Taking a modular form such that we have Fricke involution:


$\sum\_{n=1} a\_n e^{-\pi nx^2} = \frac{A}{x^k} \sum\_{n=1} a\_n e^{-\pi \frac{n}{x^2}}$ [1]


I would like to know if there exists results on possible formula with $a\_n$ coeficient which will be like the classical Poisson summation formula (it works for Dirichlet L functions with the twisted Poisson summation formula so why not for modular forms for some functions?)


More specifically, for example, does following equality holds?:


$\sum\_{n=1} a\_n e^{-(z+ \sqrt{\pi n} x)^2 } = \frac{A}{x^k} \sum\_{n=1} a\_n \widehat{f\_z}(\frac{1}{x})$ [2]


(Where $f\_z(x)=e^{-(z+ \sqrt{\pi n}x)^2}$ and Fourier transform is done on variable x: to have an iso Poisson formula)


Note that in z=0 equation [2] becomes the formula [1]. I thought this can be demonstrate by showing that both side also satisfy same partial differential equation (but partial differential equation are slightly different...).


So my question is; do you have any reference on the subject? Is there some fucntions for which an equation like [2] can hold? for example as an extension of [1]?


(Originally I would like to have asymptotic of $\sum\_{n=1}a\_n e^{-(z+ \sqrt{\pi n}x)^2}$ in x near 0.)



Possible to have Poisson Summation formula with coefficient of modular forms? (for some functions)
The functional equation of the Riemann Zeta function is equivalent to the Poisson summation formula. This should be adoptable to the setting of automorphic $L$-functions. I am not sure where the abelian Fourier Analysis should happen here, probably in higher rank though(?)


Here is a reference, which is even more general than what you ask for: [Why is the functional equation of the Riemann zeta function equivalent to the Poisson summation formula?](https://mathoverflow.net/questions/62969/why-is-the-functional-equation-of-the-riemann-zeta-function-equivalent-to-the-po)


