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I am trying to compute the temporal evolution of a coherent state $|\alpha\rangle$ using a given hamiltonian of the form:$$\hat{H}=\hbar \omega(\hat{a}^{\dagger}\hat{a}-\alpha(\hat{a}+\hat{a}^{\dagger})).$$ - My first attempt has been to find the action of the $\hat{H}$ over a Fock state $|n\rangle$ due to the fact that I can express the coherent states in terms of the Fock states: $$|\alpha\rangle=e^{\frac{-\alpha^{2}}{2}}\sum_{n=0}{\frac{\alpha^{n}}{n!}}|n\rangle,$$ and using that $f(\hat{H})|n\rangle=f(h_{n})|n\rangle$ attack the $e^{-i\hat{H}t/\hbar}$ that appears when computing the time evolution.

My problem is: I have found $\hat{H}|n\rangle$ to be: $$\hat{H}|n\rangle=\hbar \omega (n|n\rangle-\alpha\sqrt{n}|n-1\rangle-\alpha\sqrt{n+1}|n+1\rangle,$$ where I am stucked as it is not of the form $\hat{H}|n\rangle\propto |n\rangle $ so i don't know what to do with the exponential: $$e^{i\hat{H}t/\hbar}|n\rangle=e^{-i\omega t (\hat{a}^{\dagger}\hat{a}-\alpha(\hat{a}+\hat{a}^{\dagger}))}|n\rangle=?$$ I have attemped to manipulate the exponential term by using the BCH formula but I achieved nothing but a messed result

EDIT: Here I add my achieved answer using the BCH formula: $$|\alpha(t)\rangle=e^{\frac{-\alpha ^{2}}{2}}e^{\frac{-\omega^{2}t^{2}(1+\alpha ^{2})}{2}}\sum_{n=0}\frac{\alpha ^{n}}{n!}e^{-i \omega t (n-\alpha \sqrt{n+1}-\alpha \sqrt{n})}|n\rangle,$$ which seems, to my unexperienced eye, to be wrong. Thanks in advance.

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  • $\begingroup$ nice problem btw! $\endgroup$ – ZeroTheHero Mar 18 '17 at 14:05
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Yes you need to use a BCH but you can see your way if you instead recognize that your Hamiltonian can be written as $$ \hat H=\frac{\hat p^2}{2m}+\frac{1}{2}m\omega^2 (\hat x- \kappa)^2 $$ for some $\kappa$, i.e. as a shifted harmonic oscillator. What you can then do is consider the translated Hamiltonian $$ \hat H= e^{\beta(\hat a-\hat a^\dagger)} \,\hat a^\dagger \hat a\,e^{-\beta(\hat a-\hat a^\dagger)}=T(\beta)\hat a^\dagger \hat a T^{-1}(\beta) $$ for suitable choice of $\beta$ and remember that the coherent state is itself a translate of the ground state so as to get something like $$ \hat H\vert \alpha \rangle=T(\beta) \hat a^\dagger \hat a \,T^{-1}(\beta)T(\alpha)\vert 0\rangle\, . $$

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  • $\begingroup$ Firstly: Thanks for the answer! Secondly: Such a discover, that Isle Of Everywhere is delicious! I have edited the answer where my messed result with the BCH appears, I don't know where could I be wrong in my calculations. I am comfortable with the answer that you provided and I don't want to sound rude but I would like to be confident with the other method in case I can't recognize anything, so I would wait a bit more in order to close the question. $\endgroup$ – Marc C Mar 18 '17 at 15:49
  • $\begingroup$ @MarcC not so easy to do BCH since the commutator $[\hat a^\dagger \hat a,\hat a^\dagger+\hat a]$ does commute with either $\hat a^\dagger \hat a$ or $\hat a^\dagger+\hat a$... :( $\endgroup$ – ZeroTheHero Mar 18 '17 at 21:27

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