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Inverse spectral problem in QM

  1. Apr 10, 2009 #1
    I have taken QM , and i find it very interesting but my question is , we have the Hamiltonian eigenvalue problem

    [tex] i \partial _t \Psi (x,t) = \lambda _n \Psi (x,t)=(p^2+V(x))\Psi(x,t) [/tex]

    of course in general, we know the potential V(x) but my question is the inverse, if we knew how the spectrum is or for example the (approximate) value of [tex] \sum_n e^{it \lambda _n} [/tex]

    for example for Harmonic oscillator since all energies are lineal we know that

    [tex] \sum_n e^{it \lambda _n}=(1-exp(it/2))^{-1} [/tex] from this could we deduce that V(x) is a quadratic potential x*x
     
  2. jcsd
  3. Apr 10, 2009 #2

    alxm

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    I don't think you can (although I can't be bothered to try and prove it mathematically.. might be related to the v-representability problem though.).

    If you knew the eigenfunctions, then you could expand the potential on some basis and then treat it as a minimization problem. (an 'inverse Ritz variational method' if you like)
     
  4. Apr 10, 2009 #3
    Knowing energy eigenvalues is not sufficient to restore the Hamiltonian. As alxm said, you need to know all eigenfunctions too.

    Two different Hamiltonians [tex]H[/tex] and [tex]H' = UHU^{-1}[/tex] have exactly the same energy spectra if [tex]U[/tex] is any unitary operator.
     
  5. Apr 11, 2009 #4
    But H' would be most probable not of the form [tex]H' = \Delta + V(x)[/tex], so that this simple consideration is not sufficient to answer the question if it is possible to identify V(x).

    It is not possible, and I have used this as an argument against many worlds, see http://arxiv.or/abs/arXiv:0901.3262" [Broken] and the references therein.
     
    Last edited by a moderator: May 4, 2017
  6. Apr 11, 2009 #5
     
    Last edited: Apr 11, 2009
  7. Apr 11, 2009 #6
    a potential [tex] V(x,p) [/tex] that depend on momentum is not common in physics since you should modifie Newton law or E-L equation

    the idea is that sometimes you only known the spectrum but not the Eigenfunctions , perhaps using WKB approximation , you would know that eigenfunctions depend on the potential [tex] \Psi (x) = exp (i \oint (E-V(x))^{1/2}) [/tex]

    but in general you only have information about the espectrum
     
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