Question

Is the Debroglie derivation correct?
He used E=mc^2 for a photon? How can you use E=mc^2?
I thought that was only true for particles at rest, and we had to use E^2 = (mc^2)^2+ p^2c^2

So whats up!?

Answer 1

I got dis

Answer 2

Wasn't de Broglie trying to explain the wave properties of an electron?

Answer 3

Are you sure that's the way he did it?
E=mc² is only valid in the reference frame where the particle is at rest, but a photon is always moving at speed c in any reference frame.

Answer 4

Yes @gleem and @Vincent-Lyon.Fr

He was trying to explain wave property of an electron but he uses the dual nature of light to come up with it

And vincent yes..

it goes

E = mc^2 = hf
and then
mc^2 = hc/lamda
interchange
and gwall he gets lamda = h/mc

and then convienientely replaces c with v, cause v is the velocity of particles
and so
for particles lamda = h/p

Answer 5

It was't quite that simple. He used E=hv to start the logic being if the energy of a photon is related to its frequency then He Postulated that the kinetic energy of a particle might be equal to the frequency of its wave too. He then used another photon equation momentum = h/wavelength. Note that these two relationship do not use the speed of the wave.
He reasoned that the particle must have an intrinsic wave nature but at the same time it must be localized like a particle should. this can be demonstrated to be acceptable because if you take a large set of wave of frequencies that differ infinitesimally from one another you get a wave 'packet" a wave with a well localized amplitude and with a well defined velocity called the group velocity. He showed that the group velocity which he identified with the particle velocity was given by

g = c^2*p/E = particle velocity

three years later Davisson and Germer demonstrated that electron did indeed act like waves.

Answer 6

@Mashy. I think the way you describe it is not the way he worked it out.

He compared the momenta of the photon and the particle, not their energies.
What he wrote was: It is well known that a photon's momentum corresponds to a wavelength.
By p = hf/c and lambda = c/f, you get lambda = h/p

Since a massive particle also has a momentum p = mv, he said he could associate a wavelength lambda = h/p =h/mv to the massive particule.

Answer 7

Ok I understand.. but that was given in some stupid textbook.. Damn!! I hate those stupid useless authors!!

http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/02._Fundamental_Concepts_of_Quantum_Mechanics/De_Broglie_Wavelength

Even this site does the same thing..

But where did this equation come from?
p = hf/c ??

Can I use \[E = \sqrt{mc^2 + p^2c^2}\]
then put m=0 for photon
and say
\[E = pc\]
\[p= \frac{E}{c}= \frac{hf}{c}\]
\[c= f \lambda \]

giving
\[p = \frac{h}{\lambda} \]

is that the correct derivation?

Answer 8

Basically that is the way it goes and note that you did not have to arbitrarily substitute the particle velocity for the velocity of light. I think de Broglie did a little more work in his dissertation to show that a wave characteristic was consistent with particle locality.

Answer 9

Q : is that the correct derivation?
A : exactly :))

Answer 10

Thanks guys

@gleem.. son in this case, I would just say
"So a photon which is massless but has momentum can have wave characteristics associated with it. So could a particle (which has mass but that is not the point here) can also have momentum MAY obey the same equation. That was the thinking?"

If there is a more rigorous approach to convert from photon's wavelength to particles wavelength, can you provide me any links? I tried searching didn't get any.

Also
That equation works today as we know for particles. But what happens when particles are at rest? Infinite wavelength? I know macro objects have particles inside them zooming, but can't we somehow think of a micro particle being at rest (from some reference frame?). Then would that have some crazy wave property associated with it?

Answer 11

@Mashy
We cannot "find" particles at rest because we can only localize one to a distance given by
\[\Delta x > h/ \Delta p\]
Even an electron with an energy of 0.0001 eV has associated wavelength of 122.6 mu which is a wavelength in the short wave UV region.

Keep in mind that the wavelength in de Broglie's equation is not the wavelength per se of the electron but the waves that make up the electron called the pilot waves and introduced by de Broglie. I haven't found a good link but R. M. Eisberg 's "Fundamentals of Modern Physics" (1961) goes through a simplified derivation.

Answer 12

Oh yes.. good old uncertainty :D .. thank you!