Math of De-Broglie wavelength boggles me

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In summary: The frequency is the same as for photons, E/h, except that E is the total energy, not the kinetic energy. This is just a consequence of the fact that, in quantum mechanics, the total energy is the conserved quantity associated with time translation. If you want to write the wavefunction in a form that makes the E/h analogy more explicit, you have to use the phase-space wavefunction ψ(x,p), which satisfies the free-particle equationi\hbar \frac{\partial \psi}{\partial t} = \left(\frac{p^2}{2m} + V(x)\right) \psiThe analogy with the photon case is clear if you write this asi\
  • #1
etamorphmagus
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According to simple math you get [tex]\lambda=h/p[/tex] for photons. and the assumption is for everything else, whilst for everything else which is NOT restmass=0 it should be [tex]\lambda=hc/E[/tex].

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

Thank you.
 
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  • #2
etamorphmagus said:
According to simple math you get [tex]\lambda=h/p[/tex] for photons. and the assumption is for everything else, whilst for everything else which is NOT restmass=0 it should be [tex]\lambda=hc/E[/tex].
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.)
 
  • #3
Doc Al said:
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 [tex]\lambda=\frac{hc}{\sqrt{p^2c^2 + m^2c^4}}[/tex]
 
  • #4
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, [tex]E^2 = m^2 c^4 + p^2 c^2[/tex]. 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 [tex]p = \sqrt{E^2/c^2 - m^2 c^2}[/tex] So plugging into λ=h/p and squaring gives [tex]\lambda^2 = \frac{h^2}{E^2/c^2 - m^2 c^2}[/tex] which can be rearranged as [tex]E^2 = \frac{h^2 c^2}{\lambda^2} + m^2 c^4 = (hf)^2(c^2/v^2) + m^2 c^4[/tex] (assuming the equation λ = h/p still applies in relativistic QM, I'm not sure about that)
 
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  • #5
etamorphmagus said:
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.
 
  • #6
etamorphmagus said:
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 [tex]\lambda=\frac{hc}{\sqrt{p^2c^2 + m^2c^4}}[/tex]
Nope. That's only true if m = 0. Again, λ = h/p, regardless of the mass.
 
  • #7
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.
 
  • #8
JesseM said:
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"
 
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  • #9
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,

[tex]E = \frac{1}{2m}p^2[/tex]

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

[tex]E=\sqrt{(pc)^2+m^2c^4}[/tex]

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

When we apply this to the Schrodinger equation,

[tex]i\hbar\frac{\partial}{\partial t} \psi =c \sqrt{\hbar^2k^2+m^2c^2}\psi[/tex]
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?
 

FAQ: Math of De-Broglie wavelength boggles me

What is the De-Broglie wavelength equation?

The De-Broglie wavelength equation, also known as the de Broglie relation, is a fundamental equation in quantum mechanics that relates the wavelength of a particle to its momentum. It is expressed as λ = h/mv, where λ is the wavelength, h is Planck's constant, m is the mass of the particle, and v is its velocity.

How is the De-Broglie wavelength related to the wave-particle duality?

The De-Broglie wavelength is a key concept in the wave-particle duality of quantum mechanics. It suggests that particles, such as electrons and photons, can exhibit both wave-like and particle-like behavior. The wavelength is a measure of the wave-like properties of a particle, while its momentum is a measure of its particle-like properties.

Can the De-Broglie wavelength be applied to macroscopic objects?

No, the De-Broglie wavelength equation is only applicable to particles with very small masses, such as electrons and other subatomic particles. The wavelength of macroscopic objects, such as humans or cars, would be so small that it is practically impossible to observe.

How does the De-Broglie wavelength affect the uncertainty principle?

The De-Broglie wavelength is related to the uncertainty principle, which states that it is impossible to know both the position and momentum of a particle with absolute certainty. This is because the more precisely we know the position of a particle, the less precisely we can know its momentum, and vice versa. The De-Broglie wavelength is a measure of the uncertainty in the momentum of a particle.

Why is the De-Broglie wavelength important in quantum mechanics?

The De-Broglie wavelength is important in quantum mechanics because it helps us understand the wave-like behavior of particles and their interactions with matter. It also plays a crucial role in many important theories and experiments, such as the double-slit experiment, which helped confirm the wave-particle duality of light and other particles.

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