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More about wave packets

  1. Sep 17, 2007 #1
    Hello:

    I need some help with a homework problem that was taken from Quantum Physics by Gasiorowicz.

    The problem goes like this: You have a beam of electrons and know the size of the wave packet, and by the uncertainty principle you can estimate the dispersion in p at t=0. The problem is to know the size of the wave packet after the beam has crossed 10^4km in two cases: i) when the K.E. of the beam is 13.6 eV and ii) 100MeV.

    It is clear for me that the wave packet does not have a constant size since the dispersion relation does "disperse"; I also know that in the first case the problem can be treated non-relativistically, while the second case is relativistic.

    I don't want assume the wave packet to be gaussian.

    However, I do not know how to combine these ideas to solve the problem.
     
    Last edited: Sep 17, 2007
  2. jcsd
  3. Sep 17, 2007 #2
    The shape of the wave packet was not specified, so, I think, the Gaussian form assumption is justified. You are looking for a rough estimate, anyway. So, why not make your task easier?

    Eugene.
     
  4. Sep 17, 2007 #3
    If I assume the wave packet to be gaussian, the problem would be solved, since the dynamics of such a wave packet can be calculated.

    The theoretical facts I want to use are the uncertainty principle and the dispersion relation E=p^2/2m.
     
  5. Sep 17, 2007 #4

    nrqed

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    I would simply calculate the distance travelled by an electron if it had the energy given in the problem. Then I would calculate the distance travelled by an electron with momentum [itex] p_0 + \delta p [/itex] where delta p is given by the uncertainty principle. The difference between the two results is a rough estimate of the dispersion of the wave packet
     
  6. Sep 17, 2007 #5
    That's an excellent idea. Thank you very much for your help.
     
  7. Jan 13, 2009 #6
    I've been struggling with that question long time, which is what is the advantage of assuming the wavepacket as Gaussian ??please help me out:frown:
     
  8. Jan 13, 2009 #7
    In order to solve your problem you need to

    1. Assume some initial form of the wave function at t=0.
    2. Apply the non-stationary Schroedinger equation to find the wave function at t>0.

    If you choose the wave function in 1. to be Gaussian, then the solution of 2. is given by an analytically calculable integral. Moreover, Gaussian wavepackets have the advantage that it is easy to go from position representation to the momentum representation and back.
     
  9. Jan 15, 2009 #8
    appreciate your time, so you are saying such assumption lead to analytically calculable integral, by using the useful Gaussian integral relation, which make the calculation more easy. my basic understanding of the advantage of the Gaussian wavepacket is to ensure the localization and then the Normalization. how does ti sound? please enlighten me more??
     
  10. Jan 15, 2009 #9
    Of course, your wavepacket should be localized and normalized. The former condition follows from the fact that you are trying to represent a classical particle, which is localized. The latter condition is just a rule of quantum mechanics. So, localization and normalization are necessary requirements: you can't do without them. However, these requirements can be satisfied with many different functional forms of the packet. The only advantage of the Gaussian form is that it is very easy to calculate integrals with Gaussian functions, like [tex]\exp(-ar^2) [/tex].
     
  11. Jan 16, 2009 #10
    You help much appreciated........
    what du you mean by non-stationary? Is the word wavepacket means that(localization+Normalization)??
    please can you kindly elaborate this sentence by an example? cos I didn't understand it......:uhh:
    when we say Gaussian ensembles?Is this same as Gaussian wavepackets??
     
  12. Jan 16, 2009 #11
    As an example on localization and normalization, herr the bound and scattered state.....
    I s this means that the bound state is localized and normalized, and the scattered state is not?if so why is that as I think it is the opposite as the bound state usually given by exponential function and ,e.g bound particle always have an exponential wave function?so this means is not narmalised, please show me where is my confusion
     
  13. Jan 16, 2009 #12
    The non-stationary Schroedinger equation is

    [tex] i \hbar \frac{d \Psi(r,t)}{dt} = H \Psi(r,t) [/tex]

    If you know the wavefunction (or wavepacket) at t=0 you can solve this Schroedinger equation and find the wavefunction at any other time.

    By Gaussian wavepacket centered at position [tex] \mathbf{r}_0[/tex] I mean function

    [tex] \Psi(\mathbf{r}) = N \exp(-a (\mathbf{r} - \mathbf{r}_0)^2) [/tex]

    where N is a normalization constant chosen to make sure that the probability distribution associated with this wavefunction is normalized to 1. I don't know about "Gaussian ensembles".
     
  14. Jan 16, 2009 #13
    Thanks, Is the non-stationary Schroedinger equation is
    [tex]\imath\hbar\frac{\partial}{\partial t}\psi(\textbf{r},t)=\left(\frac{-\hbar^2}{2m}\nabla_r^2+V(r) \right)\psi(\textbf{r},t)[/tex]
    if so, what is the formula for stationary Schroedinger equation?
    Q2 I've read this sentence, and it is really confused me :confused:
    please enlighten me
     
  15. Jan 16, 2009 #14
    Yes, this is the non-stationary (or time-dependent) Schroedinger equation. The stationary Schroedinger equation is used to find out the energy spectrum [tex]E_n [/tex] of the system

    [tex]\left(\frac{-\hbar^2}{2m}\nabla_r^2+V(r) \right)\psi_n(\textbf{r}) = E_n \psi_n(\textbf{r})[/tex]


    About the confusing sentence you should ask its author. It is just as confusing to me.
     
  16. Jan 17, 2009 #15
    can anyone tell me what one means by these bracket [tex] \langle\ psi(r)\psi^*(r) \rangle[/tex] , is it integral or what?????and is there any idea how to calculate this integral [tex] \int\int{\mid\langle\psi(r)\psi^*(r')\rangle\mid}^2 dr dr'[/tex]
     
    Last edited: Jan 17, 2009
  17. Jan 18, 2009 #16
    Is the regular system means non-chaotic one? if so what is the irregular system, can anyone give an examples to both systems??????please help me
     
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