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De Brogli wavelength.

  1. Nov 5, 2014 #1
    according to De Brogli wavelength equation not every thing has a wavelength instead everything that moves has a wavelength right?
     
  2. jcsd
  3. Nov 5, 2014 #2

    ShayanJ

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    Gold Member

    That's right but not applicable. Because nothing is not moving in QM. There is always a motion due to zero point energy coming from Heisenberg's uncertainty principle. So every particle has a De-Broglie wavelength!
    Also you should be ready to farewell [itex] \vec p=m \vec v [/itex] when you start learning QM.
     
  4. Nov 5, 2014 #3

    bhobba

    Staff: Mentor

    Well you actually hit on a huge problem with the De-Broglie hypothesis.

    One can always jump to a frame where the particle is at rest and then issues arise such as infinite phase velocity.

    Its really just a way station to the correct quantum theory and was consigned to the dustbin of history once that was developed.

    That said its interesting to analyse it in light of the correct theory:
    http://www.gauge-institute.org/wave-particle/deBroglieP.pdf

    Thanks
    Bill
     
  5. Nov 5, 2014 #4
    what is zero point energy?
     
  6. Nov 5, 2014 #5

    ShayanJ

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    Gold Member

    I don't think we're allowed to say that. Because whenever we assign a sharp value to momentum, we're allowing the uncertainty in position to blow up. So in a frame where the particle is at rest, its actually everywhere. So you have a particle as huge as the universe itself which is at rest in the universe. I think its not strange that its De-Broglie wavelength blows up. So I think this situation isn't that much physical that we use it to judge about a formula.
    But I should say I accept that De-Broglie's formula isn't proper for actual calculations because we never associate a sharp value of momentum to a particle.

    Consider the uncertainty relation [itex] \Delta p \Delta x \geq \frac{\hbar}{2} [/itex] where [itex] \Delta p[/itex] and [itex] \Delta x [/itex] are uncertainties in particles momentum and position. We always have some vague knowledge of where the particle is. The least we can say is that its in the laboratory. So we're always confining the particle in a finite region of space. But that means [itex] \Delta x [/itex] is finite and so [itex] \Delta p [/itex] can't be zero too which means there is always some momentum associated to the particle.
     
  7. Nov 6, 2014 #6
    As you might expect, I disagree entirely with the notion that an infinite phase velocity isn't correct for a non-moving particle and in fact mathematically is exactly what suits the situation. That, just like many other seemingly explicable things in QM, may not fit comfortably with our intuitive notions. Feynman had some relevant things to say in that regard.

    Is that paper from a bonafide peer-reviewed journal by the way? Not that it's generally incompetently written, but certain statements like the one proposing that de Broglie's theory is incorrect for the reason harped on above might show a lack of full understanding and study of de Broglie's ideas,
     
  8. Nov 6, 2014 #7

    bhobba

    Staff: Mentor

    In QM you cant.

    But remember this is the De-Broglie theory we are talking about - its a mishmash of classical, quantum and relativistic ideas.

    Thanks
    Bill
     
  9. Nov 6, 2014 #8

    bhobba

    Staff: Mentor

  10. Nov 6, 2014 #9
    ok I am just curious. So becuase p=mv, and wavelength=h/p, for wavelength to be noticable, p must be small. So, to make momentum small, you can make mass a very small number so lamda is noticable. Ok, so ccould you make velocity extremely small to acheive the same effect? I mean, can u make a baseball go at like 10^-35 m/s so that in the end you generate 10nm as the wavelength?
     
  11. Nov 6, 2014 #10
     
  12. Nov 6, 2014 #11

    bhobba

    Staff: Mentor

    Sure - but for macro objects the velocity would be so small it would be way below the limits of detectability.

    Thanks
    Bill
     
  13. Nov 6, 2014 #12

    bhobba

    Staff: Mentor

    It is - see chapter 3 Ballentine - Quantum Mechanics - A Modern Development.

    However V is a QM operator.

    Thanks
    Bill
     
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