Eigenvalues of Fourier Transform on Schwartz Functions Find all the eigenvalues of the Fourier transform $\hat{f}$(viewed as an operator
acting on the class of Schwartz functions $S(R)$), i.e. all values $\lambda \in \mathbb{C}$ such that there exists a non-zero function $f \in S(R)$ with
$\hat{f}= \lambda f$.
Hint: the corresponding eigenfunctions may be found in the form $P(x)e^{−πx^2}$
, where $P(x)$ is a suitable polynomial of degree at most three.
It could be verified that $e^{−πx^2}$ is an eigenfunction with eigenvalue $\lambda = 1$. How do I find the remaining? Why should $P(x)$ be atmost degree 3. 
 A: Write the Fourier transform as $\mathcal{F}: \mathcal{S} \to \mathcal{S}$. Observe that $\mathcal{F}^{-1} \neq \mathcal{F}$ (assuming we choose the unitary definition), but that 
$$ \mathcal{F}^{-1} f(x) = \mathcal{F} f(-x) $$
Therefore we conclude that
$$ \mathcal{F}\mathcal{F}\mathcal{F}\mathcal{F} = \mathrm{Id}.$$
This tells you that the eigenvalues must be fourth roots of unity. 
For concreteness let us fix convention
$$ \mathcal{F} f(\xi) = \int_{\mathbb{R}} e^{-2\pi ix\xi} f(x) ~\mathrm{d}x $$
which seems to be the one you are using considering that you have found $\exp(-\pi x^2)$ to be an eigenfunction. 
To find the eigenfunctions, we can start with the hint.
Now, observe that 
$$ \mathcal{F}(-2\pi ix f) =  D_\xi \mathcal{F} f $$
(the rule that exchanges scalar multiplication with differentiation). Now let $\lambda$ be one of the fourth roots of unity (meaning that $\lambda \in \{1,-1,i,-1\}$), we want to solve
$$ P(-2\pi i x) \exp(-\pi x^2) = \lambda P(D_x) \exp(-\pi x^2) $$
where $P(D_x)$ is the differential operator $\sum a_n D_{x}^n$ if $P(y) = \sum a_n y^n$. 
We easily see that
$$ D_x \exp(-\pi x^2) = -2\pi x \exp(-\pi x^2) $$
so $P(y) = y$ gives us an eigenfunction for $\lambda= i$. 
Analogously you can solve for a quadratic polynomial $P(y)$ which will give you $\lambda = -1$ and a cubic polynomial $P(y)$ which will give you $\lambda = -i$. 
