= Solution
A <coherent state> is a normalized <eigenstate> of the <annihilation operator>, $a|\alpha\rangle=\alpha|\alpha\rangle$. Expanding in the <Fock state> <eigenbasis> gives a recurrence $c_{n+1}=\alpha c_n/\sqrt{n+1}$, and <wavefunction normalization> fixes
$$
|\alpha\rangle=e^{-|\alpha|^2/2}\sum_{n=0}^\infty\frac{\alpha^n}{\sqrt{n!}}|n\rangle
=D(\alpha)|0\rangle,\qquad
D(\alpha)=e^{\alpha a^\dagger-\alpha^*a}.
$$
The <displacement operator> therefore shifts the oscillator <ground state> in <phase space>. Distinct <coherent states> are not orthogonal:
$$
\langle\alpha|\beta\rangle=
\exp\left[-\frac{|\alpha|^2+|\beta|^2}{2}+\alpha^*\beta\right].
$$
The occupation probabilities form a <Poisson distribution>,
$$
P_n=e^{-N}\frac{N^n}{n!},\qquad N=|\alpha|^2,
\qquad \langle n\rangle=N,\quad (\Delta n)^2=N.
$$
For the <quantum harmonic oscillator>, the <expectation values> of position and momentum are
$$
\langle x\rangle=\sqrt{\frac{2\hbar}{m\omega}}\operatorname{Re}\alpha,\qquad
\langle p\rangle=\sqrt{2m\hbar\omega}\operatorname{Im}\alpha.
$$
The fluctuations equal those of the <ground state>,
$$
(\Delta x)^2=\frac{\hbar}{2m\omega},\qquad
(\Delta p)^2=\frac{m\hbar\omega}{2},\qquad
\boxed{\Delta x\,\Delta p=\frac{\hbar}{2}}.
$$
Thus the position <wavefunction> is a displaced <Gaussian wave packet> that saturates the <Heisenberg uncertainty principle>.
Time evolution under $H=\hbar\omega(a^\dagger a+1/2)$ preserves this form:
$$
\boxed{e^{-iHt/\hbar}|\alpha\rangle=e^{-i\omega t/2}|\alpha e^{-i\omega t}\rangle}.
$$
The center follows the classical oscillator trajectory while the packet width stays constant. Its mean <energy> is $\hbar\omega(N+1/2)$; it is generally a superposition of <energy eigenstates>, rather than a state of definite <energy>. For $N\gg1$, the relative occupation fluctuation is $1/\sqrt N$, explaining the classical appearance of a highly excited <coherent state>.
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