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

  1. May 25, 2012 #1
    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
     
  2. jcsd
  3. May 25, 2012 #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.)
     
  4. May 25, 2012 #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?
     
  5. May 28, 2012 #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.
     
    Last edited: May 28, 2012
  6. May 28, 2012 #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.
     
  7. May 28, 2012 #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.
     
  8. May 28, 2012 #7
    How does matter not consist of waves?
     
  9. May 28, 2012 #8
    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.
     
    Last edited: May 28, 2012
  10. May 30, 2012 #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]
     
    Last edited: May 30, 2012
  11. May 31, 2012 #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])
     
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