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Doubt regarding derivation of De Broglie relation

  1. Dec 14, 2015 #1
    For someone who's familiar with the de Broglie relation it's easy to say that for k=0 we have p=0 but how would we know that before deriving the result? In this image the author derives de Broglie relation by considering a wave packet in motion. As you can see where I have star-marked the author sets integration constant c=0 without giving reason. This has been bothering me since you can't just
    8PT69.jpg
     
  2. jcsd
  3. Dec 15, 2015 #2

    jfizzix

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    I think you can use symmetry considerations to understand that the integration constant [itex]c[/itex] must be zero.

    In particular, we would require that changing the sign (direction) of [itex]p[/itex] also changes the sign of [itex]k[/itex]. This is perfectly legitimate, as this is what a coordinate flip would do.

    If
    [itex]p=\hbar k + c,[/itex]
    and
    [itex](-p) = \hbar (-k) +c[/itex]
    then adding these two equations together gives
    [itex]0=2c[/itex], or [itex]c=0[/itex].
     
  4. Dec 15, 2015 #3
    I think where this proof could be wrong is because k is a scalar quantity isn't it? And scalar quantities can't flip signs like that. Correct me if I'm wrong.
     
  5. Dec 15, 2015 #4

    jfizzix

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    the wavenumber [itex]k[/itex] is a vector quantity with magnitude and direction, just like momentum. If you take all the above quantities to refer to one-dimensional propagation, this will work out just fine.

    Alternatively, if you explicitly use the vector expressions of [itex]k[/itex] and [itex]p[/itex] throughout this derivation, you should be able to show that
    [itex]\hbar = \frac{\partial p_{x}}{\partial k_{x}}=\frac{\partial p_{y}}{\partial k_{y}}=\frac{\partial p_{z}}{\partial k_{z}}[/itex].
    It gets a little tricky toward the end, but since what you derive must be true for all [itex](p_{x},p_{y},p_{z})[/itex], you can make some simplifications.
     
    Last edited: Dec 15, 2015
  6. Dec 17, 2015 #5
    While I find your proof correct now(after considering k as a vector) and satisfying what's the physical reason you would give to argue the constant must be 0?
     
  7. Dec 17, 2015 #6
    When I visualizer the situation for P=0, the wavelength becomes infinite. What does this imply? @jfizzix
     
  8. Dec 18, 2015 #7

    jfizzix

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    The physical reason I would give is that the magnitude of the wave number as well as the momentum should be constant under a reflection of coordinates.
     
  9. Dec 18, 2015 #8

    jfizzix

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    It helps to recall that the group velocity of a wave is only approximately [itex]\frac{\partial\omega}{\partial k}[/itex], and that each frequency component travels at its own velocity, so that the group velocity is a statistical average (equal to the rate of change of the mean position of the pulse).

    The derivation we're discussing here shows that the mean momentum is equal to hbar times the mean wavenumber. If the mean momentum is zero, the mean position of the pulse is stationary, but the pulse itself spreads out in time, since it will have components moving at nonzero velocities.

    If the wave's momentum components were tightly clustered near zero, then the wavepacket would spread out in opposite directions very quickly.
     
  10. Dec 19, 2015 #9
    You mean the phase component waves move out rapidly in opp. directions?
     
  11. Dec 19, 2015 #10

    jfizzix

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    I mean the positive frequency components move in the opposite direction as the negative frequency components.

    If the pulse is very narrow in space/time, then it will have to have very many wavenumber/frequency components, some of which will be moving a lot faster, simply because there are more significant frequency components making up the wave.
     
  12. Dec 19, 2015 #11
    Also when you say that group velocity is [itex]\frac{\partial\omega}{\partial k}[/itex] does this mean that if I were to find the velocity of a group packet, I'd have to evaluate this derivative at various points in space? And then sum average all the values to get an approximate number? Or just evaluate the derivatives for all values of k in the group and then sum average over k?
     
  13. Dec 20, 2015 #12

    jfizzix

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    Where a wave [itex]\Psi(x,t)[/itex]can be decomposed into an integral of plane waves, each with amplitude [itex]A(k)[/itex], we have:
    [itex]\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int dk A(k) e^{i(kx - \omega(k) t)}[/itex].
    The time derivative of the mean position [itex]\langle x\rangle[/itex] as weighted according to intensity [itex]|\Psi(x,t)|^{2}[/itex], can be given by the following formula.

    [itex]\frac{d\langle x\rangle}{dt}=\int dk |A(k)|^{2} \Big(\frac{\partial \omega(k)}{\partial k}\Big)[/itex],

    so that when [itex]\Big(\frac{\partial \omega(k)}{\partial k}\Big)[/itex] is nearly constant over the significant wavenumbers of [itex]|A(k)|^{2}[/itex], we get the approximate formula for the group velocity.

    Then, it is understood that [itex]\frac{\partial \omega(k)}{\partial k}[/itex] is to be evaluated at the peak value of [itex]k[/itex] for the pulse under consideration. The three-dimensional version of this derivation follows the same steps without a substantial increase in difficulty.

    Technically, the group velocity is the velocity of the pulse envelope (which itself only makes sense for very narrowband pulses), but this formula is how I make sense of it.
     
  14. Dec 26, 2015 #13
    Thanks for your time and answers :)
     
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