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Working on a quantum mechanics problem, I've stumbled on the problem of maximising the functional $$\int_A \varphi_m \varphi_n$$ in the limit of large $m$ and $n$, given that $n \gg m$. $\varphi_m$ is the $m$th Hermite function, defined as $$\varphi_m(x) = (2^m m! \sqrt{\pi})^{-1/2} e^{-x^2/2} H_m(x),$$ so that $\|\varphi_m\|_2 = 1$.

It's clear that this problem is equivalent to evaluating the function $$ f(m,n) = \int_{\mathbb{R}} |\varphi_m \varphi_{n}|,$$ which appears to me to be simpler, so I've concentrated myself on it. It's clear that it's not sensible to search for an exact formula, as the solution gets very complicated very quickly, so I'm only looking for an asymptotic expansion.

Only knowing if it goes to zero in this limit would be also very interesting; I have physical reason to think it does.

Using the asymptotic form of the Hermite functions I managed to prove that $$f(m,n) \sim \frac{2}{\pi} \sqrt[4]{\frac{2}{n\pi^2}}\int_{\mathbb{R}}|\varphi_m|.$$ But an asymptotic form for the rhs eludes me, so

Is there an asymptotic form for $\|\varphi_m\|_1$?

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Some (probably stupid) comments: maybe it would help if you explicitly gave the formula for $\varphi_n$, since there seem to be several different versions in use. Anyway, if you only need an upper bound, you could find an interval $[-R_n, R_n]$ where $\varphi_n$ is concentrated, and use Cauchy-Schwarz to bound the $L^1$ norm on this interval using the $L^2$ norm, since the $L^2$ norm is much easier to use. The only trick you'd need to make this work is an inequality telling you how rapidly the functions decay away from zero. Hopefully, $R_n$ would not grow too fast with $n$ for this to work. – Zen Harper May 20 '11 at 1:56
Good idea. I'm pretty sure that $R_m$ grows as $\sqrt{m}$, although I haven't seen a proof of it, so my last equation would give $f(m,n) \le \alpha \sqrt[4]{m^2/n}$, and this is enough to prove a good upper bound. Thanks! The Hermite functions decay exponentially fast, so it is easy to bound the error by restricting the interval, but I still need some formula for $R_n$. – Mateus Araújo May 20 '11 at 3:11
To determine $R_m$ if you have not already, use the representation $H_m(x)=m!\sum_{l=0}^{m/2}{(-1)^l\over l!(m-2l)!}(2x)^{m-2l}$. You can ignore the first quotient in the sum as bounded by 1, and get $|H_m(x)|\le m\cdot m!\cdot (2x)^m$ for $x>1/2$. Putting this into the above formula and taking logs, it is easy to see that $x>>\sqrt{m\log m}$ will contribute negligibly. You only have to compare $\log|H_m(x)|\sim m\log m+m\log x$ with $\log e^{-x^2}\sim -x^2$. – Junkie May 27 '11 at 6:22
You could also write $\|\varphi_m\|_1$ as an alternating sum of the integral of $\varphi_m$ between consecutive zeros (for they are all simple and $\varphi_m$ changes sign). Then try using the recurrence relations and the known informations on the zeros of Hermite polynomials. – Pietro Majer Jun 7 '11 at 5:48

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