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Wavefunction for arbitrary t

  1. Nov 19, 2014 #1
    For a particle, given the normalised eigenfunctions of the Hamiltonian, the associated energy eigenfunctions and the wavefunction describing the state of the particle at time t=0 how does one calculate the wavefunction for arbitrary t? I know you could solve the time dependent Schroedinger equation but is there not an easier way than that?
     
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  3. Nov 19, 2014 #2

    Orodruin

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    Each eigenstate of the Hamiltonian evolves independently and the evolution is trivially given by the time-dependent SE (if you know the eigenvalue, it becomes a very simple equation for the coefficient).
     
  4. Nov 19, 2014 #3

    jtbell

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    Expand the initial wave function as a linear combination of the energy eigenfunctions ##\psi_k(x)##, that is, find the coefficients ##c_k## in $$\Psi(x,0) = \sum_{k=0}^\infty c_k \psi_k(x)$$ The initial wave function then evolves as $$\Psi(x,t) = \sum_{k=0}^\infty c_k \psi_k(x) e^{-iE_k t / \hbar}$$
     
  5. Nov 19, 2014 #4
    Thanks for the quick replies. I'm seriously lacking an understanding of this. Can you tell me what the coefficient [itex]c_k[/itex] corresponds to?
     
  6. Nov 19, 2014 #5
    is it correct that [itex]c_k=\left\langle \psi_n |\psi \right\rangle[/itex]
     
  7. Nov 19, 2014 #6

    Orodruin

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    Only if n = k.
     
  8. Nov 19, 2014 #7
    silly typo there. thanks
     
  9. Nov 19, 2014 #8

    Nugatory

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    Provided that the potential is not also a function of time, right? Of course that will be the case for a very large number of very important problems.
     
  10. Nov 19, 2014 #9

    Orodruin

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    You are right of course, but based on the OP I would say we are dealing with a level where this is the case. It is also true in the adiabatic limit (where the Et in jt's post becomes an integral of E over time).
     
  11. Nov 20, 2014 #10
    So say If I'm given for example [itex]\psi_n=\sqrt{\frac{2}{L}}sin(\frac{n\pi x}{L}) [/itex] what will [itex]\left\langle \psi_n |\psi \right\rangle[/itex] be?
    Is it the same as [itex]\left\langle \psi_n |\psi_n \right\rangle[/itex]
    [tex]=\frac{2}{L}\int sin^2(\frac{n\pi x}{L})dx[/tex]

    Im not sure if this is turning into a homework help style question or not so please say if I should repost this elsewhere.
     
  12. Nov 20, 2014 #11
    oops think that last post is wrong as [itex]
    \left\langle \psi_n |\psi_n \right\rangle=1[/itex] since the wavefunction is normalized
     
  13. Nov 20, 2014 #12

    DrClaude

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    $$
    c_n = \langle \psi_n | \Psi \rangle = \int_{-\infty}^{\infty} \psi_n^*(x) \Psi(x) dx
    $$
     
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