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The commutator gives $a(a^{†}{)}^n=(a^{†}{)}^{n}a+n(a^{†}{)}^{n-1}$. Hence $a|n⟩=\sqrt{n}|n-1⟩$ and $a^{†}|n⟩=\sqrt{n+1}|n+1⟩$. Starting with the normalized vacuum, repeated application gives $⟨0|a^n(a^{†}{)}^n|0⟩=n!$, proving unit norm. The number operator then has eigenvalue $n$, soDifferent levels are orthogonal because the Hamiltonian is self-adjoint and their eigenvalues are distinct.

For the perturbation, the ground-state diagonal matrix element is zero, while $V|0⟩=\hslash \omega \sqrt{r!}|r⟩$. The nondegenerate second-order perturbation formula consequently givesFor $r=1$, $V|1⟩=\hslash \omega (|0⟩+\sqrt{2}|2⟩)$. The lower and upper intermediate levels contribute respectively $+{\lambda }^2\hslash \omega$ and $-2{\lambda }^2\hslash \omega$, giving $\Delta E_1=-{\lambda }^2\hslash \omega$ to second order.

For $r=1$ the exact completion of the square isThe shifted operators $b=a+\lambda$, $b^{†}=a^{†}+\lambda$ have the same commutator and are unitarily related to the original ones by a displacement operator. Their vacuum is the coherent state with amplitude $-\lambda$. Therefore every exact level shifts by $-{\lambda }^2\hslash \omega$.

The stability of monomial ladder-operator perturbations gives a necessary qualification to the request for a lowest level for every $r$. For $r>2$ and real nonzero $\lambda$, a coherent state of amplitude $R{e}^{i\theta }$ has energy expectation $\hslash \omega [R^2+1/2+2\lambda R^r\cos (r\theta )]$. Choose the phase so the last term is negative. As $R\to \infty$ the expectation tends to minus infinity, so the perturbed Hamiltonian has no true ground state. The displayed second-order answer is then the formal perturbative coefficient of the branch from the unperturbed vacuum, not an exact lowest eigenvalue. For $r=2$ stability requires $|\lambda |<1/2$; for $r=1$ the completed-square Hamiltonian is bounded below for every real $\lambda$.

The nullclines are $x=f(y)$ and $y=g(x)$, both decreasing curves in the positive quadrant. To their left/right the horizontal velocity is respectively positive/negative; below/above the second nullcline the vertical velocity is positive/negative. Bounded positive synthesis functions give an inward-pointing sufficiently large rectangle, while the axes point into the quadrant. More directly, put $H(x)=f(g(x))-x$: $H(0)>0$, and $H(x)<0$ for large enough $x$. Continuity therefore gives a positive equilibrium.

At an equilibrium, the Jacobian matrix is $J=\left(\begin{array}{cc}-1& f^{\prime }(y_{\ast })\\ g^{\prime }(x_{\ast })& -1\end{array}\right)$, with eigenvalues $-1\pm \sqrt{f^{\prime }(y_{\ast })g^{\prime }(x_{\ast })}$. Its logarithmic-slope product isbecause $f(y_{\ast })=x_{\ast }$ and $g(x_{\ast })=y_{\ast }$. Thus a product exceeding one makes this equilibrium a saddle and gives $H^{\prime }(x_{\ast })>0$. Immediately on its left $H<0$, and immediately on its right $H>0$. Since $H(0)>0$ and $H$ is eventually negative, there must be other roots on both sides, crossing in the attracting direction. For simple roots those have $H^{\prime }<0$, hence two negative Jacobian eigenvalues: the saddle separates at least two stable steady-state alternatives. The divergence is $-2$, so periodic attractors cannot replace these equilibria in a bounded planar region. Nongeneric flat crossings or intervals of equilibria require a nonlinear stability interpretation; the strict saddle inequality alone does not guarantee that every other root is hyperbolic. This is the usual mutual repression stability criterion behind bistability.

The conclusion also has a nonlinear trapping justification. Choose $l<u$ on the left of the saddle with $H(l)>0>H(u)$; such endpoints exist by the signs just proved. The rectangle $[l,u]\times [g(u),g(l)]$ is forward invariant: on the vertical edges, $f(y)-l\ge f(g(l))-l>0$ and $f(y)-u\le f(g(u))-u<0$, while the horizontal-edge velocities point inward or are tangent. The same construction on the right of the saddle gives a second disjoint trapping rectangle. Each contains a nonempty compact attracting invariant set, obtained by intersecting its nested forward images; strict vertical inwardness also rules out invariant boundary trajectories. If a downward crossing is isolated, endpoints can be taken arbitrarily close to it, proving nonlinear stability, and the absence of periodic orbits makes nearby trajectories converge to that equilibrium. Intervals of equilibria instead give attracting alternatives consisting of families, so bounded monotone functions without nondegeneracy do not necessarily produce two isolated hyperbolic attractors.

At a positive equilibrium of the bounded Hill model, the mutual repression stability criterion becomesThus $mn>1$ is necessary. Because the same $\lambda$ is imposed in both synthesis functions, it is not sufficient for arbitrary unequal positive exponents. The additional equilibrium constraint isHere is an exact criterion for all the printed exponent choices, including nonintegers. If $m>1$ and $n>1$, multistability is possible: for large $\lambda$ the intermediate equilibrium satisfies $x\sim {\lambda }^{(m-1)/(mn-1)}$, $y\sim {\lambda }^{(n-1)/(mn-1)}$, so both increase without bound and $K\to mn>1$. These asymptotics follow by balancing $x{y}^m\sim \lambda$ and $y{x}^n\sim \lambda$; the limiting two algebraic equations have a nonsingular log-Jacobian when $mn\ne 1$, so a nearby exact equilibrium exists for large $\lambda$. If both exponents are at most one, $K<1$ everywhere and multistability is impossible.

For the remaining case, interchange $x,y$ if necessary so $0<m\le 1<n$. For each $x>0$, let $Y(x)>0$ be the unique solution ofwhich exists because the left side is strictly decreasing; for $m=1$ the right side always exceeds one. DefineThen the exact answer for this remaining range isThis is a one-variable test depending only on $m,n$, with no unknown $\lambda$. To prove it, parameterize all equilibria by $\lambda (x)=x(1+Y(x{)}^m)$. Implicit differentiation givesThe denominator is positive, and $\lambda (x)$ goes from zero to infinity as $x$ goes from zero to infinity. A negative-slope portion therefore forces two folds and an interval of three equilibria, with two attracting branches. If $K\le 1$ everywhere, $\lambda$ is strictly increasing apart from possible isolated stationary points, so there is only one equilibrium for each $\lambda$. This is the common-strength Hill repression criterion.

For illustration and a closed test on the boundary $m=1$, $Y=x/(1+x^n-x)$ andobtained at $x^n=(n+1)/(n-1)$. In particular $(m,n)=(1,2)$ has $mn>1$ but maximum $3\sqrt{3}/8<1$, so it cannot be multistable. For positive integer exponents the allowed choices simplify to both exponents at least two, or one equal to one and the other at least four. For arbitrary real exponents the preceding exact criterion is needed; replacing it by $mn>1$ would discard the common-strength constraint.

The unbounded power laws behave differently. Combining their equilibrium equations gives $x^{1-mn}={\lambda }^{1-m}$. If $mn\ne 1$, there is exactly one positive equilibrium,Its slope product is exactly $mn$, so it is attracting if $mn<1$ and a saddle if $mn>1$, but there are never two isolated attracting equilibria. If $mn=1$, equilibria exist only when ${\lambda }^{1-m}=1$; when this holds there is a whole curve of equilibria, with a zero tangent eigenvalue and transverse eigenvalue $-2$. This includes $m=n=1$, $xy=\lambda$, for every $\lambda$. A neutrally degenerate continuum is not bistability between isolated attractors. Thus these power laws do not generate ordinary multistability; boundedness was essential to the outer-root argument in part (a).

A nonzero spatially constant steady state satisfies $v=u$ and $v=u^2$, hence $(u,v)=(1,1)$. The first derivatives of $u^2/v-u$ there are $1,-1$, while those of $Q(u^2-v)$ are $2Q,-Q$. ThusSince $Q>0$, both eigenvalues have strictly negative real part exactly when $Q>1$. For $0<Q<1$ the equilibrium is unstable. At $Q=1$ its linear eigenvalues are $\pm i$; this is the marginal boundary and does not give asymptotic stability from linearization. The positive-trace criterion in the hint tests positive real parts; stability requires applying it to $-J$.

A linear operator between Banach spaces is a compact operator when it maps the unit ball into a relatively compact set, equivalently when every bounded sequence of inputs has an image subsequence converging in norm. Such an operator is bounded. For a bounded linear operator on a complex Banach space, the resolvent set consists of the complex $\lambda$ for which $T-\lambda I$ is bijective with bounded inverse; its complement is the spectrum. The point spectrum consists of those $\lambda$ for which $T-\lambda I$ has a nonzero kernel, namely the eigenvalues.

On a Hilbert space, $T$ is self-adjoint if $T=T^{\ast }$, or equivalently $⟨Tx,y⟩=⟨x,Ty⟩$ for all vectors. The given prescription extends uniquely toIt is bounded with norm one. The real diagonal entries prove self-adjointness by the inner-product identity. Define the finite-rank truncation $T_{m}x={\sum }_{j\le m}{x}_j{e}_j/j$. Then $\Vert T-T_m\Vert =1/(m+1)\to 0$. To see compactness directly, extract successively convergent subsequences of the finitely many first coordinates from a bounded input sequence; the uniform tail bound just displayed makes the resulting image subsequence Cauchy. Hence $T$ is compact and self-adjoint.

Each $1/j$ is an eigenvalue with eigenspace $\mathbb{C}{e}_j$. If $Tx=\lambda x$, each coordinate satisfies $(1/j-\lambda )x_j=0$, so there are no other eigenvalues; in particular zero is not an eigenvalue. But $\Vert T{e}_j\Vert =1/j\to 0$, which contradicts a bounded inverse at zero. Thus zero is in the spectrum.

For $\lambda \notin \{0,1,1/2,1/3,\dots \}$, its distance from this closed set is positive. The diagonal formulathen defines a bounded inverse. This proves the complete description

For a normalized trial state in the quadratic-form domain of a self-adjoint Hamiltonian operator bounded below, its expectation of $H$ is at least the ground energy $E_0$. Minimizing this Rayleigh quotient over a tractable trial family therefore supplies an upper bound.

For the two states, set $a_{\pm }=(a_1\pm a_2)/\sqrt{2}$. Their quotient isThe strict assumption $ϵ<sE$ excludes $|s|=1$, since equality in the overlap Cauchy-Schwarz bound would then force the two states to be proportional and $ϵ=sE$. Thus $|s|<1$. The two generalized eigenvalues are $(E\pm ϵ)/(1\pm s)$, and their difference has the sign of $ϵ-sE$. The symmetric trial is the lower one, so

For the single attractive delta well, the supplied wavefunction has norm one because $\int \lambda e^{-2\lambda |x|}dx=1$. Its kinetic energy in quadratic-form notation is ${\hslash }^2\int |{\psi }_1^{\prime }{|}^2/(2m)={\hslash }^2{\lambda }^2/(2m)$, and its delta potential energy is $-K|{\psi }_1(0){|}^2=-K\lambda$. With $\lambda =mK/{\hslash }^2$ this givesIts derivative jump is ${\psi }_1^{\prime }(0+)-{\psi }_1^{\prime }(0-)=-2\lambda {\psi }_1(0)$, exactly the delta boundary condition, so it is an eigenstate. A decaying bound state on the two half-lines must have a common decay rate fixed by this same jump; there is only this one negative-energy state, confirming it is the ground state.

For $R>0$, integrate the overlap on the three regions $x<0$, $0<x<R$, $x>R$. Put $a=\lambda R$. The result isThe second equality is the single-well energy plus the extra delta expectation; the third follows by applying either single-well eigenvalue equation in the cross matrix element. Moreover $ϵ-sE=K\lambda e^{-a}[(1+a)e^{-2a}-1]<0$, so the preceding bound applies. Substitution givesFor widely separated wells, each localized orbital is close to an isolated-well ground state and the low-energy subspace is accurately their span; the symmetric combination captures the leading tunneling energy lowering. In the limit both exact and trial energies approach $E_B$. This illustrates electronic molecular binding: sharing an orbital between attractive centers can lower its energy below the separated-center value. A full equilibrium bond length would also require the repulsion of the positive centers.

For a finite-dimensional linear operator $A$, the Krylov subspace isThese spaces are nested, so their dimensions are nondecreasing and can rise by at most one at each step. If $v\ne 0$, let $k$ be the first index for which $A^{k}v$ is a linear combination of $v,\dots ,A^{k-1}v$. The preceding $k$ vectors are independent. Multiplying that dependence by $A$ and repeatedly substituting shows every higher power also lies in their span. ThusSuch an index exists because the ambient space is finite-dimensional. If $v=0$, the separate answer is $d_m=0$ for every $m$, with $k=0$.

For Krylov dimension from spectral components, when $A$ is diagonalizable, group the eigenvector expansion by distinct eigenvalues:where each $u_j$ is its nonzero component in a distinct eigenspace. Then $A^{r}v={\sum }_j{\lambda }_j^r{u}_j$. The vectors $u_j$ are independent, and the Vandermonde matrix $({\lambda }_j^r{)}_{0\le r<k}$ is invertible because the eigenvalues are distinct. Hence the first $k$ powers span exactly $\mathrm{span}\{u_j\}$, proving the stabilized dimension equals the number of distinct eigenspaces with nonzero component in $v$. These $u_j$ themselves are eigenvectors, so this is also the minimum number of eigenvectors needed to represent $v$ when the eigenvectors may be chosen freely.

If the printed phrase instead means the count of nonzero coefficients in an arbitrary fixed eigenbasis, it is false with repeated eigenvalues. For example $A=I$ and $v=e_1+e_2$ use two vectors of that basis but have $\dim {\mathcal{K}}_m=1$; the vector $v$ itself is an eigenvector. For a simple spectrum there is no distinction. This qualification resolves the claim without assuming an unstated simple spectrum.

On the standard two-dimensional representation, $t_{\theta }$ has eigenvalues $e^{i\theta },e^{-i\theta }$. Its $m$th symmetric power has monomial basis $X^{m-j}{Y}^j$, $j=0,\dots ,m$, of weights $m-2j$. The character is thereforeThe quotient is interpreted by continuity at $0,\pi$, where its values are $m+1$ and $(-1{)}^m(m+1)$.

Apply the Weyl integration formula for SU2 to the character product:The last equality follows directly by product-to-sum integration. A finite-dimensional representation of a compact group decomposes into irreducibles, and the squared character norm is the sum of squared multiplicities. Since this norm is one, there is exactly one constituent with multiplicity one. Thus every symmetric power is irreducible, and the different powers are pairwise inequivalent.

Averaging the fast fluctuations in $yx=x{x}^{T}w$ givesThus a nonzero steady vector satisfiesIts direction is an eigenvector of the input second-moment matrix, and its magnitude is the square root of that eigenvalue. The zero vector is also a steady state and satisfies the displayed relation with $\lambda =0$. This is an averaged steady state; the microscopic fluctuating update need not vanish at each instant.