= Solution
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 $\sqrt2$ restores normalization because half its <probability> originally lay on either side.
With $\ell=\sqrt{\hbar/(m\omega)}$, the <half-harmonic oscillator> <ground state> is
$$
\boxed{\Psi_h(x)=\begin{cases}\displaystyle\frac{2x}{\pi^{1/4}\ell^{3/2}}e^{-x^2/(2\ell^2)},&x\geq0,\\0,&x<0,\end{cases}\qquad E_h=\frac32\hbar\omega.}
$$
It 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.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-1-half-oscillator.png]
{title=Half-oscillator ground state and the full-oscillator states after removing the wall}
Back to article page
