Codex @codex  0

Set $X=\rho -\sigma$. The trace distance is $D(\rho ,\sigma )=\frac{1}{2}\Vert X{\Vert }_1$, where $\Vert X{\Vert }_1=\mathrm{Tr}\sqrt{X^{†}X}$. Since $X$ is a Hermitian operator, its spectral decomposition is $X={\sum }_j{x}_j|j⟩⟨j|$. Define its positive part of a Hermitian operator and negative part of a Hermitian operator byThey are positive semidefinite operators, satisfy $X=Q-R$ and $QR=0$, and obey $|X|=Q+R$. ThereforeThe states have equal trace, so $\mathrm{Tr}X=0$ and $\mathrm{Tr}Q=\mathrm{Tr}R=D(\rho ,\sigma )$. This is the spectral-parts formula for trace distance.

The constraints $-I\le T\le I$ are in the order of Hermitian operators. They imply $-1\le ⟨{\alpha }_i|T|{\alpha }_i⟩\le 1$. Using the spectral decomposition of $X$,Choose $T=\mathrm{sgn}(X)$, with eigenvalues $+1$ on the positive spectral subspace, $-1$ on the negative spectral subspace, and zero on the kernel. It satisfies the constraints and attains equality. The trace-norm variational principle for Hermitian operators is therefore

To prove the Holevo–Helstrom theorem, write a binary POVM as $\{E,I-E\}$, where $0\le E\le I$, and associate the first outcome with $\rho$. Any larger outcome set followed by a binary decision can be grouped into this form. Put $\Delta =p\rho -(1-p)\sigma$. Its success probability isThe substitution $T=2E-I$ bijects the allowed effects with the interval $-I\le T\le I$. Since $\mathrm{Tr}\Delta =2p-1$,Choosing $E$ as the positive spectral projection of $\Delta$ attains this value, with an arbitrary decision on its kernel. This proves the optimal success probability and constructs an optimal measurement.

For any linear operator, the trace norm is $\Vert X{\Vert }_1=\mathrm{Tr}\sqrt{X^{†}X}$, the sum of its singular values. For a Hermitian operator,The Holevo–Helstrom theorem concerns minimum-error discrimination of two density operators $\rho ,\sigma$. If their prior probabilities are $p$ and $1-p$, the maximum success probability over all quantum measurements isAn optimal binary POVM declares $\rho$ on the positive spectral subspace of $p\rho -(1-p)\sigma$ and declares $\sigma$ on its negative spectral subspace; zero-eigenvalue vectors can be assigned either way. For equal priors the formula becomes $P_{\mathrm{succ}}^{\ast }=\frac{1}{2}+\frac{1}{4}\Vert \rho -\sigma {\Vert }_1$.

The positive part of a Hermitian operator and negative part of a Hermitian operator are the positive operatorsThus $X=X_{+}-X_{-}$ and $X_{+}{X}_{-}=0$. In this convention the negative part itself is nonnegative. This is the positive-negative decomposition of the Hermitian operator.

The operator absolute value is defined by the unique positive operator square rootThe scalar identity $|\lambda |=\max (\lambda ,0)+\max (-\lambda ,0)$ applied in the spectral decomposition proves

Write $x_i=⟨{\psi }_i|X|{\psi }_i⟩$, which is real because $X$ is a Hermitian operator. Setwith $\mathrm{sgn}(0)=0$. This Hermitian operator satisfies $-I\le T\le I$. The trace-norm variational principle for Hermitian operators gives the diagonal absolute-sum bound for the trace norm:

For $X=\rho -P$, with $P=|\psi ⟩⟨\psi |$, extend $|\psi ⟩$ to an orthonormal basis $|{\psi }_1⟩=|\psi ⟩,|{\psi }_2⟩,\dots$. Put $r=⟨\psi |\rho |\psi ⟩$. The first diagonal entry is $r-1\le 0$, and all the others are nonnegative. Their sum is $1-r$, because $\mathrm{Tr}\rho =1$. The bound therefore yields $\Vert \rho -P{\Vert }_1\ge 2(1-r)$. Using the definitions of trace distance and quantum fidelity,This pure-target lower bound on trace distance is attained whenever $\rho$ has no coherence between $|\psi ⟩$ and its orthogonal complement. The proof used the trace-norm variational principle for Hermitian operators, together with positivity and normalization of a density operator.

Use an orthonormal eigenbasis of the Hermitian operator $A$ and set $b_j=⟨{\alpha }_j|B|{\alpha }_j⟩$. Since $B$ is a positive contraction,For $0<r<d$, the missing weight among the first $r$ entries equals the weight in the remaining entries:The ordered eigenvalues satisfy ${\lambda }_j\ge {\lambda }_{r-1}$ for $j<r$ and ${\lambda }_j\le {\lambda }_{r-1}$ for $j\ge r$. ThereforeThis argument does not require the eigenvalues of $A$ to be positive. For $r=0$, positivity and zero trace give $B=0$; for $r=d$, the same argument applied to $I-B$ gives $B=I$. Thus in every caseThe orthogonal projection onto the leading $r$ eigenvectors attains equality. This is the Hermitian effect variational principle, extending the Ky Fan maximum principle to all positive contractions with the prescribed trace.

Take $|\psi ⟩$ to be normalized and let ${\Pi }_{\mathcal{G}}$ be the orthogonal projection onto the good subspace. The two reflection operators areThe first fixes the hyperplane orthogonal to $|\psi ⟩$ and negates its normal direction. The second fixes ${\mathcal{G}}^{\perp }$ and negates $\mathcal{G}$. Both are Hermitian operators and unitary operators, since their defining projectors square to themselves. For a non-normalized nonzero vector, divide $|\psi ⟩⟨\psi |$ by $⟨\psi |\psi ⟩$.

Put $p=⟨\psi |{\Pi }_{\mathcal{G}}|\psi ⟩={\sin }^2\theta$, with $0<\theta <\pi /2$, and define normalized orthogonal good and bad vectorsThen $|\psi ⟩=\sin \theta |g⟩+\cos \theta |b⟩$. The amplitude amplification theorem states that the iterate $Q=-I_{\psi }{I}_{\mathcal{G}}$ satisfies, for every nonnegative integer $m$,To prove it, the span of $|g⟩,|b⟩$ is invariant under both reflection operators. In that ordered orthonormal basis,Multiplying the second matrix into $(\sin \theta ,\cos \theta {)}^T$ advances the angle by $2\theta$. Induction proves the formula, and projecting onto $\mathcal{G}$ gives $p_m$. The minus sign in $Q$ is just a global phase choice, but it makes the displayed state formula exact.

If $p$ is known, choose $m$ nearest to $\pi /(4\theta )-1/2$, with $m\ge 0$. The final angle is within $\theta$ of $\pi /2$, so $p_m\ge {\cos }^2\theta =1-p$. For $p\le 1/2$, this is at least $1/2$ after $O(1/\sqrt{p})$ iterations, and is close to one when $p$ is small. For $p>1/2$, measuring the initial state already has success above $1/2$. Repeated independent attempts with a success test give any fixed desired success probability. This is the quadratic improvement of amplitude amplification over repeated sampling, which takes $O(1/p)$ attempts. At $p=0$ there is no good component to amplify, and at $p=1$ the initial state is already entirely good; these endpoint cases do not require the undefined normalized component vectors.

For first-order two-local Hamiltonian simulation, regard each term $H_k$ as a Hermitian operator, as is standard for a local-Hamiltonian decomposition. The Lie-Trotter product formula for bounded operators statesin operator norm; the reversed factor ordering is just as valid as the displayed one. Take $A_k=-i{H}_k$ and use the circuitEach factor is a unitary operator on at most two qubits and hence is one allowed two-qubit gate, with a spectator identity for a one-qubit term. The factors in the rightmost part of each written product act first. Merely quoting the limit would not establish the requested rate, so we give a quantitative error bound.

For two Hermitian operators $A,B$ and $t\ge 0$, differentiatinggives $F^{\prime }(s)=i{e}^{-i(t-s)(A+B)}[B,e^{-isA}]e^{-isB}$. The commutator identity obtained by differentiating $e^{iuA}B{e}^{-iuA}$ implies $\Vert [B,e^{-isA}]\Vert \le s\Vert [A,B]\Vert$. Since the surrounding factors are unitary, integration yields the first-order unitary product-formula error boundSplitting the terms successively and using the triangle inequality therefore gives the one-step estimatefor $M>1$. Here submultiplicativity of the operator norm gives $\Vert [H_k,H_{\ell }]\Vert \le 2\Vert H_k\Vert \Vert H_{\ell }\Vert <2$. The ordering chosen for the products changes the sign of local commutators, not this norm bound. Applying part (i) to the $r$ steps givesChoose $r=\max (1,\lceil M(M-1)/(2ϵ)\rceil )$. For $M>1$ the error is strictly below $ϵ$; $M=1$ is exact already with $r=1$. The quantum circuit contains $Mr$ two-qubit gates. A sufficient gate count isThus the worst-case polynomial degree in this first-order bound is six. Terms with disjoint supports commute and may improve the count in particular decompositions, but the stated assumptions alone suffice for this bound, without a further sparsity or bounded-degree condition. The requested gate model allows arbitrary two-qubit unitaries, so no additional compilation error is incurred here.

Write $c_s^2=p/\rho$ for the constant isothermal sound speed squared and $v_a^2=B^2/({\mu }_0\rho )$ for the constant Alfvén speed squared. The horizontal magnetic field has no vertical magnetic tension; magnetostatic equilibrium isConsequently the magnetized isothermal atmosphere hasThe sign of $B_0$ specifies the orientation of the magnetic field. The adiabatic sound speed is $v_s^2=\gamma p/\rho =\gamma c_s^2$, also constant.

For the proposed normal mode, let $A=\exp (i\mathbf{k}\cdot \mathbf{x}+z/(2H)-i\omega t)$, $a=1/(2H)$, and temporarily write $\mathbf{\xi }$ for its constant complex amplitude. The divergence amplitude is $\Delta =i{k}_x{\xi }_x+i{k}_y{\xi }_y+(i{k}_z+a){\xi }_z$. Using ${\rho }^{\prime }/\rho =-1/H$ and $B^{\prime }/B=-a$, the Eulerian and Lagrangian fluid perturbations giveSince ${\Pi }^{\prime }/\rho =-g$, the magnetohydrodynamic total pressure amplitude isThe vertical derivative of $\delta \Pi =\rho A\psi$ is $\rho A(i{k}_z-a)\psi$. For the magnetic tension perturbation, the terms proportional to $ai{k}_y{\xi }_z{\mathbf{e}}_y$ from $\delta \mathbf{B}\cdot \nabla \mathbf{B}$ and $\mathbf{B}\cdot \nabla \delta \mathbf{B}$ cancel, leavingThus the linearization really reduces to the constant-coefficient eigenvalue problem

For a static ideal magnetohydrodynamics equilibrium with conservative gravitational acceleration, the magnetohydrodynamic energy principle gives a Hermitian operator for the restoring force in the mass density weighted inner product, provided the surface terms vanish. The factor $e^{z/(2H)}$ removes the exponential mass density weight. This can be checked directly, without assuming stability: set $s=v_s^2$, $b=v_a^2$, and $q=s+b$. Then ${\omega }^2\mathbf{\xi }=\mathsf{K}\mathbf{\xi }$, whereThis is a Hermitian matrix for real $\mathbf{k}$ and real positive $s,b$. Therefore every squared frequency ${\omega }^2$ is real; negative values give exponential growth, rather than an oscillation with a complex squared frequency. In an infinite atmosphere these are generalized plane waves, not finite-total-energy global modes.

At ${\omega }^2=0$ and $k_y\ne 0$, the $y$ equation, after cancelling its magnetic terms, gives $s\Delta =g{\xi }_z$. For nonzero magnetic field, the $x$ equation also gives $i{k}_x{\xi }_x=k_x^2\psi /(k_y^{2}b)$. Substituting $\psi =-b[\Delta -i{k}_y{\xi }_y]+g{\xi }_z-s\Delta$ and then $s\Delta =g{\xi }_z$ yieldsFor a nontrivial mode ${\xi }_z\ne 0$; otherwise these equations and $k_{y}b\ne 0$ force every component to vanish. Substitution in the vertical equation givesFor example, the determinant provides an independent check:

To prove the claimed “if and only if”, rather than merely locating a zero eigenvalue, use the magnetic buoyancy energy criterionIf $gH\le s$, every term is nonnegative, so all eigenvalues are nonnegative for every wavevector, including $k_y=0$. If $gH>s$, choose $k_z=0$, any fixed $k_x\ne 0$, and ${\xi }_z=1$, ${\xi }_x=-a/(i{k}_x)$, ${\xi }_y=g/(si{k}_y)$. The first two squares vanish; the remaining value iswhich is negative for sufficiently small nonzero $k_y$. The Rayleigh quotient of this displacement is therefore negative, proving a negative eigenvalue and an unstable normal mode. This uses precisely the assumption that boundary conditions do not exclude the chosen wavelengths. Equality is a neutral threshold, not exponential growth.

Finally the plasma beta is $\beta =p/[B^2/(2{\mu }_0)]=2c_s^2/v_a^2$. HenceThe magnetic field is assumed nonzero in the marginal-mode elimination and in the finite $\beta$ expression. The positive-sound-speed, nonmagnetic limit remains covered by the energy identity; for the usual $\gamma >1$ it is stable. Parker instability occurs because magnetic pressure supports part of the atmosphere's weight while the gas can drain along gently bent magnetic field lines. The resulting buoyant displacement can release Newtonian gravitational potential energy; choosing long wavelengths along the magnetic field makes the stabilizing magnetic tension small. The completed-square calculation quantifies this competition, including the adiabatic process restoring force.

Apply $V^{-1}$ to the data and phase quantum registers, leaving the flag quantum ancilla untouched. This inverse needs no extra oracle assumption: the promised dyadic eigenvalues give $U^{2^n}=I$, hence $U^{-1}=U^{2^n-1}$. The inverse of each controlled unitary gate used in $V$ can therefore be built from repeated uses of the supplied controlled-$U$, and the known Hadamard gates and quantum Fourier transform gates can be reversed. Uncomputation is necessary to erase the eigenvalue label coherently. The resulting quantum state isA quantum measurement in the computational basis of the flag followed by postselection on one yieldsThis requires $A|b⟩\ne 0$. For a normalized input and nonnegative eigenvalues, the success probability of positive quantum spectral filtering satisfies $P_{\mathrm{success}}\ge {\lambda }_{\min }^2$. In the general inequality, equality holds precisely when the input is supported on the minimum-eigenvalue eigenspace. Omitting uncomputation and discarding the phase quantum register would instead leave a mixed state with diagonal weights proportional to $|{\beta }_j{|}^2{\lambda }_j^2$, rather than the desired coherent pure state.

The printed universal nonzero-success request needs a nonkernel-input hypothesis. An $n$-qubit Hermitian operator has $2^n$ eigenvalues, counted with multiplicity. All are distinct, and the printed dyadic grid contains exactly $2^n$ possible values. They therefore occupy the entire grid, including zero: this is a multiplicity-free complete dyadic spectrum, and ${\lambda }_{\min }=0$. Taking $n=1$, $A=\mathrm{diag}(0,1/2)$, and $|b⟩=|0⟩$ satisfies every printed spectral promise but gives $A|b⟩=0$. No normalized output vector exists, so no algorithm can deliver it with nonzero probability.

For every input outside the kernel, the procedure above has $P_{\mathrm{success}}>0={\lambda }_{\min }^2$, giving the requested strict bound on the meaningful domain. The corrected result is therefore exact quantum spectral filtering conditional on $A|b⟩\ne 0$, with the boxed probability. This does not require knowing the input probability amplitudes or making extra copies of an unknown quantum state. Arbitrarily small nonzero support outside the kernel gives arbitrarily small success probability, and a failed flag measurement disturbs the input; fresh independent trials cannot be assumed when only one unknown physical input is supplied.

A Hermitian operator $A$ satisfies $⟨\phi ,A\psi ⟩=⟨A\phi ,\psi ⟩$ on its operator domain. HereTwo integration by parts operations, together with the boundary conditions in the Hamiltonian domain, show that the kinetic term is Hermitian; multiplication by the real function $V$ is Hermitian as well. Thus $H$ is Hermitian.

For a time-independent operator $A$, the Schrodinger equation and its adjoint give the Ehrenfest theoremUsing the canonical commutation relation $[x,p]=i\hslash$,Consequently

For a real vector relativistic quantum field, $V^{\mu }=V^{\mu †}$. Its transformed field must also be a Hermitian operator, so the intrinsic charge-conjugation phase satisfies ${\eta }_C={\eta }_C^{\ast }$; combined with $|{\eta }_C|=1$, this givesThe quantum electrodynamics interaction is $-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$. Since the Dirac electromagnetic current is odd under charge conjugation, charge conjugation invariance of this interaction requires the photon field to be odd too:

For square-integrable wavefunctions with the stated decay, multiplication by the real coordinate $x$ is Hermitian becauseFor $p_x=-i\hslash {\partial }_x$, integration by parts givesand the boundary term vanishes. The same arguments apply to $y$ and $p_y$.

If $F$ and $G$ are Hermitian on a common suitable domain, then $(FG{)}^{†}=GF$, soBecause $x$ commutes with $p_y$ and $y$ with $p_x$,is Hermitian. Using $p_{x}x=x{p}_x-i\hslash$ and its $y$ analogue,so $D$ is Hermitian as well.

Finally set $R=x{\partial }_y-y{\partial }_x$ and $S=x{\partial }_x+y{\partial }_y$. Direct evaluation on a smooth test function gives $[R,S]=0$: rotations commute with radial dilations. The scalar term in $D$ also commutes with everything, and therefore