De Broglie wavelength of a particle

In summary: The de Broglie wavelength has nothing to do with the speed of the particle, its related to the momentum of the particle. Thanks,.
  • #1
Harmone
2
0
The meaning of the De Broglie wavelength of a particle is entirely unclear to me.

1. To which kind of wave does the wavelength belong? To the wave that determines how likely the free particle is to show up in certain places (probabilities as squared amplitudes)?

If yes:

2. Why is the wavelength inversely related to the momentum of the particle? Why do the positions where the particle could show up get more densely spaced if the speed of the particle is higher?

3. What does such a wave that has equally high amplitues all over the universe even mean? The picture of a particle with a certain momentum (even with some probabilistic variance) and that of a probability distribution of places that says that it is equally likely that the particle shows up here or 2000 miles from here just don't fit together. (I do realize that usually wave packets are used to describe particles; it's just not clear to me what the meaning of De Broglie waves then is.)

I would be happy if someone could clear up my thoughts a little. Thanks!
 
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  • #2
Harmone said:
1. To which kind of wave does the wavelength belong? To the wave that determines how likely the free particle is to show up in certain places (probabilities as squared amplitudes)?

Yes.

2. Why is the wavelength inversely related to the momentum of the particle? Why do the positions where the particle could show up get more densely spaced if the speed of the particle is higher?

For a free particle, the probability distribution is more or less uniform. That is, you don't have alternating regions of space where the probabilty is small and large. For an idealized free particle, the wave function has the form

[tex]\psi = Ae^{i(px-Et)/\hbar}[/tex]

where A is a constant. The corresponding probability distribution is

[tex]\psi^*\psi = A^*e^{-i(px-Et)/\hbar}Ae^{i(px-Et)/\hbar} = A^*A[/itex]

3. What does such a wave that has equally high amplitues all over the universe even mean? The picture of a particle with a certain momentum (even with some probabilistic variance) and that of a probability distribution of places that says that it is equally likely that the particle shows up here or 2000 miles from here just don't fit together.

The wave function described above is an idealization. Real particles are described by wave packets which are superpositions (sums) of waves that have a range of momenta [itex]\Delta p[/itex], and are localized to a region of space [itex]\Delta x[/itex], such that [itex]\Delta p \Delta x \ge \hbar / 2[/itex] (the Heisenberg Uncertainty Principle).
 
  • #3
Thanks, jtbell!

I still have some difficulty grasping the relationship between the De Broglie wavelength and actual physical systems and why the relation between wavelength and momentum is the way it is.

If I have a real particle and if I know its momentum (to some degree), I can determine its De Broglie wavelength (to some degree). What does this wavelength tell me about the particle's future behavior? If I did not know the particle's momentum, which experiment would allow me to find out about the De Broglie wavelength of the particle?

Is there an intuitive way of understanding the inverse relationship between the De Boglie wavelength and the momentum of a particle? Here is what makes it difficult for me to think of intuitive explanations on my own: For the kind of idealized free particle described by the wavelength I cannot really imagine what the idea of momentum even means. For a real particle where it is is clear that talking about momentum makes sense, there does not seem to be a single wavelength.
 
  • #4
Harmone said:
Is there an intuitive way of understanding the inverse relationship between the De Boglie wavelength and the momentum of a particle?

Do you know "[URL principle[/URL]?


Its 4d variant is the principle by which matter propagates. Huygens principle deals with the direction in which a wave travels: In the direction of the wavefront. The same is true for matter waves. Now, light can only move with c but matter can move at any (lower) speed. It's the de Broglie wave length which determines the speed of propagation.

Consider this first:

A particle in its rest frame has zero momentum and thus has an infinite wave length. The latter simply means that the phase is equal everywhere. You might say that the phase is synchronized throughout the particle's wave function. This synchronization is only possible because different parts of the wave function "communicate" with each other.

In a moving frame the "communication delays" are shifted in a certain direction and the result is that the "synchronization" gives rise to a de Broglie wave length in the direction of motion. The de Broglie wavelength determines the momentum rather then the speed, why? Well because of the de Broglie frequency (the phase change per unit of time). A particle which has a frequency twice as high as another will have a wave length twice as short if both move at the same speed.

For the deeper insight you need a little bit of Special Relativity.

This synchronization mechanism goes very deep. If we change our speed to that of the moving particle then its de Broglie wavelength disappears! The phase appears equal throughout it's wave functions. Why? Our experience of the simultaneity of time has changed, basically so because we are build out of propagating matter waves as well.


Regards, Hans
 
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Related to De Broglie wavelength of a particle

What is the De Broglie wavelength of a particle?

The De Broglie wavelength of a particle is a concept in quantum mechanics that describes the wave-like behavior of matter. It is the wavelength associated with a particle, and it is related to the momentum of the particle. In other words, it tells us how far the particle will travel before its wave-like behavior becomes noticeable.

How is the De Broglie wavelength calculated?

The De Broglie wavelength is calculated by dividing Planck's constant (h) by the momentum of the particle (p). The equation is λ = h/p. The momentum of a particle is equal to its mass (m) multiplied by its velocity (v), so the equation can also be written as λ = h/(m*v).

What is the significance of the De Broglie wavelength?

The De Broglie wavelength is significant because it helps us understand the dual nature of particles. It shows that particles not only behave like particles with mass and position, but also like waves with a certain wavelength. This concept is essential in understanding quantum mechanics and the behavior of subatomic particles.

How does the De Broglie wavelength relate to the uncertainty principle?

The De Broglie wavelength is related to the uncertainty principle because it is one of the factors that determines the uncertainty in the position of a particle. According to the uncertainty principle, the more accurately we know the momentum of a particle, the less accurately we can know its position. This is because the De Broglie wavelength becomes smaller as the momentum becomes more precise, making it harder to pinpoint the exact position of the particle.

Can the De Broglie wavelength be measured?

Yes, the De Broglie wavelength can be measured using various experimental techniques, such as electron diffraction or neutron interference. These experiments involve passing a beam of particles through a diffraction grating, and measuring the distance between the interference fringes to determine the wavelength of the particles. The De Broglie wavelength has also been observed in larger particles, such as molecules and even small viruses, using advanced techniques and equipment.

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