The Wikipedia article about matter waves lists the De Broglie relation for the frequency $f$ of a matter wave as $$ E = hf = \hbar\omega $$ with Planck constant $h$, total energy $E$ and angular frequency $\omega$.

The article also explains phase and group velocity of the matter wave being different, so the matter wave is rather a wave packet i.e. a sum or integral of waves each having their own frequency and wave length. Approximately we can have

$$\omega(k) = \omega_0 + v_g(k - k_0)$$

as a relation between the $\omega$s of the individual waves of the wave packet and their wave vector $k$ with $v_g$ being the group velocity and $k_0$ and $\omega_0$ some constant values.

Question: Which of the $\omega(k)$ or $\omega_0$ or some combination of them appears in the De Broglie relation $E = \hbar\omega$?


1 Answer 1


The answer to your question is that the frequency that appears in the De Broglie relations is the function $\omega(k)$. Note that the Wikipedia article on group velocity that you quote says:

If the wavepacket has a relatively large frequency spread, or if the dispersion ω(k) has sharp variations (such as due to a resonance), or if the packet travels over very long distances, this assumption is not valid, and higher-order terms in the Taylor expansion become important.

In the context of matter waves, this linear approximation of the frequency as a function of wavenumber does not hold (why?) and one can actually derive a dispersion relation that is valid for matter waves. Remember that not only the relation $E=\hbar \omega$ holds, but also $p=\hbar k$, and as we know from classical mechanics, $E=\frac{p^2}{2m}$ (for a free particle). So substituting the equations for $E$ and $p$ in terms of $\omega$ and $k$ respectively, we arrive at the dispersion relation $\omega(k)=\frac{\hbar k²}{2m}$. I leave as an exersice to the OP the job of calculating and interpreting what the derivative of this expression for $\omega$ with respect to $k$.

  • $\begingroup$ You say 'the frequency ... is the function $\omega(k)$', but that could be something like $E(k)=\hbar\omega(k)$, which is a function, not the total energy $E$ featuring in the De Broglie relation. Is it $E = \int E(k)$ then? $\endgroup$
    – Harald
    Jan 2, 2022 at 10:24
  • $\begingroup$ @Harald You are confusing the hole point of the De Broglie relations, there's absoulutely nothing wrong about having $E(k)$, it is both the total energy and a function of $k$. In fact, one can write $E=\frac{\hbar^2 \omega^2}{2m}$, as I wrote in my answer, remember that $p=\hbar k$. The last expression you wrote, at least as stated, makes no sense. $\endgroup$
    – Don Al
    Jan 3, 2022 at 4:01
  • $\begingroup$ So $E=\hbar^2\omega^2/(2m)$, but then again: which $\omega$ is that. OK, lets extend to $E(k) = \hbar^2\omega(k)^2/(2m)$, but this is the energy of individual waves of the wave packet. What is the total energy of the wave packet then? $\endgroup$
    – Harald
    Jan 8, 2022 at 7:04
  • $\begingroup$ @Harald Sorry for the late reply. I made a typing mistake, the correct relation is not the one I wrote in terms of $\omega$, but it is $E=\frac{\hbar^2 k^2}{2m}$. The total energy is the expression for $E$ is just wrote. $\endgroup$
    – Don Al
    Jan 10, 2022 at 14:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.