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?
Best Answer
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.