Proof of de Broglie wavelength for electron

Question

According to de Broglie's wave-particle duality, the relation between electron's wavelength and momentum is $\lambda =h/mv$.

The proof of this is given in my textbook as follows:

1. De Broglie first used Einstein's famous equation relating matter and energy, $$E=mc^2,$$ 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, $$E=h\nu,$$ where $E =$ energy, $h =$ Plank's constant ($6.62607 \times 10^{-34}\:\mathrm{ J\:s}$), $\nu =$ frequency.

3. Since de Broglie believes particles and wave have the same traits, the two energies would be the same: $$mc^2=h\nu.$$

4. Because real particles do not travel at the speed of light, de Broglie substituted $v$, velocity, for $c$, the speed of light: $$mv^2=h\nu.$$

I want a direct proof without substituting $v$ for $c$. Is it possible to prove directly $\lambda=h/mv$ without substituting $v$ for $c$ in the equation $\lambda =h/mc$?

Answer 1

The $c$ in $mc^2$ is not the actual speed of the particle (unless we’re talking about light, but then m would be zero). $mc^2$ 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 $\frac{\hbar}{i}\frac{d}{dx}$ by finding the generator of the translation operator in the quantum mechanical way (which gives you something like $\frac{d}{dx}$ 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 $\frac{\hbar}{i}\frac{d}{dx}$, then you can just start from there.

Using $p = \frac{\hbar}{i}\frac{d}{dx}$ and the assumption that particles have a wave-like nature, you can prove that $p = \frac{h}{λ}$. Since in general, we can write the wave function of a given state of a particle as: $Ψ(x,t) = Ψ_0 e^{i(kx-ωt)}$, which gives us: $pΨ = \frac{\hbar}{i}\frac{dΨ}{dx} = \hbar k Ψ$, so $p = \hbar k = h/λ$.

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 $i \hbar \frac{d}{dt}$, 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.

Answer 2

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

Answer 3

The "proof" on the question is incorrect since it claims that the energy of the matter wave is $E=mv^2$, which is double of what it really is.

Here, I am going to demonstrate the non-relativistic case. For relativistic case, one can not prove de Broglie's hypothesis, it should be a postulate, just like Planck's hypothesis. Actually in relativistic QM, the postulate is in terms of a four-vector: $p^\mu = \hbar k^\mu$, where $p^\mu=(E/c, \mathbf{p})$ is four-momentum and $k^\mu = (\omega/c, \mathbf{k})$ is the wave four-vector.

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The group velocity for any wave is as follows: $$ v_g = \frac{\partial \omega}{\partial k} $$ where $\omega$ and $k$ are angular frequency and wave number, respevtively. We use Planck's hypothesis, $E=\hbar \omega$, where $E$ is the kinetic energy of a particle for non-relativistic case: $$ \hbar \omega = \frac{p^2}{2m} $$ where $p$ is the momentum and $m$ is the mass of the matter wave.

Since the group velocity corresponds to the actual velocity of the wave packet, then $$ v = \frac{\partial \omega}{\partial k} = \frac{1}{2m \hbar} \frac{\partial p^2}{\partial k} = \frac{p}{m \hbar} \frac{\partial p}{\partial k} $$ If you multiply both sides by $m \hbar$, then $$ \hbar mv = p\frac{\partial p}{\partial k} \\ \hbar p = p\frac{\partial p}{\partial k} \\ \hbar = \frac{\partial p}{\partial k} $$ Therefore the momentum should be proportional to the wave number: $$ p = \hbar k $$ QED.