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Wave packet

  1. Jul 1, 2003 #1

    jby

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    I read that a wave packet is really some superposition of some waves with different wave number k (just slightly different k's). While the wave packet represents the particle, is there any meaning to the individual wave? How does physicists know what to superpose?
     
  2. jcsd
  3. Jul 1, 2003 #2
    Hi jby,
    be careful since you're entering a very dangerous area of physical thinking. A wave packet will diffuse very quickly, while a particle will not. You expect a particle to be stable in time, don't you? Quantum theory tells us that the only states stable in time are the eigenstates of the Hamiltonian. And these are what you call the 'individual waves'. A wave packet, as you state correctly, always contains an ensemble of different k's, and thus an ensemble of different eigenstates, and thus an ensemble of photons. Facit: A wave packet is not a photon. Whenever there's a wave, it is made up of an ensemble of photons.
     
  4. Jul 2, 2003 #3

    Ivan Seeking

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    Re: Re: wave packet

    I was with you right up to here. Did you mean whenever there's a wave packet, it's made up of an ensemble of photons?
     
  5. Jul 2, 2003 #4

    jby

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    Re: Re: wave packet

    I don't understand.
    Let say, I have a wavefunction = sin x + sin 1.1x + sin 1.2x + sin 1.3x
    Do you mean that all four sin's, ie sin x, sin 1.1x, sin 1.2x, and sin 1.3x represents 4 different eigenstates?

    And I don't understand this: isn't that a wave packet describes a particle like one photon. We use wave packet concept because it is more localized.
     
  6. Jul 2, 2003 #5
    Re: Re: Re: wave packet

    Yes.
     
  7. Jul 2, 2003 #6
    Re: Re: Re: wave packet

    Yes.

    You can localize a photon only when it interacts (= is emitted or absorbed). There is no way of determining which path it took.
     
  8. Jul 2, 2003 #7

    pmb

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    If the particle is a free particle [I.e. the potential energy function V(x,y,z) = constant or zero] then the wavefunction can be anything you'd like, so long as the wavefunction is normalizable (i.e. the integral of |Psi(x)|^2 over all x is finite). That means that the particle can be found anywhere on the x-axis.


    If the particle is not a free particle then you can have a finite sum of eigenfunctions. But that doesn't mean that you can choose the wavefunction at will. It has to meet the boundary conditions. The eigenfunctions vanish outside the box and are sines and/or cosines inside the box - depending on where the box is.

    The meaning of the wavefunction is interpreted by the Born Postulate which says that the wavefunction represents the probability of measuring position, I.e. the probability of finding the particle in the interval x + dx is proportional to |Psi(x)|^2 dx

    Therefore: Psi(x,y,z,t) is the probability "amplitude" of the particle's presence. |Psi(x,y,z,t)|^2 is the probability "density"


    Pete
     
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