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At a critical point, a local field operator $\mathcal{O}$ has scaling dimension ${\Delta }_{\mathcal{O}}$ if $\mathcal{O}(x)\mapsto b^{-{\Delta }_{\mathcal{O}}}\mathcal{O}(x/b)$ under coarse-graining by $b$. The coupling $u$ in $\int d^{d}xu\mathcal{O}$ has renormalization-group eigenvalueA perturbation is a relevant operator when $y_u>0$, an irrelevant operator when $y_u<0$, and a marginal operator when $y_u=0$; nonlinear terms in its renormalization-group beta function decide the fate of a marginal coupling. Irrelevant microscopic interactions decay under coarse-graining, so many distinct systems approach the same fixed point and share one universality class.

Near criticality the singular free energy density is of order one per correlation volume:The heat capacity contains two derivatives with respect to temperature, soConsequently the hyperscaling relation is

Besides $g=0$, the coupling beta function vanishes atThe mass fixed-point equation isso the root continuously connected to the origin isPerturbation theory requires $b/c\ll 1$, ensuring that the fixed-point coupling is small and the omitted higher powers are suppressed.

For increasing infrared scale $s$, positive $g<g_{\ast }$ flows upward toward $g_{\ast }$ and $g>g_{\ast }$ flows downward toward it. The mass direction remains relevant, so only the tuned critical trajectory reaches the interacting fixed point; trajectories on either side leave toward the two phases.

The relevant thermal eigenvalue is obtained from the stability matrix of a renormalization-group fixed point by differentiating the mass beta function:It follows that

Let $a=B_{12}$, $b=B_{13}$, $c=B_{21}$, $d=B_{23}$, $e=B_{31}$, and $f=B_{32}$. Zero row sums put $B$ in the form used in the spectrum of a three-dimensional matrix with zero row sums. In particular, $B(1,1,1{)}^T=0$. DefineExpanding the characteristic polynomial gives $\det (\lambda I-B)=\lambda ({\lambda }^2+S\lambda +T)$, soFor the physical coefficients, $L_j{B}_{jk}=L_k{B}_{kj}$ with $L_j=M_j\sqrt{G{M}_{\star }{a}_j}$. Thus $W=\mathrm{diag}(L_j)$ gives positive diagonal symmetrization of a matrix, andConsequently the eigenvalues are real and nonpositive, and $B$ has a complete eigenvector basis. This justifies the mode expansion in part (i).

The variables $y_j=I_j{e}^{i{\Omega }_j}$ are complex inclinations, despite the question's description as eccentricities. Diagonalize the Laplace-Lagrange inclination matrix: let real eigenvectors ${\mathbf{v}}_k$ satisfy $B{\mathbf{v}}_k={\lambda }_k{\mathbf{v}}_k$, and form $V=({\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3)$. The physical matrix is diagonalizable by positive diagonal symmetrization of a matrix, as shown below.

The matrix exponential solution isWrite $c_k=I_k{e}^{i{\gamma }_k}$ with $I_k\ge 0$, and define $I_{jk}=I_k{v}_{jk}$. This givesThe eigenvalues determine nodal precession frequencies, the eigenvectors determine relative inclinations within each secular eigenmode, and the initial complex inclinations determine their amplitudes and phases. The real $I_{jk}$ can have either sign, encoding opposite nodal phases. The zero-frequency mode is a common fixed tilt.

Ignoring the distant planet at first, the inner two-by-two block has eigenvalues zero and $-(B_{12}+B_{21})$. Therefore its fast differential secular eigenmode hasFor the slow mode, strong inner coupling makes $y_1\simeq y_2=y_{\mathrm{in}}$. Add the first two evolution equations weighted by angular momenta and divide by $L_{12}=L_1+L_2$. The rigid inner pair in hierarchical secular inclination dynamics has equationswithThe two-plane matrix has eigenvalues zero and $-c(1+L_{12}/L_3)$. The architecture implies $L_3\gg L_{12}$ for comparable masses, so its leading slow frequency isThe neglected factor $1+L_{12}/L_3$ gives a relative correction of order $\sqrt{\beta }$; imperfect locking of the inner pair supplies higher-order corrections in its weak coupling to the outer planet.

A bipartite density operator ${\rho }_{AB}$ has positive partial transpose when ${\rho }_{AB}^{T_B}\ge 0$ in a product basis. Positivity is independent of the choice of local bases, although the matrix representing the partial transpose changes.

If the state is a separable quantum state, write ${\rho }_{AB}={\sum }_j{p}_j{\rho }_j^A\otimes {\rho }_j^B$. Taking its partial transpose givesIndeed, the matrix transpose preserves the nonnegative eigenvalues of each local density operator; a tensor product of positive semidefinite operators is positive, and so is their nonnegative sum. This proves the necessary direction of the positive partial transpose criterion.

The distinct values in one period imply that $r$ is the least positive period, $r\mid N$, and $|Y|=r$. Write $x=j+tr$, where $0\le j<r$ and $0\le t<N/r$. Use the positive-exponent quantum Fourier transform convention already used in the concept article. The amplitude of first-register value $y$ isBy the root-of-unity filter, the inner sum vanishes unless $y=mN/r$. DefineThe distinct $f(j)$ give an orthonormal basis of the second register, and the transformed state isReindexing $j+l$ modulo $r$ showsConsequently measurement in this common eigenbasis returns each $m$ with probability $1/r$; the corresponding eigenvalue of $U_1$ is a uniformly sampled $r$th root of unity. A single measured phase $-m/r$ reveals the denominator $r/\gcd (m,r)$, which need not be $r$. Repeated samples permit exact period recovery from a Fourier sample, for example by taking the least common multiple of their reduced denominators. The simultaneous eigenspaces and their probabilities depend on $r$, not on the particular labels $f(j)$.

The usual sufficient input promises for the HHL algorithm are efficient quantum state preparation of the nonzero vector $b$, efficient access to a sparse matrix $A$ with at most $\mathrm{poly}(n)$ nonzero entries per row, and a condition number $\kappa ={\lambda }_{\max }/{\lambda }_{\min }=\mathrm{poly}(n)$ after an efficiently known normalization. Both the positions and values of the nonzero matrix elements must be computable coherently in polynomial time. More generally, efficient Hamiltonian simulation can replace the sparsity promise. We assume $A$ is invertible; if zero eigenvalues occur, one must instead restrict $b$ to their orthogonal complement and specify the desired inverse on that support.

With $a={\lambda }_{\max }$ and a known bound $\kappa$, choose $c=a/\kappa \le {\lambda }_{\min }$. The HHL controlled reciprocal rotation then succeeds with probabilityThis is the required inverse-polynomial lower bound. In the usual approximate algorithm, inverse-polynomial requested error also gives polynomial runtime under these input promises. The original HHL paper states the dependence on sparsity, conditioning, and accuracy.

There is a qualification for a literally exact version. Representability of the eigenvalues in $n$ bits does not itself make exact quantum phase estimation efficient: using $e^{2\pi iA}$ requires controlled powers corresponding to evolution times as large as $2\pi 2^{n-1}$. Under the paper's idealization we may describe their exact action, but polynomial runtime for that exact circuit additionally requires efficient implementations of those controlled powers, or equivalent efficient exact spectral access. Assuming an operation executes exactly does not bound its cost. This distinction is recorded in cost of exact phase estimation on a dyadic spectrum.

Use the weighted Neumann heat kernel with constant drift. The spatial operator has the self-adjoint formso its homogeneous Neumann boundary conditions give a regular Sturm-Liouville problem in the weighted inner productThe constant eigenfunction is ${\varphi }_0=1$, with eigenvalue ${\lambda }_0=0$ and squared norm $N_0=(e^{\beta L}-1)/\beta$. For $k\ge 1$, put $q_k=k\pi /L$. Substitution of $\varphi =e^{-\beta x/2}v$ in $\mathcal{L}\varphi =-\lambda \varphi$ gives $v^{\prime \prime }+(\lambda -{\beta }^2/4)v=0$ and the Robin boundary conditions $v^{\prime }=\beta v/2$ at both endpoints. HenceIndeed,The Sturm-Liouville eigenfunction expansion is complete and orthogonal. Its heat kernel, relative to the measure $e^{\beta y}dy$, isFor a coefficient $a_k(t)=⟨{\varphi }_k,u(\cdot ,t){⟩}_{\beta }/N_k$, two applications of integration by parts yieldSolving these scalar linear differential equations and summing gives the Neumann boundary-forcing formula:This depends only on the given data. In particular, the zero eigenvalue retains the changing weighted mean:The endpoint derivatives of the boundary integral are understood as interior limits. Differentiating each homogeneous Neumann eigenfunction at the endpoint before summing would incorrectly discard the prescribed boundary forcing: the short-time heat kernel makes that interchange invalid. The stated compatibility conditions give the classical boundary traces.

For distinct eigenvalues, the formula holds for $n=0,1$. If the lower-left entry of $L^n$ is $b_n=({\lambda }_1^n-{\lambda }_2^n)/({\lambda }_1-{\lambda }_2)$, multiplication by $L$ gives $b_{n+1}={\lambda }_1{b}_n+{\lambda }_2^n=({\lambda }_1^{n+1}-{\lambda }_2^{n+1})/({\lambda }_1-{\lambda }_2)$. This proves by mathematical induction thatSumming the absolutely convergent matrix exponential series givesFor the repeated eigenvalue ${\lambda }_1={\lambda }_2=\lambda$, the continuous limit is $A(t)=e^{\lambda t}\left(\begin{array}{cc}1& 0\\ t& 1\end{array}\right)$. This is also the exponential of a two-dimensional Jordan block.

For real ${\lambda }_1,{\lambda }_2$, computeThus $L$ is a non-normal matrix. For distinct eigenvalues, takeThese satisfy $L{v}_j={\lambda }_j{v}_j$ but have inner product $v_1^T{v}_2=1$. The eigenvectors are therefore nonorthogonal. When the eigenvalues coincide, only the direction $v_2$ remains as an eigenvector; $L$ is a defective Jordan block.

Differentiate the quadratic energy:For $x_1\ne 0$, put $r=x_2/x_1$. Immediate energy growth occurs precisely when ${\lambda }_2{r}^2+r+{\lambda }_1>0$. With both eigenvalues negative, such directions exist iff $1-4{\lambda }_1{\lambda }_2>0$. Writing $\delta =\sqrt{1-4{\lambda }_1{\lambda }_2}$, the growing initial conditions are the two opposite open wedgesIf $x_1=0$, energy decreases. No energy growth at any time, for any initial condition, is possible exactly whenIndeed, this makes the symmetric part of a matrix $H$ a negative semidefinite matrix, so $\dot{E}\le 0$ along every trajectory. If the product is smaller, the wedges already supply counterexamples at $t=0$. This is the instantaneous energy-growth criterion for a linear system.

At each fixed time, the maximum over nonzero initial conditions is the optimal energy amplification of a linear system:Write $a=e^{{\lambda }_{1}t}$, $d=e^{{\lambda }_{2}t}$ and $b=(a-d)/({\lambda }_1-{\lambda }_2)$. The symmetric matrix $A^{T}A$ has trace $\tau =a^2+b^2+d^2$ and determinant $a^2{d}^2$. Its largest eigenvalue isAn optimal initial condition is a right singular vector of $A(t)$ associated with its largest singular value.

If ${\lambda }_2>{\lambda }_1$, let $\alpha =1/({\lambda }_2-{\lambda }_1)$. Then $A(t)/e^{{\lambda }_{2}t}\to \left(\begin{array}{cc}0& 0\\ \alpha & 1\end{array}\right)$, so the precise long-time asymptotic statement isThe Cauchy-Schwarz inequality shows that the unit initial conditions achieving this leading factor areThis is an adjoint eigenvector, since $L^T(\alpha ,1{)}^T={\lambda }_2(\alpha ,1{)}^T$. It differs from the right eigenvector $v_2=(0,1{)}^T$: the initial condition maximizes the projection onto the slow mode, whereas the eventual state aligns with that right eigenvector. For ${\lambda }_2<0$, even the optimal energy ultimately decays to zero; the prefactor describes enhanced excitation of the slow mode.

Replace coordinate inequalities by the Loewner order and coordinate sums by matrix trace. The maximizing semidefinite program isIts feasible set is the fantope. In an orthonormal eigenbasis of $X$, its objective is ${\sum }_i{\lambda }_i(X)Y_{ii}$, and the diagonal entries of $Y$ belong to the capped simplex. Choosing $Y$ as the orthogonal projection matrix onto the largest $k$ eigenvectors attains the sum of the largest eigenvalues, as in the Ky Fan maximum principle.

The minimizing semidefinite program isFor example, its Lagrangian arises by assigning $S⪰0$ to $I-Y⪰0$, keeping $Y⪰0$ in the domain, and using free $t$ for $\mathrm{tr}Y=k$. Finiteness of the supremum over $Y$ requires $S+tI-X⪰0$.

Equality follows directly by choosing $S$ to have eigenvalues $({\lambda }_i(X)-t{)}_{+}$ in the same orthonormal eigenbasis, with $t$ between the $k$th and $(k+1)$st eigenvalues, or below the smallest one for $k=n$. This is the threshold semidefinite program for the largest eigenvalues, and it also verifies the endpoint case $k=n$.

The mass matrix is a positive-definite matrix, and integration by parts with periodic boundary conditions gives $C^T=-C$. HenceThis is energy conservation for semidiscrete Galerkin advection: $U^{T}MU=\Vert u_h{\Vert }_{L^2(0,1)}^2$. Thus the semidiscretization is stable and conserves its finite element L2 norm.

For a mesh-independent comparison with the grid norm, the Fourier symbol of $M$ is $h(2+\cos \theta )/3$, between $h/3$ and $h$. ConsequentlyThe conserved energy therefore gives a uniform bound in $\sqrt{h{\sum }_m|U_m{|}^2}$ as well. Equivalently, each Fourier mode evolves with the purely imaginary exponent $\sigma (\theta )=3i\sin \theta /[h(2+\cos \theta )]$. This concerns the semidiscrete method; a chosen time integrator must separately control those imaginary eigenvalues.