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.


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 $\lambda = -i$.

  • $\begingroup$ 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 '15 at 14:20
  • $\begingroup$ The equation that the Fourier transform is $\lambda$ times itself? $\endgroup$ – Willie Wong Apr 15 '15 at 15:19
  • $\begingroup$ 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 '15 at 15:21
  • 1
    $\begingroup$ 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 $$ P(D_x) \exp(-2\pi x^2) = \lambda P(-2\pi i x) \exp(-2\pi x^2) $$ $\endgroup$ – Willie Wong Apr 16 '15 at 7:47
  • $\begingroup$ 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 $P$. $\endgroup$ – Willie Wong Apr 16 '15 at 7:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.