Poisson Summation Formula appears to fail when applied to Hermite Functions (why?) [duplicate]

Question

I came across an odd circumstance where it appears as though the poisson summation formula fails to yield a correct answer (involving Hermite Functions), and I don't quite understand why this happens.

Define the family of Hermite Functions by: $$ \psi_n(x) = \frac{(-1)^n}{\sqrt{2^n n! \sqrt{\pi}}}e^{\frac{x^2}{2}}\frac{d^n}{dx^n}e^{-x^2} $$ And the Fourier Transform by: $$ \hat{f}(\xi) =\mathscr{F}\{f\}(\xi) =\frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} {f(x)e^{- i x \xi}dx} $$ It's known that Hermite Functions are eigenfunctions of the fourier transform: $$ \mathscr{F}\{\psi_n\}(x) = (-i)^n \psi_n(x) $$ Now the Poisson Summation formula states that: $$ \sum_{n = -\infty}^{\infty}f(n) = \sum_{n=-\infty}^{\infty}\hat{f}(n) $$ However, if we apply this to Hermite Functions, we arrive at: $$ \sum_{n = -\infty}^{\infty} \psi_k(n)=\sum_{n = -\infty}^{\infty} (-i)^k\psi_k(n) = (-i)^k \sum_{n = -\infty}^{\infty} \psi_k(n) $$ We know the sum converges for any choice of $k$, given that $\psi_k(x) \in S(\mathbb{R})$, where $S(\mathbb{R})$ denotes Schwartz space on the real number line (space of rapidly decreasing functions). The problem is introduced when $k = 4m+2$ for some $m \in \mathbb{Z}$. We should arrive at: $$ \sum_{n=-\infty}^{\infty} \psi_k(n) = - \sum_{n = -\infty}^{\infty} \psi_k(n) \\ \implies \sum_{n = -\infty}^{\infty} \psi_{4m + 2}(n) = 0 $$ But manually summing up (for example) $ \psi_2(n) = (4\pi)^{\frac{-1}{4}}(2n^2-1)e^{\frac{-n^2}{2}}$ appears to generate a non-zero number (around $1.331$ for this example), leading to a contradiction. Now my question is, what am I missing here? Is there some restriction on the poisson summation formula I'm not considering, or did I just make an algebra error? Any help would be appreciated.

Answer 1

Your mistake is that you are not using the correct form of the Poisson summation formula based on your choice of how to define the Fourier transform. There are multiple conventions on how to define the Fourier transform in terms of where you stick factors of $2\pi$ (in the exponent or as a coefficient) and this leads to multiple versions of the Poisson summation formula, one for each convention.

Here is my suggestion: for the rest of your life, define the Fourier transform of a function $f$ in $S(\mathbf R)$ to be $(Ff)(y) = \int_{\mathbf R} f(x)e^{-2\pi ixy}\ dx$. (I can't write this as $\hat{f}(y)$ or as $(\mathscr F f)(y)$ because you decided to use both of those notations for something else.) By using $2\pi$ in the exponent in this way, you will never again be misled about the Poisson summation formula because it has a clean expression: $$ \sum_{n \in \mathbf Z} f(n) = \sum_{n \in \mathbf Z} (Ff)(n). $$ If you translate this back into your convention on what the Fourier transform is then this summation formula becomes an uglier expression: $$ \sum_{n \in \mathbf Z} f(n) = \sqrt{2\pi}\ \sum_{n \in \mathbf Z} \hat{f}(2\pi n). $$ (When you unravel the notation, these two summation formulas say exactly the same thing: $\sum_n f(n) = \sum_n (\int_{\mathbf R} e^{-2\pi inx}f(x)\ dx)$. For functions $\mathbf R \rightarrow \mathbf C$ there is just one Poisson summation formula but multiple notations for it in terms of "the" Fourier transform. If you did not have a factor of $1/\sqrt{2\pi}$ in your definition of the Fourier transform then you wouldn't need to multiply on the right side by $\sqrt{2\pi}$ but you'd still have to sum on the right side over Fourier transform values on $2\pi\mathbf Z$ instead of Fourier transform values on $\mathbf Z$. The simplest way to avoid factors of $2\pi$ in awkward places from a choice of how to define the Fourier transform is to place $2\pi$ in the exponent when defining the Fourier transform because when you unravel the notation that is exactly where it appears.)

Let's check how this looks with $f(x)$ being your $\psi_2(x) = (2x^2-1)e^{-x^2/2}/\sqrt[4]{4\pi}$. Since $\hat{\psi_2} = -\psi_2$, the corrected Poisson summation formula for $\psi_2$ should say $$ \sum_{n \in \mathbf Z} \psi_2(n) = -\sqrt{2\pi}\sum_{n \in \mathbf Z} \psi_2(2\pi n). $$ When I calculate the sum on the left side and right sides over $-15 \leq n \leq 15$ I get the same number to lots of decimal places: $$1.3313348084818105674376253882631445973\ldots,$$ which you had found to be around $1.331$ on the left side. (The truncated sums on both sides over $-1 \leq n \leq 1$ are not close, being $0.113162\ldots$ on the left and $1.3313348\ldots$ on the right. The sum on the right converges much more rapidly because $\psi_2(2\pi n)$ has the factor $e^{-2\pi^2n^2} = e^{-19.73\ldots n^2}$ while on the left side $\psi_2(n)$ has factor $e^{-n^2/2} = e^{-.5 n^2}$. Truncations of the sum on the right already achieve the digits in the number displayed above when summing over $-2 \leq n \leq 2$, while truncations of the sum on the left don't reach that accuracy until you sum over $-13 \leq n \leq 13$.)

Because of the extra factors of $2\pi$ in the ugly version of Poisson summation connected to your definition of the Fourier transform, there is no reason to think the sum of $\psi_2$ over $\mathbf Z$ should be $0$ from $\psi_2$ having its negative as its Fourier transform by your convention for the Fourier transform.

Now I'll show you a genuinely vanishing sum from Poisson summation. For $a > 0$ and a nonnegative integer $m$, the $m$th derivative of $e^{-ax^2}$ is a polynomial multiple of $e^{-ax^2}$. Set $$ \frac{d^m}{dx^m}(e^{-ax^2}) = (-1)^mH_{a,m}(x)e^{-ax^2}. $$ For example, $H_{a,0}(x) = 1$, $H_{a,1}(x) = 2ax$, and $H_{a,2}(x) = 4a^2x^2 - 2a$. Here is a "general" Fourier transform formula involving the functions $H_{a.m}(x)e^{-(1/2)ax^2}$: $$ \int_{\mathbf R} H_{a,m}(x)e^{-(1/2)ax^2}e^{-iaxy}\ \frac{dx}{\sqrt{2\pi/a}} = (-i)^mH_{a,m}(y)e^{-(1/2)ay^2}. $$ I like to use $a = 2\pi$, giving us $$ \int_{\mathbf R} H_{2\pi,m}(x)e^{-\pi x^2}e^{-2\pi ixy}\ dx = (-i)^mH_{2\pi,m}(y)e^{-\pi y^2}. $$ So by my preferred definition of the Fourier transform, denoted $F$, the function $G_m(x) = H_{2\pi,m}(x)e^{-\pi x^2}$ has $F(G_m) = (-i)^mG_m$. Then by the cleaner form of Poisson summation associated to the definition of the Fourier transform that I use,
$$ \sum_{n \in \mathbf Z} G_m(n) = (-i)^m\sum_{n \in \mathbf Z} G_m(n). $$ Therefore if $m$ is not a multiple of $4$ the sum on the left has to be $0$: $$ \sum_{n \in \mathbf Z} H_{2\pi,m}(n)e^{-\pi n^2} = 0. $$ When $m = 1$ this says $$ \sum_{n \in \mathbf Z} 2(2\pi)ne^{-\pi n^2} = 0, $$ which is obvious since terms at $n$ and $-n$ cancel. When $m = 2$ we have $H_{2\pi,2}(x) = 4\pi(4\pi x^2-1)$, so after dividing by $4\pi$ we get
$$ \sum_{n \in \mathbf Z} (4\pi n^2 - 1)e^{-\pi n^2} = 0, $$ and it is not obvious that sum should be $0$. But if we calculate the sum over $-2 \leq n \leq 2$ it is already 0 to 9 decimal places, and the sum over $-4 \leq n \leq 4$ is 0 to over 30 decimal places. When $m = 3$ we get a boring sum equal to $0$ since $H_{2\pi,3}(x)$ is an odd polynomial (of degree $3$), and the case $m=5$ is uninteresting for a similar reason, but when $m = 6$ we get something not obvious again: $$ H_{2\pi,6}(x) = 64\pi^3(64\pi^3x^6 - 240\pi^2x^4 + 180 \pi x^2 - 15), $$ so we must have $$ \sum_{n \in \mathbf Z} (64\pi^3n^6 - 240\pi^2n^4 + 180 \pi n^2 - 15)e^{-\pi n^2} = 0. $$ Let $s(N)$ be the partial sum for the left side over $-N \leq n \leq N$. Then I compute $s(1) = -.637\ldots$, $s(2) = -.000001324\ldots$, $s(3)$ is $0$ to 14 decimal places, and $s(4)$ is $0$ to over 25 decimal places.

Having tried to show why the version of Poisson summation using the definition of the Fourier transform that I prefer is nicer, I'll admit that the version of the Fourier transform that you prefer also admits a symmetric form of the Poisson summation formula. For $f \in S(\mathbf R)$ and $t > 0$, let $f_t(x) = f(tx)$. Then $\widehat{f_t}(y) = (1/t)\widehat{f}(y/t)$, so the ugly Poisson summation formula with $f_t$ in place of $f$ becomes $$ \sum_{n \in \mathbf Z} f_t(n) = \sqrt{2\pi}\ \sum_{n \in \mathbf Z} \hat{f_t}(2\pi n), $$ which says $$ \sum_{n \in \mathbf Z} f(tn) = \frac{\sqrt{2\pi}}{t}\ \sum_{n \in \mathbf Z} \hat{f}(2\pi n/t). $$ To make this look nice we want $t = 2\pi/t$, so $t = \sqrt{2\pi}$. Using that value for $t$, the previously ugly form of Poisson summation becomes the nicer-looking formula $$ \sum_{n \in \mathbf Z} f(\sqrt{2\pi}n) = \sum_{n \in \mathbf Z} \hat{f}(\sqrt{2\pi}n). $$ If you use $f(x) = \psi_2(x) = (2x^2-1)e^{-x^2/2}/\sqrt[4]{4\pi}$, for which $\widehat{\psi_2} = -\psi_2$ then we must have
$$ \sum_{n \in \mathbf Z} \psi_2(\sqrt{2\pi}n) = 0, $$ so that is the vanishing sum you should have found instead of thinking $\sum_{n \in \mathbf Z} \psi_2(n)$ is $0$. And this vanishing sum is something we saw already: $\psi_2(\sqrt{2\pi}x) = (4\pi x^2-1)e^{-\pi x^2}/\sqrt[4]{4\pi}$, which up to scaling by $\sqrt[4]{4\pi}$ is $H_{2\pi,2}(x)$ from above, so this vanishing sum is up to scaling the earlier vanishing sum $\sum_{n \in \mathbf Z} (4\pi n^2-1)e^{-\pi n^2}$.