Rayleigh-Ritz variational principle

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