I'm stuck on some simple mathematics in finding the time evolved coherent state for a single-mode field from Gerry and Knight, Introductory Quantum Optics page 51.

The Hamiltonian is given by $\hat{H} = \hbar \omega \left(\hat{n} + \frac{1}{2}\right)$, and the time evolution is found as follows:

$$ \lvert \alpha, t \rangle \equiv \exp(-i\hat{H}t/\hbar) \lvert \alpha\rangle = e^{-i\omega t/2}e^{-i\omega t \hat{n}}\lvert\alpha\rangle = e^{-i\omega t/2}\lvert\alpha e^{-i\omega t}\rangle, $$

however I do not understand this last step; why the term $e^{-i\omega t \hat{n}}$ can be brought into the ket like that as a phase change of the annihilation operator eigenvalue $\alpha$.

Here, $\hat{a} \lvert\alpha\rangle = \alpha \lvert\alpha\rangle$, and $\hat{n} = \hat{a}^\dagger\hat{a}$.

Thanks for any help!


1 Answer 1


\begin{align} e^{-i\omega t/2}e^{-i\omega t \hat n}\vert\alpha\rangle&= e^{-i\omega t/2}\sum_{N}e^{-i\omega t \hat n}e^{-\vert\alpha\vert^2/2} \frac{\alpha^n}{\sqrt{n!}}\vert n\rangle\\ &=e^{-i\omega t/2}\sum_{N}e^{-\vert\alpha\vert^2/2} \frac{e^{-i\omega nt}\alpha^n}{\sqrt{n!}}\vert n\rangle\\ &=e^{-i\omega t/2}\sum_{N}e^{-\vert\alpha\vert^2/2} \frac{(e^{-i\omega t}\alpha)^n}{\sqrt{n!}}\vert n\rangle\, ,\\ &=e^{-i\omega t/2}\vert e^{-i\omega t}\alpha\rangle \end{align}

  • $\begingroup$ Thank you! It hadn't occurred to me to use the definition in terms of the number states for some reason, I guess it was too obvious for the book to state $\endgroup$ Apr 8, 2017 at 10:39
  • $\begingroup$ It's a classic. If you don't "see" what to do you'll never get it. $\endgroup$ Apr 8, 2017 at 13:12
  • $\begingroup$ What do you mean by that? Like the book won't spell things out for you? Thankfully the lecture course doesn't require me to memorise this definition of the coherent state $\endgroup$ Apr 10, 2017 at 13:47
  • 1
    $\begingroup$ The answer above by ZeroTheHero is the correct way to do it, just be aware that a 1/2 term is missing in the exponential of the e term of the coherent state. $\endgroup$
    – alonoid
    Nov 5, 2020 at 13:58
  • 1
    $\begingroup$ @alonoid fixed thanks. $\endgroup$ Jul 27, 2021 at 1:27

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