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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 $E = \hbar\omega$?

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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 $k$.

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  • $\begingroup$ 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

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