# Why the name 'displacement' operator?

I'm studying coherent states of the harmonic oscillator and I have learned about the so called displacement operator, which is the operator defined as $$D(\alpha) = e^{\alpha a^\dagger -\alpha^* a}$$ whose action over the fundamental state $|0\rangle$ is to produce a coherent state $|\alpha\rangle$. Now, why is it called displacement operator? In which sense does it displace the state $|0\rangle$?

A coherent state is characterized by a complex number $\alpha \in \mathbb C$. Applying the displacement operator $D(\beta)$ to $|\alpha\rangle$ translates $\alpha$ in the complex plane by $\beta$, in the following sense: $$D(\beta) |\alpha\rangle \sim |\alpha+\beta\rangle . \tag 1$$
Here, $\sim$ means "up to a phase". The precise relation is: $${D}(\beta){D}(\alpha) = e^{(\beta\alpha^*-\beta^*\alpha)/2} {D}(\alpha + \beta) , \tag 2$$ the exponential is just a phase factor.
Note 1: (1) follows from (2) because $|\alpha\rangle = D(\alpha) |0\rangle$ - multiply (2) from the right with $|0\rangle$ to get (1).
Note 2: That also makes sense for the ground state $|0\rangle$, because it is equal to the coherent state with $\alpha=0$: $$D(\alpha)|0\rangle = |0+\alpha\rangle .$$