# De Broglie relations use which frequency or frequencies of a wave packet?

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 $$E = \hbar\omega$$?

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:
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 $$k$$.
• 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? Jan 2, 2022 at 10:24
• @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. Jan 3, 2022 at 4:01
• 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? Jan 8, 2022 at 7:04
• @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. Jan 10, 2022 at 14:49