Rigorous derivation of DeBroglie wavelength

Question

I've scoured the internet as much as I can, and I've yet to find a rigorous derivation of the DeBroglie wavelength. They all go something like this:

$$E=\gamma mc^2, \ \gamma \approx 1 \ \therefore\ E=mc^2$$ $$E=\hbar \omega=\frac{2 \pi \hbar c}{\lambda}$$ And then they all make the leap that because massive particles can't go the speed of light, the energy equations above must also satisfy $$E=mv^2$$ $$E=\frac{2 \pi \hbar v}{\lambda}$$ $$\frac{2 \pi \hbar v}{\lambda}=mv^2$$ $$\frac{2 \pi \hbar v}{|\vec{p}|v}=\lambda$$ $$\frac{2 \pi \hbar}{|\vec{p}|} \equiv \lambda$$ But I find this "derivation" very frustrating because we are using relationships that are unique to light, such as $E=\hbar\,\omega$, and just extrapolating the relationship to massive particles with (what seems like) very little basis other than "this should work". Is there any proof/derivation out there that states a few acceptable and consistent assumptions at the start and then just does math from there on?

Answer 1

There is nothing to "derive". The de Broglie wave length is always $\lambda =h/|\vec{p}|$. More precisely, the momentum $\vec{p}$ and the wave vector $\vec{k}$ are related by $$\vec{p}= \hbar \vec{k}, \tag{1} \label{1}$$ where $|\vec{k}|=2\pi/\lambda$.

The energy-momentum relation $$E=c\sqrt{\vec{p}^2+(mc)^2} \tag{2} \label{2}$$ holds for massive ($m\ne 0$) as well as massless ($m=0$) particles like the photon. In the latter case, you simply have $E= c |\vec{p}|$. In the case of a nonrelativistic (nr) particle ($|\vec{v}|<\!\!<c$, or, equivalently $|\vec{p}|<\!\! <mc$) it is sometimes convenient to subtract the rest energy $mc^2$ defining $$E_{\rm nr}= E- mc^2 =\vec{p}^2/2m = m \vec{v}^2/2. \tag{3} \label{3}$$ In any case, you obtain the dispersion relation $\omega= \omega(\vec{k})$ by inserting \eqref{1} into \eqref{2} and using $E= \hbar \omega$ arriving at $$\omega(\vec{k})= c \sqrt{\vec{k}^2+(mc/\hbar)^2}, \tag{4}$$ or, in the nonrelativistic case, $$\omega_{\rm nr}(\vec{k}) = \frac{\hbar \vec{k}^2}{2m}.\tag {5}$$

Answer 2

In addition to the correct answer: wavelength is not a good physical quantity for manipulation. Yes, we are more conformable with it than wave-number, but it's worth the effort to use $\vec k$ instead of $\lambda$ and $\hat n=\vec k/||k||$, esp since wavelength doesn't transform like the space component of a geometric object.

In relativity, the wave 4-vector is:

$$ k_{\mu} \equiv (\omega/c, \vec k) $$

and the four momentum is

$$ p_{\mu} \equiv (E/c, \vec p) = \hbar k_{\mu} $$

which contains all the deBrogile relationships you need.

Moreover, when working with waves, the fact that phase:

$$ \phi = \vec k \cdot \vec x - \omega t = k_{\mu}x^{\mu} $$

is a Lorentz scalar will come in handy. Wavelength only muddles the situation.

Sorry: back to your question about "hassles photons". Note that

$$ k_{\mu} \equiv (\omega/c, \vec k) $$

does not depend on speed of propagation. For light:

$$ k_{\mu} = p_{\mu}/\hbar $$ $$ k_{\mu} = (E/c, \vec p)/\hbar $$ $$ k_{\mu} = (p, \vec p)/\hbar $$ $$ k_{\mu} = (k, \vec k)=(\omega/c, \vec k)$$

so we immediately see light is dispersionless:

$$ v_{ph} = \omega/k = c $$ $$ v_{gr} = \frac{d\omega}{dk} = c $$

while massive particles are dispersive, since:

$$ c^2p_{\mu}p^{\mu} = (\hbar c)^2k_{\mu}k^{\mu}$$ $$ E^2-(pc)^2 = \hbar^2(\omega^2 - (kc)^2) = (mc^2)^2 $$

from which you can work out dispersion relations. Note there is a cut-off minimum frequency for $k=0$:

$$ \omega_0 = \frac{m c^2}{\hbar} $$

Final note: this is why ppl use natural units with $\hbar = c = 1$.