Wave-particle duality as an uncertainty relation for the average confidence width

Abstract

We introduce the average confidence width $\hat{I}{”}_{a}x={\int }_0^1\hat{I}{”}_{c}x(\hat{I}{¸}_x)d\hat{I}{¸}_x$: the confidence width $\hat{I}{”}_{c}x(\hat{I}{¸}_x)$ -- the smallest position interval carrying a fraction $\hat{I}{¸}_x$ of the probability -- averaged over all levels. It is the first moment of the decreasing rearrangement of $|\ddot{I}ˆ{|}^2$, an $L^1$ mean-absolute-deviation measure of localization, so the product $\hat{I}{”}_{a}x\hat{I}{”}_{a}p$ is dilation invariant and obeys $\hat{I}{”}_{a}x\hat{I}{”}_{a}p\ge c\hslash$. Reading $1/\hat{I}{”}_{a}x$ as a particle character and $1/\hat{I}{”}_{a}p$ as a wave character, this lower bound on combined spread is identically an upper bound on combined particle-and-wave character: uncertainty and wave-particle duality are two faces of one inequality. A mean-entropy argument with the Bialynicki-Birula-Mycielski relation gives the rigorous $c\ge \ddot{I}€/e$, while the achievable constant $c^{\ast }$ is set by the ground state of the Fourier-invariant operator $|x|+|p|$, $c^{\ast }\le E_0^2\approx 1.217$. Hence $\ddot{I}€/e\le c^{\ast }\le E_0^2<4/\ddot{I}€$: the optimal state is sub-Gaussian, so the Gaussian -- optimal for the Heisenberg and entropic relations -- is not the duality optimum.

7 pages, 4 figures