Math of De-Broglie wavelength boggles me

[etamorphmagus]

According to simple math you get \lambda=h/p for photons. and the assumption is for everything else, whilst for everything else which is NOT restmass=0 it should be \lambda=hc/E.

So why is it referred to as simple \lambda=h/p when clearly it is not true mathematically. It would be true for p>>m but why not give the general form \lambda=hc/E for electrons and "matter"?

Thank you.

 

[Doc Al]

According to simple math you get \lambda=h/p for photons. and the assumption is for everything else, whilst for everything else which is NOT restmass=0 it should be \lambda=hc/E.

I think you have it backwards. λ = h/p is the more general statement; p = E/c is only true for photons. (And what do you mean by 'according to simple math'? λ = h/p was de Broglie's hypothesis.)

 

[etamorphmagus]

I think you have it backwards. λ = h/p is the more general statement; p = E/c is only true for photons. (And what do you mean by 'according to simple math'? λ = h/p was de Broglie's hypothesis.)

THIS MATH: http://hyperphysics.phy-astr.gsu.edu/hbase/debrog.html
As you can see m=0 for photons, and the hypothesis is that this is the same for "matter particles" AS IF THEY ARE m=0. Doesn't make sense.

The general "matter wavelength" should be \lambda=\frac{hc}{\sqrt{p^2c^2 + m^2c^4}}

 

[JesseM]

E=hf applies for photons (Planck's law)
f=c/λ applies for photons (in general for a wave moving at v, f=v/λ)
So, for photons, E=hc/λ, which means λ=hc/E
For all particles, E^2 = m^2 c^4 + p^2 c^2. For photons m=0 so this reduces to E=pc. So plugging that into the wavelength equation gives λ=hc/pc=h/p

de Broglie's hypothesis was that this last equation λ=h/p would work for all particles, not just photons. What was it that gave him the intuition that it should be this equation, rather than E=hf or λ=hc/E, that should work for particles other than photons? I'm not really sure, anyone have any ideas? In any case his hypothesis was shown to be correct experimentally, while the other two don't match experimental data for particles other than photons. If you want an equation relating energy to wavelength of frequency for other particles, for massive particles we have p = \sqrt{E^2/c^2 - m^2 c^2} So plugging into λ=h/p and squaring gives \lambda^2 = \frac{h^2}{E^2/c^2 - m^2 c^2} which can be rearranged as E^2 = \frac{h^2 c^2}{\lambda^2} + m^2 c^4 = (hf)^2(c^2/v^2) + m^2 c^4 (assuming the equation λ = h/p still applies in relativistic QM, I'm not sure about that)

 

[JesseM]

THIS MATH: http://hyperphysics.phy-astr.gsu.edu/hbase/debrog.html
As you can see m=0 for photons, and the hypothesis is that this is the same for "matter particles" AS IF THEY ARE m=0. Doesn't make sense.

Why not? Are you assuming that some other equation like E=hf does apply for massive particles? That's just as arbitrary. And remember that λ=hc/E, which you want to apply to massive particles, was itself derived by plugging f=c/λ into E=hf, but f=c/λ only applies in the case of a wave moving at c, for a wave moving at some smaller v the frequency/wavelength relation would be f=v/λ.

 

[Doc Al]

THIS MATH: http://hyperphysics.phy-astr.gsu.edu/hbase/debrog.html
As you can see m=0 for photons, and the hypothesis is that this is the same for "matter particles" AS IF THEY ARE m=0. Doesn't make sense.

I don't know how you are drawing that conclusion. The only use of 'm = 0' is when deducing that p = E/c for a photon.

The general "matter wavelength" should be \lambda=\frac{hc}{\sqrt{p^2c^2 + m^2c^4}}

Nope. That's only true if m = 0. Again, λ = h/p, regardless of the mass.

 

[etamorphmagus]

Right right, thanks a bunch Doc Al and JesseM.
JesseM, it might be that he liked λ=h/p because it is simple. Dirac said the math must be beautiful. But the choice does seem arbitrary.

 

[cesiumfrog]

de Broglie's hypothesis was that this last equation λ=h/p would work for all particles, not just photons. What was it that gave him the intuition that it should be this equation, rather than E=hf or λ=hc/E, that should work for particles other than photons? I'm not really sure, anyone have any ideas?

He pretty much proved it theoretically, up to a point (the choice of h). https://www.physicsforums.com/showthread.php?t=249679#7"

 

[ytuab]

For photon case, the problem has been solved, because m=0;
But in the electron case, what is the frequency ?

If we don't include the rest mass energy, the energy of the free particle becomes,

E = \frac{1}{2m}p^2

But when we include the big rest mass energy, the energy becomes very big,

E=\sqrt{(pc)^2+m^2c^4}

So in this case, frequency of the electron becomes much bigger than the upper case.

When we apply this to the Schrodinger equation,

i\hbar\frac{\partial}{\partial t} \psi =c \sqrt{\hbar^2k^2+m^2c^2}\psi
But I heard this equation doesn't satisfy relativistic causality.
(Of course, this equation is not Lorentz-covariant.)
Because this Schroedinger equation doesn't include the antiparticle (minus energy solution) like Dirac equation.

So I have one question.
The frequency of the Dirac wavefunction means real electron's matterwave?
And what is the electron's frequency?