIn particular, the odd resonance of a strongly repulsive symmetric double-delta potential hasThe negative imaginary part identifies a decaying scattering resonance. Ifthen, to leading nonzero order,Its time-dependent amplitude is proportional toso its survival probability decays as $e^{-\Gamma t/\hslash }$ and its lifetime isPhysically, this is the lowest odd standing wave trapped for a long time between two strong repulsive barriers and leaking out by quantum tunnelling.

Conjugating the reflection operator by a unitary operator reflects about the conjugated state:Thus one may take $|\psi ⟩=|{\psi }_{\mathrm{in}}⟩=U|00⟩$.

Introduce a quantum ancilla with basis $|i⟩$ and defineThe completeness relation gives $V^{†}V=I$, so $V$ is an isometry and extends to a unitary operator on a sufficiently large system-plus-ancilla space. Prepare the ancilla in a fixed state, apply that unitary, and perform the projective measurement $\{I\otimes |i⟩⟨i|\}$. Outcome $i$ has probability $\Vert M_i|\psi ⟩{\Vert }^2$ and leaves the system in the normalized state $M_i|\psi ⟩$. By linearity the same holds for mixed states, implementing the generalized measurement.

Discarding the outcome of the complete projective measurement givesFor $m$ projectors, let $\omega =e^{2\pi i/m}$ and $U={\sum }_j{\omega }^j{P}_j$. Then the pinching identity isThe Concavity of Von Neumann entropy and its invariance under unitary operators implyEquality holds exactly when every conjugate in the average is the same, equivalentlyThus equality holds when the input already has no coherence between distinct measurement subspaces.

The no-cloning theorem for two pure states states that, for distinct nonorthogonal states $|c_0⟩$ and $|c_1⟩$, there is no unitary operator $U$ and fixed blank state $|0⟩$ such that

To derive this from state discrimination, letand suppose such a cloner existed. Applying it repeatedly would produce $N$ copies. The inner product of the two possible $N$-copy states has magnitude $s^N$, so their optimal success probability under the Helstrom-Holevo bound would beAlready for $N=2$,because $0<s<1$ implies $s^4<s^2$. But cloning followed by the two-copy Helstrom measurement for two pure states would itself be a quantum procedure applied to the original single state, contradicting the optimal one-copy bound $P_1$. Hence the assumed unitary cloner cannot exist.

Equivalently, clone-assisted asymptotic state discrimination would give $P_N\to 1$, although two nonorthogonal states are not perfectly distinguishable. Thus the Helstrom--Holevo theorem implies the no-cloning theorem.

Let $X=C([0,T];L^2(\mathbb{R}))$ and define the map suggested by the variation-of-constants formula:The strong continuity of the evolution family and the continuity of $f$ imply that $\Phi$ maps $X$ into itself. Because $U(t,s)$ is unitary and $\Vert f(t,u){\Vert }_2\le C$,

Equip $X$ with the exponentially weighted supremum normThis equivalent norm makes $X$ a Banach space. Using that $f$ is a globally Lipschitz function and that the unitary operator $U(t,s)$ preserves the norm,Choose $\alpha >L$. Then $\Phi$ is a contraction mapping, so the Banach fixed-point theorem gives a unique fixed point $u\in X$. This fixed point is exactly the required mild solution of an abstract Cauchy problem:The weighted-norm argument works on the whole prescribed finite interval, so no subdivision of $[0,T]$ is needed.

Put $a={\lambda }_P$ and $b={\lambda }_Q$. The Pauli operators are Hermitian and satisfy $P^2=Q^2=I$ and $PQ+QP=0$, soThus $V$ is a unitary operator. For every Pauli operator $R$, according as $R$ commutes or anticommutes with $P$ and $Q$, expansion of $VRV$ gives one of $\pm R$ or $\pm RPQ$, up to the phase that makes it Hermitian. It is therefore another Pauli operator, so $V$ normalizes the Pauli group and is a Clifford operation. Finally,which is the normalized projection onto the eigenvalue-$b$ eigenspace of $Q$. Hence $V$ maps the eigenvalue-${\lambda }_P$ eigenspace of $P$ onto the eigenvalue-${\lambda }_Q$ eigenspace of $Q$.

For hermitian $x$ with $\Vert x\Vert \le 1$, its spectrum lies in $[-1,1]$. Continuous functional calculus therefore shows that $1-x^2$ is positive. LetIt commutes with $x$, andsatisfies $u^{\ast }u=u{u}^{\ast }=x^2+v^2=1$. Thus $u$ is unitary and

For positive trace-class covariance operators $C_i=L_i{L}_i^{\ast }$, their Procrustes distance between covariance operators iswhere $\mathcal{O}(H)$ is the group of unitary operators on the real Hilbert space $H=L^2[0,1]$. Expanding the square gives the equivalent formula

For each fixed $\alpha$, right multiplication by $x^{-\alpha }$ is a bijection of $G$, whose inverse is right multiplication by $x^{\alpha }$. Therefore $D_x$ permutes the displayed orthonormal basis of the tensor-product space. It is consequently a unitary operator, with

The spectral norm of an operator $O$ is the induced operator normEvery Pauli X gate $X_i$ is unitary, so $\Vert X_i\Vert =1$. The triangle inequality for the norm therefore givesIn fact equality holds, as the product of $+1$ eigenvectors of the $X_i$ has eigenvalue $n$ under $J_X$.

The HHL algorithm requires coherent, efficient and repeatable preparation of the normalized state $|b⟩$, normally through a known preparation circuit and its inverse; possession of a single unknown physical specimen does not supply that access. The component of $b$ on any discarded or unresolved small-eigenvalue subspace must also be negligible. Here $U$ is a unitary operator, so it is invertible and all its singular values equal one, giving condition number $\kappa =1$.

Standard HHL is stated for a Hermitian matrix with an efficient sparse-access or block encoding oracle. A non-Hermitian $U$ can be embedded in the Hermitian block matrixpart (a) supplies efficient access to $U^{†}$. With inverse-polynomial target precision, $m=O(\log n)$ phase bits, and an efficient preparation oracle for $|b⟩$, the runtime is $\mathrm{poly}(n)$. The output is the normalized quantum state proportional to the solution $x$, rather than a classical list of all its amplitudes.

Let $L_n=\mathrm{span}\{e_1,\dots ,e_n\}$ and let $P_n$ be its orthogonal projection. For $z\in \mathbb{C}$, defineBecause $A$ is unitary,The displayed finite Hermitian matrix uses only finitely many matrix entries of $A$, and its least eigenvalue can be approximated by an arithmetic algorithm in the SCI hierarchy.

As $L_n$ increases densely,A unitary operator is normal, so the spectral theorem for normal operators on a separable Hilbert space identifies the limit asThe functions ${\gamma }_n$ are continuous and decrease to a continuous function on the compact unit circle. The Dini theorem therefore gives uniform convergence there.

Take successively finer rational meshes $G_n$ around $\mathbb{T}$. From the finitely computed values of ${\gamma }_n$, retain the mesh minima in the comparison neighborhoods whose radii are ${\gamma }_n(z)$; equivalently, use the standard local-minimum construction for a decreasing approximation to a distance function. Call the resulting finite set ${\Gamma }_n(A)$. Uniform convergence and the shrinking mesh implyin Hausdorff distance. Every operation at stage $n$ is finite and arithmetic, so $({\Gamma }_n)$ is the required one-limit sequence.

There is no finite-stage certificate that the whole output has the correct Hausdorff error: the convergence of ${\gamma }_n$ has no uniform computable rate over all unitary operators, and unseen matrix entries can still reveal a missing spectral component. A small computed residual can certify that an individual output point lies near the spectrum, but it cannot verify that ${\Gamma }_n(A)$ covers all of $\mathrm{Sp}(A)$. Thus the full finite-stage output is not verifiable without additional information.

Since $\gcd (x,N)=1$, multiplication by $x$ modulo $N$ is a bijection of ${\mathbb{Z}}_N$, with inverse multiplication by the modular inverse $x^{-1}$. Hence $U_x$ permutes the standard orthonormal basis. It preserves all inner products and satisfiesso $U_x$ is a unitary operator.

The number operator $N$ is self-adjoint. Alsoafter $\mathbf{q}\mapsto -\mathbf{q}$. Thus both exponents are anti-Hermitian and ${\mathcal{P}}_1,{\mathcal{P}}_2$ are unitary. Their product obeysand likewise for creation operators. Both generators annihilate the vacuum, so the product leaves it invariant. Substitution in the free-field mode expansion yieldsHence $\mathcal{P}={\mathcal{P}}_1{\mathcal{P}}_2$ implements parity with intrinsic parity ${\gamma }_p$.