Inverse spectral problem in QM

  • #1
391
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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
alxm
Science Advisor
1,842
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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)
 
  • #3
1,744
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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.
 
  • #4
674
83
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.
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.
 
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  • #5
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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).
QUOTE]

If you choose a unitary operator [tex] U [/tex] commuting with momentum, then the transformed Hamiltonian is of the form [tex]H' = \Delta + V(x,p)[/tex]. I don't see anything wrong with having potential V that depends on both position and momentum of the particle.
 
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  • #6
391
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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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