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Constructing a wave packet requires adding (superposing) many (if not infinite) plane waves of different wavevectors. A single plane wave has a well-defined wavelength, and hence, from de Broglie $$p = \frac{h}{\lambda}$$ a well-defined momentum. Here $h$ is Planck's constant. Now the wave packet has many wavelengths and hence many wavevectors/momenta in it. This simply means that the momentum is defined over some $\Delta k$ range.

Now the wave packet has a group velocity $v_g = d\omega/dk$. Here begins my confusion. If we know the group velocity of the wave packet (of the particle), wouldn't this mean that the momentum of the particle, $p = m~ v_g$, and also its de Broglie wavelength are known? So what about the uncertainty of the wavelength and the wavevectors mentioned earlier?

What I am thinking is, that we have two "kinds" of de Broglie momenta in this situation. One is that associated with each of the plane waves constituting the wave packet, and also with the so-called "phase velocity", while the other is that of the wave packet as a whole and associated with the group velocity.

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I found a problem (on some 4-year-old university homework paper, unfortunately I don't know its source.) says:

An electron has a de Broglie wavelength of $1.5 \times 10^{-12}$ m. (i) Find its kinetic energy and (ii) the group and phase velocities of its matter waves.

The solution has to go in one way only. The kinetic energy is $$T = E - E_0 = \sqrt{c^2~p^2 + m_0^2~c^4} - m_0~c^2$$ To find $p$, we have to use the above wavelength of de Broglie.

Then it is asking about finding $v_g$ and $v_p$, which are both related to the wave packet describing the electron.

Therefore, the conclusion is: A moving wave packet, that represents some quantum particle, does have a de Broglie wavelength of its own.

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  • $\begingroup$ I think you should read this first en.wikipedia.org/wiki/Wave_packet, and see the animation on how the Gaussian wave packet propagates. $\endgroup$
    – hyportnex
    Oct 22, 2023 at 0:50
  • $\begingroup$ There are no answers here, not even a single mention of 'de Broglie' in this Wikipedia article. $\endgroup$ Oct 22, 2023 at 1:03
  • $\begingroup$ @flippiefanus A quantum free particle is dispersive: the dispersion relation is $\omega = \hbar k^2/2m$, which implies that the phase velocity is half the group velocity. $\endgroup$
    – march
    Oct 22, 2023 at 6:14
  • $\begingroup$ I think you should understand what group velocity v. phase velocity is, de broglie or no broglie... and this is why I suggested the wikipedia article and you should also should take @march's admonition into heart: free-space is dispersive for de Broglie waves. $\endgroup$
    – hyportnex
    Oct 22, 2023 at 7:28
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    $\begingroup$ The de Broglie wavelength $\lambda = h/p$ is associated with a quantum state that is a momentum eigen-state in free space, represented by a plane wave in a state whose momentum is exactly $p$. Every other quantum state that is not a momentum eigen-state has no "wavelength" for it is a wave packet, a linear combination of plane waves of various wavelengths and of various directions. A wave packet has an average wavelength, a wave packet also has an average velocity, a rms velocity etc., but does not have a single velocity, or wavelength or "momentum". $\endgroup$
    – hyportnex
    Oct 22, 2023 at 14:02

1 Answer 1

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What I am thinking is, that we have two "kinds" of de Broglie momenta in this situation. One is that associated with each of the plane waves constituting the wave packet, and also with the so-called "phase velocity", while the other is that of the wave packet as a whole and associated with the group velocity.

This is largely correct.

The momentum associated with the group velocity is the expectation value of the momentum, i.e., the mean value of momentum that will average out of measuring momentum on an ensemble of particles that have been identically prepared in the wave packet.

The momenta associated with each of the plane waves within the wave packet are the allowed values of momentum that can form part of that ensemble of measurements.

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  • $\begingroup$ Thanks. So the wave packet is an ensemble of particles? I have read that a single particle is described with a wave packet and in this case we have no ensemble but that of the plane waves. So is the momentum (wavelength) of the wave packet now the average of the momenta of all the plane waves? $\endgroup$ Oct 22, 2023 at 11:59

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