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For the dimensionless quantum harmonic oscillator $H=-d^2/d{x}^2+x^2$, use the compactly supported trial ${\psi }_{\gamma }=\cos (\gamma x)$ on $|x|\le \pi /(2\gamma )$ and zero elsewhere, with $\gamma >0$. It is continuous and belongs to the oscillator energy quadratic form domain: its weak derivative is square-integrable even though its ordinary derivative jumps at the endpoints. Exact integration givesThe minimum occurs at ${\gamma }^4=({\pi }^2-6)/12$, giving the displayed upper bound, approximately $1.13572$, compared with exact ground state value $1$. This illustrates why Rayleigh-Ritz variational principle trials need the energy form domain rather than the smaller operator domain.

For the dimensionless quantum harmonic oscillator $H=-d^2/d{x}^2+x^2$, the normalized compact trial function is ${\psi }_a=\sqrt{15/(16a^5)}(a^2-x^2)$ on $|x|\le a$ and zero elsewhere. Its kinetic and potential contributions to the energy quadratic form are $5/(2a^2)$ and $a^2/7$. Minimizing their sum gives $a^4=35/2$ and the displayed upper bound, approximately $1.19523$, compared with exact ground state eigenvalue $1$. The weak derivative is square-integrable because the trial is continuous at its endpoints. Jumps in the ordinary first derivative do not invalidate the Rayleigh-Ritz variational principle: it requires the energy form domain, rather than the smaller operator domain.

With vanishing surface pressure/mass density and regular admissible displacements, integration by parts makes the axisymmetric adiabatic displacement operator symmetric in the mass density-weighted inner product. Completing the pressure square gives the displayed energy form. Nonnegative form for every admissible displacement excludes exponentially growing axisymmetric adiabatic modes in the Cowling approximation. A negative trial Rayleigh quotient proves negative spectrum by the Rayleigh-Ritz variational principle. Nonnegative stratification and epicyclic coefficients give a simple sufficient stability condition.

For the dimensionless half-harmonic oscillator $-{\psi }^{\prime \prime }+x^2\psi =\lambda \psi$ on $x>0$ with $\psi (0)=0$, the trial family $x{e}^{-\alpha x}$, $\alpha >0$, is admissible for the Rayleigh-Ritz variational principle. The moments ${\int }_0^{\infty }{x}^n{e}^{-2\alpha x}dx=n!/(2\alpha {)}^{n+1}$ give norm squared $1/(4{\alpha }^3)$, derivative energy $1/(4\alpha )$, and potential energy $3/(4{\alpha }^5)$. Its Rayleigh quotient is ${\alpha }^2+3/{\alpha }^2$, minimized at $\alpha =3^{1/4}$. The resulting upper bound $2\sqrt{3}$ exceeds the exact lowest eigenvalue three; the exact ground-state shape is $x{e}^{-x^2/2}$.

For $H=-d^2/d{x}^2+x^2+ϵx^4$, use the normalized real wavefunction ${\psi }_{\alpha }=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$, $\alpha >0$. The Gaussian integral gives $⟨x^2⟩=1/(2\alpha )$ and $⟨x^4⟩=3/(4{\alpha }^2)$, while the kinetic contribution is $\alpha /2$. Thus the Rayleigh quotient is the displayed expression. For positive $ϵ$, its stationary width satisfies ${\alpha }^3-\alpha -3ϵ=0$. At $ϵ=2$ the unique positive solution is $\alpha =2$, giving $E_0\le 13/8$ by the Rayleigh-Ritz variational principle. For $ϵ<0$, taking $\alpha \downarrow 0$ makes the negative quartic contribution dominate, so the energy is unbounded below even within this normalized family. The positive-coupling bound and negative-coupling instability concern different signs of the same variational expression.

For $H=p^2/(2m)+k{x}^4$ with $k>0$, the normalized even trial wavefunction $f_{\alpha }=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$ has Rayleigh quotientThe Gaussian moment recurrence gives the potential term, and integration by parts gives the kinetic term. Its minimum is at ${\alpha }^3=6mk/{\hslash }^2$, proving the displayed upper bound by the Rayleigh-Ritz variational principle.

An odd trial family $g_{\beta }=(4{\beta }^3/\pi {)}^{1/4}x{e}^{-\beta x^2/2}$ is orthogonal to the exact even ground state. The Sturm oscillation theorem identifies the lowest odd state as the first excited state, so minimizing within this family bounds $E_1$:Orthogonality to an approximate ground trial alone would not justify this excited-state bound; odd parity ensures orthogonality to the exact ground state. Both bounds exhibit the characteristic quartic-oscillator energy scale $({\hslash }^{4}k/m^2{)}^{1/3}$.

For $E(a)=a/2-V_0\sqrt{a/(1+a)}$ with $a,V_0>0$, differentiation gives $E^{\prime }(a)=1/2-V_0/[2\sqrt{a}(1+a{)}^{3/2}]$. The positive stationarity equation has a unique solution because its left side increases strictly. Moreover $E^{\prime \prime }(a)=V_0(1+4a)/[4a^{3/2}(1+a{)}^{5/2}]>0$, so it is the global minimum. Substituting the stationarity equation gives the displayed negative energy. The only positive zero is $(\sqrt{1+16V_0^2}-1)/2$; $a=0$ is only a limiting zero, since the limiting constant trial is not a square-integrable function. The Rayleigh-Ritz variational principle therefore proves a negative-energy bound state for every positive well depth. For weak depth, the simpler choice $a=V_0^2$ yields an explicit upper bound $E_0\le -V_0^2/2+\epsilon$ withThis is a variational upper bound, not an equality for the exact weak-depth coefficient.

A confined ground state requires $k>0$; take $\alpha >0$ so the trial wavefunction is normalizable. Its normalized form isThe Gaussian moment recurrence gives $⟨x^2⟩=1/(2\alpha )$ and $⟨x^4⟩=3/(4{\alpha }^2)$. Also, integration by parts gives $⟨p^2⟩={\hslash }^2\int |f_{\alpha }^{\prime }{|}^{2}dx={\hslash }^2\alpha /2$. The Rayleigh quotient for the quartic oscillator is consequentlyDifferentiation yields ${\mathcal{E}}^{\prime }(\alpha )={\hslash }^2/(4m)-3k/(2{\alpha }^3)$, so the unique minimum is at ${\alpha }_{\ast }=(6mk/{\hslash }^2{)}^{1/3}$. Since the quotient diverges at both endpoints $0$ and $\infty$, this is the global minimum in the trial family. The Rayleigh-Ritz variational principle givesFor the first excited energy eigenstate, choose the odd trial familyThe even confining potential has an even, nodeless ground state, by the nodeless theorem for a one-dimensional ground state. By the Sturm oscillation theorem, its first excited state is odd with one node. The proposed family has the same parity operator eigenvalue and one node, and is orthogonal to the exact ground state, not merely to its Gaussian trial approximation. Minimization in the odd linear subspace therefore bounds the first excited energy from above. For completeness, its Rayleigh quotient is $3{\hslash }^2\beta /(4m)+15k/(4{\beta }^2)$, minimized at ${\beta }^3=10mk/{\hslash }^2$, giving $E_1\le (9{\hslash }^2/8m)(10mk/{\hslash }^2{)}^{1/3}$. These are Gaussian variational bounds for a quartic oscillator.

If $k=0$, there is no normalizable free particle ground state on the whole line, though the spectral infimum is zero. If $k<0$, the Hamiltonian operator is not bounded below, so a ground-energy bound of the requested kind does not exist.

Normalize the trial quantum state as $(|{\varphi }_0⟩+c|{\varphi }_1⟩)/\sqrt{1+c^2}$. With the phases used above, the off-diagonal position matrix element is $x_{01}=-32a/(9{\pi }^2)$. DefineThe diagonal perturbation matrix elements vanish by parity, so the Rayleigh quotient isThe stationarity equation is $W+\Delta c-W{c}^2=0$. The minimum continuously connected to the unperturbed ground energy eigenstate hasEquivalently, diagonalize the Hamiltonian operator matrix in this two-dimensional trial space. Its lower eigenvalue isThereforeThis retains exactly the $k=1$ term of the full ground-state second-order nondegenerate perturbation theory sum. Every omitted term is negative, so the full quadratic correction is more negative than this restricted one. This is consistent with the Rayleigh-Ritz variational principle: minimizing over a two-state subspace gives an upper bound on the exact ground energy. The bound applies to the unexpanded two-state minimum; a truncated series is to be compared order by order.

Use $⟨O⟩=⟨G|O|G⟩$. The given state is a two-species uniform boson number state: only the $q=0$ mode of each species is occupied. Applying an annihilation operator to an unoccupied mode gives zero, while $A_0^{†}{A}_0$ has eigenvalue $n_A$. ThusThe two species have independent fixed occupation numbers, so mixed expectation values factorize. The inverse discrete Fourier transforms yieldIn the momentum-space Hamiltonian operator, only $p=q=r=t=0$ contributes: each annihilation operator must remove an occupied boson, and each creation operator must restore it to the same mode. HenceThere is no replacement of $n_B$ by $n_B+1$ in this energy: in the real-space calculation the relevant neighboring-site commutator is zero. Both expressions assume distinct neighbouring sites, as in ii.

To match the prescribed spin length in the mean, impose $n_A+n_B=2sL=:N$. At fixed integer $N$, the energy is minimized by maximizing $n_A{n}_B=[N^2-(n_A-n_B{)}^2]/4$. Consequently the best choices within this relaxed family areWhen $N$ is even, $n_A=n_B=sL$ and $⟨H⟩=-J{s}^{2}L$. When $N$ is odd, either assignment of $(N-1)/2,(N+1)/2$ gives $⟨H⟩=-J(N^2-1)/(4L)$. Without fixing $N$, this auxiliary boson Hamiltonian operator has no lower bound: taking $n_A=n_B\to \infty$ sends its expectation value to $-\infty$.

Next, $S_i^x=(a_i^{†}{b}_i+b_i^{†}{a}_i)/2$. Each term changes the two species' occupation numbers, producing a state orthogonal to $|G⟩$. ThereforeFor distinct sites $i,j$ modulo $L$, expand the product of the two spin operators. The terms $S_i^{+}{S}_j^{+}$ and $S_i^{-}{S}_j^{-}$ again have zero expectation value. The remaining terms giveHere $⟨b_i{b}_j^{†}⟩=n_B/L+{\delta }_{ij}$ and the Kronecker delta vanishes at distinct sites. ConsequentlyThe same holds for the $y$ components. At the balanced choice and as $L\to \infty$, each transverse two-point correlation function tends to $s^2/2$ independently of separation, whereas the one-point transverse magnetization is zero. This is transverse spin order in a number eigenstate: the state has a definite total $S^z=(n_A-n_B)/2$ and retains the global U(1) symmetry of rotations about $z$, yet has the transverse correlations of an ordered state with no selected common azimuth. The sum of the $x$ and $y$ correlations tends to $s^2$, describing ferromagnetic long-range order within this approximation. The thermodynamic limit is understood with both the separation and $L$ growing, without identifying the two sites periodically.

Fixed-spin qualification. The proposed state is not actually in the physical fixed-$s$ Hilbert space for $s>0,L>1$. It satisfies the local spin constraint in the Schwinger boson representation only in expectation value. For example, with $L=2,s=1/2,n_A=n_B=1$,The nonzero component $a_0^{†}{b}_0^{†}|\mathrm{vac}⟩/2$ has local total occupation numbers $(2,0)$ and hence local spins $(1,0)$, instead of $(1/2,1/2)$. More generally the local total number operator $N_n$ has varianceEach species contributes a binomial distribution at a given site, with success probability $1/L$, and the two contributions are independent. In the physical subspace that variance must be zero. The proposed state even hasby the Casimir operator identity from a. Thus the calculated energy is a relaxed mean-field theory estimate, not a Rayleigh-Ritz variational principle bound for the stated fixed-$s$ quantum XY chain. Projecting onto $N_n=2s$ at every site would give a physical trial state and generally change its expectation values; in the two-site example the projected state has energy $-J$, whereas the unprojected value is $-J/2$. The calculated long-range order characterizes the supplied approximation and does not establish long-range order in the exact one-dimensional ground state.

For fixed electron-electron interaction and electron number, a nondegenerate ground-state electron density determines the external potential up to an additive constant, and hence determines the Hamiltonian and ground-state observables.

Suppose two potentials $v,v^{\prime }$ differing by more than a constant have different ground states $\Psi ,{\Psi }^{\prime }$ but the same density $\rho$. Strict Rayleigh-Ritz variational principle inequalities giveAdding produces the contradiction $E+E^{\prime }<E+E^{\prime }$. This proves the first Hohenberg-Kohn theorem under the stated nondegeneracy assumptions. A constant potential shift changes energy by $N$ times that constant without changing the density.^[ref]

Work at fixed enclosed mass $m$, so the perturbations are fluid Lagrangian perturbations. The equilibrium acceleration vanishes. Varying the factor $r^{-4}$ in the gravitational force contributes $4Gm\delta r/(4\pi r^5)$, while varying the factor multiplying the acceleration gives no first-order term because the background acceleration is zero. With $\delta r=r\xi e^{-i\omega t}$, the linearized momentum equation isConservation of the mass of each displaced shell gives its fractional density change as minus the fractional volume change. ThusThe adiabatic relation is $\delta p=\gamma p\delta \rho /\rho =-\gamma p\chi$. Use $dm/dr=4\pi r^2\rho$ and the stellar hydrostatic equation $p^{\prime }=-Gm\rho /r^2$ to turn the mass-coordinate momentum equation into $(\delta p{)}^{\prime }-{\omega }^2\rho r\xi +4p^{\prime }\xi =0$. It follows that the radial stellar pulsation equation isThe stellar adiabatic exponent may vary with radius, so it must remain inside the derivative.

Expand the derivative and multiply by $r^3$. The self-adjoint differential equation becomesHence $f(r)=\gamma p{r}^4$ and $g(r)=r^3[(3\gamma -4)p{]}^{\prime }+{\omega }^2\rho r^4$ in the requested form. Multiplying by a real eigenfunction and integrating gives the boundary term $[\gamma p{r}^4\xi {\xi }^{\prime }{]}_0^R$. It vanishes for a regular centre and bounded admissible surface displacement and derivative, since $p(R)=0$. Therefore the weighted stellar pulsation Rayleigh quotient isReal coefficients and the symmetric endpoint conditions give the Sturm-Liouville problem; the Rayleigh-Ritz variational principle identifies its minimum with the lowest squared frequency. For complex amplitudes the corresponding quadratic forms use modulus squares, and the real eigenfunctions may be chosen for this calculation.

Integrate the second term of $K$ by parts. Its boundary term $-[(3\gamma -4)p{r}^3{\xi }^2{]}_0^R$ again vanishes. The remaining integrand isSince $\chi =a+3\xi$, one has $3{\xi }^2+2a\xi =({\chi }^2-a^2)/3$. Substitution proves the positive-square radial pulsation energy identityIf $\gamma \ge 4/3$ throughout the positive-pressure interior, both terms are nonnegative for every admissible displacement. No negative squared frequency is then possible. Hence radial dynamical instability requires $\gamma <4/3$ somewhere inside the star.

For the opposite necessary condition, test the quotient with a nonzero constant $\xi =C$. It has $\chi =3C$ andIf $\gamma \le 4/3$ everywhere and is strictly below $4/3$ on a positive-pressure interval, this is negative. The lowest eigenvalue is at most this negative trial quotient, proving instability. Strict dynamical stability requires the stronger weighted conditionand in particular $\gamma >4/3$ somewhere. This is the pressure-weighted radial instability criterion viewed as a necessary test for strict stability.

The word “stability” needs a marginal qualification: if $\gamma =4/3$ throughout, $\xi =C$ solves the pulsation equation with $\omega =0$. The star is marginally stable in this radial sense, so the strict $\gamma >4/3$ requirement applies to positive fundamental squared frequency, not to a definition that includes a zero-frequency neutral mode. There is no exponentially growing mode in this equality case, but the zero restoring force also permits a secular solution linear in time.

The conditions are useful as rapid tests, but local threshold crossings are not a complete criterion for a mixed exponent profile. If $\gamma \ge 4/3$ everywhere and exceeds it on a positive-pressure interval, the square identity forces $K>0$ for every nonzero eigenfunction: zero energy would first force constant $\xi$ and then force that constant to vanish. If $\gamma \le 4/3$ everywhere with strict inequality on an interval, the constant trial proves instability. When both types of region occur, $\gamma <4/3$ somewhere does not alone prove instability, and $\gamma >4/3$ somewhere does not alone prove stability. A negative pressure-weighted average proves instability, while a positive one tests only the homologous displacement; another radial shape may still have negative energy. Determining the mixed case requires minimizing the full weighted stellar pulsation Rayleigh quotient or solving the radial stellar pulsation equation. These are radial adiabatic statements and do not rule out nonradial, convective or nonadiabatic instabilities.

For a self-adjoint Hamiltonian operator bounded below, the Rayleigh-Ritz variational principle says $E[\psi ]=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩\ge E_0$. Expanding in energy eigenstates proves this, since the quotient is a weighted average of energies. Optimize a tractable trial family to obtain an upper bound on the ground energy.

If $H{\psi }_0=E_0{\psi }_0$, a regular trial error $\psi ={\psi }_0+ϵ\eta$ gives the exact cancellationThe linear terms vanish because $(H-E_0){\psi }_0=0$. Here $\eta$ is a fixed finite-energy direction, or is uniformly bounded in the quadratic-form norm. A Hilbert-space norm bound alone is insufficient for an unbounded Hamiltonian operator: a norm-$ϵ$ admixture of an eigenstate with energy of order ${ϵ}^{-2}$ can have order-one energy error. The usual quadratic-error claim therefore presupposes a controlled-energy perturbation.

For the first excited energy minimize the quotient on states orthogonal to the exact ground state. Alternatively diagonalize $H$ in a trial subspace and use the min-max principle: its second Ritz eigenvalue is an upper bound for the second exact eigenvalue. An arbitrary trial function without either orthogonality or a min-max construction need not bound the first excited level.

The one-dimensional even potential has even ground state and odd first excited state, by the Sturm oscillation ordering. Choose the automatically orthogonal odd trial function ${\psi }_a(x)=Nx{e}^{-a{x}^2/2}$, $a>0$. Integration by parts gives the kinetic numerator $\int |{\psi }_a^{\prime }{|}^2$, and the given Gaussian moments giveFor example, the kinetic factor is $(I_0-2a{I}_2+a^2{I}_4)/I_2=3a/2$, while the potential factors are $I_4/I_2$ and $I_6/I_2$. Differentiation gives the unique positive minimizer $a^3-a-5=0$. ThusThis is an upper estimate, not an exact eigenvalue. Improve it with odd polynomial-Gaussian trial functions such as $(x+b{x}^3)e^{-a{x}^2/2}$, or a larger odd Ritz basis, and optimize the width. The parity restriction keeps all such states orthogonal to the even ground state.

For a nonzero normalized or unnormalized trial function in the Hamiltonian's quadratic-form domain, the Rayleigh-Ritz variational principle gives $E_0\le ⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$. Expanding in energy eigenstates, or using the spectral measure, makes the quotient a weighted average of spectral energies, all at least $E_0$. For an even one-dimensional potential, the ground state is even and the first excited bound state is odd by the nodal ordering theorem. Restricting to odd trial functions therefore gives an upper bound on the lowest odd energy and on $E_1$ when that bound state exists; a negative odd quotient also certifies its existence.

Here ${\hslash }^2/(2m)=1$. For the Gaussian trial function, integration by parts gives kinetic quotient $\int |{\psi }^{\prime }{|}^2/\int |\psi {|}^2=a/2$, and the potential quotient is $-V_0\sqrt{a/(1+a)}$. ThereforeFor $a>0$, its zero obeys $a(1+a)=4V_0^2$, giving the unique positive zeroAlso $E^{\prime }(a)=1/2-V_0/[2\sqrt{a}(1+a{)}^{3/2}]$, so its stationary points satisfy $a(1+a{)}^3=V_0^2$. The left side increases strictly from zero to infinity, proving that there is exactly one stationary point, a minimum.

As $a\downarrow 0$, $E(a)=-V_0\sqrt{a}+O(a)$ approaches zero from below; as $a\to \infty$, it tends to infinity. The sketch and variational inequality therefore give a negative energy for every $V_0>0$. Since this decaying potential has essential spectrum beginning at zero, at least one bound state exists for every positive well depth.

For small $V_0$, the minimizer is $a_{\ast }=V_0^2-3V_0^4+O(V_0^6)$. Substitution gives $E(a_{\ast })=-V_0^2/2+V_0^4/2+O(V_0^6)$. Meanwhile the potential is strictly greater than $-V_0$ except at a single point and the kinetic energy is nonnegative, so a normalized bound-state expectation is strictly greater than $-V_0$. Thus

Let primes denote radial derivatives of the equilibrium and put $\Delta u=u^{\prime }+2u/r$. The linearized continuity, adiabatic pressure and Poisson equation areTake the time derivative of the Poisson equation and substitute continuity. Integration in radius givesCentre regularity excludes a changing point mass at the origin, so $C(t)=0$ andThe perturbed self-gravity must be retained; this calculation does not make the Cowling approximation.

The radial linear momentum equation is $\rho {\partial }_{t}u=-{\partial }_r\delta p-\delta \rho {\Phi }^{\prime }-\rho {\partial }_r\delta \Phi$. Differentiate it in time and use the preceding equations:The equilibrium has $p^{\prime }=-\rho {\Phi }^{\prime }$ and ${\Phi }^{\prime \prime }+2{\Phi }^{\prime }/r=4\pi G\rho$. The remaining terms simplify asConsequently the radial stellar oscillation equation isThe derivative acts on the full variable coefficient $\gamma (r)p(r)$; discarding ${\gamma }^{\prime }$ would change the result.

For $u=r\xi (r)e^{-i\omega t}$, $\Delta u=(3\xi +r{\xi }^{\prime })e^{-i\omega t}$. Substitute and multiply by $-r^3$. Combining the two terms proportional to ${\xi }^{\prime }$ into a total derivative gives the radial stellar pulsation equation,

Set $P=\gamma p{r}^4$, $W=\rho r^4$ and $Q=r^3[(4-3\gamma )p{]}^{\prime }$. The Sturm-Liouville operator isFor regular physical perturbations, $\xi$ is finite at the centre and $P{\xi }^{\prime }\to 0$ at the free surface. For a nonzero-frequency mode the fluid displacement is $u/(-i\omega )$, so vanishing Lagrangian pressure perturbation is equivalent to $\gamma p(3\xi +r{\xi }^{\prime })\to 0$. With $p(R_s)=0$ and finite $\xi$, this gives $P{\xi }^{\prime }\to 0$. Zero-frequency modes use the same displacement boundary condition directly. Assume bounded $\gamma$, positive $\rho ,p,\gamma$ in the interior and the usual finite-energy endpoint domain. Integration by parts givesThus this physical self-adjoint differential operator has real eigenvalues ${\omega }^2$. The endpoint domain is essential: the fact that $p$ vanishes does not by itself allow arbitrary singular trial functions. The regular free-surface realization of the Sturm-Liouville problem is the one used here.

The Rayleigh-Ritz variational principle gives the fundamental squared frequency as the infimum of the weighted stellar pulsation Rayleigh quotient,The infimum is over admissible finite-energy functions satisfying the physical endpoint conditions. If this quotient is nonnegative for every such function, all radial frequencies are real and there is no exponentially growing radial mode. A negative value for even one trial function proves a negative eigenvalue and an exponentially growing solution, because ${\omega }^2<0$. A zero lowest value is marginal and needs separate treatment of neutral motion.

Choose the homologous trial function $\xi =1$, which corresponds to radial velocity proportional to $r$. Its gradient contribution is zero, andThe denominator $\int \rho r^{4}dr$ is positive. HenceThis pressure-weighted radial instability criterion is sufficient, not necessary: another trial function can detect instability even if this one does not. For a constant stellar adiabatic exponent it recovers instability below $4/3$. At constant $\gamma =4/3$, the homologous mode is neutral. For constant $\gamma >4/3$, $Q=(4-3\gamma )r^3{p}^{\prime }>0$ in a normally stratified hydrostatic equilibrium, so the quotient is positive and the star is radially stable.

Let $a_{\xi }=\mathbf{\xi }\cdot \nabla \Psi$, $d_{\xi }=\nabla \cdot \mathbf{\xi }$ and ${\rho }_{\Psi }=d\rho /d\Psi$. Since $\nabla p=-\rho \nabla \Psi$, the pressure and mass density perturbations are $\delta p_{\xi }=\rho a_{\xi }-\gamma p{d}_{\xi }$ and $\delta {\rho }_{\xi }=-\rho d_{\xi }-{\rho }_{\Psi }{a}_{\xi }$. Integrating the pressure-gradient term by parts in the mass density-weighted inner product givesThe boundary term vanishes for regular admissible displacements because $p,\rho$ vanish there and $\delta p=\rho a-\gamma pd$. Completing the pressure square giveswhereThis effective-potential stratification coefficient is real. All coefficients of the bilinear form are real, so $⟨\eta ,\mathcal{F}\xi ⟩=⟨\mathcal{F}\eta ,\xi ⟩$: the operator is symmetric on the stated boundary domain, giving the usual self-adjoint realization of the stellar normal-mode problem.

Set $\eta =\xi$ and use $\mathcal{F}\xi =-{\omega }^2\xi$. The Cowling energy principle for a rotating barotropic star isThe Rayleigh quotient $Q[\xi ]/\int \rho |\xi {|}^2$ tests stability. If $Q\ge 0$ for every admissible displacement, no mode has ${\omega }^2<0$, so there is no exponentially growing mode. If an admissible trial displacement has $Q<0$, the Rayleigh-Ritz variational principle puts negative spectrum below zero; in the usual discrete stellar mode problem this gives a mode with $\omega =i\sigma$, $\sigma >0$. Equality allows neutral modes, rather than establishing strictly positive frequencies. A locally negative coefficient alone is not a complete instability proof: a trial function must also control the pressure and other positive terms.

Let $\beta =\mu [A]+\mu [B]$ and $\gamma =\mu [A+B]$, where all three matrices are symmetric. Submultiplicativity and the triangle inequality giveand similarly the other commutator term is bounded by $2\Vert B{\Vert }_2{e}^{x\beta }$. Apply part (a) also to $A+B$ in the integral from part (b). The outer factor one-half cancels these twos, leavingThe exponential divided difference evaluates this integral. Thuswhen $\beta \ne \gamma$, andwhen they coincide. The second expression is both the direct equal-exponent integral and the continuous limit of the first. The Rayleigh-Ritz variational principle also gives $\gamma \le \beta$, although the integral computation does not require a strict inequality. These are valid coarse norm bounds; the cancellation between the two products can make the actual small-step error substantially smaller.

By the Rayleigh-Ritz variational principle, $\mathrm{tr}({\Pi }_{0}D)$ is maximized by taking ${\Pi }_0$ to project onto the positive-eigenvalue eigenspace of $D$. Put $|⟨{\alpha }_0|{\alpha }_1⟩|=\cos \theta$. On $S$, the characteristic polynomial of $D$ isso its positive eigenvalue isPart (b) then gives the Helstrom-Holevo bound