= Solution
The <spatial translation operator> is $U(\alpha)=e^{-i\alpha\widehat p/\hbar}$. From $[\widehat x,\widehat p]=i\hbar$ and the preceding central-commutator result,
$$
\boxed{[\widehat x,U(\alpha)]=\alpha U(\alpha).}
$$
Thus $\widehat xU|x\rangle=U\widehat x|x\rangle+\alpha U|x\rangle=(x+\alpha)U|x\rangle$, consistent with $U|x\rangle=|x+\alpha\rangle$. Since $\langle x|U=\langle x-\alpha|$, the translated <wavefunction> is $\boxed{\psi(x-\alpha)}$.
For the <quantum harmonic oscillator>, the <creation and annihilation operators> give $\widehat p=i\sqrt{m\hbar\omega/2}(a^\dagger-a)$. Put $s=\alpha\sqrt{m\omega/(2\hbar)}$. The central <commutator> of $sa^\dagger$ and $-sa$ is $s^2$, so
$$
U=e^{s(a^\dagger-a)}=e^{-s^2/2}e^{sa^\dagger}e^{-sa}.
$$
As $a|0\rangle=0$ and $(a^\dagger)^n|0\rangle=\sqrt{n!}|n\rangle$,
$$
U|0\rangle=e^{-s^2/2}\sum_{n=0}^\infty\frac{s^n}{\sqrt{n!}}|n\rangle.
$$
Taking position <wavefunctions> proves
$$
\boxed{\psi_0(x-\alpha)=e^{-m\omega\alpha^2/(4\hbar)}\sum_{n=0}^\infty\left(\frac{m\omega}{2\hbar}\right)^{n/2}\frac{\alpha^n}{\sqrt{n!}}\psi_n(x).}
$$
The coefficient squares sum to $e^{-s^2}\sum_n s^{2n}/n!=1$, checking normalization of this <coherent state>.
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