# Inverse spectral problem in QM

## 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

$$i \partial _t \Psi (x,t) = \lambda _n \Psi (x,t)=(p^2+V(x))\Psi(x,t)$$

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 $$\sum_n e^{it \lambda _n}$$

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

$$\sum_n e^{it \lambda _n}=(1-exp(it/2))^{-1}$$ from this could we deduce that V(x) is a quadratic potential x*x

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alxm
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)

Knowing energy eigenvalues is not sufficient to restore the Hamiltonian. As alxm said, you need to know all eigenfunctions too.

Two different Hamiltonians $$H$$ and $$H' = UHU^{-1}$$ have exactly the same energy spectra if $$U$$ is any unitary operator.

Knowing energy eigenvalues is not sufficient to restore the Hamiltonian. As alxm said, you need to know all eigenfunctions too.

Two different Hamiltonians $$H$$ and $$H' = UHU^{-1}$$ have exactly the same energy spectra if $$U$$ is any unitary operator.
But H' would be most probable not of the form $$H' = \Delta + V(x)$$, 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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But H' would be most probable not of the form $$H' = \Delta + V(x)$$, so that this simple consideration is not sufficient to answer the question if it is possible to identify V(x).
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If you choose a unitary operator $$U$$ commuting with momentum, then the transformed Hamiltonian is of the form $$H' = \Delta + V(x,p)$$. I don't see anything wrong with having potential V that depends on both position and momentum of the particle.

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a potential $$V(x,p)$$ 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 $$\Psi (x) = exp (i \oint (E-V(x))^{1/2})$$

but in general you only have information about the espectrum