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?

  • 2
    $\begingroup$ 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. $\endgroup$ Commented Dec 12, 2013 at 16:39
  • $\begingroup$ @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. $\endgroup$
    – Abhinav
    Commented Dec 12, 2013 at 17:20
  • 1
    $\begingroup$ 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 $p$. $\endgroup$ Commented Dec 12, 2013 at 17:24
  • $\begingroup$ 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? $\endgroup$
    – Abhinav
    Commented Dec 12, 2013 at 17:27
  • $\begingroup$ Related: physics.stackexchange.com/q/60655/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Feb 5, 2014 at 8:21

1 Answer 1


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.

  • $\begingroup$ I'm sorry, can you elaborate why $e^{-\alpha^* a} | 0 \rangle = | 0 \rangle $ ?I don't see that. If I expand the expontential function into a taylor series, then this whole term should just compute to zero. $\endgroup$ Commented Aug 26, 2017 at 13:05
  • $\begingroup$ @Quantumwhisp when you expand it the zeroth order term (in $a$) of the exponential series is the identity, which gives $| 0 \rangle $. $\endgroup$
    – Abhinav
    Commented Aug 27, 2017 at 20:23
  • $\begingroup$ The expectation value of, say the number operator, in a coherent state $| \alpha>$ turns out to be $|\alpha|^2$. If we choose $\alpha = |\alpha| e^{I \theta}$, then what are the typical values for $|\alpha|$ and $\theta$? $\endgroup$
    – W. Voltera
    Commented Jul 22, 2018 at 18:40

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