I'm curious where the de Broglie relation $p=\frac{h}{\lambda}$ comes from?

I know that for light (which has no rest mass), the following is true:

$E=pc$ and $E=hf$


$$pc=hf \Rightarrow p=\frac{hf}{c}=\frac{h}{\lambda}$$

But how is the expression $p=\frac{h}{\lambda}$ obtained for a massive particle where $E\neq pc$? I've read some people claim that the expression can be derived, and others saying it's an experimentally verified relationship.


2 Answers 2


This is covered nicely in the first (historically oriented) chapter of Weinberg's first book on quantum field theory. Here, he explains the motivation behind De Broglie's hypothesis as follows:

Of course, the main clue was provided by the analogy with radiation. However, there was more: If we want to try to implement wave-particle duality, it must be possible to describe particles my means of a wave with a phase which depends on position and time as:

$$\varphi(\vec x,t)=2\pi (\vec k \cdot \vec x -\nu t)$$

where $\vec k$ is the wave-vector and $\nu$ the frequency of the wave. Now, the key is Lorentz invariance. For this phase to stand a chance to be Lorentz invariant, we must have that $\vec k$ and $\nu$ transform under boosts like $\vec x$ and $t$, and hence like $\vec p$ and $E$. This forces them to be proportional, with an identical proportionality constant $\alpha$. From the Einstein relation $E=h\nu$ it is then natural to guess at $\alpha^{-1}=h$, and indeed this gives the correct De Broglie relations, including the ones you quote. Once we accept this, everything follows directly.

As we see, the De Broglie hypothesis can be made plausible, but it remains a hypothesis that needs to be verified experimentally (and cannot be proven).

  • $\begingroup$ I guess that the argument consists in substance to realize that both $\vec{k}$ (from the shape of its contribution to the phase) and $\vec{p}$ (because it is the generator of the Lie spacial translation group) perform a translation of the system by a vector $\vec{x}$; thus they have to be proportional. I am not sure, invoking special relativity is necessary here, or at least I wonder whether it is really necessary. $\endgroup$
    – gatsu
    Oct 10, 2014 at 10:22
  • $\begingroup$ Shouldn't $\vec k$ incorporate the $2\pi$ factor in your formula? $\endgroup$
    – Ruslan
    Oct 10, 2014 at 12:01

The de Brogile supposes that a particle is moving at some speed v, where "E = pv"

In other words, the de Brogile wave is not moving at the speed of light, but at the speed of the particle itself. This is apparent from $\lambda f$ which gives the speed of the particle.


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