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For a self-adjoint Sturm-Liouville operator $L_{ϵ}=L_0+ϵL_1$ with common boundary conditions and an $L^2$-normalized lowest eigenfunction $U_0$ of $L_0$, evaluating the Rayleigh quotient on $U_0$ gives the displayed upper bound. More generally the lowest eigenvalue is the infimum of affine functions of $ϵ$, so it is concave in that parameter. The trial state must satisfy the boundary requirements of the variational domain; changing the operator without changing the trial function's normalization can still be legitimate when the weight is fixed.

For real smooth coefficients with $p,w>0$, the Rayleigh quotient $\Lambda [y]=\int [p(y^{\prime }{)}^2+q{y}^2]dx/\int w{y}^{2}dx$ has the displayed first variation. Its stationary points solve the Sturm-Liouville equation on an admissible class whose boundary variations make $[p{y}^{\prime }\eta {]}_a^b=0$. Homogeneous Dirichlet boundary conditions do this by setting $\eta =0$ at the ends; homogeneous Neumann boundary conditions do it through $y^{\prime }=0$. More general self-adjoint boundary conditions can require a boundary contribution to the quadratic form.

For an eigenfunction, the weaker identity $[py{y}^{\prime }{]}_a^b=0$ suffices to give $\lambda =\Lambda [y]$ by integration by parts, but it does not by itself imply the condition on every variation. An explicit counterexample uses $p=q=w=1$ on $[0,1]$ and $y=\cos \pi x+\sin \pi x$. This solves $-y^{\prime \prime }+y=(1+{\pi }^2)y$ and has $[y{y}^{\prime }{]}_0^1=0$. Define a nearby familyIt keeps $y_{\epsilon }(0)=1$, $y_{\epsilon }^{\prime }(0)=\pi$, $y_{\epsilon }(1)=-1+\epsilon$, and $y_{\epsilon }^{\prime }(1)=\pi /(-1+\epsilon )$, so $[y_{\epsilon }{y}_{\epsilon }^{\prime }{]}_0^1=0$ for small $\epsilon$. Nevertheless, $\eta ={\partial }_{\epsilon }{y}_{\epsilon }{|}_0$ has $\eta (0)=0$, $\eta (1)=1$, and ${\int }_0^1{y}^{2}dx=1$, giving $\delta \Lambda =2[y^{\prime }\eta {]}_0^1=-2\pi \ne 0$. Boundary-energy cancellation alone therefore cannot replace a specified variational domain.

For positive-definite within-group scatter $W$ and between-group scatter $B$, directions maximizing $a^{T}Ba/(a^{T}Wa)$ solve the displayed generalized eigenvalue problem. Put $u=W^{1/2}a$; the quotient becomes the Rayleigh quotient of $W^{-1/2}B{W}^{-1/2}$. Its ordered orthonormal eigenvectors give directions orthogonal in the $W$ inner product. There are at most $g-1$ nonzero directions, where $g$ is the number of groups. These directions summarize separation for linear discriminant analysis.

In atomic units, use the radial trial wavefunction $\psi (r)=\max (1-r/\beta ,0)$ with $\beta >0$. Its radial norm integral is $\int r^2{\psi }^{2}dr={\beta }^3/30$, its gradient contribution is $\int r^2{\psi }^{\prime 2}dr=\beta /3$, and its Coulomb contribution is $-2\int r{\psi }^{2}dr=-{\beta }^2/6$. In the hydrogen radial Rayleigh quotient, this gives the displayed energy, minimized at $\beta =4$. It is above the exact hydrogen ground-state energy $-1/2$, as required by the quantum variational principle. The derivative jumps at $r=\beta$, but the function has a square-integrable weak first derivative and is an admissible trial for the energy quadratic form. One must use that form rather than assume that a classical second derivative exists at the join.

Conic optimization using the completely positive cone. For a real symmetric matrix $Q$, the trace-normalized program $\min \{⟨Q,X{⟩}_F:X\in {\mathrm{CP}}_n,\mathrm{tr}X=1\}$ minimizes a weighted average of nonnegative-unit-vector Rayleigh quotients. The weights are the squared norms of the factors in $X={\sum }_j{x}_j{x}_j^T$. Hence a minimizing rank-one factor attains the same value as the original orthant minimum.

Conic optimization using the copositive cone. The constraint $Q-\lambda I\in {\mathrm{COP}}_n$ supplies an exact reformulation of a Rayleigh quotient minimum on the nonnegative orthant. Exact conic formulation does not itself provide an efficient membership algorithm.

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.

Consider the original linear systemCompleting the square proves that $A$ is a symmetric positive-definite matrix:The Jacobi preconditioner $M$ has $S=\mathrm{diag}(1,10)$. Symmetric positive-definite preconditioning givesIts eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$ have eigenvalues $51/50$ and $49/50$. HenceThe characteristic polynomial of $A$ is ${\lambda }^2-101\lambda +2499/25$. Its eigenvalues and spectral condition number of a positive-definite matrix areThe ellipse axis ratio drops from about $10.0020$ to $1.0202$. The transformation multiplies the second coordinate by ten, bringing the two energy scales together.

Both panels show the same energy level sets, expressed in error coordinates $e=x-x_{\ast }$ and $\eta =M^{1/2}e$. Their coordinate axes have equal scales. The marked displacement $(1,0.1{)}^T$ becomes $(1,1{)}^T$; its energy remains $e^{T}Ae/2=1.02$. These are exact contours of the specified quadratic form.

Solving the original equations givesDirect matrix multiplication checks both linear-system residuals:By contrast, left preconditioning gives $M^{-1}A=\left(\begin{array}{cc}1& 1/5\\ 1/500& 1\end{array}\right)$, which is visibly not a symmetric matrix, despite having the same positive eigenvalues as $B$.

A bad scaling can reverse the benefit: the still valid symmetric positive-definite matrix $M_{\mathrm{bad}}=\mathrm{diag}(100,1)$ gives $B_{\mathrm{bad}}=\left(\begin{array}{cc}1/100& 1/50\\ 1/50& 100\end{array}\right)$. Its Rayleigh quotients on the coordinate unit vectors are $0.01$ and $100$, so ${\kappa }_2(B_{\mathrm{bad}})\ge 10000$. Cheapness and positive definiteness do not guarantee improved conditioning.

Let $A$ be a real positive-definite matrix and let the columns of $U$ be an orthonormal basis of a two-dimensional linear subspace in three-dimensional Euclidean space. Points in that central plane are $x=Uq$, with $|x{|}^2=|q{|}^2$. Restricting the ellipsoid $x^{T}Ax=1$ gives $q^T(U^{T}AU)q=1$. The compressed matrix $U^{T}AU$ is a positive-definite matrix, and its extreme eigenvalues give the distance formulas in the title by the Rayleigh quotient. The two extremal axes are the images under $U$ of the corresponding eigenvectors. This restriction reduces a problem with two Lagrange multipliers to a two-dimensional quadratic form.

For positive diffusivity $D(x)$, a diffusion equation with potential-driven drift has diffusive flux $J=-D(c_x+\beta V^{\prime }c)$. On a finite interval with zero flux at both endpoints, the total amount is conserved. The substitution $c=e^{-\lambda t}{e}^{-\beta V}y$ converts the decay eigenvalue problem to the Sturm-Liouville problemThe Rayleigh quotient is $\lambda =\int p|y^{\prime }{|}^2/\int w|y{|}^2\ge 0$. Its zero eigenfunction is constant, giving the stationary solution $c_{\mathrm{eq}}=C{e}^{-\beta V}$. This is a spatial Boltzmann distribution; $D$ changes relaxation rates but cancels out of the zero-flux equilibrium profile. For a sufficiently regular positive $D$ and bounded $V$, the constant eigenfunction is the unique zero mode up to scale.

For a static base state with zero incremental body force, homogeneous incremental tractions or homogeneous Dirichlet boundary conditions eliminate the boundary term in integration by parts. The displayed Rayleigh quotient then gives real squared frequencies because of major symmetry. Positive moduli give strictly positive squared frequencies for modes with nonzero displacement gradient. Pure dead loading leaves constant translations at zero; fixed displacement conditions remove those modes.

Consider the symmetric matrix and initial unit vectorIts orthonormal eigenbasis consists of $u_1=(1,1{)}^T/\sqrt{2}$ and $u_2=(1,-1{)}^T/\sqrt{2}$, with eigenvalues $3$ and $1$. Since $x_0=(u_1+u_2)/\sqrt{2}$, direct multiplication givesThe first unnormalized vectors are $(1,0{)}^T$, $(2,1{)}^T$, $(5,4{)}^T$, and $(14,13{)}^T$. If ${\theta }_k$ is the acute angle from the line spanned by the dominant eigenvector, then $\tan {\theta }_k=3^{-k}$. Every application divides the unwanted-to-wanted component ratio by three: normalization changes lengths, while the spectral gap selects the direction.

The right panel compares the power iteration direction error, Rayleigh quotient error, and eigenpair residual. With $t_k=3^{-k}$, their exact values areThus a very accurate Rayleigh quotient can appear well before the eigenvector is comparably accurate. This is an example of quadratic Rayleigh-quotient improvement for the power method.

In a two-level system, let the Hamiltonian operator have matrix $H=\mathrm{diag}(0,2)$ in the orthonormal basis $|0⟩,|1⟩$. The normalized real trial quantum states and their Rayleigh quotient areThe excited-state probability amplitude is $\sin \theta$, so its probability is ${\sin }^2\theta$. Thus $E(\theta )\ge 0=E_0$, with equality if and only if the excited-state probability amplitude vanishes: $\theta \in \pi \mathbb{Z}$. The remaining quantum state is $|0⟩$ up to a global phase. At $\theta =\pi /4$ the trial energy is $1$; at $\theta =\pi /2$ it is $2$. These are upper bounds on the ground-state energy, even though neither is a minimizing trial.

The real trial family contains the ground state, so it attains the exact minimum despite not covering every complex quantum state. More generally, a family covering all normalized vectors in a two-dimensional Hilbert space gives the exact lowest eigenvalue of any Hermitian matrix by minimizing its Rayleigh quotient. For a complex trial $c_0|0⟩+c_1|1⟩$ with $|c_0{|}^2+|c_1{|}^2=1$, the same calculation gives $E=2|c_1{|}^2$: a relative quantum phase cannot change this diagonal Hamiltonian energy. This finite example is the weighted mean proof of the variational method with just two energy eigenstates.

Unrestricted minimization of a Rayleigh quotient seeks the ground state. To bound the $k$th excited energy from above, restrict normalized trial wavefunctions to the orthogonal complement of the exact lower-energy eigenspaces. In an exact orthonormal eigenbasis, all remaining expansion coefficients multiply energies at least $E_k$, so their weighted average is at least $E_k$. A known symmetry sector can supply some of these exclusions automatically. Orthogonality only to inaccurate approximate lower states does not automatically retain the same rigorous excited-state upper bound.

For $\omega >0$ on $x>0$, use the regular finite branch at zero, decay at infinity, and a finite quadratic-form energy. For trial functions with $[x{y}^2{]}_0^{\infty }=0$, integration by parts givesThe Rayleigh quotient is at least $\omega$, with equality exactly for a nonzero multiple of $e^{-\omega x}$. Direct substitution verifies that this is an eigenfunction with eigenvalue $\omega$, so the lower bound is attained. The zero coefficient $p(0)=0$ makes this a singular-endpoint Sturm-Liouville problem; regularity replaces a prescribed zero endpoint value. Continuous piecewise differentiable trials of finite form energy are allowed in the Rayleigh-Ritz method, even if their derivative has a jump away from the endpoint.

For $-(p{y}^{\prime }{)}^{\prime }=\lambda wy$ on $[0,1]$ with $p=(1+x)e^{2x}$, $w=e^{2x}$, $y(0)=0$ and $y^{\prime }(1)=0$, the Rayleigh quotient is $\int p{y}^{\prime 2}/\int w{y}^2$. The trial $y_1=x{e}^{-x}$ gives numerator $5/12$ and denominator $1/3$, hence $0<{\lambda }_{\min }\le 5/4$. Every displayed $y_n$ obeys the same mixed boundary conditions; so does $\sin [(m+1/2)\pi x]$. Adding any of these to the trial space cannot worsen the optimized upper bound. Matrix elements $A_{ij}=\int p{y}_i^{\prime }{y}_j^{\prime }$ and $M_{ij}=\int w{y}_i{y}_j$ give a finite-dimensional generalized eigenvalue problem $Av=\lambda Mv$. The case $n=1$ repeats the original trial and adds no variational freedom.

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 $\kappa >0$ and the radial exponential trial state $e^{-\alpha r}$, $\alpha >0$, put $a={\hslash }^2/(m\kappa )$. Direct radial integration gives the Rayleigh quotientThe kinetic expectation is ${\hslash }^2{\alpha }^2/(2m)$; the potential expectation follows by splitting its radial integral at $R$. The common prefactor cancels from the stationary condition, giving $a\alpha =(1+2\alpha R)e^{-2\alpha R}$. Its positive solution is unique, since the left side increases and the right side decreases. For $\epsilon =R/a\to 0$, it satisfies $a\alpha =1-2{\epsilon }^2+O({\epsilon }^3)$.

The quantum variational principle makes $E(\alpha )$ an upper bound for the ground energy. The explicit choice $\alpha =1/a$ gives $E_0(R)\le (\kappa /a)[1/2-\{1-(1+\epsilon )e^{-2\epsilon }\}/\epsilon ]$, whose first correction to $-\kappa /(2a)$ is $2\kappa R^2/(3a^3)$. The true energy is no lower than the point-Coulomb value because capping makes the potential less attractive.

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.

Consider $-y^{\prime \prime }-(n-1)y^{\prime }/x+x^{2}y=\lambda y$ for a positive integer $n$, with regular radial data at zero and square-integrable decay at infinity in the measure $x^{n-1}dx$. Its Sturm-Liouville form has $p=w=x^{n-1}$ and $q=x^{n+1}$. For the Gaussian trial $y=e^{a{x}^2}$, $a<0$, the Rayleigh quotient isThe Gamma function recurrence evaluates the ratio, and the minimum is $n$ at $a=-1/2$. It is the exact lowest eigenvalue: after the endpoint term $[x^n{y}^2{]}_0^{\infty }$ vanishes, integration by parts givesEquality gives the displayed positive Gaussian eigenfunction. For $n=1$, the radial regularity condition is the Neumann boundary condition at zero, not a vanishing value there. Changing the boundary condition changes the lowest eigenvalue and whether this trial is admissible.