# How to derive the displacement operator from the relation $|\alpha\rangle=e^{\alpha(a^\dagger-a)}|0\rangle$?

I've been studying coherent states, an encountered a problem. My teacher told me, one can derive displacement operator via complexifying the parameter $$\alpha$$ in this formula: $$|\alpha\rangle=e^{\alpha( a^\dagger - a)} |0\rangle$$ ($$|\alpha\rangle$$ denotes a coherent state, and $$|0\rangle$$ denotes the ground state.)

For more formally speaking, I should say this is actually performing the "analytic continuation".

Then we get this: $$D(\alpha) = e^{\alpha a^\dagger -\alpha^* a}$$ I know $$\alpha$$ denotes a complex number, but I still have no idea how to get final answer, by my own calculation I can't get that complex conjugate term. Can some one please show me the correct whole process？

Coherent state is defined as: \begin{align} a|\alpha\rangle=\alpha|\alpha\rangle \end{align} 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.

@Varun Govind's answer is correct. I'll just offer another way to get to the result which is basis independent, requires less computation, and motivates the definition of $$D$$. $$D(\alpha)$$ is unitary since the argument of the exponential is antihermitian. Making $$\alpha(t)$$ depend on a real parameter, noted $$t$$, you can view it as the evolution operator and the real parameter as time. In particular, to see how the operators evolve, you can write out Heisenberg's equations of motion and solve them. For the creation/annihilation operators, writing $$a_t:=D(\alpha)^\dagger aD(\alpha)$$: $$\frac{d}{dt} a_t = [\dot \alpha^* a_t-\dot \alpha a_t^\dagger,a_t] \\ =\dot \alpha$$ so you get using the inititial condition $$a_0 = a$$: $$D(\alpha)^\dagger aD(\alpha)= a+\alpha\\ D(\alpha)^\dagger a^\dagger D(\alpha)= a^\dagger+\alpha^*$$ This is the mathematical translation of $$D$$ being the displacement operators, it acts as a translation on the creation/annihilation operators. In particular, you get: $$aD(\alpha)|0\rangle = D(\alpha)(a+\alpha)|0\rangle \\ = \alpha D(\alpha)|0\rangle$$ which proves that $$D(\alpha)|0\rangle$$ is a coherent state. Note that there is no canonic choice of a coherent state, these are defined up to a global phase.

Note that it also shows that $$D(\alpha)$$ can be used to express the time evolution operator of a forced harmonic oscillator. This is how you physically obtain coherent states: start with the ground state of a harmonic oscillator and add an excitation.

Hope this helps.