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For the dimensionless half-harmonic oscillator $-{\psi }^{\prime \prime }+x^2\psi =\lambda \psi$ on $x>0$ with $\psi (0)=0$, the trial family $x{e}^{-\alpha x}$, $\alpha >0$, is admissible for the Rayleigh-Ritz variational principle. The moments ${\int }_0^{\infty }{x}^n{e}^{-2\alpha x}dx=n!/(2\alpha {)}^{n+1}$ give norm squared $1/(4{\alpha }^3)$, derivative energy $1/(4\alpha )$, and potential energy $3/(4{\alpha }^5)$. Its Rayleigh quotient is ${\alpha }^2+3/{\alpha }^2$, minimized at $\alpha =3^{1/4}$. The resulting upper bound $2\sqrt{3}$ exceeds the exact lowest eigenvalue three; the exact ground-state shape is $x{e}^{-x^2/2}$.

A half-harmonic oscillator has $V=m{\omega }^2{x}^2/2$ on $x>0$ and an infinite wall at $x=0$. Its wavefunction vanishes at the wall. Only the odd full-oscillator energy eigenstates satisfy this condition; restricting them to the positive half-line and multiplying by $\sqrt{2}$ restores normalization. Thus $E_r=\hslash \omega (2r+3/2)$ for $r=0,1,\dots$. With $\ell =\sqrt{\hslash /(m\omega )}$, its normalized ground state isIt has no interior node on its allowed half-line, and energy $3\hslash \omega /2$.

The boundary conditions on the positive half-line, absent from the transcription, select the half-harmonic oscillator. The trial state $x{e}^{-\alpha x}$ with $\alpha >0$ satisfies both endpoints and has finite norm and energy. Integration by parts, with vanishing endpoint term, givesUsing ${\int }_0^{\infty }{x}^n{e}^{-2\alpha x}dx=n!/(2\alpha {)}^{n+1}$, the three required integrals areThus the exponential trial state for a half-harmonic oscillator gives $\Lambda (\alpha )={\alpha }^2+3/{\alpha }^2$. Its derivative vanishes at ${\alpha }^4=3$, and the arithmetic-geometric mean inequality confirms the global minimum:For comparison, the exact lowest half-line state is proportional to $x{e}^{-x^2/2}$ and has eigenvalue three. The exponential trial family gives a valid upper bound rather than an exact eigenvalue.

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.

Suddenly remove the wall of a half-harmonic oscillator in its ground state, leaving the same harmonic potential on the whole line. The sudden approximation keeps its wavefunction unchanged during removal. The new normalized ground state is ${\psi }_0={\pi }^{-1/4}{\ell }^{-1/2}{e}^{-x^2/(2{\ell }^2)}$, soThe integration is over the initially occupied half-line. The final first excited state instead has probability $1/2$, since the initial half-line state equals $\sqrt{2}$ times its positive-half restriction. This also checks that removing the wall does not simply create the full oscillator's odd state.