# 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?

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Note that (1) does not follow very directly from (3) because squeezed states are also minimal uncertainty states but need not be eigenstates of $a$. However, there will always be some annihilation operator $\hat b=u\hat x+iv\hat y$ for which they are eigenstates. – Emilio Pisanty Dec 12 '13 at 16:39
@EmilioPisanty, that's why (3) has \delta X = \delta P (in the choice of units where m= \hbar = \omega = 1). Squeezed states do not satisfy the equality. – Abhinav Dec 12 '13 at 17:20
You can definitely do that, but note that setting $\omega=1$ is in no way a natural unit. You can set $-i[x,p]=\hbar=1$, but specifying $\omega$ and $m$ does a bit of violence to phase space. You should either say $\Delta x\Delta p=\frac12\Rightarrow\exists a:\ a|\alpha\rangle =\alpha|\alpha\rangle$, or state the minimal uncertainties you're setting in terms of the constants that define $a$ in terms of $x$ and $p$. – Emilio Pisanty Dec 12 '13 at 17:24
Hmm, I like the term 'violence to phase space'. Or instead can (3) be defined to look like $\Delta X$ = $\sqrt{\frac{\hbar}{2m\omega}}$, $\Delta P$ = $\sqrt{\frac{\hbar m \omega}{2}}$ ? In this case, will this relation too be satisfied by squeezed states? – Abhinav Dec 12 '13 at 17:27
Related: physics.stackexchange.com/q/60655/2451 and links therein. – Qmechanic Feb 5 '14 at 8:21

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.