Is the vacuum state a coherent state? I'm asking because I got introduced to the state $|0\rangle$ as a fock-state. Nevertheless:
$$
\hat{a} |0\rangle = 0 |0 \rangle
$$
It is an eigenstate of $\hat{a}$ with eigenvalue $0$, and it can be obtained the same way any other coherent states are obtained via the displacement operator with parameter 0:
$$
\hat{D}(\alpha=0)|0 \rangle = e^{0 \hat{a}^\dagger - 0 \hat{a}}|0\rangle = |0\rangle
$$
Would one consider the vacuum state a coherent state?
 A: The coherent state $\vert \alpha\rangle$ is just a vacuum state $\vert 0\rangle$ translated in $x$ and $p$ space so $\alpha=x_0+ip_0$.  Thus the vacuum state is a coherent state that has not been displaced, i.e. $x_0=p_0=0$.
In fact, a nice way to see this is in the Wigner function formalism.  The vacuum state is just a Gaussian sitting at the centre of $(x,p)$ space whereas a coherent state is the same state displaced to another point.  This is illustrated in the figures below, taken from this site: on the left is the Wigner function of the vacuum state, and on the right that of a coherent state.
 
Note also that the Wigner function for the coherent state is everywhere positive, and positivity of the Wigner function is sometimes taken as a marker of classicality so in this sense coherent states (and the vacuum state) are "classical states".
A short movie illustrating the time evolution of the Wigner function of a coherent state can be  found on the coherent state wikipage; it shows the Wigner function does not deform and remains non-negative at all times  Of course since the vacuum state is an eigenstate of the Hamiltonian and lies at the centre of $(x,p)$,  its Wigner function would actually remain there at all times.
A: Coherent state is a superpostition of states with different particle number with a weight of Poisson distribution. In other words, 
$$|\alpha\rangle =e^{-{|\alpha|^2\over2}}\sum_{n=0}^{\infty}{\alpha^n\over\sqrt{n!}}|n\rangle $$ where $|n\rangle$ is a state with $n$ number of particles.
Thus when you say that my system is in coherent state, you mean that your system does not have a definite particle number. This is why if you remove a particle from that state with 
$$a|\alpha\rangle=\alpha|\alpha\rangle$$
you dont change the state.  That's why coherent state is an eigenstate of annihilation operator. 
But the reason that vacuum is an eigenstate of annihilation op is; it does not have any particle so there is nothing to annihilate. that's why it does not change.
So in this sense, I would not say vacuum is a coherent state because it has a definite particle number $0$. and the reason of this similarity under the act of annihilation operator is due to different reasons.
A: The ordinary coherent states may be generated in several ways: (1) as minimum uncertainty states, (2) as eigenstates of the annihilation operator, and (3) as states displaced from the ground state via the operator $D (\alpha ) =e^{\alpha a^{\dagger}-\alpha^{*} a}$, where $\alpha$ is a complex number. The states generated by these three methods are equivalent. I will show how they are equivalent, and by their equivalence, it is enough to show that the ground state $|0\rangle$ is a minimum uncertainty state.
Proof:
\begin{align}
x^{2} &=\frac{\hbar}{2 m \omega}\left(a+a^{\dagger}\right)^{2} \\
p^{2} &=-\frac{m \omega \hbar}{2}\left(a-a^{\dagger}\right)^{2}
\end{align}
Using the second definition of Coherent state (as eigenstates of the annihilation operator),
\begin{align}
a|\alpha\rangle=\alpha|\alpha\rangle
\end{align}
which implies $\left\langle\alpha\left|a^{\dagger} a\right| \alpha\right\rangle=|\alpha|^{2}$. It is trivial to check that this indeed defines a minimal wave-packet
\begin{align}
\begin{array}{r}
\left\langle\alpha\left|\left(a+a^{\dagger}\right)\right| \alpha\right\rangle=\left(\alpha+\alpha^{\star}\right) \\
\left\langle\alpha\left|\left(a-a^{\dagger}\right)\right| \alpha\right\rangle=\left(\alpha-\alpha^{\star}\right) \\
\left\langle\alpha\left|\left(a+a^{\dagger}\right)\left(a+a^{\dagger}\right)\right| \alpha\right\rangle=\left(\alpha+\alpha^{\star}\right)^{2}+1 \\
\left\langle\alpha\left|\left(a-a^{\dagger}\right)\left(a-a^{\dagger}\right)\right| \alpha\right\rangle=\left(\alpha-\alpha^{\star}\right)^{2}-1
\end{array}
\end{align}
from which follows
\begin{align}
\left\langle(\Delta x)^{2}\right\rangle_{\alpha} &=\left\langle x^{2}\right\rangle_{\alpha}-\langle x\rangle_{\alpha}^{2}=\frac{\hbar}{2 m \omega} \\
\left\langle(\Delta p)^{2}\right\rangle_{\alpha} &=\left\langle p^{2}\right\rangle_{\alpha}-\langle p\rangle_{\alpha}^{2}=\frac{\hbar m \omega}{2}
\end{align}
and accordingly
\begin{align}
\left\langle(\Delta x)^{2}\right\rangle_{\alpha}\left\langle(\Delta p)^{2}\right\rangle_{\alpha}=\frac{\hbar}{4}
\end{align}
So the states $|\alpha\rangle$ satisfy the minimum uncertainty relation.
In the $|n\rangle$ base the coherent state looks like:
\begin{align}
|\alpha\rangle=\sum_{n} c_{n}|n\rangle=\sum_{n}|n\rangle\langle n \mid \alpha\rangle
\end{align}
Since
\begin{align}
|n\rangle=\frac{\left(a^{\dagger}\right)^{n}}{\sqrt{n !}}|0\rangle
\end{align}
we have
\begin{align}
\langle n \mid \alpha\rangle=\frac{\alpha^{n}}{\sqrt{n !}}\langle 0 \mid \alpha\rangle
\end{align}
and thus
\begin{align}
|\alpha\rangle=\langle 0 \mid \alpha\rangle \sum_{n=0}^{\infty} \frac{\alpha^{n}}{\sqrt{n !}}|n\rangle
\end{align}
The constant $\langle 0 \mid \alpha\rangle$ is determined by normalization as follows:
\begin{align}
1=\sum_{n}\langle\alpha|n\rangle \langle n| \alpha\rangle=|\langle 0 \mid \alpha\rangle|^{2} \sum_{m=0}^{\infty} \frac{|\alpha|^{2 m}}{m !}=|\langle 0 \mid \alpha\rangle|^{2} e^{|\alpha|^{2}}
\end{align}
solving for $\langle 0 \mid \alpha\rangle$ we get:
\begin{align}
\langle 0 \mid \alpha\rangle=e^{-\frac{1}{2}|\alpha|^{2}}
\end{align}
up to a phase factor. Substituting, we obtain the final form:
\begin{align}
|\alpha\rangle=e^{-\frac{1}{2}|\alpha|^{2}} \sum_{n=0}^{\infty} \frac{\alpha^{n}}{\sqrt{n !}}|n\rangle
\end{align}
Now,
\begin{align}
\sum_{n=0}^{\infty} \frac{\alpha^{n}}{\sqrt{n !}}|n\rangle=\sum_{n=0}^{\infty} \frac{\alpha^{n}}{n !}\left(a^{\dagger}\right)^{n}|0\rangle=e^{\alpha a^{\dagger}}|0\rangle
\end{align}
which implies
\begin{align}
|\alpha\rangle=e^{-\frac{1}{2}|\alpha|^{2}} e^{\alpha a^{\dagger}}|0\rangle .
\end{align}
Due to $a|0\rangle=0$, we have $e^{-\alpha^{*}a}|0\rangle=|0\rangle$. Thus, the above equation can be written as
\begin{align}
|\alpha\rangle=e^{-\frac{1}{2}|\alpha|^{2}} e^{\alpha a^{\dagger}} e^{-\alpha^{*} a}|0\rangle .
\end{align}
Using the Baker-Campbell-Hausdorff $(\mathrm{BCH})$ formula for any two operators $A$ and $B$ which commute with the commutator of $A$ and $B$
$$
e^{-\frac{1}{2}[A, B]} e^{A} e^{B}=e^{A+B}
$$
and the relation
\begin{align}
\left[\alpha a^{\dagger}, \alpha^{*} a\right]=|\alpha|^{2},
\end{align}
we arrive at,
\begin{align}
|\alpha\rangle = e^{\alpha a^{\dagger}-\alpha^{*} a}|0\rangle = D(\alpha)|0\rangle
\end{align}
and $|\alpha \rangle $ is a “displaced vacuum state”. $D(\alpha)$ is the displacement operator.
So, we showed $(2) \equiv (1) $ and $(2) \equiv (3)$.
Now to show that the ground state $|0\rangle$ is a minimum uncertainty state:
Since
\begin{align}
\left\langle 0\left|\left(a+a^{\dagger}\right)\left(a+a^{\dagger}\right)\right| 0\right\rangle &=\left\langle 0\left|a a^{\dagger}\right| 0\right\rangle=1 \\
\left\langle 0\left|\left(a-a^{\dagger}\right)\left(a-a^{\dagger}\right)\right| 0\right\rangle &=-\left\langle 0\left|a a^{\dagger}\right| 0\right\rangle=-1
\end{align}
it follows that
\begin{align}
\left\langle x^{2}\right\rangle_{0}\left\langle p^{2}\right\rangle_{0}=-\frac{\hbar^{2}}{4} 1(-1)=\frac{\hbar^{2}}{4}
\end{align}
and finally, since $\langle x\rangle_{0}=\langle p\rangle_{0}=0$, it follows that
\begin{align}
\left\langle(\Delta x)^{2}\right\rangle_{0}\left\langle(\Delta p)^{2}\right\rangle_{0}=\frac{\hbar^{2}}{4}
\end{align}
But note that $|n\rangle$ is not a minimum uncertainty state.
\begin{align}
\left\langle n\left|\left(a+a^{\dagger}\right)\left(a+a^{\dagger}\right)\right| n\right\rangle=\left\langle n\left|a a^{\dagger}+a^{\dagger} a\right| n\right\rangle=\left\langle n\left|2 a^{\dagger} a+\left[a, a^{\dagger}\right]\right| n\right\rangle=2 n+1
\end{align}
and similarly
\begin{align}
\left\langle n\left|\left(a-a^{\dagger}\right)\left(a-a^{\dagger}\right)\right| n\right\rangle=-(2 n+1)
\end{align}
which implies
\begin{align}
\left\langle(\Delta x)^{2}\right\rangle_{n}\left\langle(\Delta p)^{2}\right\rangle_{n}=\frac{\hbar^{2}}{4}(2 n+1)^{2}
\end{align}
so $|n\rangle$ is not minimal!
