# De broglie frequrency

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

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Suraj M
Gold Member
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?

ChAshutosh
bhobba
Mentor
Remember De-Broglies ideas are a mishmash of quantum and relatvistic ideas. Best to forget about them - they have long since been consigned to the dustbin of history. But since its relatvistic E is not the classical 1/2 mv^2:
http://hyperphysics.phy-astr.gsu.edu/hbase/debrog.html

Thanks
Bill

ChAshutosh
Suraj M
Gold Member
Bhobba, is there noway of achieving E = mv² by adding the energy due to SHM of each particle along the wave, to the KE?

bhobba
Mentor
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

Suraj M
Gold Member
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?

bhobba
Mentor
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

Suraj M
Gold Member
The regular SHM terms! ##\omega## = 2##\pi /T## ; A = amplitude of the wave

bhobba
Mentor
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