= Solution
The <sudden approximation> keeps the wavefunction unchanged at the instant of switching: a finite step of potential creates no delta-function impulse in the <Schrödinger equation>. Just before switching the old <ground state> differs from the printed real wavefunction only by an overall phase, which does not affect probabilities.
Let $\widetilde\psi_0(x)=(\pi\hbar)^{-1/4}e^{-(x-d)^2/(2\hbar)}$, $d=qE$, be the new ground state. Its overlap with the initial state is, up to that phase,
$$
A_0=(\pi\hbar)^{-1/2}\int_{-\infty}^\infty
e^{-[x^2+(x-d)^2]/(2\hbar)}\,dx.
$$
Since $x^2+(x-d)^2=2(x-d/2)^2+d^2/2$, the <Gaussian integral> gives $A_0=e^{-d^2/(4\hbar)}$. The <Born rule> and the <ground-state overlap after sudden oscillator displacement> thus give
$$
\boxed{P_0=|A_0|^2=e^{-q^2E^2/(2\hbar)}.}
$$
This is an overlap probability, not a claim that the original wavefunction instantaneously relaxes to the new ground state. It becomes one when $qE=0$.
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