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Quantum mechanics - a free particle

  1. Mar 29, 2008 #1
    Hello everyone!

    If we measure the position of a particle in a free space, and say we find that it is at x0,

    what is the wavefunction right after the measurement in x representation?

    shouldn't it be delta (x-x0), because delta functions are the eigenfunctions of the position operator?

    Another question is how do I find the eigenfunctions of the Hamiltonian of a free particle in x representation?

    Thanks in advance:blushing:
     
  2. jcsd
  3. Mar 29, 2008 #2

    Tom Mattson

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    Staff Emeritus
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    Hello! :)

    Yes, up to a normalization constant I think. I don't have my quantum book handy so you might want to check on this. But the wavefunction will at least be proportional to a delta function.

    You solve the Schrodinger equation with [itex]V(x)=0[/itex].
     
  4. Apr 10, 2008 #3
    thank you!:smile:

    I solved the Schrodinger equation and got the function:
    Aexp(ikx)+Bexp(-ikx).

    This means that any private case (like A=0 and B=1) is an eigenfunction of the Hamiltonian. Can I conclude that the Hamiltonian has an infinite number of eigenfunctions?

    I also notice that there are only two basic "types" here - exp(ikx) and exp(-ikx).
    Does it mean that these two form a basis of the Hamiltonian eigenfunctions space? is it considered as a Hilbert space?
     
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