 A very short abstract on de Broglie wavelength Date: Sunday, August 01, 2004 @ 10:02:42 GMTTopic: Science I would like to send you a possible picture of vacuum oscillation to explain de Broglie wavelength as follows: Let's look at the general Shroedinger equation for a free electron not under any potential field: -(h/4 Pi.m) d^2Psi/dx^2 = i dPsi/dt. (a) If we view space as a medium with some travelling space wave that has the speed c (c : light speed in vacuum), we shall have: dx= cdt, then dt = dx/c. (1). Replace (1) into (a) , we have: -(h/4 Pi.mc) d^2Psi/dx^2 = i dPsi/dx (b) Now we can see the Compton wavelength as h/mc appears in (b) with m as the mass of an electron. From the above, we probably can see the picture of de Broglie wavelength as follows: Let's suppose a particle is moving with the speed c. This particle will "see" a space wave with the space wavelength as lambda_0 = h/mc that is equal to Compton wavelength of an electron. If an electron is moving with a speed v, the electron will see the space wavelength as: lambda_e = lambda_0/(v/c) = (h/mc) / v/c = h/mv. (2) (2) is de Broglie wavelength ! The space probably has a wavelength that is equal to the Compton wavelength of an electron. That explains why a moving electron has de Broglie wavelength. Please check this! Best, Sinh.

This article comes from ZPEnergy.com
http://www.zpenergy.com

The URL for this story is:
http://www.zpenergy.com/modules.php?name=News&file=article&sid=856