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Choose a shift $\mu$ which is not an eigenvalue. Inverse iteration applies power iteration to $B=(A-\mu I{)}^{-1}$, implemented by a linear system solve rather than by forming an explicit matrix inverse:A single LU decomposition of $A-\mu I$ can be reused for all the solves at this fixed shift. An eigenpair $A{u}_j={\lambda }_j{u}_j$ transforms intoThus an eigenvalue ${\lambda }_{\ast }$ uniquely closest to $\mu$ becomes the dominant eigenvalue of $B$. For a diagonalizable matrix, the argument in the preceding subpart gives convergence of diagonalizable fixed-shift inverse iteration, provided the starting coefficient in the target eigenspace is nonzero, with power iteration direction error $O(q_{\mu }^k)$, whereFor any prescribed simple eigenvalue, taking a sufficiently close but unequal shift makes the numerator small while the denominator stays close to its positive separation from the remaining eigenvalues. This makes $q_{\mu }$ small and gives very fast convergence per iteration. The Rayleigh quotient of $v_k$ estimates ${\lambda }_{\ast }$ directly. If ${\beta }_k$ estimates the associated eigenvalue of $B$, the corresponding estimate is $\mu +{\beta }_k^{-1}$.

The selection principle also applies when the other eigenvalues have nontrivial blocks in the Jordan normal form: their contributions acquire powers of $k$, multiplying strictly smaller geometric factors. A simple uniquely closest eigenvalue still attracts the direction when the Riesz projection of $v_0$ onto its generalized eigenspace is nonzero. For nonreal target eigenvalues, allow complex shifts and arithmetic; a real shift cannot distinguish a conjugate pair of equal distances. Setting $\mu ={\lambda }_{\ast }$ makes the solve singular, and an exactly zero starting target component prevents convergence to that target in exact arithmetic. When ${\lambda }_{\ast }<\mu$ is real, account for the sign behavior of fixed-shift inverse iteration.

Take a smooth variation $u+\epsilon v$ with $v=0$ on the boundary. Differentiating under the integral gives the first variationIntegration by parts removes the derivatives of $v$, and the boundary contribution vanishes. The fundamental lemma of the calculus of variations therefore givesFor a regular constraint $J[u]=c$, the Lagrange multiplier rule says that an extremal satisfies $\delta I=\lambda \delta J$ for a constant $\lambda$. Equivalently, apply the preceding Euler–Lagrange equation to $F-\lambda G$. Regularity means that $\delta J$ is not identically zero as a linear functional on the admissible variations. For the quadratic constraint below this holds at every nonzero admissible $u$; a zero constraint leaves only $u=0$.

In polar coordinates the constrained integrand isIts Euler–Lagrange equation is $-2\lambda ru-{\partial }_r(2r{u}_r)-{\partial }_{\theta }(2u_{\theta }/r)=0$, orThe field is single-valued and periodic in $\theta$. Substitution $u=f(r)\cos (n\theta )$ with $n\in \{0,1,2,\dots \}$ givesThe centre is part of the disk, so one must also select the regular radial solution: for $n\ge 1$, smoothness gives $f(r)=O(r^n)$ and the regular derivative behaviour; for $n=0$, $f$ is bounded and $f^{\prime }(0)=0$. A singular solution at zero does not define an admissible smooth field on the disk.

For fixed $n$, let $c_n={\int }_0^{2\pi }{\cos }^2(n\theta )d\theta$, so $c_0=2\pi$ and $c_n=\pi$ for $n\ge 1$. The angular derivative term has integral $n^2{c}_n$, including zero when $n=0$. ConsequentlyThe one-dimensional Euler–Lagrange equation for $\hat{I}-\lambda \hat{J}$ isexactly the same radial equation. The common factor $c_n$ cancels.

For two regular solutions with the same $n$ and different multipliers, multiply the equation for $f_1$ by $f_2$, multiply that for $f_2$ by $f_1$, subtract and integrate. The result isAt $a$, both functions vanish; at zero, regularity makes the boundary term tend to zero. Since the multipliers are distinct,This is weighted Sturm–Liouville orthogonality with weight $r$.

Finally, multiply one radial equation by $f$ and integrate by parts. The same endpoint conditions yieldMultiplication by $c_n$ provesFor any nonzero admissible solution the energy on the left before subtraction is positive, so its multiplier is the positive Rayleigh quotient $\lambda =\hat{I}/\hat{J}$.

Use the normalized Rayleigh quotient $R(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$. Its real directional derivative in an arbitrary complex direction $\eta$ isIt vanishes for every direction precisely when $(H-R)\psi =0$, since testing $\eta =(H-R)\psi$ gives its squared norm. Conversely an eigenvector makes every derivative zero. Stationarity over all states is stronger than stationarity in a restricted trial family.

Normalize the Gaussian as ${\psi }_{\alpha }=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$, with $\alpha >0$. Its moments are $⟨X^2⟩=1/(2\alpha )$, $⟨X^4⟩=3/(4{\alpha }^2)$ and $⟨P^2⟩={\hslash }^2\alpha /2$. The printed potential energy consequently givesPut $a=\alpha \hslash /(m\omega )$. Then $\mathcal{E}/(\hslash \omega )=a/4+1/(4a)+15/(256a^2)$, whose second derivative is $1/(2a^3)+45/(128a^4)>0$. It diverges at both positive endpoints, so its sole stationary point is the unique global minimum. Its equation isThe variational principle therefore yields the optimized upper boundThe printed minimizer is inconsistent with the printed quartic coefficient. At $a=3/2$ the derivative is $5/48>0$, so this is not the stationary Gaussian for that potential energy. It is nevertheless a legitimate trial vector and gives the simple, weaker bound $\boxed{E_0\le 109\hslash \omega /192}$. If the quartic coefficient were instead $5m^2{\omega }^3/(16\hslash )$, four times the printed value, then $a=3/2$ would indeed minimize the family and its energy would be $31\hslash \omega /48$. This possible coefficient correction is identified explicitly rather than silently used. The reusable calculation is Gaussian trial for a quadratic-plus-quartic oscillator.

For a nonzero trial vector use the Rayleigh quotient $\mathcal{E}(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$. Its first variation in any allowed direction $\eta$ isVanishing for all complex directions implies $(H-\mathcal{E})\psi =0$; an eigenvector conversely makes it vanish. The ground energy is the infimum of this quotient for a lower-bounded Hamiltonian, so minimizing over a trial family provides an upper bound to it, exact when that family contains a ground eigenvector.

Set $a=⟨\varphi ,H\varphi ⟩$, $b=⟨\varphi ,I\varphi ⟩$ and $c=⟨\varphi ,HI\varphi ⟩$. All three are real: $H,I$ are self-adjoint and commute. Also $I^{†}I=I^2=1$. Expansion of the numerator and norm givesDifferentiate the quotient, retaining only nonzero trial vectors:Thus the stationarity equation is $\boxed{(t^2-1)(c-ab)=0}$. If $c=ab$, the quotient is identically $a$ wherever defined. Otherwise $t=\pm 1$ are the two extrema, withIn this nonconstant case $|b|<1$, so both denominators are positive. If $c-ab>0$, the plus value is the maximum and the minus value the minimum; the assignments reverse when $c-ab<0$. One can also see global extremality by resolving the trial vector into the orthogonal $I=\pm 1$ sectors: its energy is a weighted mean of their two fixed projected energies.

The null-vector exception matters. If $|b|=1$, then $I\varphi =\pm \varphi$, the covariance $c-ab$ vanishes and one of $\varphi \pm I\varphi$ is zero. That value has no Rayleigh quotient and is not an extra stationary physical state.

For particle exchange, change of integration variables shows that $I\varphi (x_1,x_2)=\varphi (x_2,x_1)$ is unitary and $I^2=1$, hence self-adjoint. Exchanging coordinates interchanges the two kinetic derivatives and leaves the symmetric potential unchanged, so $[I,H]=0$. The vectors $\varphi \pm I\varphi$ lie in its symmetric and antisymmetric sectors. Their normalized nonzero versions are respectively bosonic and fermionic spatial trial wavefunctions.

Because the energy of a general trial state is the weighted average of its sector energies, at least one nonzero symmetry projection has energy no greater than the original trial. Thus minimizing can be performed in a definite exchange-symmetry sector. This is a variational approximation within the trial subspace; it does not assert that projecting an arbitrary trial vector produces an exact eigenvector of $H$. For particles with a prescribed statistics sector one minimizes within that physical sector.

Let $E=f_H(\psi )$. For any admissible variation $\eta$, differentiation with respect to a real parameter givesStationarity for all variations, including their imaginary multiples, forces $(H-E)\psi =0$; conversely this eigenvector equation makes the derivative vanish. Thus stationary Rayleigh quotients are exactly eigenvectors, with their eigenvalues as the quotient values.

Put $\kappa =e^2/(4\pi {ϵ}_0)$ and assume $\alpha >0$. The squared norm of the radial trial function is $4\pi {\int }_0^{\infty }{r}^2{e}^{-2\alpha r}dr=\pi /{\alpha }^3$. Its kinetic expectation, computed through $\int |\nabla \psi {|}^2$, is ${\hslash }^2{\alpha }^2/(2m)=\kappa a{\alpha }^2/2$. Splitting the potential integral at $R$ givesFor example, the first integral is $[1-e^{-2s}(1+2s+2s^2)]/(4{\alpha }^3)$ and the second is $e^{-2s}(R/(2\alpha )+1/(4{\alpha }^2))$, where $s=\alpha R$. ThereforeIts derivative is $\kappa [a\alpha -(1+2\alpha R)e^{-2\alpha R}]$. The actual stationary condition is consequentlyThe extra Coulomb factor in the printed stationary equation is an error: the common prefactor cancels on differentiation. The corrected equation is dimensionless on both sides. Its left side increases from zero, while its right side decreases from one, so it has one positive solution and gives the minimum in this trial family.

Let $\epsilon =R/a$ and $y=a\alpha$. Expansion of $(1+2s)e^{-2s}$ gives $1-2s^2+O(s^3)$. ThusThere is no linear correction in $R/a$. By the quantum variational principle, any normalized trial state gives an upper bound on the true ground energy. Choosing $\alpha =1/a$ already gives the exact boundIts small-radius expansion isUsing the optimized $\alpha$ improves this bound starting at order ${\epsilon }^4$, and hence gives the same leading finite-size correction. The true energy is also at least the point-nucleus value $-\kappa /(2a)$, since the finite-radius potential is nowhere more attractive.

For a nonzero trial vector use the Rayleigh quotient $\mathcal{E}(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$. Its first variation in any allowed direction $\eta$ isVanishing for all complex directions implies $(H-\mathcal{E})\psi =0$; an eigenvector conversely makes it vanish. The ground energy is the infimum of this quotient for a lower-bounded Hamiltonian, so minimizing over a trial family provides an upper bound to it, exact when that family contains a ground eigenvector.

Set $a=⟨\varphi ,H\varphi ⟩$, $b=⟨\varphi ,I\varphi ⟩$ and $c=⟨\varphi ,HI\varphi ⟩$. All three are real: $H,I$ are self-adjoint and commute. Also $I^{†}I=I^2=1$. Expansion of the numerator and norm givesDifferentiate the quotient, retaining only nonzero trial vectors:Thus the stationarity equation is $\boxed{(t^2-1)(c-ab)=0}$. If $c=ab$, the quotient is identically $a$ wherever defined. Otherwise $t=\pm 1$ are the two extrema, withIn this nonconstant case $|b|<1$, so both denominators are positive. If $c-ab>0$, the plus value is the maximum and the minus value the minimum; the assignments reverse when $c-ab<0$. One can also see global extremality by resolving the trial vector into the orthogonal $I=\pm 1$ sectors: its energy is a weighted mean of their two fixed projected energies.

The null-vector exception matters. If $|b|=1$, then $I\varphi =\pm \varphi$, the covariance $c-ab$ vanishes and one of $\varphi \pm I\varphi$ is zero. That value has no Rayleigh quotient and is not an extra stationary physical state.

For particle exchange, change of integration variables shows that $I\varphi (x_1,x_2)=\varphi (x_2,x_1)$ is unitary and $I^2=1$, hence self-adjoint. Exchanging coordinates interchanges the two kinetic derivatives and leaves the symmetric potential unchanged, so $[I,H]=0$. The vectors $\varphi \pm I\varphi$ lie in its symmetric and antisymmetric sectors. Their normalized nonzero versions are respectively bosonic and fermionic spatial trial wavefunctions.

Because the energy of a general trial state is the weighted average of its sector energies, at least one nonzero symmetry projection has energy no greater than the original trial. Thus minimizing can be performed in a definite exchange-symmetry sector. This is a variational approximation within the trial subspace; it does not assert that projecting an arbitrary trial vector produces an exact eigenvector of $H$. For particles with a prescribed statistics sector one minimizes within that physical sector.

Let $E=f_H(\psi )$. For any admissible variation $\eta$, differentiation with respect to a real parameter givesStationarity for all variations, including their imaginary multiples, forces $(H-E)\psi =0$; conversely this eigenvector equation makes the derivative vanish. Thus stationary Rayleigh quotients are exactly eigenvectors, with their eigenvalues as the quotient values.

Put $\kappa =e^2/(4\pi {ϵ}_0)$ and assume $\alpha >0$. The squared norm of the radial trial function is $4\pi {\int }_0^{\infty }{r}^2{e}^{-2\alpha r}dr=\pi /{\alpha }^3$. Its kinetic expectation, computed through $\int |\nabla \psi {|}^2$, is ${\hslash }^2{\alpha }^2/(2m)=\kappa a{\alpha }^2/2$. Splitting the potential integral at $R$ givesFor example, the first integral is $[1-e^{-2s}(1+2s+2s^2)]/(4{\alpha }^3)$ and the second is $e^{-2s}(R/(2\alpha )+1/(4{\alpha }^2))$, where $s=\alpha R$. ThereforeIts derivative is $\kappa [a\alpha -(1+2\alpha R)e^{-2\alpha R}]$. The actual stationary condition is consequentlyThe extra Coulomb factor in the printed stationary equation is an error: the common prefactor cancels on differentiation. The corrected equation is dimensionless on both sides. Its left side increases from zero, while its right side decreases from one, so it has one positive solution and gives the minimum in this trial family.

Let $\epsilon =R/a$ and $y=a\alpha$. Expansion of $(1+2s)e^{-2s}$ gives $1-2s^2+O(s^3)$. ThusThere is no linear correction in $R/a$. By the quantum variational principle, any normalized trial state gives an upper bound on the true ground energy. Choosing $\alpha =1/a$ already gives the exact boundIts small-radius expansion isUsing the optimized $\alpha$ improves this bound starting at order ${\epsilon }^4$, and hence gives the same leading finite-size correction. The true energy is also at least the point-nucleus value $-\kappa /(2a)$, since the finite-radius potential is nowhere more attractive.

Multiply the Sturm-Liouville equation by $y$ and integrate. The endpoint values and integration by parts giveSince $I[y]=1$,Thus the multiplier equals the stationary value of $J$ at the extremal, as in stationary values of a quadratic functional with a quadratic constraint. It is also the corresponding Rayleigh quotient.

Let $y$ be a smooth constrained minimizer. For any smooth variation $h$ vanishing at the endpoints, use the normalized admissible curve $y_{\epsilon }=(y+\epsilon h)/\sqrt{J[y+\epsilon h]}$. Since $J[y]=1$, differentiating at zero givesIn particular this holds for every compactly supported test function $h$ in $(0,1)$. Integration by parts and the fundamental lemma of the calculus of variations givewhich is the required Bessel differential equation with $\boxed{\lambda =I[y]}$. The lemma used here says that a continuous function whose integral against every compactly supported smooth test function is zero must itself vanish. No multiplier-existence theorem is needed: the explicit normalization produces the variation directly.

The endpoint $0$ causes no divergent energy for these variations: smoothness and $y(0)=h(0)=0$ imply $y,h=O(x)$. Also $J[v]>0$ for every nonzero admissible $v$, and $v/\sqrt{J[v]}$ has unit $J$ and energy $I[v]/J[v]$. Thus minimizing $I$ with $J=1$ is equivalent to minimizing the scale-invariant Rayleigh quotient $I/J$. The zero function is excluded whenever this quotient is written.

The trial function $y=x(1-x)$ givesIts Rayleigh quotient is $5k^2+10$, yielding the upper bound. Conversely, $1/x\ge x$ on $(0,1)$ and the derivative energy is nonnegative, so $I[y]\ge k^{2}J[y]$ for every admissible $y$. Part (b) therefore gives

The dispersion relation shows that an infinite rod has unstable long-wave modes. A finite rod excludes sufficiently long wavelengths, so stability depends on its length and the precise end conditions. The fourth spatial derivative requires two conditions at each end; fixing transverse positions alone does not specify whether slopes are constrained.

First take displacement-fixed ends with free rotation. Setting $\varphi (0)=\varphi (l)=0$ leaves slope variations free; the bending boundary term in the first variation then gives the corresponding natural boundary conditions on the slope, ${\varphi }_{xx}(0)={\varphi }_{xx}(l)=0$. The spatial normal modes are $\sin (n\pi x/l)$, andThe first mode destabilizes first. ConsequentlyAt equality, the first mode has zero restoring force and is marginal, with a static component and a possible linear-in-time component in the linear theory. Every integer $n<l\sqrt{\alpha /\beta }/\pi$ gives an exponentially growing mode.

If "fixed" instead means clamped, the clamped boundary condition is $\varphi ={\varphi }_x=0$ at both ends. The critical compression is thenFor example, the static critical shape $1-\cos (2\pi x/l)$ obeys the four clamp conditions and the zero-frequency equation at this load. To establish that it is the first threshold, write $v={\varphi }_x$: it has zero endpoints and zero mean, and periodic extension gives ${\int }_0^l{v}_x^{2}dx\ge (2\pi /l{)}^2{\int }_0^l{v}^{2}dx$ by the zero-mean periodic Poincaré inequality, with equality at $v\propto \sin (2\pi x/l)$. This proves the critical Rayleigh quotient. Both cases are instances of the compressed Euler-Bernoulli beam and its endpoint-dependent buckling threshold. The source does not specify the endpoint rotation, so both interpretations are stated explicitly.

For each unit vector iterate compute the Rayleigh quotient ${\rho }_k=x_k^{\ast }A{x}_k$ and the eigenpair residual $r_k=A{x}_k-{\rho }_k{x}_k$. Use the same multiplication $A{x}_k$ for both diagnostics and the next normalized iterate. For any complex matrix, this Rayleigh quotient minimizes $\Vert A{x}_k-\mu x_k{\Vert }_2$ over complex scalars $\mu$, because $x_k^{\ast }{r}_k=0$. Under power iteration convergence for a diagonalizable matrix, ${\rho }_k\to {\lambda }_1$ and $\Vert r_k{\Vert }_2\to 0$, even when the vectors alternate signs or phases.^[ref]

A dimensionless stopping quantity iswhen the denominator is positive; $\Vert A{\Vert }_2$ is the operator norm, and a known upper bound may replace it if computing it is costly. A small eigenpair residual certifies a small backward error of an approximate eigenpair, but does not establish that the eigenvalue found is dominant.

For a Hermitian matrix with orthonormal eigenbasis $u_j$, write $x_k={\sum }_j{a}_j{u}_j$. Then the independently useful identityshows that some eigenvalue lies within $\Vert r_k{\Vert }_2$ of ${\rho }_k$. If a specific target eigenvalue ${\lambda }_1$ is separated from all the others by ${\delta }_k={\min }_{j>1}|{\lambda }_j-{\rho }_k|>0$, it also gives ${\epsilon }_k\le \Vert r_k{\Vert }_2/{\delta }_k$. This separation is measured from the computed Rayleigh quotient; it differs from the dominant-modulus spectral gap. For a nonnormal matrix, the same orthogonal expansion is unavailable and a small eigenpair residual alone need not imply a comparably small distance to the spectrum.

A tied largest modulus can prevent even direction convergence. With $D=\mathrm{diag}(2,-2)$ and $x_0=(1,1{)}^T/\sqrt{2}$,The two directions form a cycle; neither is an eigenvector. There is no dominant-modulus spectral gap. A tied modulus is not a universal failure: an initial eigenvector remains on its line, and $A=2I$ preserves every direction. Subspace iteration addresses a separated cluster by targeting a dominant invariant subspace instead of one eigenvector.

An orthogonal start can fail despite a strict spectral gap. For $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ and $x_0=(1,-1{)}^T/\sqrt{2}$, the dominant component is zero, so $A{x}_0=x_0$ and every iterate equals $x_0$. The Rayleigh quotient is exactly $1$ and the eigenpair residual is zero, although the dominant eigenvalue is $3$. Exact arithmetic never creates a missing component.^[ref]

The third panel contrasts these failures with $N=\mathrm{diag}(-3,1)$, $x_0=(1,1{)}^T/\sqrt{2}$, for whichHere the horizontal eigenvector direction converges with a strict spectral gap, while the normalized vectors alternate signs. This is the successful direction convergence described under sign and phase in power iteration.

For a nonorthogonal eigenbasis, even the word “orthogonal” must be used carefully. The diagonalizable matrix $C=\left(\begin{array}{cc}3& 2\\ 0& 1\end{array}\right)$ has right eigenvectors $v_1=(1,0{)}^T$, $v_2=(-1,1{)}^T$, and dominant left eigenvector $w_1=(1,1{)}^T$. For $x_0=(a,b{)}^T$, the dominant coefficient is $c_1=w_1^T{x}_0=a+b$. The starting vector $v_2$ is nonorthogonal to $v_1$ but stays at the smaller eigenvalue $1$; $(0,1{)}^T$ is orthogonal to $v_1$ but has $c_1=1$ and converges to the dominant direction. The left eigenvector test, not the right-eigenvector inner product, controls the outcome.

A random vector drawn from a continuous probability distribution with a probability density function on the full real or complex vector space has $c_1\ne 0$ with probability one. This avoids the exact missing-component hyperplane, but does not guarantee a large component or fast convergence. Floating-point arithmetic may seed a missing component, yet a structurally preserved zero can remain zero, so accidental perturbations are not a convergence guarantee.

For a fixed circular elastic membrane, the Rayleigh-Ritz variational principle applied to $f(r)=1-(r/a{)}^p$, $p>0$, gives ${\lambda }_1\le (p+3+2/p)/a^2$. The numerator of its Rayleigh quotient is $p/2$, and the denominator is $a^2{p}^2/[2(p+1)(p+2)]$. Minimization gives $p=\sqrt{2}$, ${\lambda }_1\le (3+2\sqrt{2})/a^2$, and the displayed upper bound for the ordinary frequency ${\nu }_1=c\sqrt{{\lambda }_1}/(2\pi )$. Every $p>0$ has finite radial Dirichlet energy, which is the required variational admissibility condition.

For a real symmetric matrix with $|{\lambda }_1|>|{\lambda }_2|$ and a starting vector having nonzero component in the leading eigendirection, power-method direction errors arewhereas the Rayleigh quotient error isThe improvement occurs because orthogonality cancels terms linear in the direction error.

The quantum variational principle estimates the lowest energy by choosing a trial wavefunction and computing its expectation value, without solving the Schrödinger equation exactly. Minimizing over a manageable family of trial wavefunctions gives an upper bound; enlarging that family can improve the estimate.

Let $H$ be a self-adjoint operator bounded below, and put $E_{\ast }=\inf \sigma (H)$. For $\Vert \psi \Vert =1$ in its operator domain $D(H)$,Trial states may also lie in the form domain $Q(H)=D((H-bI{)}^{1/2})$, where $b\le E_{\ast }$. Then use the quadratic form of a positive quantum Hamiltonian shifted by $b$:This extends the expectation value beyond $D(H)$ without requiring $H\psi$ to exist. Equality for a normalized state holds exactly in the energy eigenspace at $E_{\ast }$, if that eigenvalue exists. ^[ref].

A short proof makes the direction of the bound memorable. If $H$ has a complete orthonormal basis of energy eigenstates ${\varphi }_n$, writeThe expectation value is a weighted average of allowed energies:Equality requires every occupied energy eigenstate to have energy $E_{\ast }$. ^[ref]. With continuous spectrum, the spectral theorem replaces the sum by an integral against the scalar spectral measure; the same nonnegative-difference argument works.

For an unnormalized trial wavefunction, divide by its squared norm: $R_H[\psi ]=q_H[\psi ]/\Vert \psi {\Vert }^2$ is the form version of the Rayleigh quotient. The quantum variational principle supplies an upper bound on $E_{\ast }$; it supplies neither a lower bound on $E_{\ast }$ nor, by itself, a guarantee that a bound state exists.

To see the method at work, choose the real Gaussian wave packetThe Gaussian integral gives wavefunction normalization and the two useful position averages:Thus its position probability density is a normal distribution with standard deviation $1/\sqrt{2a}$: large $a$ means narrow confinement. In units $\hslash =m=1$, integration by parts and ${\psi }_a^{\prime }=-ax{\psi }_a$ give the kinetic energyThese smooth, rapidly decaying trial wavefunctions belong to the operator domains of both polynomial Quantum Hamiltonians below. Narrowing the Gaussian wave packet increases its kinetic energy; widening it increases the potential energy in either confining potential.

For the quantum harmonic oscillator, take $\hslash =m=\omega =1$:The trial values at $a=1/2,1,2$ are respectively $0.625,0.5,0.625$. The derivative $E_{\mathrm{ho}}^{\prime }(a)=(1-a^{-2})/4$ changes sign at $a=1$, where kinetic energy and potential energy are both $1/4$. Here the trial family contains an exact ground state:The zero-point energy remains positive even at the best width. For comparison, removing the potential leaves $E(a)=a/4\to 0$ as $a\to 0$: the free Quantum Hamiltonian on the line has spectral bottom zero but no normalizable zero-energy energy eigenstate. A sequence of ever broader trial wavefunctions need not approach an actual bound state.

For a pure quartic oscillator, keep $\hslash =m=1$ and set the quartic coefficient to one:The trial values at $a=1,2,3$ are respectively $1,0.6875,0.833333\dots$. Setting the derivative to zero givesThe sign of this derivative changes from negative to positive at the optimum, giving the smallest energy within this Gaussian wave packet family:At this width, $⟨T⟩\approx 0.454280$ and $⟨V⟩\approx 0.227140$, so $2⟨T⟩=4⟨V⟩$, as expected from the power-law quantum virial theorem for a scale-optimized trial wavefunction. The number $0.681420$ is an upper estimate of the ground-state energy, not an asserted exact energy eigenvalue. Indeed,depends on $x$ for every $a>0$, so no member of this trial family is an energy eigenstate. Nevertheless, a single width parameter already turns a difficult Schrödinger equation into a useful, explicit energy estimate.

For axisymmetric heat conduction in a unit disk, separated modes satisfy $-(r\kappa R^{\prime }{)}^{\prime }=\nu rR$, with weight $r$, outer Dirichlet boundary condition $R(1)=0$, and zero radial flux at the origin. Integration by parts gives the displayed Rayleigh quotient. Its Euler-Lagrange equation under fixed $\int r{R}^2$ recovers the mode equation; the natural endpoint term is $r\kappa R^{\prime }\delta R$. The Rayleigh-Ritz variational principle says that every admissible nonzero trial profile gives an upper bound on the bottom of the spectrum, and on the lowest eigenvalue when this bottom is attained by a fundamental diffusion mode. Finite weighted energy, rather than pointwise existence of a second derivative at the origin, is the relevant trial admissibility requirement.

A bounded spherically symmetric solution of the Helmholtz equation in a ball with zero value at radius $a$ satisfies $-(r^2{\psi }^{\prime }{)}^{\prime }=k^2{r}^2\psi$. The substitution $u=r\psi$ gives $u^{\prime \prime }+k^{2}u=0$ with $u(0)=u(a)=0$, so the displayed modes have $n\ge 1$; the apparent singularity at zero is removable. Their Rayleigh quotient is ${\int }_0^a{r}^2{\psi }^{\prime 2}dr/{\int }_0^a{r}^2{\psi }^{2}dr$. The trial profile $1-r^2/a^2$ gives the upper bound $21/(2a^2)$, compared with the exact lowest value ${\pi }^2/a^2$. Boundedness at zero eliminates the $1/r$ solution and makes the endpoint term in the Sturm-Liouville problem vanish.

For a radial stellar oscillation, put ${\xi }_r=r\eta$. The self-gravitating adiabatic equation is $({\Gamma }_{1}P{r}^4{\eta }^{\prime }{)}^{\prime }+r^3[(3{\Gamma }_1-4)P{]}^{\prime }\eta +\rho r^4{\omega }^2\eta =0$. It is a Sturm-Liouville problem whose Rayleigh quotient tests radial stability. The familiar constant-exponent threshold is ${\Gamma }_1=4/3$; a variable exponent requires the full integral test.

A fixed circular elastic membrane of radius $a$ has radial normal modes satisfying the displayed Sturm-Liouville problem, with weight $r$, $R(a)=0$ and regularity at the origin. The regular solutions are Bessel functions of the first kind $J_0(\sqrt{\lambda }r)$. Integration by parts gives the positive Rayleigh quotient $\lambda ={\int }_0^{a}r|R^{\prime }{|}^{2}dr/{\int }_0^{a}r|R{|}^{2}dr$. Distinct modes are orthogonal in the weighted inner product ${\int }_0^{a}r\overline{R}Sdr$.