= Solution
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:
$$
\langle0|\Psi_h\rangle=\int_0^\infty\frac{2x}{\sqrt\pi\ell^2}e^{-x^2/\ell^2}\,dx=\frac1{\sqrt\pi},
$$
where $\int_0^\infty xe^{-x^2/\ell^2}dx=\ell^2/2$. Therefore
$$
\boxed{P_{\mathrm{new\ ground}}=\frac1\pi\simeq0.318.}
$$
The <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 $1/\sqrt2$, 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.
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