Wave-particle duality as an uncertainty relation for the average confidence width
이 뉴스, 어떠셨어요?
한 번의 탭으로 반응을 남겨요 · 로그인 불필요
Abstract
We introduce the average confidence width $\Delta_a x=\int_0^1 \Delta_c x (\theta_x) d \theta_x$: the confidence width $\Delta_c x(\theta_x)$ -- the smallest position interval carrying a fraction $\theta_x$ of the probability -- averaged over all levels.
It is the first moment of the decreasing rearrangement of $|\psi|^2$, an $L^1$ mean-absolute-deviation measure of localization, so the product $\Delta_{a} x\,\Delta_{a} p$ is dilation invariant and obeys $\Delta_{a} x\,\Delta_{a} p\ge c\,\hbar$.
Reading $1/\Delta_{a} x$ as a particle character and $1/\Delta_{a} p$ 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 $c\ge\pi/e$, 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 $\pi/e\le c^\ast\le E_0^2<4/\pi$: the optimal state is sub-Gaussian, so the Gaussian -- optimal for the Heisenberg and entropic relations -- is not the duality optimum.