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 $\alpha=q+ip$
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.
