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I Width of a stationary wave packet as a function of time

  1. Apr 6, 2016 #1
    Trivial question: If a wave-packet is spreading with time, should the width of the wave-packet grow linearly with time?
     
  2. jcsd
  3. Apr 6, 2016 #2

    blue_leaf77

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    Not so trivial as the answer depends on the form of the wavepacket itself as well as the Hamiltonian operator.
     
  4. Apr 6, 2016 #3
    i am talking about a time dependent behaviour of a gaussian wave-packet. [itex] -\frac{h^2}{2m}\frac{∂^2}{∂x}+Vx[/itex]
     
  5. Apr 6, 2016 #4

    blue_leaf77

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    What does ##Vx## mean?
     
  6. Apr 6, 2016 #5
    I meant V(x) that is the potential. Part of the hamiltonian of the time dependent shrodinger equation
     
  7. Apr 6, 2016 #6

    blue_leaf77

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    If ##V(x)## is unknown, it's impossible to determine the way the wavepacket evolves in time. Even if it is given, only very few forms allow us to perform the calculation analytically.
     
  8. Apr 6, 2016 #7
    Ok so if V(x) is zero?
     
  9. Apr 6, 2016 #8

    blue_leaf77

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    ##V(x)=0## corresponds to a free particle and it can be solved analytically in momentum space. Suppose your initial wavepacket is ##|\psi_0\rangle##. The wavepacket at time ##t## is ##|\psi(t)\rangle = \exp(\frac{-iHt}{\hbar}) |\psi_0\rangle## where ##H = p^2/(2m)##. Then use the completeness relation in momentum basis
    $$
    |\psi(t)\rangle = \int dp' \hspace{2mm} e^{\frac{-iHt}{\hbar}} |p'\rangle \langle p'|\psi_0\rangle
    $$
    Now just plug in the initial Gaussian wavepacket in momentum space ##\psi(p,0) =\langle p|\psi_0\rangle## and compute the integral. From this point on, it's up to you whether you want to express the final state in position or in momentum basis. The momentum basis representation is easy, you just need to calculate
    $$
    \langle p|\psi(t)\rangle = \int dp' \hspace{2mm} \langle p|e^{\frac{-iHt}{\hbar}} |p'\rangle \langle p'|\psi_0\rangle
    $$
    Since ##e^{\frac{-iHt}{\hbar}} |p'\rangle = e^{\frac{-i\hat{p}^2t}{2m\hbar}} |p'\rangle = e^{\frac{-ip'^2t}{2m\hbar}} |p'\rangle##, the above equation becomes
    $$
    \langle p|\psi(t)\rangle = \int dp' \hspace{2mm} e^{\frac{-ip'^2t}{2m\hbar}} \langle p|p'\rangle \langle p'|\psi_0\rangle = e^{\frac{-ip^2t}{2m\hbar}} \langle p|\psi_0\rangle
    $$
     
    Last edited: Apr 6, 2016
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