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B Particle or wave?

  1. Mar 16, 2017 #1
    How anything can be a particle or wave at the same time.
  2. jcsd
  3. Mar 16, 2017 #2
    Quantum scale objects (electrons say), can be one or the other depending on how you measure them.
    I think it's best to just call them quantum objects.
  4. Mar 16, 2017 #3
    Yeah but at a particular time object or a Quantum object behave like a particle or wave but not simultaneously both
  5. Mar 16, 2017 #4
  6. Mar 16, 2017 #5
    Feymann has a good view on this.They are acting just like "Quantum Mechanical Way".This a problem which turning physics to english as he claimed.I agree with Feymann
  7. Mar 17, 2017 #6
    Wave-particle duality is an outdated concept, so don't bother too much with that.
  8. Mar 17, 2017 #7
    I just want to know that waves we talk about of particles, are they electromagnetic waves ??
    ,and when someone is talking about wavelike property of macroscopical objects are they talking about our oscillation up and down ie. Our wave function otherwise how can someone think of us as a wave.
  9. Mar 17, 2017 #8


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    Staff: Mentor

    No, the wave we're talking about are not electromagnetic waves.
    It can't. A quantum object is neither a particle nor a wave, although it has some properties that we usually associate with waves and some properties that we usually associate with classical particles.
  10. Mar 17, 2017 #9
    Which waves are these?
  11. Mar 17, 2017 #10
  12. Mar 17, 2017 #11
    I mean what type of wave are they actually are?
  13. Mar 17, 2017 #12


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    As several posters have noted before, there is no wave-particle dualism in "modern quantum mechanics" (the today still valid version of QM has been developed independently in 1925/26 by Heisenberg+Jordan+Born, Dirac, and Schrödinger).

    In non-relativistic quantum mechanics one associates a wave function with the "state" of a single particle like an electron, ##\psi(t,\vec{x})##. It takes values in ##\mathbb{C}## (for scalar particles, spin ##s=0##) or ##\mathbb{C}^{2s+1}## (for particles with spin ##s \in \{1/2,1,\ldots \}##). The meaning of the wave function is that
    is the probability distribution for the position ##\vec{x}## of the particle at time ##t## (Born's Rule).

    All the "quantum weirdness" like "wave-particle duality" and similar ideas of "old quantum theory" vanish, as soon as you accept this probabilistic meaning of the wave function (or more generally any kind of quantum state).
  14. Mar 17, 2017 #13
    ... "probability waves" ...
  15. Mar 17, 2017 #14
    As I understand, in modern quantum mechanics, the wave-particle duality was "outdated." There is no duality, there are only quantum objects, or better to say: vectors in the Hilbert space of infinite dimensions. What we perceive as a "wave" is only a "name" just to give a name to something, that "in essence" is a mathematical function, like sin and cos.
    And what is sin and cos? Trigonometric functions ...
  16. Mar 17, 2017 #15
    Partially true, but I think there is more to it than that, so I also partially disagree, but the full analysis would have to go on a separate thread. Roughly and briefly three points here:
    1. Structure of wave as (amplitude function)×f(x-vt, y, z) [f being sin or cos etc. type function, or combination etc.], or F(x-vt, y, z) in general etc., or includes different types of modulation ...
    2. Thus not limited only to sin, cos ... , unless perform Fourier alalysis (expansion).
    3. Besides waves there is also wave-packets, very useful and common even in QM, and these are not sin, cos only (nothing but ... , sometimes, unless Fourier ...).

    Thus it's a big issue.
  17. Mar 17, 2017 #16


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    Any reason for separating the scalar particles out of the ##\mathbb{C}## formula, please? Is this OK:

    In non-relativistic quantum mechanics one associates a wave function with the "state" of a single particle like an electron, ##\psi(t,\vec{x})##. It takes values in ##\mathbb{C}^{2s+1}## (for particles with spin ##s \in \{0,1/2,1,\ldots \}##?

  18. Mar 17, 2017 #17
    I remember there were three kinds of interpretations of a measurement on a particle.

    1. realist position, which implies that the particle was there before measurement. Because you didn't do measurement, you did not know it was there. This implies a loophole in quantum mechanics that there is hidden variable to determine position of the particle. This is what Einstein believed in.

    2. orthodox position, the particle is not anywhere, but observation created a form of measurement (forced wavefunction of the particle to collapse into one determinate state) The method of observation deeply affects the form of measurement you get.

    3. agnostic position, refuse to answer such question because it is nutty to ask for position of a particle before measurement.

    Bell's inequality proved that the statistic of realist position and orthodox position are different. This eliminated agnostic position. And the result is that orthodox position is correct (This is why Einstein was wrong).

    Now back to your question. If you measure position of a particle, you get position of a particle, which is a particle. Now if you measure momentum of the particle, which is related to wavelength by de broglie formula, you get wavelength, which is a wave. Because position and momentum do not commute each other, you cannot get both information accurately. When position wavefunction collapse, you get measurement of a particle, but the momentum measurement is lost. Vice versa, this shows that a free particle has position and momentum, which is particle and wave property. Measurement method will change their property, which is different from classical mechanics. In classical mechanics, properties are static, particle is particle, wave is wave.
  19. Mar 17, 2017 #18


    Staff: Mentor

    These descriptions seem oversimplified to me. They also seem to assume a collapse interpretation of QM; there are also no collapse interpretations.
  20. Mar 17, 2017 #19
    maybe I used wrong words. Three opinions maybe better?
  21. Mar 17, 2017 #20
    I agree, then the "wave particle duality", or whatever you want to call it, is not outdated
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