MEST-The quantum space-time explain some questions of quantum mechanics

The probability of displacement and period of wave are the quantum space-time. The paper explain of the two-slit interference and the uncertainty relation. (1) $S=P(r)=P(\lambda)={f^2}$. According to the Benford's law, (2) $T=P(t)=ln(1+\frac{1}{t})={\nu}$. Among it, S: the quantum space, f: the amplitude, r: the displacement, T: the quantum time, t: the period, $\nu$: the frequence, $\lambda$: the wavelength, P(x): the probability function. (3) $E'{\psi}=i{\hbar}\frac{\partial{\psi}}{{\partial}t}$. (4) $m'{\psi}=i{\hbar}\frac{{\partial}{\psi}{\partial}t} {{(\partial}x)^2}$, equation (3) over equation (4), substituting equation (1) and (2) into it, (5) $E'{\psi}=m'{\psi}c'^2=m'{\psi}\frac{({\partial}f^2)^2}{({\partial}\nu)^2}$, getting the energy-wave and mass-wave equation, (6) $E'=i{\hbar}\frac{{\partial}f^2}{{\partial}\nu}$. (7) $m'=i{\hbar}\frac{{{\partial}\nu}}{{\partial}f^2}$. (8) $\Delta{E'}\Delta{\nu}=\Delta{E'}\Delta{t}=i{\hbar}\Delta{f}^2, (\Delta{f}^2\geq\frac{1}{2})$. (9) $\Delta{p'}\Delta{f^2}=\Delta{p'}\Delta{\lambda}=i{\hbar}\Delta{f}^2, (\Delta{f}^2\geq\frac{1}{2})$. Among it, $E'$: the energy of wave, $m'$: the mass of wave, $c'$: the velocity of light, ${\psi}$: the Wave Functions, $f^2$: the probability. Here, the equation (8) and (9) are new uncertainty relation. In the two-slit interference, because (10) $ c'=\frac{\lambda}{t}=\frac{f^2}{t}$, so (11) $ f^2={\lambda}$(the wavelength), so (12) ${\lambda}\geq{d}$( the width of the slits). Measuring the time of light at one slit(1) and its energy at other silt(2) together. And the measuring probability, if (13) $f_1^2\geq\frac{1}{2}, f_2^2\geq\frac{1}{2}, f_1^2=f_2^2, (f_1^2+f_2^2)\leq1$, then (14) $f_1^2=f_2^2=\frac{1}{2}$.