Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 10

Write the energy form as $Q[\psi ]={\int }_{\alpha }^{\beta }(p|{\psi }^{\prime }{|}^2+q|\psi {|}^2)dx$ and the weighted squared norm as $N[\psi ]={\int }_{\alpha }^{\beta }w|\psi {|}^{2}dx$. For real coefficients and appropriate homogeneous boundary conditions, the Sturm-Liouville problem is the stationary-point equation for their Rayleigh quotient $Q/N$.

One suitable choice is homogeneous Dirichlet boundary conditions, $\psi (\alpha )=\psi (\beta )=0$, with variations vanishing there. Assume real $q$ and the stated positive $p,w$. Introduce a Lagrange multiplier $\mu$ for $N[\psi ]=1$. For real variations, integration by parts givesThe boundary term vanishes, so the fundamental lemma of the calculus of variations yieldsMultiplying this equation by $\psi$, integrating, and using the boundary conditions gives $Q[\psi ]=\mu N[\psi ]=\mu$. Thus the multiplier, the eigenvalue, and the constrained stationary value coincide. Conversely every normalized eigenfunction makes the first variation vanish for all tangent variations satisfying $\delta N=0$.

Homogeneous Neumann boundary conditions, $p{\psi }^{\prime }=0$ at both endpoints, or mixed Dirichlet and Neumann conditions also work with this energy form. General Robin boundary conditions require the corresponding boundary terms in the form; they cannot simply be inserted into the displayed unmodified integral.

For a regular self-adjoint problem with a lowest eigenvalue and a complete orthonormal eigenbasis in the weighted inner product, expand an admissible trial function as $\psi ={\sum }_j{c}_j{\psi }_j$. ThenThe Rayleigh-Ritz variational principle therefore provides the upper boundChoose a family satisfying the essential boundary conditions and finite energy, evaluate the quotient, and minimize over its parameters. Enlarging the trial family can only improve or preserve the bound; equality is attained if it includes an exact ground eigenfunction. For a finite linear trial space, this becomes a generalized matrix eigenvalue problem.

Here $p=w=1$ and $q=x^2$ on the whole real line. The quantum harmonic oscillator energy form is $Q[\psi ]={\int }_{\mathbb{R}}(|{\psi }^{\prime }{|}^2+x^2|\psi {|}^2)dx$. Admissible trials have a square-integrable weak derivative and finite potential energy; they need not have a classical second derivative everywhere.

Put $C_a=\sqrt{15/(16a^5)}$, with $a>0$. Direct integration confirms the normalization:The kinetic energy and potential energy terms in the dimensionless Rayleigh quotient areThus ${\lambda }_0\le 5/(2a^2)+a^2/7$. Differentiating with respect to $a$ gives $-5/a^3+2a/7=0$, so the unique minimizing width is $a=(35/2{)}^{1/4}$. The two energy contributions are equal there, and the compact parabolic trial bound for the quantum harmonic oscillator isThe trial is continuous and zero at $\pm a$, so its weak first derivative is $-2C_{a}x$ inside and zero outside. The jumps of that derivative do not add delta terms to the first derivative, and its squared norm is finite. The energy-form calculation is therefore valid even though the trial does not belong to the operator domain of the harmonic oscillator. Applying the second-derivative operator naively and omitting its endpoint distributions would be incorrect.

For any nonzero constant $C$, differentiating ${\psi }_0=C{e}^{-x^2/2}$ gives ${\psi }_0^{\prime \prime }=(x^2-1){\psi }_0$. HenceIt is indeed the lowest eigenvalue, since the factorization $H=(-d/dx+x)(d/dx+x)+1$ makes $⟨\psi ,H\psi ⟩\ge \Vert \psi {\Vert }^2$, with equality for this Gaussian. The variational bound exceeds the exact value by $\sqrt{10/7}-1\simeq 0.195229$, or about $19.52\%$.

The PDF's prefactor $1/\sqrt{2\pi }$ does not normalize this wavefunction: its squared norm is $1/(2\sqrt{\pi })$. The normalized ground state has prefactor ${\pi }^{-1/4}$. This normalization error does not affect the eigenvalue, because the equation is homogeneous in the wavefunction.