quantum mechanics - How are the definitions of a coherent state equivalent?

I am trying to understand coherent states. As far as I could find there are three equivalent definitions and in general many sources start from a different one, still I fail to see their equivalence. I restate the definitions and their equivalences given in the Wikipedia page:

1. Eigenstate of annihilation operator: $$a|\alpha \rangle =\alpha|\alpha \rangle$$


2. Displacement operator of the vacuum: $$|\alpha \rangle =e^{\alpha a^{\dagger}-\alpha^{*} a}|0 \rangle$$


3. State of minimal Uncertainty: $$\Delta X= \Delta P = \frac{1}{\sqrt{2}}$$

I fail to see how they are the same! Can someone please explain how to derive these from one another?

Answer

Refer to the nice complement on coherent states in the book by Cohen-Tannoudji, Diu and Laloë, volume 1. It starts off defining coherent states as neither of the ones you mention, and then derives all properties.

To answer the question, if you start with definition 2, you can easily show 1, and then from 2, 3. First expand the exponential using Baker-Campbell-Hausdorff formula: $$e^{\alpha a^\dagger -\alpha^* a}=e^{\alpha a^\dagger}e^{-\alpha^* a}e^{\frac{-1}{2}|\alpha|^2[a^\dagger,-a]}$$ and let it act on the vacuum state $|0\rangle$ to get $$|\alpha \rangle = e^{-|\alpha|^2/2}e^{\alpha a^\dagger}e^{-\alpha^* a}|0\rangle \\ = e^{-|\alpha|^2/2}e^{\alpha a^\dagger}|0\rangle \\ = e^{-|\alpha|^2/2} \sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}} |n\rangle$$ Now that you have the expression for $|\alpha \rangle$ in terms of states you already know, you can operate $a$ on it to find that it is indeed an eigenstate of the lowering operator, showing that definition 2 implies definition 1.

Property 3 follows from finding $\langle X^2 \rangle$ and $\langle P^2 \rangle$ for this state, by expressing the operators in terms of $a$ and $a^\dagger$, a fairly standard exercise.