Past exam of the physics A course of the University of Cambridge / 2018 / Paper 1 / Section B / B7

The one-dimensional time-independent Schrödinger equation isHere $m$ is the particle mass, $V(x)$ its potential energy, $E$ the stationary energy eigenvalue, and $\hslash =h/(2\pi )$ the reduced Planck constant. The spatial wavefunction is $\Psi$; the full stationary state is $\Psi (x)e^{-iEt/\hslash }$. Physical bound states satisfy the boundary conditions and wavefunction normalization $\int |\Psi {|}^{2}dx=1$. The operator on the left is the Hamiltonian operator, the sum of kinetic and potential energies.

Assume $a>0$ and define $\omega =\sqrt{2a/m}$ so that the potential is that of a quantum harmonic oscillator. The two trial forms have derivativesSubstitution into the stationary Schrödinger equation cancels the $x^2$ terms precisely when $a=2{\hslash }^2{C}^2/m$. Normalizability selects the positive root,The remaining constants giveFor later overlaps choose positive normalization constants. The Gaussian integral gives $A=(2C/\pi {)}^{1/4}$, while its second-moment integral gives $B=2\sqrt{C}A$. Thus both wavefunctions have unit norm and are orthogonal by their opposite parity.

The real Gaussian is positive everywhere and has no nodes; the odd state has exactly one node at $x=0$. The Sturm oscillation theorem orders one-dimensional bound states by their node count. Consequently these are the ground state and first excited state, respectively, also agreeing with the oscillator spectrum $E_n=(n+1/2)\hslash \omega$.

The infinite wall imposes $\boxed{\Psi (0)=0}$, and the wavefunction vanishes for $x<0$. The even full-oscillator ground state fails this boundary condition; the lowest admissible state is the positive-half restriction of the odd first excited state. Multiplying by $\sqrt{2}$ restores normalization because half its probability originally lay on either side.

With $\ell =\sqrt{\hslash /(m\omega )}$, the half-harmonic oscillator ground state isIt rises from zero at the wall, reaches its wave-amplitude maximum at $x=\ell$, and decays to zero as $x\to \infty$. It is positive throughout the interior of the allowed region, with no interior node, as required for its ground state. The wall need not make the derivative continuous with the forbidden half-line.

For removal of a half-oscillator wall, the sudden approximation keeps the state unchanged during the change of potential. Its probability in the new full-oscillator ground state ${\Psi }_0={\pi }^{-1/4}{\ell }^{-1/2}{e}^{-x^2/(2{\ell }^2)}$ is the squared overlap:where ${\int }_0^{\infty }x{e}^{-x^2/{\ell }^2}dx={\ell }^2/2$. ThereforeThe integral runs only over the initially occupied half-line. The initial state is not the odd full-line eigenstate: it is zero, rather than negative, on the left. Indeed its overlap with that final first excited state is $1/\sqrt{2}$, giving probability $1/2$, and the remaining probability lies in higher states. Subsequent evolution under the new fixed Hamiltonian operator changes the quantum phases of its energy components without changing these energy-measurement probabilities.