Maximum of a series of integrals of Hermite functions

2011-05-27T02:31:54Z

Given the function $$f(A) := \sum_{n=1}^{\infty}\left( \int_A \varphi_0\varphi_n \right)^2,$$ where $A$ is any measurable subset of $\mathbb{R}$, and $\varphi_n$ is the $n$th Hermite function, I want to know for which sets $f$ attains its maximum.

I've already proven that $$f(A)\le \frac{1}{4}$$ for all $A$, and that $f(\mathbb{R}^+) = 1/4$. But it is crucial for my problem to find other sets that would also maximise $f$ or prove that none exists, and for doing that I am at a loss. Even finding local maxima would be interesting to me.

The proof I already obtained is somewhat long and very indirect, so I won't include it here; also, I'm very interested to see how a mathematician would approach this problem.

Answer by Piotr Migdal for Maximum of a series of integrals of Hermite functions

2011-05-27T10:09:48Z

Write your quantity as: $$f(A)=\hbox{Tr}\left[ P_A|0\rangle\langle0|P_A(\mathbb{1}-|0\rangle\langle0|) \right],$$ where $P_A$ is the projection on A, and $|0\rangle\langle0|$ is the projection on $\varphi_0$. Note that then you need only to investigate properties of the $\varphi_0$, not every Hermite function (as they form an orthonormal basis).

With the properties of Tr and projection operators you get $$f(A)= \hbox{Tr}[|0\rangle\langle0|P_A|0\rangle\langle0|]-\hbox{Tr}[(|0\rangle\langle0|P_A|0\rangle\langle0|)^2]$$ $$=\lambda-\lambda^2.$$

So:

indeed, $\max f(A) = \frac{1}{4}$,
$f(A)=\frac{1}{4}$ iff $\int_{A} \varphi_0^2(x)dx=\frac{1}{2}$.

Depending what you need the formula for, but if it is about the filtering of the higher-order modes, here (sec. 6.) is a numerical remark.

Maximum of a series of integrals of Hermite functions - MathOverflow

Maximum of a series of integrals of Hermite functions

Answer by Piotr Migdal for Maximum of a series of integrals of Hermite functions