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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.

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

Let $H$ be the one-dimensional Hamiltonian operator and letbe its discrete energy eigenvalues. Expanding a normalized trial wavefunction in an orthonormal energy basis,givesThus the Rayleigh-Ritz variational principle givesfor every nonzero admissible trial wavefunction. One chooses a parameterized family and minimizes the quotient.

When $V(x)=V(-x)$, parity commutes with $H$. In one dimension the nondegenerate ground state is even and the first excited state is odd. Restricting the trial family to odd wavefunctions makes every trial state orthogonal to the ground state, soThis is the odd-state variational principle for an even potential.

Sinceexcept at the single point $x=0$, and the kinetic-energy expectation is nonnegative, every normalized wavefunction satisfiesTherefore

For the upper bound, use the admissible choiceThenThe Rayleigh-Ritz variational principle now givesIndeed, the displayed calculation provides the sharper remainder $O(V_0^4)$.

For any nonzero state, the Rayleigh quotient isExpand $|\psi ⟩={\sum }_n{c}_n|{\psi }_n⟩$ in normalized energy eigenstates. ThenBecause the ground state is unique, equality holds exactly when all coefficients except $c_0$ vanish. Thus the Rayleigh-Ritz variational principle gives its minimum at the ray of $|{\psi }_0⟩$.

For the proposed exact state, write $\tilde{\psi }=e^{-\beta x^n}$. Its logarithmic derivatives giveThe stationary Schrödinger equation, after multiplication by $2m/{\hslash }^2$, isMatching the highest power requires $2n-2=6$, so $n=4$, and cancellation of $x^6$ requires $16{\beta }^2=1$. Normalizability selects $\beta =1/4$. Then $12\beta =3$ also cancels the quadratic term, leaving $\epsilon =2$. ThereforeThis is the exact ground state of a solvable sextic potential candidate $e^{-x^4/4}$.

For the Gaussian trial state ${\psi }_{\alpha }=e^{-\alpha x^2/2}$, normalization cancels from the quotient. With respect to the probability density proportional to $e^{-\alpha x^2}$,HenceThe stationary equation isWriting $z={\alpha }^2$, the quadratic has exactly one positive root,Since the quotient tends to infinity as $\alpha \downarrow 0$ or $\alpha \to \infty$, this is the unique global minimizer. ThusUsing the stationary equation to replace $15/(8{\alpha }_{\ast }^3)$ by ${\alpha }_{\ast }/6+1/(2{\alpha }_{\ast })$ gives the best estimateThis is the Gaussian variational estimate for a solvable sextic potential.

The exact eigenfunction $e^{-x^4/4}$ is positive and has no nodes. The nodeless theorem for a one-dimensional ground state therefore identifies it as the true ground state, with dimensionless energy $2$. The variational value is consistent: since ${\alpha }_{\ast }^2>3/2$,so $E_0^{\ast }>{\hslash }^2/m=E_0$, as every trial-state upper bound must satisfy.

For $\psi ={\sum }_n{\alpha }_n{\varphi }_n$ and an orthonormal trial set,Stationarity subject to ${\alpha }^{†}\alpha =1$ givesThe finite-subspace variational method, equivalently the Rayleigh-Ritz variational principle, says that the minimum quotient is the smallest eigenvalue of $\mathcal{H}$. It is therefore the optimal upper bound on $E_0$ available from this span.

Set $T=A-zI$. The squares of the singular values of $T{P}_n^{\ast }$ are the eigenvalues of the finite-dimensional positive operatorThe Rayleigh-Ritz variational principle and its min-max characterization show that, as the trial space grows, its $(j+1)$-st eigenvalue counted upward cannot increase. Therefore, for each fixed $j\ge 0$,once $n>j$. Being nonnegative, it has a limitThe core assumption ensures that these Ritz limits are the min-max values of $T^{\ast }T$, rather than values for a smaller closed restriction.

For a normal operator, the spectral theorem gives three possibilities.

vanish, while ${\sigma }_{\inf +m}(T)>0$.

For $k\in \mathbb{N}$, defineIn the first case $H_k=0$ for all sufficiently large $k$. In the second case its first $m$ summands tend to one and all remaining summands eventually vanish, so $H_k\to m$. In the third case $H_k=k+1\to +\infty$. These are exactly the three values in the definition of $h$, and hence