# Proof of de Broglie wavelength for electron

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 $\lambda =h/mc$?

• This doesn't actually prove what you're stating. Which textbook is this? Commented Jul 16, 2018 at 18:48
• Direct proof starting with what axioms? One could take the de Broglie relationship as an axiom. Commented Jul 16, 2018 at 19:02
• Commented Jul 16, 2018 at 20:33

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.

• Please refer to this link for learning how to format math in your question: math.meta.stackexchange.com/questions/5020/… Commented Jul 16, 2018 at 19:13
• Thanks, this was my first comment here, and I was a bit too lazy to look up how to write the math properly. It’s basically latex, right? Commented Jul 18, 2018 at 8:18

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.

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.

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.

• "which is double of what it really is" More precisely, twice the total minus rest energy for the non-relativistic case. Commented Jul 16, 2018 at 21:50
• There is no rest energy in non-relativistic case. So, your statement is maybe overprecise :) Commented Jul 16, 2018 at 21:51
• Your statement is a contradictio in terminis. @Oktay Commented Jul 16, 2018 at 21:53
• OK. Why is that? Commented Jul 16, 2018 at 21:55
• Rest energy is the energy of a system at rest, which is the extreme non-relativistic case. The Schrödinger equation does not contain the rest energy, because for non-relativistic energy differences it is unnecessary. This does not mean that a non-relativistic electron does not have a rest energy. Commented Jul 16, 2018 at 21:56