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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.

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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 $\lambda = -i$.

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    $\begingroup$ How did you come up with $P(-2\pi i x) \exp(-\pi x^2) = \lambda P(D_x) \exp(-\pi x^2)$? $\endgroup$
    – saurav90
    Apr 15, 2015 at 14:20
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    $\begingroup$ The equation that the Fourier transform is $\lambda$ times itself? $\endgroup$ Apr 15, 2015 at 15:19
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    $\begingroup$ But we have to prove $F(P(x)exp(-\pi x^2)) = \lambda P(y)exp(-\pi y^2)$ before using it? $\endgroup$
    – saurav90
    Apr 15, 2015 at 15:21
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    $\begingroup$ You want to solve $$\mathcal{F}(f) = \lambda f $$ Make the guess that $$f = P(-2\pi i x) \exp (-\pi x^2)$$ Using the properties of the Fourier transform you know that for this particular $f$ you have $$\mathcal{F}(f) = P(D_x) \exp(-2\pi x^2)$$ So plugging into the equation you want $$ P(D_x) \exp(-2\pi x^2) = \lambda P(-2\pi i x) \exp(-2\pi x^2) $$ $\endgroup$ Apr 16, 2015 at 7:47
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    $\begingroup$ Maybe this would help: the equation you are quoting is not a general identity. You are trying to solve that equation for the correct polynomial $P$. $\endgroup$ Apr 16, 2015 at 7:48

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