TITLE: Eigenstates of Fourier transformation
QUESTION [7 upvotes]: Let $\gamma$ be defined on $\mathbb R^n$ by $\gamma (x)=e^{-π x^2}$. With $\mathcal F$ standing for the Fourier transformation defined on the Schwartz space by
$$
(\mathcal F u)(\xi)=\int e^{-2iπ x\cdot \xi} u(x) dx,
$$
we have 
$
\mathcal F\gamma =\gamma.
$
We can also define $\mathcal F$ for the tempered distribution ($\mathscr S'$)
with the duality formula
$$
\langle \widehat T,\phi\rangle_{\mathscr S',\mathscr S}=\langle T,\widehat{\phi}\rangle_{\mathscr S',\mathscr S}.
$$
For instance, the Poisson summation formula is $\widehat S=S$ with 
$S=\sum_{k\in \mathbb Z^n}\delta_k$. Finally the question:
determine all the tempered distributions $T$ such that
$$
\mathcal F T=T.
$$

REPLY [4 votes]: In case the specific distribution-theoretic argument is not clear... as it hadn't really been overtly mentioned in comments or answers, and is not really suggested by the classical argument as in Titchmarsh and such:
E.g., for Poisson summation on $\mathbb R$: observe that the distribution "sum over integers" is annihilated by multiplication by $e^{2\pi ix}-1$, and is translation invariant. Observe that these two conditions are interchanged by Fourier transform. Show that the space of such distributions is one-dimensional: the multiplication annihilation shows that any such distribution is of order $0$ and supported at integers. By classification of distributions supported at a point, it is a sum of Dirac deltas at integers. By translation-invariance, it is (a scalar multiple of) sum-over-evaluation-at-integers. A just-slightly-more-complicated version applies in $n$-dimensions.