Rayleigh quotient ($R[\psi ]$)

For a self-adjoint operator $H$, the Rayleigh quotient isFor a symmetric matrix this reduces to $x^{T}Ax/(x^{T}x)$.

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 $E_j$.

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 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