= Solution
Writing $a=cx+idp$, where $c=\sqrt{m\omega/(2\hbar)}$ and $d=1/\sqrt{2m\omega\hbar}$, the <canonical commutation relation> gives
$$
[a,a^\dagger]=-2icd[x,p]=2cd\hbar=\boxed{1}.
$$
Consequently $[a^\dagger a,a]=-a$ and $\boxed{[H,a]=-\hbar\omega a}$. If $a|\phi_0\rangle$ were nonzero, this <commutator> would make it an <energy eigenstate> with <energy> $E_0-\hbar\omega$, contradicting the lowest-energy property. Therefore \b[$a|\phi_0\rangle=0$].
The <commutator product rule> and induction give
$$
[a,(a^\dagger)^{n+1}]
=[a,(a^\dagger)^n]a^\dagger+(a^\dagger)^n[a,a^\dagger]
=(n+1)(a^\dagger)^n.
$$
Acting on the vacuum and shifting the sum then shows that the <unnormalized bosonic coherent state> $|\widetilde\psi\rangle=e^{\alpha a^\dagger}|0\rangle$ satisfies $a|\widetilde\psi\rangle=\alpha|\widetilde\psi\rangle$.
\b[The displayed series lacks its <wavefunction normalization> factor.] Since $(a^\dagger)^n|0\rangle=\sqrt{n!}|n\rangle$, its squared norm is $\sum_n|\alpha|^{2n}/n!=e^{|\alpha|^2}$. The normalized <coherent state> needed for the stated <expectation values> is
$$
|\alpha\rangle=e^{-|\alpha|^2/2}\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}|n\rangle,
\qquad a|\alpha\rangle=\alpha|\alpha\rangle.
$$
Use $\langle a\rangle=\alpha$, $\langle a^\dagger\rangle=\alpha^*$, $\langle a^\dagger a\rangle=|\alpha|^2$ and $\langle aa^\dagger\rangle=|\alpha|^2+1$. For $x=x_0(a+a^\dagger)$, $x_0=\sqrt{\hbar/(2m\omega)}$,
$$
\langle x\rangle=x_0(\alpha+\alpha^*),\qquad
\langle x^2\rangle=x_0^2\left[(\alpha+\alpha^*)^2+1\right],
\qquad \boxed{\Delta x=\sqrt{\frac{\hbar}{2m\omega}}}.
$$
The displacement changes the mean position but not its ground-state uncertainty. Unnormalized bras and kets give the same physical <expectation values> only after division by their squared norm.
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