Overlap of nth QHO excited state and momentum-shifted QHO ground state

[HBHSU]

$\newcommand{\ket}[1]{|#1\rangle}$
$\newcommand{\bra}[1]{\langle#1|}$
I have a momentum-shifting operator $e^{i\Delta p x/\hbar}$ acting on the ground state $\ket{0}$ of the QHO, and I want to compute the overlap of this state with the n$^{th}$ excited QHO state $\ket{n}$. Given $\hat{a}^\dagger=\sqrt{\dfrac{m\omega}{2\hbar}}x-ip\sqrt{\dfrac{1}{2m\omega\hbar}}$, I let $c=i\Delta p\sqrt{\dfrac{2}{m\omega\hbar}}$ and obtain $e^{c\hat{a}^\dagger}\ket{0}=\sum_{m=0}^\infty\dfrac{c^m(\hat{a}^\dagger)^m}{m!}\ket{0}=\sum_{m=0}^\infty\dfrac{c^m}{\sqrt{m!}}\ket{m}$. Then $\bra{n}e^{c\hat{a}^\dagger}\ket{0}=\bra{n}e^{\frac{i\Delta px}{\hbar}+\frac{p\Delta p}{m\omega\hbar}}\ket{0}=\bra{n}e^{\frac{i\Delta px}{\hbar}}e^{\frac{p\Delta p}{m\omega\hbar}}e^{\frac{-(\Delta p)^2}{\hbar m\omega}}\ket{0}=\dfrac{(i\Delta p\sqrt{\frac{2}{m\omega\hbar}})^n}{\sqrt{n!}},$ using the Baker-Campbell-Hausdorff formula. Now I write $\bra{n}e^{c\hat{a}^\dagger}\ket{0}=\dfrac{(i\Delta p\sqrt{\frac{2}{m\omega\hbar}})^n}{\sqrt{n!}}e^{-\frac{p\Delta p}{m\omega\hbar}}e^{\frac{(\Delta p)^2}{\hbar m\omega}}$. Is this valid? I have concerns about moving the exponential of the momentum operator out of the inner product. Might I instead write $\ket{n}$ in terms of coherent states in order to find the overlap?

 

[vanhees71]

I think, it's much simpler to use yhour equation for the coherent state and just pick out the coefficient of $|n \rangle$ in its expansion in terms of number eigenstates, which you already have!