= Coherent state after shifting an oscillator equilibrium
{title2=$\alpha_0=-\sqrt{m\omega/(2\hbar)}\,x_0$}
Suddenly shifting a <quantum harmonic oscillator> equilibrium from zero to a signed position $x_0$ leaves the old <ground state> unchanged at the instant of the <quantum quench>. The new <annihilation operator> is $a'=a-\lambda$, with $\lambda=\sqrt{m\omega/(2\hbar)}x_0$, so $a'|0\rangle=-\lambda|0\rangle$. In the translated <energy eigenstates> $|n'\rangle=e^{-ipx_0/\hbar}|n\rangle$, the expansion is
$$
|0\rangle=e^{-\lambda^2/2}\sum_{n=0}^\infty\frac{(-\lambda)^n}{\sqrt{n!}}|n'\rangle.
$$
The sign follows by translating the old vacuum by $-x_0$ before expanding it in the old <Fock states>. A <Baker--Campbell--Hausdorff formula> with central <commutator> separates the resulting exponential into creation and annihilation parts. Evolution with the shifted <Hamiltonian> preserves this <coherent state>, with $\alpha(t)=\alpha_0e^{-i\omega t}$. Its mean occupation is $|\alpha_0|^2$ and its mean energy above the new potential minimum is $\hbar\omega(|\alpha_0|^2+1/2)$. Omitting $(-\lambda)^n$ in favour of $|\lambda|^n$ changes the state when $x_0>0$.
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