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!
Best Answer
\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}