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Let $\hat{P}$ project onto a leading unit eigenvector of $H+W$, with $H$ as in the Davis-Kahan curvature lemma. Its maximizing Rayleigh quotient gives $\mathrm{tr}(H(P-\hat{P}))\le \mathrm{tr}(W(\hat{P}-P))$. Duality of the operator norm and trace norm, together with the rank-two bound $\Vert \hat{P}-P{\Vert }_{\ast }\le \sqrt{2}\Vert \hat{P}-P{\Vert }_F$, proves the estimate. No gap condition on the perturbed matrix is needed.

The Rayleigh quotient measures the quadratic form of a self-adjoint operator per unit squared norm. For a nonzero vector $x$ and a finite-dimensional Hermitian matrix $A$,For a real symmetric matrix, replace the conjugate transpose $x^{\ast }$ by $x^T$. More generally, for a self-adjoint operator $H$ on a Hilbert space, use $R_H(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$ for nonzero $\psi$ in its operator domain. On an eigenvector it returns the exact eigenvalue. Multiplying the vector by any nonzero scalar leaves the quotient unchanged, so it measures a direction rather than a chosen length.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ from the finite-dimensional spectral theorem example,The first value is not an eigenvalue: it is an average of the two directional responses. To see every direction, write the unit vector $q(\theta )=\cos \theta e_{+}+\sin \theta e_{-}$ in the orthonormal eigenbasis. ThenThe largest value is $3$ along $e_{+}$; the smallest is $1$ along $e_{-}$. A direction halfway between them has quotient $2$. In the left panel, the color on the unit circle represents the quotient; in the right panel, the same values are plotted against the angle measured from $e_{+}$.

The Hermitian Rayleigh quotient range explains why this is useful: evaluating one trial vector gives an upper bound on the smallest eigenvalue and a lower bound on the largest one, without solving a characteristic polynomial. At an extreme eigenvector, a small angular error changes the quotient only quadratically. Near $e_{+}$ in this example, $3-R_A(q(\theta ))=2{\sin }^2\theta =2{\theta }^2+O({\theta }^4)$. This is the geometric reason for quadratic Rayleigh-quotient improvement for the power method and motivates the Rayleigh-Ritz method.

There is a concrete historical link with normal modes. In The Theory of Sound (1877), Volume I, Sections 88–89, Lord Rayleigh studies the period of a constrained trial vibration. He shows that the period has a stationary point at a normal mode and that a trial type bounds the longest natural period. These energy ratios are a mechanical setting for the modern Rayleigh quotient and generalized Rayleigh quotient; the book describes the result through vibration periods, rather than today's matrix notation.^[ref]

Multiply the narrow-gap centrifugal stability equation by the complex conjugate mode and use integration by parts with zero endpoint radial velocities. This gives the displayed real Rayleigh quotient. Its denominator is positive. Positive Rayleigh discriminant throughout the gap gives positive ${\omega }^2$ for every nonzero mode, excluding exponential axisymmetric growth; nonnegativity permits marginal cases. In the alternative convention $\Phi =-{\kappa }_{\mathrm{ep}}^2$, the numerator is $-k^2\int \Phi |u{|}^{2}dr$, so the stable sign is $\Phi <0$. For rigid rotation, ${\kappa }_{\mathrm{ep}}^2=4{\Omega }^2$ gives ${\omega }_n^2=4{\Omega }^2{k}^2/[k^2+(n\pi /d{)}^2]$.

For a nonzero trial function with $y(0)=y(L)=0$ and $y(L-x)=y(x)$, substituting $L-x$ for $x$ proves ${\int }_0^{L}x{y}^{2}dx=(L/2){\int }_0^L{y}^{2}dx$. This gives the displayed Rayleigh quotient. The Poincaré inequality on an interval bounds it below by $2{\pi }^2/L^3$, with equality for a nonzero multiple of $\sin (\pi x/L)$. Thus this sine is the best reflection-symmetric trial for $-y^{\prime \prime }=\lambda xy$, while the unrestricted lowest eigenfunction need not be reflection-symmetric. The polynomial $x(L-x)$ gives $20/L^3$, only a factor $10/{\pi }^2$ above the sine's value. Both are upper bounds on the unrestricted lowest eigenvalue by the Rayleigh-Ritz method.

On $(0,1)$, integration by parts gives the displayed Fourier sine series. For normalized Dirichlet eigenfunctions $y_i=\sqrt{2}\sin [(i+1)\pi x]$, indexed from $i=0$, it has the form $A[y_0+{\sum }_{i\ge 1}{c}_i{y}_i]$, with $A=4\sqrt{2}/{\pi }^3$ and $c_i=[1-(-1{)}^{i+1}]/[2(i+1{)}^3]$. Its Rayleigh quotient for $-d^2/d{x}^2$ is $10$, since ${\int }_0^1(u^{\prime }{)}^{2}dx=1/3$ and ${\int }_0^1{u}^{2}dx=1/30$. Reflection symmetry forces $c_1=0$, because that first excited eigenfunction is antisymmetric about $1/2$. Thus the first excited state's contribution to the error is exactly zero; the first nonzero correction is $c_2=1/27$.

On $(0,L)$ with zero Dirichlet boundary conditions, the normalized trial $\sqrt{2/L}\sin (\pi x/L)$ has kinetic Rayleigh quotient ${\pi }^2/L^2$ and mean position $L/2$ by reflection symmetry. The Rayleigh-Ritz variational principle therefore gives the displayed upper bound for a real linear potential. For $ϵ\ge 0$ the lower bound ${\lambda }_1\ge {\pi }^2/L^2$ follows from the Poincaré inequality. For $ϵ\ne 0$ the trial is not an exact eigenfunction, so the upper bound is strict.

A scalar field solving the Helmholtz equation inside a sphere of radius $R$ and Laplace's equation outside has a regular interior mode $b{j}_{\ell }(\sqrt{s}r)P_{\ell }(\cos \theta )$ and a decaying exterior mode $a{r}^{-\ell -1}{P}_{\ell }(\cos \theta )$. Continuity of the field and its radial derivative gives, for $x=\sqrt{s}R>0$,The Spherical Bessel function recurrence turns this into $j_{\ell -1}(x)=0$, not $j_{\ell }(x)=0$. This elimination does not require division by a possibly vanishing boundary value.

For $\ell =0$, $j_0(x)=\sin x/x$ and $j_{-1}(x)=\cos x/x$, so the first positive root is $\pi /2$. For $\ell =1$, the matching condition is $j_0(x)=0$, whose first positive root is $\pi$. The distinction matters when a scalar model permits the monopole sector but a physical multipole model excludes it.

To compare angular sectors, put $u=r{j}_{\ell }(\sqrt{s}r)$. Its equation and boundary conditions are $-u^{\prime \prime }+\ell (\ell +1)u/r^2=su$, $u(0)=0$ and $u^{\prime }(R)+\ell u(R)/R=0$. Its Rayleigh quotient isBoth added terms are nonnegative and increase with $\ell$. Thus the lowest scalar eigenvalue is ${\pi }^2/(4R^2)$ in the $\ell =0$ sector; if only $\ell \ge 1$ is admitted, the lowest is ${\pi }^2/R^2$ in the $\ell =1$ sector. Zeros at $x=0$ caused by the vanishing normalization of $j_{\ell }$ are not additional eigenvalues.

For a self-adjoint operator with two normalized orthogonal eigenvectors $u_0,u_1$ and $E_0<E_1$, the Rayleigh quotient of ${\psi }_t=u_1+t{u}_0$ is $(E_1+t^2{E}_0)/(1+t^2)$. It is stationary at $t=0$ but has second derivative $2(E_0-E_1)<0$ there. Thus an exact excited eigenfunction can be a maximum within a trial family. A constrained minimum requires exclusion of lower states, as in an excited-state variational constraint.

Impose zero Dirichlet boundary conditions on an interval. The isoperimetric constraint $I[y]=1$ is regular because $D{I}_y[y]=2I[y]=2\ne 0$. The Lagrange multiplier rule applied to $J-\lambda (I-1)$ gives the Sturm-Liouville equationMultiplying by $y$ and using integration by parts gives $J[y]=\lambda I[y]=\lambda$. Thus the multiplier is the stationary value of the original functional, or its Rayleigh quotient. This calculation does not require $w$ to have one sign; positive-weight spectral conclusions require additional hypotheses.

If two stationary functions have different multipliers, multiplying their equations crosswise and subtracting gives $({\lambda }_n-{\lambda }_m)\int w{y}_n{y}_m=0$. The endpoint terms vanish. This proves the weighted orthogonality of Sturm-Liouville eigenfunctions. Higher stationary values need not be local minima of the constrained functional.

For a nonrotating self-gravitating star, let $p_1,{\rho }_1,{\varphi }_1$ be Eulerian perturbations and $\Delta p=-\gamma p\nabla \cdot \mathbf{\xi }$ the adiabatic Lagrangian perturbation. Integration of the linear momentum equation against ${\mathbf{\xi }}^{\ast }$, using hydrostatic balance and the perturbed Poisson equation, yields the displayed Rayleigh quotient. The gravitational integral is over all space and the other integrals over the star. The compression denominator is $\gamma p$, not $\gamma \rho$. Surface terms vanish for compactly supported trials or the physical free-surface and exterior potential conditions.

A second-order differential equation in the displayed divergence form. Under the usual regular spectral assumptions, $p>0$, $w>0$, the coefficients have suitable regularity, and separated self-adjoint boundary conditions specify a Sturm-Liouville problem. Multiplying by $y$ and using zero endpoint values gives the Rayleigh quotient $\lambda =\int (p{y}^{\prime 2}+q{y}^2)dx/\int w{y}^{2}dx$ for a nonzero real eigenfunction. Sign conventions for the potential and eigenvalue vary; they must be read from the stated equation.

The weighted operator $L=W^{-1}[-(P\xi \prime )\prime +Q\xi ]$, with real coefficients and $P,W>0$ in the interior, defines a Sturm-Liouville problem. Its realization as a self-adjoint differential operator requires a domain whose endpoint conditions annihilate the boundary form $[P(\eta {\prime }^{\ast }\xi -{\eta }^{\ast }\xi \prime )]$. Its Rayleigh quotient uses the weight $W$ in the denominator.

Let $s=r(1-r)$ and $p=r(1-r^2)$. Under reflection $r\mapsto 1-r$, $s$ is symmetric and $p=(3/2)s+a$, where $a=r(1-r)(r-1/2)$ is antisymmetric. Their cross terms vanish in both the $L^2$ inner product and the Dirichlet energy form: each integrand is odd about the midpoint. Their separate Rayleigh quotients are $R[s]=10$ and $R[a]=42$. Thus every quotient on their two-dimensional span is a positive weighted average of ten and forty-two, so the minimum remains ten. Adding a direction with the wrong symmetry need not improve the ground-state estimate.

The symmetric quartic $s^2=r^2(1-r{)}^2$ does add a useful direction. On the span of $s,s^2$, the generalized eigenvalue problem has stiffness matrix $K=\left(\begin{array}{cc}1/3& 1/15\\ 1/15& 2/105\end{array}\right)$ and mass matrix $M=\left(\begin{array}{cc}1/30& 1/140\\ 1/140& 1/630\end{array}\right)$. Its least eigenvalue is $56-4\sqrt{133}\simeq 9.86974962$, close to the exact Dirichlet ground value ${\pi }^2$. This is an explicit useful enlargement of a polynomial Rayleigh-Ritz trial space for a one-dimensional box.

A nonconstant smooth periodic roll obeys $\mu ⟨T_0^{\prime 2}⟩-⟨T_0^{\prime \prime 2}⟩-⟨T_0^{\prime 4}⟩=0$. The orthogonal-roll perturbation raises the increasing-gradient functional for poorly conducting convection by the displayed amount. The variance is strictly positive: a nonconstant periodic function has a derivative that vanishes somewhere and is nonzero somewhere. The Hessian of $V$ therefore has a positive direction. Its self-adjoint linearized evolution operator has a positive Rayleigh quotient and hence a positive eigenvalue, proving transverse linear instability wherever a nonconstant roll exists.

A trial function is a chosen candidate used to approximate a solution or evaluate a variational principle. For an eigenvalue upper bound from a Rayleigh quotient, it must be nonzero, have finite numerator and denominator, and lie in the appropriate variational domain. In a Dirichlet problem it satisfies the fixed endpoint values. Natural boundary conditions need not always be imposed on a trial in the quadratic-form domain, but a candidate violating an essential boundary condition does not give the stated bound.

If a trial quantum state is dominated by two energy eigenstates of distinct energies $E_a<E_b$, then its Rayleigh quotient is approximately their probability-weighted mean. Rearranging that mean gives the displayed relative admixture. Neglected states must make small contributions to the energy numerator as well as to wavefunction normalization; small probabilities alone need not suffice if their energies are large.

For $-{\psi }^{\prime \prime }+\cos (\pi x)\psi =\lambda \psi$ on $(0,1)$ with zero Dirichlet boundary conditions, use $\psi =\sin (\pi x)+\alpha \sin (2\pi x)$. Its Rayleigh quotient is $({\pi }^2+\alpha +4{\pi }^2{\alpha }^2)/(1+{\alpha }^2)$. In the normalized sine basis the restricted operator has matrix $\left(\begin{array}{cc}{\pi }^2& 1/2\\ 1/2& 4{\pi }^2\end{array}\right)$. Its smaller eigenvalue gives the displayed upper bound, attained within this trial space at $\alpha =3{\pi }^2-\sqrt{9{\pi }^4+1}$. The other stationary parameter gives the larger restricted eigenvalue. This is a finite-dimensional Rayleigh-Ritz method for a regular Sturm-Liouville problem with a sign-changing potential.

The variational method estimates the lowest energy of a Quantum Hamiltonian by choosing a trial wavefunction, evaluating its energy, and optimizing the trial parameters. It uses the quantum variational principle, the Rayleigh-Ritz variational principle expressed for quantum states. Let $H$ be a self-adjoint operator bounded below, and define $E_0=\inf \sigma (H)$, the bottom of its spectrum. Every normalized admissible trial quantum state satisfiesWhen $E_0$ is an energy eigenvalue, it is the ground-state energy; equality holds precisely for a quantum state in the ground-state subspace. Otherwise the bottom of the spectrum need not be attained by any normalizable wavefunction.

Admissible means finite form energy. Choose a real $b\le E_0$ and use the shifted quadratic form of a positive quantum Hamiltonian:The form domain $Q(H)$ is independent of this choice of $b$ and can be larger than the operator domain $D(H)$. For $\psi \in D(H)$, $E_{\psi }=⟨\psi ,H\psi ⟩$ is the usual expectation value; for nonzero unnormalized trials, divide $q_H[\psi ]$ by $\Vert \psi {\Vert }^2$ to get the form version of the Rayleigh quotient. See the quantum variational principle for the general domain justification.

Thus each trial energy is an upper bound on $E_0$, and minimizing within a trial family improves that upper bound. It is not a lower bound on $E_0$. The Gaussian variational energy for the harmonic oscillator reaches $E_0=1/2$ because its family contains an exact ground state. The exact variational solution in a two-level system makes the equality condition visible as vanishing excited-state probability amplitude.

The method is useful when solving the Schrödinger equation exactly is difficult but integrals of a trial wavefunction are tractable. Enlarging a trial family cannot increase its optimal energy, although an optimum need not be attained. The beauty is that even an imperfect trial gives a rigorous bound, provided its form energy is computed exactly: numerical integration errors require separate control. The Gaussian variational estimate for a solvable sextic potential provides an existing example beyond a quantum harmonic oscillator.

A short proof explains the direction. With a complete orthonormal eigenbasis of energy eigenstates, write $\psi ={\sum }_n{c}_n{\varphi }_n$, where $H{\varphi }_n=E_n{\varphi }_n$ and ${\sum }_n|c_n{|}^2=1$. ThenThis is a weighted mean of nonnegative energy differences; equality requires every occupied energy eigenstate to have energy $E_0$. ^[ref]. For continuous spectrum, the spectral theorem replaces the sum by an integral against the scalar spectral measure of $\psi$, with the same nonnegative integrand. This proves the bound without assuming an orthonormal eigenbasis exists.

The historical setting predates quantum mechanics: Rayleigh's The Theory of Sound, volume I, appeared in 1877, and Ritz's paper on solving variational problems of mathematical physics appeared in volume 135 of Crelle's journal, catalogued by its publisher as 1909, pp. 1–61. These are original works associated with the Rayleigh-Ritz method; the dates here identify the publications rather than assert priority. ^[ref]; ^[ref].

Let $H$ be a lower-bounded self-adjoint Hamiltonian operator and $I$ a commuting self-adjoint operator with $I^2=1$. For a normalized trial vector $\varphi$, define $a=⟨H{⟩}_{\varphi }$, $b=⟨I{⟩}_{\varphi }$, $c=⟨HI{⟩}_{\varphi }$. The Rayleigh quotient on $\varphi +tI\varphi$ is $[(1+t^2)a+2tc]/[1+t^2+2tb]$, with derivative $2(c-ab)(1-t^2)/(1+t^2+2tb{)}^2$. Unless $c=ab$, its extrema are the normalized nonzero vectors $\varphi \pm I\varphi$, with energies $(a\pm c)/(1\pm b)$. Their difference is $2(c-ab)/(1-b^2)$, identifying the maximum and minimum.

The orthogonal projections $(1\pm I)/2$ split any trial energy into a weighted mean of the two sector energies, so one nonzero sector projection never raises the trial energy. If $|b|=1$, one projected vector is zero and its quotient is undefined; the remaining family has constant energy. Exchange symmetry makes these sectors bosonic and fermionic spatial wavefunctions. Stationarity within this two-dimensional trial space does not generally make a vector an exact eigenvector of $H$.

For quantum witness embedding $J$, verifier unitary $U$ and accepting projector $\Pi$, the witness acceptance operator is the positive contraction $F=J^{†}{U}^{†}\Pi UJ$. The maximum acceptance over normalized quantum witnesses is ${\lambda }_{\max }(F)$ by the Rayleigh quotient. Restricting to basis quantum witnesses only tests its diagonal entries and need not find that maximum.