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Free particle normalization

  1. Jul 23, 2010 #1
    why it is not possible to normalize the free-particle wawe functions over the whole range of motion of the particles?
     
  2. jcsd
  3. Jul 23, 2010 #2

    tom.stoer

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    The whole range of motion is from minus infinity to plus infinity (no restricted range, otherwise the particle would not be free).
    The free particle is described by a plane wave u(x) = exp(ikx).
    The normalization means to integrate |u(x)|² = 1; the integral will certainly divergy
     
  4. Jul 23, 2010 #3

    DrDu

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    One should add that exp(ikx) is obviously not the only possible form for the wavefunction of a free particle. Rather the wavefunction may have almost any form, especially normalizable.
    However, there are no normalizable wavefunctions which are eigenfunctions of the Hamiltonian.
     
  5. Jul 23, 2010 #4

    tom.stoer

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    Let me see if I understood you correctly.

    You propose to use (nearly) arbitrary wave packets constructed from plane waves (Fourier modes) and let these wave packets evolve in time using the free Hamiltonian. That means you construct normalizable wave functions, but they are no longer eigenstates of the free Hamiltonian.

    Yes, of course you are right. I assumed that "free particle" means "eigenstate of the free Hamiltonian H", but that need not be the case.

    @alimehrani: what was your intention?
     
  6. Nov 30, 2011 #5
    yes
    yes
    exactly
    thank you very much
    i think you are very good teacher
     
  7. Nov 30, 2011 #6
    thank tom.stoer and others
    you greatly help me
     
  8. Nov 30, 2011 #7
    this problem exist also about a particle in a square potential barrier!????
    what is the justification here?

    another question:
    we assume that wave function is zero at infinity while we couldn't observe such a thing in the cases of free particle and particle in a square potential barrier?
     
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