= Solution
The <annihilation operator> associated with the new coordinate $X=x-x_0$ is
$$
a'=\sqrt{\frac{m\omega}{2\hbar}}(x-x_0)+\frac{i}{\sqrt{2m\hbar\omega}}p=a-\lambda.
$$
As the initial state is the old vacuum,
$$
\boxed{a'|\Psi(0)\rangle=-\lambda|\Psi(0)\rangle,\qquad
\alpha_0=-\lambda=-\sqrt{\frac{m\omega}{2\hbar}}x_0}.
$$
Equivalently, applying $a'$ to the number-state expansion in part (c) and shifting its summation index returns $-\lambda$ times the same state. It is therefore a <coherent state> of the displaced oscillator.
The new <Hamiltonian> is $H=U_{\min}+\hbar\omega(a'^\dagger a'+1/2)$. In the <Schrödinger picture> its evolution gives
$$
|\Psi(t)\rangle=e^{-i(U_{\min}/\hbar+\omega/2)t}
e^{-|\alpha_0|^2/2}\sum_{n=0}^\infty\frac{(\alpha_0e^{-i\omega t})^n}{\sqrt{n!}}|n'\rangle.
$$
Consequently
$$
\boxed{a'|\Psi(t)\rangle=\alpha(t)|\Psi(t)\rangle,\qquad
\alpha(t)=\alpha_0e^{-i\omega t}}.
$$
The displayed prefactor is only a global phase. Equivalently, in the <Heisenberg picture> $a'(t)=a'(0)e^{-i\omega t}$. The mean occupation $|\alpha(t)|^2=\lambda^2$ is constant, and $\hbar\omega(\lambda^2+1/2)$ reproduces the energy above the new minimum found in part (b). Its mean position and momentum likewise reproduce that part's oscillating wavepacket centre.
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