= Solution
A normalized oscillator <coherent state> is an <eigenstate> of the <annihilation operator>, $a|\alpha\rangle=\alpha|\alpha\rangle$. Expanding it in <Fock states> gives
$$
|\alpha\rangle=e^{-|\alpha|^2/2}\sum_{n=0}^\infty\frac{\alpha^n}{\sqrt{n!}}|n\rangle.
$$
Since $x=\sqrt{\hbar/(2M\omega)}(a+a^\dagger)$, $\langle a^2\rangle=\alpha^2$, and $\langle aa^\dagger\rangle=|\alpha|^2+1$, the <variance> is $\langle x^2\rangle-\langle x\rangle^2=\hbar/(2M\omega)$. Hence
$$
\boxed{\Delta x=\sqrt{\frac{\hbar}{2M\omega}},}
$$
independently of the complex amplitude $\alpha$.
For $A=\alpha a^\dagger$ and $B=-\alpha^*a$, the <commutator> is the scalar $[A,B]=|\alpha|^2I$. The <Baker--Campbell--Hausdorff formula> therefore terminates and gives the <displacement operator>
$$
D(\alpha)=e^{\alpha a^\dagger-\alpha^*a}
=e^{-|\alpha|^2/2}e^{\alpha a^\dagger}e^{-\alpha^*a}.
$$
Because $a|0\rangle=0$, applying it to the <Fock vacuum> yields exactly the <Fock state> expansion above:
$$
\boxed{|\alpha\rangle=D(\alpha)|0\rangle.}
$$
For a general complex $\alpha$, its <complex conjugate> is required in the annihilation term. The expression with $-\alpha a$ printed in the paper is valid when $\alpha$ is real.
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