De broglie frequrency

  • Thread starter ChAshutosh
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  • #1
As we know de broglie wavelength is λ = h/p and f = E/h
λ = h/p
v/f = h/mv as f*λ=v

then

f = mv^2/h

but it should f = E/h and E=1/2*mv^2
 

Answers and Replies

  • #2
Suraj M
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When you say $$E= \frac{1}{2} mv^2 $$ you are considering only kinetic energy, do you think that, it is the only type of energy the body has, whose wave motion is also being is considered?
 
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  • #4
Suraj M
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Bhobba, is there noway of achieving E = mv² by adding the energy due to SHM of each particle along the wave, to the KE?
 
  • #5
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Bhobba, is there noway of achieving E = mv² by adding the energy due to SHM of each particle along the wave, to the KE?
I don't know what you mean by that.

Like most I have read about the De-Broglie theory and seen the equations. They are an inconsistent mish-mash eg what's the wavelength of a stationary particle - and by frame jumping you can always go to a frame where its stationary. Because of that I tend to not get worried about manipulations involving them - it's wrong anyway.

In QM the KE of a free particle is 1/2 mV^2 where V is the velocity operator.

Interestingly one can actually prove that from symmetry in QM - but that is another story.

Thanks
Bill
 
  • #6
Suraj M
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I don't know what you mean by that.
I mean the particles would have an extra amount of energy of ##½m \omega^2 A^2## can we use that?
 
  • #7
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I mean the particles would have an extra amount of energy of ##½m \omega^2 A^2## can we use that?
You lost me.

Can you explain what your terms mean?

Thanks
Bill
 
  • #8
Suraj M
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The regular SHM terms! ##\omega## = 2##\pi /T## ; A = amplitude of the wave
 
  • #9
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I think you are trying to introduce concepts from the quantum harmonic occilator used in QFT. QFT is a whole new ball game not compatible with ordinary QM.

The energy in QM is actually based on the Galilaen transformations while QFT is based on the Lorentz transformations.

Thanks
Bill
 

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