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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.

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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
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1 Answer

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

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.

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Very cool. Seems to be a good example of mathoverflow.net/questions/40005 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
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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
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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
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