Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 10

For an eigenfunction in the operator domain, $H{\psi }_n=E_n{\psi }_n$. Substitution into the Rayleigh quotient givesThe cancellation only needs a finite nonzero norm; orthonormalization makes both denominator and norm equal to one.

Assume $H$ is self-adjoint on its specified operator domain, and the admissible variations preserve that domain and any boundary conditions, so no boundary terms spoil self-adjointness. Let $N=⟨\varphi ,\varphi ⟩>0$. The first variation of the Rayleigh quotient isAt $\varphi ={\psi }_n$, the residual vanishes, so $\boxed{\delta F[{\psi }_n]=0}$. More explicitly, for a normalized ${\psi }_n$ and $\varphi ={\psi }_n+\eta$,For a fixed admissible direction scaled by a small parameter, the change is second order. This is stationarity, and is not necessarily a minimum for an excited state.

Conversely, vanishing first variation for every admissible complex direction yields the stationary equationAt the original stationary point, this is $(H-F[{\psi }_n]){\psi }_n=0$. The printed equation with an arbitrary perturbed $\varphi$ in the final factor is not generally true. Taking $\varphi ={\psi }_n+ϵ{\psi }_m$ with $E_m\ne E_n$ gives $(H-E_n)\varphi =ϵ(E_m-E_n){\psi }_m\ne 0$, even though the first-order change in $F$ is zero. Thus the valid claim is the stationary equation at the extremal state, not an exact eigenvalue equation for every variation.

Assume a complete orthonormal eigenbasis and a self-adjoint operator bounded below, with the trial state in its operator domain or associated energy-form domain. The expansion and orthogonality giveConsequentlyIt is a weighted average of eigenvalues, all at least $E_0$. This is the Rayleigh-Ritz variational principle; equality holds precisely when the trial state lies in the ground-state subspace. A continuous part of the spectrum would require the corresponding spectral integral in place of part of the sum.

Use the intended infinite square well interpretation: the kinetic-energy operator on $(-a,a)$ with zero Dirichlet boundary conditions. The variational energy is most safely evaluated through its quadratic form,The boundary term vanishes because the trial function vanishes at both walls. Its zero extension has derivative jumps there, so the form interpretation avoids assigning an unrestricted full-line second derivative to it.

Set $u=x/a$ and $b=\lambda a^2$. Apart from an irrelevant factor $a^2$, the trial function is $q_b(u)=(1-u^2)(1+b{u}^2)$. Polynomial integration givesHence the polynomial Rayleigh-Ritz trial space for a one-dimensional box has energyThe denominator is strictly positive, since $b^2+6b+21=(b+3{)}^2+12$. Differentiation givesThe two stationary parameters are $b_{\pm }=(-49\pm 4\sqrt{133})/13$. The smaller is a local maximum; the larger is a local minimum. Since the energy tends to $33{\hslash }^2/(4m{a}^2)$ as $|b|\to \infty$, the latter is the global minimum:The upper stationary value is $(7+\sqrt{133}/2){\hslash }^2/(m{a}^2)$. Exact roots give a slightly tighter bound than substituting the rounded hints.

There are two qualifications to the printed operator and comparison energy. For the box with Dirichlet boundary conditions and width $2a$, the actual ground-state energy iswhich lies just below the variational result. The coefficient $6/5$ printed as an exact eigenvalue is inconsistent with this box problem. Also, merely declaring the differential expression to be zero outside the interval does not impose hard-wall confinement or specify a self-adjoint domain. Read literally on a space admitting exterior-supported states, it has zero-energy states outside and does not have the claimed positive ground energy. The boxed variational result is the well-defined answer for the intended confined particle; a different full-line operator would need its domain and matching conditions supplied.