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In his book Introduction to Quantum mechanics, Griffiths gives a diagram of a person giving a sharp jerk to a rope creating a bump travelling down the rope. Here's the image:enter image description here

I understand that the wave has a well defined position but an ill defined wavelength (it doesn't seem periodic)
To bring about a transition to the Uncertainty principle, Griffiths writes that a spread in wavelength corresponds to a spread in momentum, which is what I don't quite understand. I say so because apart from those on the bump, all the particles have a zero momentum (a well-defined zero momentum). As for particles on the bump, what makes their momentum different from a 'regular transverse wave,' except that there's an additional rightward velocity along with downwards/upwards, depending on whether it's on the left or right part of the bump? Seems pretty well defined to me.

hft
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Ambica Govind
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1 Answers1

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Simplest answer:

The momentum and the wavelength of quantum particles are related by de Broglie’s equation:$$p=\hbar k,$$ so the statement of Griffith is almost trivial in this sense.

In addition to this, I give a brief explanation about this equation.

First of all, it must be pointed out that the momentum of “waves” is different fro the momentum of particles which consist of a bump. In fact, the particles in the rope are actually only moving in a direction perpendicular to the direction of the wave. (For example, this comedy-video may be helpful. In this movie, each audience just stands and sits, but they don't move all together in a horizontal direction, do they?) In other words, the momentum of the wave and the momentum of the particles on the rope are completely different, and especially particles in the rope don’t have the rightward velocity. (e.g. see this post)

Then, what is a momentum of waves? We already know the answer from the calsssical wave theory: the wave, $\exp\big({i(\omega(k) t-k{x})}\big)$, have the group velocity $v=\frac{d\omega}{dk}$, and this group velocity corresponds to the velocity of the center of waves.

On the other hand, if we regard the bump as a quantum mechanical particle, the velocity of the particle is the same as the velocity of the wave $$v=\frac{d\omega}{dk},$$ and Einstein's equation holds $$E=h\omega,$$ so we can conclude that: \begin{split}p&=mv=m\frac{d\omega}{dk}= \frac{m}{\hbar}\frac{dE}{dk}\\&= \frac{p}{\hbar}\frac{dp}{dk}\\ \therefore p&=\hbar k.\end{split} This is the derivation of de Broglie's equation for the simplest case.

The important point is that we consider the wave as particles, and that the motion of a wave is different from the motion of an individual particle.

Siam
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