A very short abstract on de Broglie wavelength
Date: Sunday, August 01, 2004 @ 10:02:42 GMT Topic: 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.

