How much does the de Broglie relation depend on Special relativity?

In summary: And again, the wave function is simply a convenient way of representing the probability that a particle will be found at a particular location. It has nothing to do with the actual matter itself.In summary, de Broglie's theory invokes both Special and Galilean Relativity. His theory starts with energy and momentum relations that are based in Galilean Relativity, but the wave function is simply a representation of probability that a particle will be found at a particular location.
  • #1
lalbatros
1,256
2
I went back to my old introductory books on Quantum Mechanics and noticed the derivation of the de Broglie relation.
Therefore I have now this question:

How much does the de Broglie relation (p = hb k) depend on Special relativity?
Would the dependence be strong enough that one may consider the de Broglie relation and its experimental confirmations as further support for Special Relativity?

Thanks for your comments
___
You can find the original derivations by de Broglie in these documents:

http://tel.archives-ouvertes.fr/docs/00/04/70/78/PDF/tel-00006807.pdf
http://ebookbrowse.com/de-broglie-kracklauer-pdf-d277824038

More readable derivations:

http://quantummechanics.ucsd.edu/ph130a/130_notes/node66.html
http://en.wikipedia.org/wiki/Theore...for_the_Schrödinger_equation#de_Broglie_waves
 
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  • #2
I wouldn't say the De Broglie wavelength has anything to do with SR, other than the obvious fact that momentum doesn't always fit into the classical sense. Most introductory QM doesn't involve SR, though a framework for combining QM and SR has been worked out quite well. (Before anyone gets mad at me for saying QM and Relativity are combined, it's QM and General Relativity that are incompatible.)
 
  • #3
Thanks whovian.

I have seen two kinds of derivations related to Special Relativity.
The first was obtained by relating wave group velocity with particle velocity, which is a more "visual" explanation.
The second, which is simpler, is based on the phase velocity since it starts by assuming that the phase must be an invariant:

Since this expression is invariant:

img701.png


only this can be the expression for the phase in a matter waver:

img702.png


Therefore, I get the impression that SR is really needed to derive the de Broglie equation.
Could we come to the same conclusion in Galilean relativity, or by other means without invoking SR?
 
  • #4
Hi lalbatros,

If you're serious about studying de Broglie's work, his inexpensive book "Introduction to the Study of Wave Mechanics" is indispensable. He does begin the derivation of his theory with a few observations that come from both SR and the theory of the photon. I agree with you though, that it may be possible to derive his theory from Galilean relativity if certain observations are incorporated. That's a study that I've postponed but now may be an appropriate time to take it up.

An interesting thing is that Schroedinger at first expected his wave equation to be relativistic since the de Broglie relations, on which his equation is based, are said to be Lorentz invariant. De Broglie theory seems to strangely walk the borderline between SR and Galilean relativity.
 
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  • #5
I feel a different puzzlement can be focussed on with respect to deBroglie's lambda =hp. I thin the very strangest assumption is that matter consist of waves. Once you take this odd fact, it makes more sense that momentum is like wavelength. Once you believe that, and the probabilistic aspect, then the uncertainty principle is a geometric fact about x and d/dx. Not being able to know x and d/dx is true for any random variable and it's. So the really odd thing about quantum mechanics is that matter is a wave. A lot of other things will follow from that hypothesis. For instance, shorter wavelength implies higher frequency implies higher energy and momentum, makes sense for any wave.
 
  • #6
The very heart of de Broglie theory is that there must exist both particle and wave associated with the particle. (Matter is associated with the wave but doesn't necessarily consist of it). De Broglie parted company with nearly all of the early QM heavyweights on that point. Because of the incontrovertible experimental evidence of de Broglie's "wave", that part of his theory needed to be acquired in all QM interpretations. But because few could accept his theory in totality you see his theory watered down and only semi-incorporated into QM.
 
  • #7
PhilDSP said:
Matter is associated with the wave but doesn't necessarily consist of it

How does matter not consist of waves?
 
  • #8
algebrat said:
How does matter not consist of waves?

The particle, in de Broglie theory, is what carries the mass aspect of matter. That the wave is in some sense secondary is expressed by the fact that refraction or dispersion of the wave is controlled by the mass term of the particle.

Zitterbewegung is an example of how a particle can move in ways that correlate to a wave function.
 
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  • #9
Back to the original question: There are 2 essential invocations of relativity in de Broglie theory. The theory starts with the "relativistic energy relations"

[itex]E = \frac{m_0 c^2}{\sqrt{1 - \beta^2}} \ \ \ \ \ \ \ \ \ \ p = \frac{m_0 v}{\sqrt{1 - \beta^2}}[/itex]

The phrase "relativistic" is no doubt apt. The values for energy and momentum vary when velocity is varied. But we ought to notice that these are expressions of Galilean Relativity. They are invoked, and de Broglie specifically says so in his book "Non-Linear Wave Mechanics", from within a Galilean reference frame. Indeed it can be demonstrated that those equations express solutions of the Maxwell equations, to first order approximation, of a free charged particle with uniform recti-linear motion (entirely without recourse to the Lorentz transformation).

From there de Broglie develops the connections to wave-length, frequency, group velocity and phase velocity. The second invocation of relativity occurs when he bases the relationships between those values in different inertial frames on the invariance of the form of the wave equation:

[itex]\Psi = a_0 e^{2 \pi i f't'} \ \ \ [/itex] for the non-transformed frequency and time period of the wave for observer A

[itex]\Psi = a_0 e^{2 \pi i f(t - \frac{x}{v_{phase}})} \ \ \ [/itex] for the transformed frequency and time period of the wave for observer B (from the LT)

That can be verified by substitution using the transformations
[itex]f = \frac{f'}{\sqrt{1 - \beta^2}} \ \ \ \ \ \ \ \ v_{phase} = \frac{c}{\beta} = \frac{c^2}{v}[/itex]

Since [itex]f = \frac{E}{h}[/itex] and [itex]E = \frac{pc^2}{v_{phase}} \ \ \ [/itex] the second wave equation becomes

[itex]\Psi = a_0 e^{\frac{i}{\hbar}(Et - px)} \ \ \ [/itex] just as you quoted earlier

What I think we want to do to understand how the transformation between frames looks when evaluated in terms of wave-lengths and frequencies, is to develop a covariant 3-vector plus t form for the transformation, along the lines of

[itex]\Psi = a_0 e^{2 \pi i ft} \ \ \ [/itex] and [itex]\ \ \ \Psi = a_0 e^{2 \pi i f'(t' + \frac{x'}{v_{phase}})} \ \ \ [/itex] so that correlated with the first 2 wave equations, we can find how x -> x' relates [itex]\lambda[/itex] -> [itex]\lambda'[/itex]
 
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  • #10
There's a misprint in one equation above. The relationship is [itex]E = \frac{pc^2}{v} \ \ \ [/itex] ([itex]v[/itex] can be called [itex]v_{group}[/itex])
 

1. How does the de Broglie relation relate to Special Relativity?

The de Broglie relation is a fundamental equation in quantum mechanics that describes the relationship between the wavelength and momentum of a particle. It is based on the principles of Special Relativity, which states that the laws of physics should be the same for all observers regardless of their relative motion.

2. What is the de Broglie wavelength and how is it calculated?

The de Broglie wavelength is a fundamental property of matter that describes the wave-like behavior of particles. It is calculated using the de Broglie relation, which states that the wavelength (λ) is equal to Planck's constant (h) divided by the momentum (p) of the particle (λ = h/p).

3. How does Special Relativity affect the de Broglie wavelength of particles?

Special Relativity has a significant impact on the de Broglie wavelength of particles. It predicts that the wavelength of a moving particle will be shorter than that of a stationary particle, and the degree of shortening is dependent on the particle's velocity.

4. Can the de Broglie relation be applied to all particles, including photons?

Yes, the de Broglie relation can be applied to all particles, including photons. Even though photons are massless particles, they still have momentum and can exhibit wave-like behavior, making the de Broglie wavelength applicable to them.

5. How does the de Broglie relation support the concept of wave-particle duality?

The de Broglie relation is one of the key pieces of evidence that supports the concept of wave-particle duality. It demonstrates that particles, which were previously thought to only exhibit particle-like behavior, also have wave-like properties. This duality is a fundamental aspect of quantum mechanics and has been experimentally verified numerous times.

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