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Not a real wave

  1. Jun 19, 2004 #1
    why can't we describe a probability wave as a real wave like a sound wave? what are the inconsistencies that arise?
     
  2. jcsd
  3. Jun 19, 2004 #2

    chroot

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    You can't actually detect a wavefunction; when you make measurements, you will always detect the particle in a specific "crisp" location, or as having a specific "crisp" momentum, etc.

    - Warren
     
  4. Jun 19, 2004 #3
    you can't detect it directly but you can certainly infer its existence and calculate its shape...an analogy would be a sound wave which you can't see but you can detect in other ways...it seems to me that the collapse of the wave function wouldn't necessarily prohibit our thinking of it as an actual wave...
     
  5. Jun 19, 2004 #4

    jcsd

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    For a start the quantum mechanical wavefunction is an essientially complex quantity, secondly it is possible for the wavefunction to have different values at 2 different points in space, yet the probabilty of finding a particle at those points may be the same.

    Mianly for these two reasons it is impossible to identify the wavefunction with any physically observable quantity.
     
  6. Jun 19, 2004 #5

    ZapperZ

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    Other than the fact that QM wavefunction is a COMPLEX quantity, the wavefunction also exits in a CONFIGURATION space, not REAL space (other than for a 1-particle system). For more than 1-particle, you have a multidimensional space. This is one of the common misconception of QM that has been listed.[1]

    Zz.

    [1] D. Styer, Am. J. Phys., v.64, p.31 (1996).
     
  7. Jun 19, 2004 #6

    selfAdjoint

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    And the kicker for Scroedinger himself (who wanted to believe in the physical reality of his wave) was that the wave function of each separate particle exists in a space separate from that of every other particle's wave function. The particles all exist together in our spacetime, but their wave functions can't; they exist somewhere else and are locally mapped into spacetime by the operators.
     
  8. Jun 20, 2004 #7

    reilly

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    Multiparticle Systems

    ****************


    Not so at all. The coordinates used in multiparticle wavefunctions, just as those used in multiparticle classical physics, all refer to the same space-time -- the one we inhabit. This can be confirmed in any book on QM. Examples include the physics of the hydrogen atom, Fermi-Thomas calculations of the properties of heavy atoms, all of solid state physics, and all of the various forms of quantum field theory.

    If what you say is true, what, then is the significance of the coloumb potential between two charged particles? If what you say is true, how can we possibly describe scattering experiments?

    Regards,
    Reilly Atkinson
     
  9. Jun 21, 2004 #8
    okay, i thought the wavefunction value would be a measurement of the probability - what is the value you're refering to? thanks...

    _____________

    i don't think the fact that the wavefunction is a complex quantity prohibits our viewing it as something analagous to a wave that simply has some unusual properties, it isn't 'real' in the usual sense... and the interesting question of what sort of space does the wave inhabit, it must inhabit 4 dimensional spacetime at some point or we wouldn't see interference would we?
     
  10. Jun 21, 2004 #9

    selfAdjoint

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    I read this in a respectable history of QM. I have just checked my copy of Inward Bound, and it's not there. Could I have seen it in Jammer's Conceptual Foundations? I no longer have my copy of that book - lost in a move - so could someone who has it check? It would be in a discussion of Schroedinger's early realist views of his wave function.
     
  11. Jun 22, 2004 #10
    As I recall, Schröddinger viewed his wave function like some kind of charge density.

    However this problem has been long solved, hasn't it? I thought Born's probabilistic interpretation was accepted. That is, that the square of the wave function is the probability density. I guess that's why they call the wave function a wave, in analogy to em waves, whose square is the intensity, which we can measure.

    I read somewhere that Schrödinger's early interpretation of the wave function is correct in a limited number of cases. For example, in Bose-Einstein condensates where all the particles share the same wave function, it IS incidentaly a charge density, or a density of something.

    When I started studying QM this semester this puzzled me greatly. Why on earth is the solution of Schrödinger's equation called a wave function??? The differential equation in question is NOT a wave equation, since it's only first order in time because it comes from the Hamilton-Jacobi equation.
     
  12. Jun 22, 2004 #11

    reilly

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    The Schrodinger Eq came to be known as a wave eq. because, among other things, deBroglie waves and electron diffraction needed explanation in the 1920s. The Schrodinger Eq. did just that, and Schrodinger nailed the hydrogen spectrum as well -- a truly astonishing result then and now.. For a free particle, the solutions of the Schrodinger Eq. are standard waves -- exponential of i(kx-Et). It's the form of the solutions that determines a wave equation, not its order.

    |psi|**2 as a charge density is not always true. The relativistic Klein-Gordan equation has a probability/charge density that involves d(psi)/dt. Relativity makes QM a little tricky.

    Regards,
    Reilly Atkinson
     
  13. Jun 23, 2004 #12
    Yeah, the solutions for free particles are plane waves, but there is a nasty trick there. The imaginary exponential, as you know, comes from the imaginary unity on the right side of the equation. If it were not there, in the time dependence we would get the typical solution of heat transfer problems, because it is in fact an eq of diffusion.

    A few months ago I read Schrödinger's original papers, and it is interesting to point out that in the first pages he already was comparing the solutions he found to those of a vibrating cord with two points fixed. It was something like we obtain the hydrogen energies much like we obtain the normal modes of vibration in a cord.
    I didn't really understand all the details in the article because the mathematics was kinda hard core, but I recommend them for people who get a bit startled when they first see it on a blackboard with no further information.

    How the heck did he come about thinking up that change of variable
    S=kln(psi)!!!??
     
  14. Jun 23, 2004 #13
    It would be nice to read a book where the Schrödinger equation were not a postulate, but a derivation. Of course there would still have to be a postulate regarding time dependence, but I would perfer a more fundamental one.

    Any one fell up to the task?If you do or already have done it please do post it!!!
     
  15. Jun 23, 2004 #14

    selfAdjoint

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  16. Jun 24, 2004 #15
    Way out of my league. But thanks!
     
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