PDA

View Full Version : Compton wavelength vs DeBroglie wavelength.



Garvs
2007-Feb-16, 09:56 AM
Could some please correct, confirm or elaborate on my understanding of these two quantum properties.

My understanding is that the Compton wavelength of a particle is based on it's rest mass, and the DeBroglie wavelength is related to a particles momentum. Is it then fair to say that the DeBroglie wavelength of a particle with zero momentum, ie at rest, is equal to its Compton wavelength.

Thanks in advance.

tusenfem
2007-Feb-16, 01:06 PM
Compton wavelength: LC = h / (m0 c), where m0 is the rest mass
de Broglie wavelength (non-rel): LB = h / sqrt(2 m0 K) , where K is kinetic energy
de Broglie wavelength (rel): LB = h c / sqrt(2 E0 K + K2)

A little algebra shows that:
LB = (h / m0 c) / sqrt( 2 (K/E0) + (K/E0)2 )
where the numerator is LC

So one finds that:

LB / LC = 1 / sqrt( 2 (K/E0) + (K/E0)2 )

which goes to infinity as K goes to zero.

Ken G
2007-Feb-16, 08:22 PM
So tusenfem is saying the answer is no, the deBroglie wavelength of a particle at rest is infinite, not the Compton wavelength. That's true-- highly nonrelativistic particles can show wave properties over scales much larger than their Compton wavelength. A better way to think of the comparison is that as the deBroglie wavelength becomes shorter than the Compton, relativistic effects become very important.

Garvs
2007-Feb-16, 11:26 PM
Thanks folks.