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A very short abstract on de Broglie wavelength
Posted on Sunday, August 01, 2004 @ 10:02:42 GMT by vlad

Science VacuumZeroPoint writes: 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.


 
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