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Bound states as a solution of free particles?

  1. May 19, 2015 #1
    It came to me just now that because we can always take the Fourier transform of a well-behaved function, this means we can think of any such state as a superposition of free-particle momentum eigenstates. E.g., the Hermite polynomial eigenfunctions of the harmonic oscillator. They have a Fourier transform (whatever it is) and can therefore be thought of as superpositions of functions e^(ipx) . These are free momentum eigenstates, even though individually they are not solutions to the harmonic oscillator Schrodinger equation.

    From a math point of view, this is trivial. But does it have any theoretical significance?
  2. jcsd
  3. May 19, 2015 #2
    Sure, this gives you the momentum space wave function.
  4. May 19, 2015 #3
    Right. Of course. Thank you! The Fourier transform (momentum space wave function) will be time-dependent in this case. Unlike the free particle case. But it has the same physical meaning.

    But does it still work if you have a velocity dependent potential? Classically, that changes the form of the canonical momentum. Mathematically you can still take the Fourier transform. I'm just wondering if the result in that case would have the same physical meaning: momentum space wave function.
  5. May 19, 2015 #4
    In the case of small perturbations to a free Hamiltonian (even velocity-dependent ones), it makes sense to cast the perturbation term into this momentum basis, and compute the matrix elements of it between various free states. This is known as the Interaction Picture, and is a convenient way to set up a lot of scattering problems.
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