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The sudden approximation keeps the wavefunction unchanged at the instant of switching: a finite step of potential creates no delta-function impulse in the Schrodinger 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 ${\tilde{\psi }}_0(x)=(\pi \hslash {)}^{-1/4}{e}^{-(x-d{)}^2/(2\hslash )}$, $d=qE$, be the new ground state. Its overlap with the initial state is, up to that phase,Since $x^2+(x-d{)}^2=2(x-d/2{)}^2+d^2/2$, the Gaussian integral gives $A_0=e^{-d^2/(4\hslash )}$. The Born rule and the ground-state overlap after sudden oscillator displacement thus giveThis is an overlap probability, not a claim that the original wavefunction instantaneously relaxes to the new ground state. It becomes one when $qE=0$.