RayleighRitz method
The RayleighRitz method is an algorithm for obtaining approximate solutions to eigenvalue ODEs. It can be neatly summarized as follows:

1.
Choose an approximate form for the eigenfunction with the lowest eigenvalue (the ground state wavefunction, in the language of quantum mechanics). Include one or more free parameters^{}.

2.
Find the expectation value of the eigenvalue with respect to the trial eigenfunction.

3.
Minimize the resulting equation with respect to the free parameter(s), hence finding a value for the free parameter.

4.
Substitute this new eigenfunction back into the expectation value.

5.
The expectation value obtained is an upper bound for the actual eigenvalue of the true eigenfunction.
1 Example
Consider the Schrödinger equation for a onedimensional harmonic oscillator potential:
$$\left(\frac{{\mathrm{\hslash}}^{2}}{2m}\frac{{\partial}^{2}}{\partial {x}^{2}}+\frac{1}{2}{m}^{2}{\omega}^{2}\right)\psi =E\psi $$ 
where $m$ is the mass of the particle in the well, and $\omega $ is the angular velocity a classical particle would move with in the well. This equation can be solved exactly using Frobenius’ method, and leads to eigenfunctions of the form of Hermite polynomials^{} multiplied by Gaussians, and halfinteger eigenvalues of the form ${E}_{n}=(n+1/2)\mathrm{\hslash}\omega $. Since the solutions are known, it is a good test case. We choose the ground state wavefunction of the infinite potential well as our trial eigenfunction:
$$\psi =\frac{\mathrm{cos}(\frac{\pi x}{2a})}{\sqrt{a}}$$ 
with $a$ as our free parameter. We now find the expectation value:
$$\u27e8E\u27e9=\u27e8\psi \widehat{H}\psi \u27e9={\int}_{a}^{a}{\psi}^{*}\widehat{H}\psi \mathit{d}x$$ 
Evaluating the integral^{}, we find
$$\u27e8E\u27e9=\frac{{\mathrm{\hslash}}^{2}{\pi}^{2}}{8m{a}^{2}}+m{\omega}^{2}{a}^{2}\left(\frac{1}{6}\frac{1}{p{i}^{2}}\right)$$ 
We now minimise this with respect to $a$ to obtain:
$$2m{\omega}^{2}a\left(\frac{1}{6}\frac{1}{{\pi}^{2}}\right)=\frac{{\mathrm{\hslash}}^{2}{\pi}^{2}}{4m{a}^{2}}$$ 
Hence:
$$a=\pi {\left(\frac{3}{4({\pi}^{2}6)}\right)}^{\frac{1}{4}}{\left(\frac{\mathrm{\hslash}}{m\omega}\right)}^{\frac{1}{2}}$$ 
Substituting this into the expecation value $\u27e8E\u27e9$ we obtain
$$\u27e8E\u27e9=\frac{1}{2}{\left(\frac{{\pi}^{2}6}{3}\right)}^{\frac{1}{2}}\mathrm{\hslash}\omega $$ 
$$\u27e8E\u27e9\approx 0.568\mathrm{\hslash}\omega $$ 
The analytical value is of course $0.5\mathrm{\hslash}\omega $. Considering the crudeness of the approximation used, the result is impressive.
Title  RayleighRitz method 

Canonical name  RayleighRitzMethod 
Date of creation  20130322 17:52:18 
Last modified on  20130322 17:52:18 
Owner  invisiblerhino (19637) 
Last modified by  invisiblerhino (19637) 
Numerical id  8 
Author  invisiblerhino (19637) 
Entry type  Definition 
Classification  msc 65L60 