Past exam of the options II course of the University of Oxford / 2012 / J / J2

Since $d/dx=\sqrt{m\omega /\hslash }d/d\xi$, dividing the Schrödinger equation by $\hslash \omega /2$ yieldsFor $\psi =f{e}^{-{\xi }^2/2}$, differentiation givesThe ${\xi }^{2}f$ terms cancel, leaving

Differentiate the power series twice and align the coefficient of ${\xi }^k$. The equation becomesThusEven coefficients are generated from $a_0$, and odd coefficients from $a_1$; the two parity sequences are independent.

A nonzero polynomial solution of degree $n$ has $a_n\ne 0$ but $a_{n+2}=0$. The recurrence therefore forces $2n+1-ϵ=0$. Since $ϵ=2E/(\hslash \omega )$,Conversely, for this value of $ϵ$, choose only the parity sequence containing degree $n$ and set the other initial coefficient to zero. The recurrence terminates at $n$, producing a nonzero polynomial times the decaying Gaussian. Thus every displayed energy occurs under the supplied normalizability criterion.

The Born probability is the integral of $|{\psi }_n{|}^2$ over the indicated interval. Changing variables to $\xi$ cancels the dimensional normalization factors and givesSince $\frac{d}{d\xi }(\xi e^{-{\xi }^2})=e^{-{\xi }^2}-2{\xi }^2{e}^{-{\xi }^2}$, integration givesAs $\ell \to \infty$, the correction term tends to zero and $P_0(\ell )\to 1$ by normalization. Hence $\boxed{P_1(\ell )\to 1}$. This is expected because the expanding interval exhausts the whole position space and the state is normalized: the particle is somewhere on the line with total probability one.