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Wave formula question

  1. Sep 17, 2005 #1
    why is the wave formula equal to zero when x equals postive/negative infinity?
     
  2. jcsd
  3. Sep 17, 2005 #2

    quasar987

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    Hi,

    I think is a requirement that follows from the probabilistic interpretation. If the integral

    [tex]\int_{-\infty}^{\infty}|\Psi|^2dx[/tex]

    is to converge, psi is to go to zero at infinity
     
    Last edited: Sep 17, 2005
  4. Sep 18, 2005 #3

    reilly

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    Generally. it's not zero. Yes for bound states. But, scattering particles obey the Sommerfeld radiation conditions, outgoing waves go as exp(ikx), incoming go as exp(-ikx). To see this in action in a relatively simple way, see scattering from a square well potential in any QM text.
    Regards,
    Reilly Atkinson
     
  5. Sep 19, 2005 #4
    thank you very much! :)
     
  6. Sep 19, 2005 #5

    dextercioby

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    Scattering states are typically eigenstates of unbounded operators (we usually call them "plane waves" and "spherical waves"), like the free particle's Hamiltonian and momentum operators. Ortonormalizing such eigenstates (which are NOT in [itex] L^{2}\left(\mathbb{R}^{3}\right) [/itex] ) is a tricky problem. See Bogoliubov's text on axiomatical QFT.

    Daniel.
     
  7. Sep 19, 2005 #6

    reilly

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    Kemble's QM text from the late 1930s show that the problems of unbounded operators, and scattering have been quite well understood for a long time, at least enough to do physics. Also, Arnold Sommerfeld deals with wave equations, primarily for electromagnetic wave problems, in unbouded space, and does so with a fair amount of rigor (cf Partial Differential Equations in Physics, 1949) Then there's a more sophisticated treatment treatment by Hille and Phillips in their Functional Analysis and Semi-Groups (1957)-- they show that single particle position operators and momentum operators have eigenvalues and eigenfuctions, for example. Then there's the quite exhaustive discussion of the use of wave packets by Goldberger and Watson in their Collision Theory(1964) The set of physics mathematics driven by physical intuition that has been proved wrong by mathematicians is close to one of measure zero.(Think of how useful delta functions are in working with Green's Functions.)

    There's plenty of mathematical rigor to back up common pratice in physics. As Goldberger and Watson putit,

    "Mathematics is an interesting intellectual sport, but it should not be allowed to stand in the way of obtaining sensible information about physical processes.This attitude may offend some purists, who should feel free to consult the mathematical literature whenever the mood strikes them." Their manifold contributions to physics validate their statments, at least in my opinion.

    And, I do not suggest that rigorous mathematics should be eschewed by physicists; it's a matter of taste and interest. The good Professor Bogoliubov was a master in both the inuitive and and rigorous worlds.
    Regards,
    Reilly Atkinson
     
  8. Sep 21, 2005 #7
    wow! that's interesting~
     
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