Proof of de Broglie wavelength for electron According to de Broglie's wave-particle duality, the re

Jaeden Weaver

Jaeden Weaver

Answered question

2022-05-14

Proof of de Broglie wavelength for electron
According to de Broglie's wave-particle duality, the relation between electron's wavelength and momentum is λ = h / m v
The proof of this is given in my textbook as follows:
1.De Broglie first used Einstein's famous equation relating matter and energy,
E = m c 2 ,
where E = energy, m = mass, c = speed of light.
2.Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation,
E = h ν ,
where E = energy, h = Plank's constant ( 6.62607 × 10 34 J s ), ν = frequency.
3.Since de Broglie believes particles and wave have the same traits, the two energies would be the same:
m c 2 = h ν .
m c 2 = h ν .
4.Because real particles do not travel at the speed of light, de Broglie substituted v, velocity, for c, the speed of light:
m v 2 = h ν .
I want a direct proof without substituting v for c. Is it possible to prove directly λ = h / m v without substituting v for c in the equation λ = h / m c?

Answer & Explanation

Bentyrchjurvk

Bentyrchjurvk

Beginner2022-05-15Added 15 answers

De Broglie proposed that the relation p = h / λ would not only hold for photons but also for massive particles. This inspired Schrödinger to propose his famous equation.
Peia6tvsr

Peia6tvsr

Beginner2022-05-16Added 4 answers

The c in m c 2 is not the actual speed of the particle (unless we’re talking about light, but then m would be zero). m c 2 is simply the energy the particle has at rest. I don’t know exactly how De Broglie did it, but you can prove it like this:
First you can prove that the momentum operator should be i d d x by finding the generator of the translation operator in the quantum mechanical way (which gives you something like d d x as the generator) and the classical way (which gives you the momentum as the generator), and simply state that both results should be equivalent. If this sounds unfamiliar, then I suggest you look into Noether’s theorem (it’s one of the coolest theorem’s in maths/physics, so I would suggest it anyway). But if you’re okay with just assuming that the momentum operator is equal to i d d x , then you can just start from there.
Using p = i d d x and the assumption that particles have a wave-like nature, you can prove that p = h λ . Since in general, we can write the wave function of a given state of a particle as: Ψ ( x , t ) = Ψ 0 e i ( k x ω t ) , which gives us: p Ψ = i d Ψ d x = k Ψ , so p = k = h / λ
You can do the same thing to prove Planck’s theorem by first finding the generator of time translation and prove that the operator of energy should be i d d t , and then letting this operator act on Ψ again.
N.B.: In the most general case Ψ ( x , t ) should be a superposition of wave functions with different Ψ 0 , k and/or ω , but then you can’t be sure what the momentum of your particle is anymore.

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