Rayleigh quotient ($R_A(x)=\frac{x^{\ast }Ax}{x^{\ast }x}$)

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]

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.

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.

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 a nonzero real solution of ${\varphi }^{\prime \prime }+(C-k^2)\varphi =0$ on a finite interval, integration by parts gives $\mathcal{R}=\int (C{\varphi }^2-{\varphi }^{\prime 2})dz/\int {\varphi }^{2}dz$ in the displayed form. A Neumann boundary condition only at the lower endpoint does not remove the upper boundary term. Equality with $k^2$ additionally requires $\varphi (z_2){\varphi }^{\prime }(z_2)=0$. With the lower condition, nonconstant oscillatory solutions satisfy this when their phase advance is an integer multiple of $\pi /2$; affine solutions must be constant, while nonzero hyperbolic-cosine solutions fail the condition. Infinite intervals additionally require convergence of both integrals.

For a finite-dimensional Hermitian matrix $A$, the Rayleigh quotient has precisely the range $[{\lambda }_{\min },{\lambda }_{\max }]$, where the endpoints are its extreme eigenvalues. The finite-dimensional spectral theorem writes a nonzero vector as $x={\sum }_j{c}_j{e}_j$ in an orthonormal eigenbasis, givingThus the quotient is a convex combination of eigenvalues. The bounds follow immediately, and equality at either endpoint means that $x$ lies in the corresponding extreme eigenspace, including any repeated eigenvalue. Every intermediate value occurs: for orthonormal extreme eigenvectors, $x=\sqrt{1-t}{e}_{\min }+\sqrt{t}{e}_{\max }$ gives $R_A(x)=(1-t){\lambda }_{\min }+t{\lambda }_{\max }$ for $0\le t\le 1$. If all eigenvalues are equal, the quotient is constant.

The Hermitian hypothesis matters. The real matrix $B=\left(\begin{array}{cc}1& 4\\ 0& 1\end{array}\right)$ has only the eigenvalue $1$, but $(1,1)B(1,1{)}^T/2=3$. Its quotient is governed by the symmetric part of a matrix $(B+B^T)/2$, not by a real orthonormal eigenbasis of $B$. A nonsymmetric matrix therefore need not obey the displayed eigenvalue bounds.

For a real symmetric matrix $A$ and nonzero real vector $x$, the quotient rule and the quadratic form gradient give the displayed gradient. It vanishes exactly when $Ax=R_A(x)x$, so the stationary points are eigenvectors. Equivalently, the Lagrange multiplier condition for optimizing $x^{T}Ax$ on the unit sphere is $Ax=\lambda x$. The smallest and largest eigenvalues give the global extrema; other eigenvectors can be stationary without being extrema, as a direction mixing larger and smaller eigenvalues changes the quotient in opposite ways.

An odd trial state is orthogonal to the even oscillator ground state. Its Rayleigh quotient therefore bounds the first excited energy from above, unlike an arbitrary trial state. The kinetic quadratic form permits continuous compactly supported trials with derivative jumps.

For symmetric $A$ and positive-definite $B$, the generalized Rayleigh quotient is $x^{T}Ax/(x^{T}Bx)$. Its stationary values are the generalized eigenvalues of $Av=\lambda Bv$, and its maximum is the largest such eigenvalue.

Starting from a unit vector $v_k$, Rayleigh quotient iteration sets ${\mu }_k=R[v_k]$, solvesand normalizes $v_{k+1}=w_k/\Vert w_k{\Vert }_2$. The shift ${\mu }_k$ is updated from the new vector.

Let a real symmetric matrix have simple eigenpairs $({\lambda }_i,w_i)$ and letThen its Rayleigh quotient isAfter one shifted inverse solve and normalization, the relative coefficients areThus the eigenvector direction error is cubed at each sufficiently close iterate.

The extremal Rayleigh quotients of a self-adjoint operator are the extremal spectral values. In particular, a Hamiltonian with a discrete spectrum satisfies $R[\psi ]\ge E_0$, with equality precisely on its ground-state eigenspace. More generally, minimizing over states orthogonal to the first $j$ eigenspaces gives an upper bound on $E_j$.

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.

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.

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.

For the Dirichlet boundary conditions on $(-a,a)$, take the two-dimensional even trial space spanned by $(a^2-x^2)$ and $x^2(a^2-x^2)$. Using the kinetic-energy quadratic form gives, for $b=\lambda a^2$,The minimum occurs at $b=(-49+4\sqrt{133})/13$ and equals $7-\sqrt{133}/2\simeq 1.2337187027$. It is slightly above the exact infinite square well coefficient ${\pi }^2/8\simeq 1.2337005501$. The zero extension of a trial function has derivative jumps at the walls, so the appropriate admissibility statement uses the Dirichlet kinetic-energy form; it need not belong to the unrestricted full-line second-derivative operator domain.

Choose a nonzero smooth bounded profile $\varphi (s)$ supported in $0\le s<1$, constant near zero, and set $R_{\epsilon }(r)=\varphi (r/\epsilon )$. Its radial diffusion Rayleigh quotient isFor $n>2$ this tends to zero, proving that there is no positive spectral gap. Zero is not an eigenvalue with the outer Dirichlet boundary condition: zero energy would force $R^{\prime }=0$ on $0<r<1$, hence $R=0$. Thus the bottom is not attained, and a lowest exponentially decaying fundamental diffusion mode must not be assumed merely because the spatial region is bounded.

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.

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.

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 a one-dimensional even potential, the ground state is even and the first excited state is odd. Every normalized odd trial wavefunction is therefore orthogonal to the ground state and gives

Forand the trial state ${\psi }_a(x)=e^{-a{x}^2/2}$,This is negative for some $a>0$ for every $V_0>0$, proving that the attractive well has a bound state. For small $V_0$, taking $a=V_0^2$ gives

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.

For an orthonormal trial set ${\varphi }_1,\dots ,{\varphi }_N$, the Rayleigh quotient on their span isIts minimum is the smallest eigenvalue of $\mathcal{H}$, which is therefore the optimal variational upper bound available in that trial subspace.

For the first two sine modes in a width-$a$ infinite well with
$V(x)=9{\hslash }^{2}x/(m{a}^3)$, the Hamiltonian in units ${\hslash }^2/(m{a}^2)$ isIts lower eigenvalue gives the upper bound