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Wavepacket Dispersion and how this links to particle behaviour

  1. Nov 27, 2011 #1
    Hi,

    I'm trying to get my head round modelling particles in free space in quantum mechanics. I appreciate that we can "build" wavepackets by superposing many plane waves with different k-numbers (i.e. with different frequencies & momentums & energies I think). The greater the number of phase waves making up the packet, the greater the localisation too giving a narrower packet.

    I wonder if anyone could clarify some issues regarding dispersion of this packet please. I understand that the phase velocities will vary for each component wave and this spread causes dispersion, whilst the envelope of the packet travels at the group velocity,
    v_g = d\omega / dk

    My questions arise in terms of relating this to the original particle of mass m and travelling at velocity v, so momentum, p = mv.
    1) Is the average k-number, k_av of the component waves in the packet related to the particle's momentum by p= h-bar * k_av? So therefore, is it right to say that the particle's momentum doesn't affect the rate of dispersion.
    2) We can relate particle energy to momentum by E = (p^2) / 2m and the planar wave energy is given by E = (h-bar * k)^2 / 2m. How do we relate wavepacket energy to the particle energy and will this affect dispersion?

    Any help would be much appreciated.
     
  2. jcsd
  3. Nov 27, 2011 #2

    kith

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    For wave packets, neither position x nor momentum p of the particle are well-defined. Only the average momentum is given by <p>=hbar<k>.
    Energy too is not well-defined. The equation <E> = hbar²<k>²/2m holds again only for average values.
     
  4. Nov 27, 2011 #3

    Bill_K

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    But these are not equal. <E> = ħ2<k2>/2m, and for a wave packet <k2> ≠ <k>2.
     
  5. Nov 27, 2011 #4

    kith

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    Yes, of course you're right.
     
  6. Nov 28, 2011 #5

    Rap

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    If you have a free particle, its momentum wavefunction f(k) will not change in time. The expectation of momentum will be [itex]\langle p \rangle=\int (\hbar k) f(k)f^*(k)dk[/itex]. The expectation of energy will be [itex]\langle E \rangle= \int \frac{(\hbar k)^2}{2m}f(k)f^*(k)dk[/itex], so you can see that [itex]\langle E \rangle=\frac{\langle p^2 \rangle}{2m}[/itex], not [itex]\frac{\langle p \rangle^2}{2m}[/itex].

    The uncertainty in momentum will be [itex]\Delta^2 p=\langle p^2 \rangle-\langle p\rangle^2[/itex] so that [itex]\langle E\rangle=\frac{\langle p\rangle^2+\Delta^2p}{2m}[/itex]. The expectation of the energy will be larger than the square of the expected momentum divided by 2m.
     
  7. Nov 28, 2011 #6
    Ok, thanks. I see how energy and momentum are not well defined. Therefore, could we say that the dispersion of a wavepacket is only determined by it's width and not by the particle's momentum and energy?

    Since if a wavepacket has greater average energy (using the definitions you've given above) then the average wavenumber is greater but the spread of k-number is unaffected assuming the packet width is the same?

    Thanks for your help!
     
  8. Nov 28, 2011 #7

    Rap

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    I'm not sure what you mean by the dispersion of the wave packet. I will assume you mean the rate at which the position wave function increases its width or uncertainty. The width of the positional wave packet, lets call it g(x,t), increases in time, while the width of the momentum wave packet f(k) does not. The rate of increase of the width of g(x,t) is due to the width of f(k). Its not determined by the width of g(x,t). You can see this even in classical physics. If you are not too sure of the momentum, and you calculate the position of a particle over time, the spread in its position as time goes by will increase. The spread in its position is not what causes that spread to increase in time.
     
  9. Nov 28, 2011 #8
    Thanks Rap. I think I see now, so the spreading of the wavefunction g(x,t) with time is due to the spectral distribution f(k) only. Therefore, this is completely independent of the particle's properties for which the wavepacket is modelling, since we arbitrarily choose f(k) and generally choose a Gaussian spectrum?

    Also, a linked but slightly divergent query I wonder if anyone could shed any light on please... with regard to Heisenberg's Uncertainty Principle I've seen both \Delta(x) \Delta(p) >= h-bar and \Delta(x) \Delta(p) >= h-bar/2 & in cases when makes estimates for confinement energies in atoms, it seems common just to use \Delta(x) \Delta(p) EQUALS h-bar. Is there a reason for the inclusion of the factor of 2 in some cases but not others?

    Thanks
     
    Last edited: Nov 28, 2011
  10. Nov 28, 2011 #9

    Rap

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    h-bar/2 is correct. hbar alone is not. I think if you have a confined particle with an exact energy, then the equality is good. But in the case of a free particle, the momentum wave function has a fixed width, while the position wave function's width increases in time, and then the equality does not hold. For bound particles which have been localized inside the boundaries, their g(x,t) will spread also, and the equality will not hold.
     
  11. Nov 28, 2011 #10
    Thanks Rap. That makes sense.

    Just one final question - is it right to say that the particle's properties (velocity, momentum etc) are completely independent of the wavepacket shape and the rate at which its width increases since we arbitrarily choose f(k) (often to be Gaussian)?
     
  12. Nov 28, 2011 #11

    Rap

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    No. It is not correct to speak about a particle as if it had an "actual" position or momentum, because eventually you will come up with logical contradictions and inconsistencies. This is the "hidden variable" attempt to interpret QM and it doesn't work. The wave function tells you what the probabilities are that you will get a particular result if you make a particular measurement, and that's it. The properties of the particle are completely described by the wave function. The particle does not "have" a momentum, it has an expected value of momentum, and that is completely determined by the wave function. It has an uncertainty in momentum, which is also completely determined by the wave function. Critics of this way of looking at things call it the "shut up and calculate" approach. It's not really, it's the "shut up with the dumb interpretations and come up with a better logically consistent interpretation, and when you do, let us know, and until then, calculate"
     
  13. Nov 28, 2011 #12

    kith

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    What do you mean when you say that we choose f(k) arbitrarily? If I have a particle in an unknown wavepacket state, I can measure f(k) by performing reapeated momentum measurements at identically prepared particles.
     
  14. Nov 29, 2011 #13
    Thanks, got it I think!
     
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