Past exam of the quantum theory course of the University of Oxford / 2014 / 2

The dimensionless coordinate for the quantum harmonic oscillator gives $d/dx=\sqrt{m\omega /\hslash }d/d\xi$. The time-independent Schrödinger equation therefore becomesFor $\psi =e^{-{\xi }^2/2}f$, direct differentiation yieldsCancelling the nonzero exponential reduces the equation to the Hermite differential equation

Substitute the power series and collect each power of $\xi$:Matching the power series coefficients gives the recurrence relationThe even and odd coefficients form separate sequences, determined by $a_0$ and $a_1$, respectively. For example, $a_2=(1-ϵ)a_0/2$ and $a_3=(3-ϵ)a_1/6$. These are the two independent power series solutions of the Hermite differential equation.

The issue is the square-integrable condition on the whole real line, not just decay at one end. Because the quantum harmonic oscillator potential is even, the even and odd parts of a square-integrable wavefunction are themselves square integrable and satisfy the same equation. It is therefore enough to examine each parity sequence separately.

For a nonterminating sequence, put $b_j=a_{2j+p}$, where $p=0$ or $1$. Its recurrence relation givesSince $ϵ$ is real for an energy eigenvalue, the ratio is eventually positive. After multiplying by one constant phase, the tail coefficients are all positive. Choose $0<\delta <1/2$. For sufficiently large $j$, the ratio is at least $(1-\delta )/(j+1)$, so for some $K>0$ the tail coefficients obey $b_j\ge K(1-\delta {)}^j/j!$. Consequently, for $\xi >0$, the magnitude of $f$ eventually exceeds a positive multiple of ${\xi }^p{e}^{(1-\delta ){\xi }^2}$ after subtraction of a fixed polynomial. Multiplication by $e^{-{\xi }^2/2}$ leaves exponential growth like $e^{(1/2-\delta ){\xi }^2}$, incompatible with the square-integrable condition. Thus each nonzero parity series must terminate. The parity argument also excludes cancellation between the two nonterminating sequences at both infinities. This proves Normalization forces Hermite-series termination.

If the terminating polynomial has degree $n$, its top nonzero coefficient requires $2n+1-ϵ=0$. Conversely, when $ϵ=2n+1$, choosing the seed of parity $n$ nonzero makes the recurrence relation terminate exactly at degree $n$. The opposite seed must be zero, since that sequence does not terminate. ThereforeThe resulting polynomials are proportional to the physicists' Hermite polynomials, and ${\psi }_n\propto H_n(\xi )e^{-{\xi }^2/2}$ is normalizable because any polynomial times this Gaussian function is square integrable. The two parity seeds describe the full solution space of the second-order differential equation; the preceding argument therefore exhausts the normalizable stationary states. Each eigenspace has dimension one. General time-dependent quantum states can of course be superpositions of these energy eigenstates.

For a normalized wavefunction, the expectation value of a function of position isprovided the integral exists. For an unnormalized state, divide this expression by $\int |\psi {|}^{2}dx$. Put $b=m\omega /\hslash >0$. The ground state has probability density $\sqrt{b/\pi }{e}^{-b{x}^2}$, which is even. ThusThe zero mean position ${\mathbb{E}}_{{\psi }_0}(x)=0$ expresses symmetry, whereas this positive expectation value measures the mean distance from the centre.