# Find $\psi\in L^2$ so that the overlap $\langle \psi(x+\delta),\psi(x-\delta)\rangle$ is as small as possible

Is there a way to find / characterise functions $\psi\in L^2$ which make the value of this integral small?

$$\text{Re}\int_{\mathbb R} \overline{ \psi (x+\delta)}\,\psi(x-\delta)\;\text d x$$

where $\psi$ is normalised, $\langle\psi,\psi\rangle=1$, and $\delta$ is a itsy-bitsy constant compared to the width of $\psi$, defined as $$\sigma = \sqrt{\langle\psi,X^2\psi\rangle-\langle\psi,X\,\psi\rangle^2}$$ where $X$ is the usual multiplication operator $X\,\psi(x) = x\,\psi(x)$.

Apparently the considered integral is very close to $1$. My goal is find functions $\psi$ which preferably make the integral small. My goal is not to find the minimizer of the integral but some characteristics of $\psi$ to make the integral smaller. I think here I fail to find the right mathematical notions and concepts.

Things I tried are the following

• Consider the integral as a functional $f_\delta[\psi]=\text{Re}\int_{\mathbb R} \overline{ \psi (x+\delta)}\,\psi(x-\delta)\;\text d x$ and try to find $\psi$ which minimise this functional. Unfortunately I am not very confident in dealing with functionals but my result was something like a function with hard edges (like a square pulse).
• Assume that $\psi$ is differentiable and make an expansion like $$\text{Re}\int_{\mathbb R} \overline{ \psi (x+\delta)}\,\psi(x-\delta)\;\text d x = 1 - \underbrace{\text{Re}\int \bigg(\overline{\psi'(x)}\,\psi'(x)-\overline{\psi''(x)}\,\psi(x)\bigg)\;\text d x}_{\text J} \;\cdot\frac{\delta^2}{2} + \mathcal O(\delta^4)$$ Now I have to find $\psi$ which make the integral $\text J$ big. But how?
• I looked at the integral as the expectation value of the translation operator $T=\exp(- 2\text i \delta P)$ where $P=-\text i \frac{\text d}{\text d x}$: $$\text{Re}\int_{\mathbb R} \overline{ \psi (x+\delta)}\,\psi(x-\delta)\;\text d x = \text{Re}\langle \psi,T\psi\rangle$$ but I can't see how it could help.
• oh, sorry. Its just the multiplication operator $X\,\psi(x) = x\psi(x)$. This is the usual notation in quantum mechanics... – thyme Jun 8 '14 at 20:07

Since the translation operator is continuous in $L^2$, we have $$\text{Re}\int_{\mathbb R} \overline{ \psi (x+\delta)}\,\psi(x-\delta)\;\text d x \to \langle \psi,\psi\rangle = 1$$ for every $\psi$. So, to achieve smallness, we need $\psi$ for which translation is not very continuous (has poor modulus of continuity). This matches your "hard edges" description. For example, we could divide $[0,2]$ into subintervals of length $2\delta$ and let $\psi=1$ on every other subinterval (and $0$ elsewhere). Then $\overline{ \psi (x+\delta)}\,\psi(x-\delta)\equiv 0$.
Another, related source of near-orthogonality are quick oscillations. Recall that the sine wave is orthogonal to its shift by $1/4$ of period. You can take any slowly varying function $\phi$ and multiply it by $\sin (\pi x/(8\delta))$ (or by $\exp( \pi i x/(8\delta))$. This product, $\psi$, will have the desired near-orthogonality for that particular $\delta$.