# 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 $|0\rangle$?

## 1 Answer

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 $\alpha=0$: $$D(\alpha)|0\rangle = |0+\alpha\rangle .$$

• Furthermore, if you look at quasi-probability distributions in phase space (e.g. Wigner function), they are simply displaced by the displacement operator, no matter what the initial state is. Another way to see it, is that the displacement operator shifts the average values of the quadratures, but does not change the higher order moments. – Frédéric Grosshans May 17 '17 at 14:53