MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
The Hermite functions are $\phi_n(x)={(-1)^n\over\sqrt{2^nn!\sqrt{\pi}}}e^{x^2/2}{\partial^n\over\partial x^n}(e^{-x^2})$. – Junkie May 27 '11 at 4:40
You can also note that $\int_x^\infty \phi_0\phi_n={H_{n-1}(x)\over\sqrt{n!2^n}}{e^{-x^2}\over\sqrt\pi}$, which helps to reduce the set $A$ via endpoints. Here $H_n$ is the $n$th Hermite polynomial, via the physicists counting, so $H_1(x)=2x$. – Junkie May 27 '11 at 5:17
Letting the set be $A=[-t,t]$, there is a local maximum somewhere around $t=1/2$, but it seems to be somewhat beneath it, and I don't know what the value is. This might be known, though. – Junkie May 27 '11 at 5:52
up vote 3 down vote accepted

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 $$f(A)= \hbox{Tr}[|0\rangle\langle0|P_A|0\rangle\langle0|]-\hbox{Tr}[(|0\rangle\langle0|P_A|0\rangle\langle0|)^2]$$ $$=\lambda-\lambda^2.$$


  • 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.

share|cite|improve this answer
Very cool. Seems to be a good example of as this works for any orthonormal basis, not just for the Hermite functions. Also there's nothing special with $|0\rangle$, any other vector works. But I would make your expressions a bit simpler: first express $f$ as $$f(A) = \langle 0|P_A(1-|0\rangle \langle 0|)P_A|0\rangle$$, then by linearity $$f(A) = \langle 0|P_A|0\rangle - (\langle 0|P_A|0\rangle)^2$$ I need this formula to study violations of a Bell inequality in a quantum optical system. So, the reference is a bit related, but not much. – Mateus Araújo May 27 '11 at 15:20
Actually, $f(A)=\int_A\phi_0^2 -\left(\int_A\phi_0^2\right)^2$ is just the Parseval identity. – Pietro Majer May 27 '11 at 18:41
@Pietro: Could you clarify? By 'Perseval identity' I know $|\psi|_2=|U \psi|_2$ for unitary $U$ acting on a vector $\psi$ (especially when $U$ is the Fourier transform). – Piotr Migdal May 28 '11 at 11:38
I think Pietro is referring to the identity $$\|x\|^2 = \sum_n |\langle x, \varphi_n\rangle|^2,$$ setting $x = \varphi_0$ and passing the term $|\langle \varphi_0, \varphi_0\rangle|^2$ from the rhs to the lhs. Doing the inner product with weight $P_A$, of course. It is interesting to notice that the result is more general than I thought; however I prefer the direct proof. – Mateus Araújo May 28 '11 at 23:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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