Decomposing a coherent state? Can you decompose a coherent state $|\alpha\rangle$ into $p|p\rangle+q|q\rangle$, where $|p\rangle$ and $|q\rangle$ are eigenstates of the $P$ and $Q$ operators respectively with eigenvalues p and q?
The above relation feels "natural", but I wonder if it is well-known?
Where $\alpha=q+ip$
 A: You are effectively asking for a change of basis from oscillator number states to position eigenstates $|x\rangle$, essentially the wavefunction of the Schrödinger wavepacket. (Momentum eigenstates are expressible in terms of the latter, so they are overkill.)
You already know that
$$
\langle  n|  x \rangle= \frac{1}{\pi^{1/4}\sqrt{2^n~n!}}~ (x-\partial_x)^n~ e^{-x^2/2} ~.
$$
Consequently
$$
|\alpha\rangle = e^{-|\alpha|^2/2} \sum_{n=0}^\infty  {\alpha^n\over \sqrt{n!} }|n\rangle \\ = \int \!\!dx ~ |x\rangle 
e^{-|\alpha|^2/2} \sum_{n=0}^\infty  {\alpha^n\over \sqrt{n!} }\langle x|n\rangle \\
= \int \!\!dx ~ |x\rangle 
 \frac{e^{-|\alpha|^2/2}}{\pi^{1/4}}\sum_{n=0}^\infty  {(\alpha/\sqrt{2 })^n\over  n!  } ~ (x-\partial_x)^n~ e^{-x^2/2} \\ 
=\int \!\!dx ~ |x\rangle ~
 \frac{e^{-|\alpha|^2/2}}{\pi^{1/4}} e^{  {\alpha\over \sqrt{2 }}  ~ (x-\partial_x) }  ~ e^{-x^2/2} \\
=\int \!\!dx ~ |x\rangle ~
 \frac{e^{-|\alpha|^2/2}}{\pi^{1/4}} e^{   -\alpha^2/4     } e^{  {\alpha\over \sqrt{2 }}  ~  x  }e^{ - {\alpha\over \sqrt{2 }}  ~  \partial_x  } ~ e^{-x^2/2} \\
=\int \!\!dx ~ |x\rangle ~
 \frac{e^{(\alpha^2-|\alpha|^2)/2}}{\pi^{1/4}}    ~ e^{-\left(x-\sqrt{2 }\alpha\right )^2/2}  .
$$
Compare to the standard wavefunction of the coherent state.

*

*You should be able to easily confirm this is an eigenstate of $(\hat x + i\hat p)/\sqrt{2}$      with eigenvalue α.

